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Subject : General English

Subject Code : FIS 6305

Credits : 2

Semester : 3 (one)

Lecturer : Drs. IBP. Mardana, M.Si.

PHYSICS EDUCATION DEPARTMENT

FACULTY OF MATEMATHICS AND NATURAL SCIENCES

GANESHA UNIVERSITY OF EDUCATION

YEAR 2011

Standard Competency:

Students able to understand the fundamental concept of special relativity and quantum theory

as basic to explain the physics phenomenon in the range of sub-atomic and light velocity.

COURSE 1

I. Basic Competency:

Able to analyze the relativity theory and its consequences in physics phenomenon

II. Indicators:

1.1 Can explain the principle of relativity

1.2 Can explain the inertial frame of reference

1.3 Can formulate the Galileo transformation

1.4 Can explain the essential of the zero experiment of ether of Michelson-Morley

1.5 Can explain the postulate Einstein whose based of the special theory of relativity

1.6 Can formulate the Lorentz transformation

III. Subject Matter

1. Rationale

The theory developed until at the end of the nineteenth century had been very successful

in explaining a wide range of natural phenomena. Newton mechanics beautifully explained

the motion of objects on earth, furthermore, it form the basis for successful treatment of

fluids, wave motion, and sound. Kinetic Theory, on the other hand, based on Newton Laws,

explained the behavior of gases and other materials, and Maxwell’s theory of

electromagnetism, not only brought together and explained electric and magnetic phenomena,

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but it predicted the existence of EM waves that would behave in every way just like light, and

vice versa whose speed 3x108 m/sc. Physics as it was mentioned above is referred to as

classical physics. The new physics that grew out of the great revolution at the turn of the

twentieth century is now called modern physics featured by (1) the special theory of relativity

and (2) the theory of quantum. The questions are what the special theory of relativity is? , and

what is its contribution in modern physics?

It’s difficult to say that an object is at motion or at rest without respect to any reference

called a frame of reference. The velocities of object, therefore, observed by two people with

different frame of reference are difference. Let’s study the problem: A railroad car is moving

along straight, level tracks at a constant velocity of 3.0 m/s to the north. In the car a woman is

walking up the aisle with a constant velocity of 1.0 m/s also to the north. What is the woman

velocity? Two possible answers are: (1) with respect to the railroad car (observer (O) at rest in

railroad car), her velocity is 1.0 m/s, and (2) with respect to the railroad tracks (observer (O’)

at rest on the ground), her velocity is 1.0 m/s. Refer to the problem can be concluded that

motion any object at nature is relative. The next problem is which frame of reference can be

used to observe relative motion of object.

Based on the first law of Newton, an object will move straight with constant velocity if

there is no force exerts on it. Object or place at rest or at motion straight with constant

velocity, in which obey the first law of Newton can be used as frame of reference called

inertial frame of reference. In all inertial reference frames, all laws of physics have the same

mathematical form. It is called principle of relativity. Newton Laws, however, implicitly

assume that time is absolute at every inertial frame of reference. Under the Galileo

transformation we can transform the observation result between two difference inertial frame

of reference O and O’.

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y y’

*A

u

x=x’

O (x,y,z,t) O’ (x’,y’,z’,t’)

z z’

Galileo transformation formulae as shown below:

(1.1)

(1.2)

(1.3)

(1.4)

Verify that the velocity transform to x-axis:

(1.5)

The inverse Galileo transformation, that is:

(1.6)

(1.7)

(1.8)

(1.9)

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(1.10)

Problem 1.1

A river of width L moves uniformly with a speed v. A canoe is to be paddled at a constant

speed c with respect to the river. Calculate the time needed to paddle directly across the river

and back, and compare it to the time needed to paddle the same distance upstream and back.

Ignore the turn around and start-up times.

Solution: (time for across the river and back)

(time for upstream and back)

2. The Null Experiment of Michelson-Morley

The electromagnetic theory of Maxwell, one side, has succeeded to explain wave

phenomena, include light, thoroughly. However, the other side, it appeared two problem,

namely (1) every wave in its propagation need medium, and what the medium of sun light

propagates toward the earth?, and (2) which reference the speed of light (3x108 m/s) is

measured from?. At that time, ether, a hypothetic medium, was proposed as medium for light

propagation. Absolutely, ether is regard at rest in solar system from where the speed of light

(3x108 m/s) was observed.

Michelson Morley was interested to study the existence of ether in our nature. They

developed simply optic apparatus as shown in figure 1.1.

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Figure 1.1 Michelson-Morley Experiment Equipment

The difference in times for rays L// and L will be

(Hint: )

3. Postulate of The Special Theory of Relativity

The resolution of the difficulties in explaining the null experiment of Michelson-

Morley, Albert Einstein (1905 proposed two postulate, namely:

1. The principle of relativity: All laws of physics have the same mathematical form in all

inertial reference frames (invariance).

2. The constancy of the light speed: the speed of light at vacuum has the same value of

c=3x108 m/s in all inertial systems.

The first postulate essentially asserts us that there is no experiment by which we can

measure velocity with respect to absolute space, that’s why, the result, is the relative speed of

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two inertial system. The second postulate merely asserts that the speed of light is the same for

all observers at every inertial frame of reference. In the other word, the speed of light is

absolute value at every inertial frame of reference. It will change our understanding of the

time. If we adapt the absolute of the speed of light, the implication, that the time is relative for

observers at every inertial of reference.

4. Lorentz Transportation

Since c is constant for all observer in both O and O’ and is the same in all direction, all

observer in both frames of reference must detect a spherical wave front expanding from their

origin, as shown in figure 1,2. Light in free space is regard isotropic. The spherical equation

can be written:

1.11

1.12

1.13

Figure 1.2. The isotropic of light

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The transformation matched with the postulate Einstein, especially respect to the isotropic of

light is Lorentz transformation. Lorentz proposed set of transformation, that is:

1.14

1.15

1.16

1.17

The Inverse Lorentz transformation, that is ,

1.18

1.19

1.20

1.21

where ; and

By using above transformation, we can relate the result of observation an event at inertial

frame of reference O’(x’,y’,z’,t’) to another O(x,y,z,t). The coordinate system of Lorentz

transformation is well-known as 4-dimensional space (x,y,z) and time (t) coordinate system.

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y y’

*A

u

x=x’

O (x,y,z,t) O’ (x’,y’,z’,t’)

z z’

Lorentz transformation formulae as shown below:

1.22

1.23

1.24

1.25

Verify that the velocity transform to x-axis:

1.26

The inverse Lorentz formulae

1.27

1.28

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1.29

1.30

1.31

Problem 1.2

Observers on an asteroid measure a spaceship to zip by at speed of 0.60c, as figure 1.3. All

bridge and asteroid clocks are started when the bridge of the ship passed, the asteroid

observers measure a laser flash to occur at a position having coordinates (3.0, 0.5, -0.2) km

respect to them. At what position and time does the spaceship captain measure that the laser

flash occurred?

Solution:

Fix O in the asteroid, and fix O’ in the bridge of the spaceship. Then ,

;

The asteroid observer are in O, so ; ;

; . The spaceship captain is in O’. By using Lorentz coordinate

transformations. It yield:

; ; ; and

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TASK-1

1. Explain in your words what is meant by term “relativity”. Are there different theories

of relativity?

2. Does the Michelson-Morley experiment shows that the ether does not exist or that it is

merely unnecessary?

3. Explain in your own words the terms time dilation and length contraction.

4. How fast must an object move before its length is contracted to one-halt its proper

length.

5. Several spacecraft leave a space station at the same time. Relative to an observer on

the station, A travel at 0.60c in the x direction, B at 0.50c in the y direction, c at 0.50c in

the negative x direction, and D at 0.50c at 450 between the y and negative x direction.

Find the velocity components, direction, and speed of B, C, and D as observer from A.

COURSE 2

I. Basic Competency:

Able to analyze the relativity theory and its consequences in physics phenomenon

II. Indicators:

1. Can apply the length contraction of relativistic

2. Can apply the dilation time of relativistic

3. Can apply the mass defect of relativistic

4. Can analyze the equivalence mass and energy

5. Can compute the energy and momentum based on the view of special theory of

relativistic

6. Can explain the Doppler effect in the special theory of relativity

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III. Subject Matter

1. The Consequences of The Special Theory of Relativity

As we accept the Einstein’s Postulate, that the speed light are the same for all inertial

observers, the consequences that time run slow, and moving object are shortened. The length

of an object that measured in the direction of motion become increasing smaller, and time

intervals on the object become increasingly larger as the object moves at higher speeds

respect to one inertial frame of reference. Our mind-set paradigm about space and time are

altered for rapidly moving objects. It’s due to relativistic effect, the classical definition of

mass, energy and momentum must be changed to be relativistic one. In addition, the concept

of mass and energy is not separated as known in classical physics; however, mass and energy

are equivalent in relativistic point of view.

2. Time Dilation

The time at which an event occurs depends not only on the frame of reference chosen,

but also on the position of the event. Therefore it seems reasonable that the intervals would be

measured differently by observers moving at different speeds. These time interval may be the

period of a repetitive atomic process, the length of a biological process, the time between the

ticks of a mechanical clock, or other difference time between two events.

Suppose an observer at rest in O and a clock is moving by in O’. The clock is at rest in

O”, but has speed v with respect to O. The clock ticks once at time t 2 and again at time t2’, as

shown in figure 1.4.

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Figure 1.4 . Time dilation

Both ticks occur at the same position in O’: x2=x2’. Applying the Lorentz coordinate

transformation can be obtained:

Subtracting gives , remember that . Since the clock at rest in O’, let’s

call the time interval (proper interval time). The zero refers to zero speed.

1.43

t’ is always greater than t for v geater than zero but less than c. The measured increase in

time intervals of a moving object is called time dilation.

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Problem 1.3

Mouns are created in a process near the top of a mountain 4630 m above sea level. The mean

muon traveling at v=0.99c downward with =7.1 will decay at sea level. What is the mean

muon lifetime when is at rest?

Solution:

In our frame of reference, the mean muon travels 4630 m at 0.99c=2.97x108 m/s in a time of

about 4630m/(2.97x108 m/s)=1.56x10-5s=15.6 s. Since the muon is moving in our frame of

reference, 15.6 s=t, then to=t/=2.2 s

That is, to=2.2 s is the mean lifetime when the muon is at rest.

Problem 1.4

Jack and Jill are 25-year-old twins. Jack must stay on earth with a head injury, but astronaut

Jill travels at 0.98c to a star 24.5 light years away and return immediately. Ignoring the end-

point acceleration times, find the twin’s ages when she returns. (One light year =1c. year, the

distance light travels in one year).

Solution:

From the earth-bound frame of reference, Jill travels a total of 49 light years (out and back)

at 0.98c in a time interval of 50 years=49c x years /0.98c. Therefore 50 years of earth time

have passed, so Jack is (25+50) years=75 years old. This 50 years is dilated time, however,

Jill’s time interval is t0. Since =5.0 for v=0.98c, t0=t/=50 years/5.0=10 years. Jill

therefore has only aged 10 years, so he is (25+10) years old. She is 40 years younger that her

twin!

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Question that might come to the problem 1.4, since the choice of frame of reference is

relative, why don’t we place Jill in O? She then sees the earth move away and returns, and

therefore it is Jack who has traveled out and back at 0,98c. He should be the one who is 40

years younger. Since them both can’t be 40 years younger. Each twin expects the other to be

younger. Does this prove relativity wrong? This apparent contradiction is called the twin

paradox. Recall, however, that are dealing with the special theory of relativity, which refers to

inertial reference frames. In the twin paradox, the earth is an approximately inertial reference

frame, but Jill’s spaceship isn’t. It must be accelerated to start the trip, be decelerated when it

turning to earth. The choice of frames of reference is relative in special Theory of relativity

only if the frames of reference are all inertial. Therefore an attempt to use the special theory in

a no inertial frame of reference invites incorrect result. Jack does age more rapidly than Jill.

Experiments confirm this prediction.

3. Length Contraction

Suppose that an observer at rest in O, and want to measure the length of an object

moving past him. First, fix O’ in the object. Then, find the distance to the two object ends of

the object, x2 and x1. Correctly, x2 and x1 are measured at the same time t2=t1, as show in

figure 1.5.

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X1

X2

O O’

Applying the Loretz coordinate transformation,

and

Subtracting, yield:

Note that is the length as measured in O’. Since the object is at rest with respect to

O’, let’s call this length Lo , where the subscript “zero” refer to zero speed (proper length),

then represent the length L of the moving object measured by O. This give:

1.44

Equation 1.19 tells that L is always less than Lo. The decrease in the measured lengths of

moving objects is called length contraction.

Problem 1.4.

A 0.125 m3 cubical box is placed in the cargo hold of a spaceship, which then flies past us at

0.80c. If some edges of the box are parallel to the motion of the spaceship, determine the

box’s dimensions as we’d measure them.

Solution:

A 0.125 has sides of (0.125 m3)1/3=0.5. Also, v=0.80c gives =0.60=1/. The lengths of

those edges perpendicular to the direction of motion would be changed. The length of those

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edges parallel to the motion would be L= (0.50)(0.60)=0.30 m. (With the y and z

dimensions unchanged and the dimension contracted, the volume would be (0.50 m)2(0.30

m)=0.075 m3. Of course a person in the cargo measures the volume of the box 0.125 m3, since

the box at rest with respect to that person.

4. Relativistic Mass and Momentum

Suppose that the following situation: Observer in S1 and S2are given identical balls that

will make a perfectly elastic collision with each other. When the observers move past one

another, each will throw a ball with vx =0 and vy’ that each measures to have the same

magnitude but opposite directions.

After a completely symmetric collision, the balls will rebound with the opposite velocity, -vy.

Both S1 and S2 will have a relativistic relative speed (v close to c) but v y will be classically

(vy<<c). The observer in S1 throws a ball that is practically at rest in S1, so it will call its

mass mo. The ball thrown by the observer in S2 is measured to be moving at a relativistic

speed in S1. so it is called m. The observer in S1 throws the ball upward at speed vy. The y

component of the other ball’s velocity as measured in S2 is –vy. This velocity must be

transformed to S1 using the inverse Lorentz transformation, and becomes:

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In the symmetric collision, the momentum is equal and opposite, so:

This gives:

1.45

Equation (1.45) is the relativistic mass transformation. It tells that measurement of the mass

of an object gives a mass that increases as relative speed v increase. Since and m

increase with v, linier momentum is no longer directly proportional to velocity, as shown in

equation (1.46).

1.46

Problem 1.5

If a spaceship were to move past the earth at 0.60c, what would the crew measure for the

standard kilogram (which is at rest on earth)?

Solution:

When at rest, the standard kilogram has a mass of exactly 1 kg by definition. The relative

speed of the standard kilogram with respect to the spaceship is 0.60C, which gives

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==1.25. Therefore m=mo=(1.25)(1 kg)=1.25 kg. The mass mo is called rest

mass. It is the smallest mass an object has because as the object speeds up, the mass increase.

Problem 1.6

What is the momentum of a proton moving at speed of v=0.86c?

Solution:

=

=

5. The Equivalent Mass and Energy

Work may be done on a body to increase it kinetic energy (Ek) In classical physics,

. In relativistic physics, that expression is not generally correct, even if m is the

relativistic mass. To obtain the correct expression let’s start an object from rest with a net

external force F in the +x direction. Then the work done by F will be stored in the form of

kinetic energy,

=

=

=

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1.47

1.48

1.49

1.50

where is the relativistic mass increase.

Equation 1.49 tells us that when an object at rest in can be assigned a rest energy Eo, when an

object is in motion, it will have a total energy:

The formulae is well-known the mathematical statement of the

equivalence of mass and energy.

Problem 1.7

A 0 meson (mo=2.4x10-28 kg) travels at a speed v=0.80c. What is its kinetic energy?

Compare to a classical calculation.

Solution:

The mass of the 0 meson at v=0.8c is

kg

Thus its Ek is

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6. The Relativistic Law of Conservation of mass-energy

The total relativistic energy is given by equation:

1.51

In interaction of particles at relativistic speeds, can be replaced classical principle of energy

with one based on the total relativistic energy. The relativistic law of conservation of mass-

energy, namely:

In an isolated system of particles, the total relativistic energy

remains constant.

Beside that, the relationship between relativistic energy (E) and momentum (P) can be

analyzed as below:

(square both side)

(multiplying by c4)

It is yield :

1.52

Equation 1.52 is a useful mnemonic device for remembering the relation among the total

energy, momentum, and rest energy.

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Problem 1.8

Show that the kinetic energy (Ek) of a particle of rest mass m0 is related to its momentum p by

the equation

Solution:

Let’s star from the equation:

TASK-2

1. The proper life time of a certain particle is 100 ns. (1) How long does it live in the

laboratory if it moves a v= 0.96c?, (2) How far does it travel in the laboratory during the

time, and (3) How far does it travel in its own frame reference?(nomor 5)

2. An electron and a proton are each accelerated through a potential difference of 10.0

million volts. Find the momentum (in MeV/c) and the kinetic energy (in MeV/c) of each,

and compare with the result of using the classical formulas.(nomor 7)

3. Prove that the relativistic expression (nomor 8)

9. Find the kinetic energy of an electron moving at speed of (a) v=1.00x10-4c;

(b) v=1.00x10-2c; (c)v=0.300c; (d) v=0.999c.

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10. A helium nucleus (alpha particle) contains two proton and two neutrons and has a

mass of 4.001506u. (a) What is the binding energy of a helium nucleus? (b) What is the

mass difference in kg between a helium nucleus and its constituents?

Course 3

I. Basic Competency:

Students are able to understand the theory of quantum and its implication in physics.

II. Indicators:

1. Able to derive e/m of electron base on J.J Thomson experiment.

2. Able to derive electron charge unit (e) based on the oil–drop experiment of Millikan.

3. Can explain the quantum property of particle.

4. Able to explain the approach of Raleigh-Jeans in blackbody radiation.

5. Able to explain the postulate Max Planck about blackbody radiation.

6. Able to derive Max Planck theory about blackbody radiation

III. Subject Matter

At the 18th and 19th century, the scientist paid full attention toward the validity of the

powerful of Newton’s Laws and theory of Electromagnetic Wave to describe the behavior of

the universe. Many questions appeared, such as: (1) What is the fundamental thing as

constituent of the matter?, (2) Does the matter has continuum property absolutely?, (2) Why

are the scientists difficult to explain the spectrum of light emitted by hot object. As we known

that empirically all object emit radiation whose total intensity is proportional to the fourth

power the Kelvin temperature ( ). It is called the power fourth of Kelvin temperature

of Stefan-Boltzman. The spectrum of light emitted by a hot dense object as shown in figure

2.1 for an idealized blackbody (a body who absorb all radiation falling on it and the radiation

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it emits). Beside that, it is found that the wavelength at the peak spectrum whose bring the

maximum energy shifted to the short wavelength. The multiplication of the wavelength at the

peak with it’s the Kelvin temperature related by equation . It is

known as Wien’s displacement law. Mayor problems will cover in this section are in related

with the quantum property of matter and wave as a based to explain the dynamic of particle in

the level of atomic.

1. The Quantum Property of Matter

1) J.J Thomson of e/m electron

J. J. Thomson, using a device similar to a cathode ray tube discovered the electron

and measured the ratio of its electric charge (e) to its mass (m). Thomson's experiment was

concerned with observing the deflection of a beam of particles in a combined electric and

magnetic field. Its result established: 1)the existence of the electron; 2) the fact that the

electron has a mass (me); 3) the fact that the electron has a charge (e); 4) that both charge and

mass are quantized; 5) the ratio of e/m. Eliminating the time between the equations of motion

in the x- and y-directions yields the equation for the parabolic trajectory between the plates:

2.1

Beyond the plates, the trajectory is a straight line because the charge is then moving in a field

free space. The total deflection of the beam is sum of the deflections in regions y1 and y2 and

can be shown to be:

2.2

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Figure 2.1 J.J Thomson of e/m electron apparatus

Thomson measures q/m for cathode rays based on equation 2.2. and found a unique

value for this quantity that was independent of the cathode material and the residual gas in the

tube. This independence indicated that cathode corpuscles are a common constituent of all

matter. The modern accepted value of q/m is (1.758796 ± 0.000010) x 1011 coulombs per

kilogram. Thus Thomson is credited with the discovery of the first subatomic particle, the

electron. Because it was shown later that electrons have a unique charge e, the quantity he

measured is now denoted by e/me. He also found that the velocity of the electrons in the beam

was about one-tenth the velocity of light, much larger than any previously measured material

particle velocity.

2) The oil–drop experiment of Millikan

Figure 2.1 The Oil-drop experiment of Millikan Apparatus

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Millikan’s measurement of the charge on the electron is one of the few truly crucial

experiments in physics and, at the same time, one whose simple directness serves as a

standard against which to compare others. With no electric field, the downward force is mg

and the upward force is bv. The equation of motion is

where b is given by Stokes’ law:

and where h is the coefficient of viscosity of the fluid drop. The terminal velocity of the

falling drop vf is

When an electric field E is applied, the upward motion of a charge qn is given by

Thus the terminal velocity vr of the drop rising in the presence of the electric field is

In this experiment, the terminal speeds were reached almost immediately, and the drops

drifted a distance L upward or downward at a constant speed.

2.3

Where, is the fall time, and is the rise time.

Based on the equation 2.3, Millikan succeeded to determine the value of electron, that is,

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e=1.60217733x10-19 C

It can be concluded that the matter is formed by quanta of elementary particle whose charge

1.60217733x10-19 C, and mass 9.1x10-31 kg.

2. Blackbody Radiation and Its Empirical Laws

A blackbody is a body that absorbs or emits all the radiation falling on it. Empirically,

the spectrum of light emitted by a hot dense object is shown by figure 2.1. The spectrum

contains a continuous range of frequencies. One of an ideal blackbody in our nature is sun.

Sun radiates energy to the earth over empty space, in the form of electromagnetic wave. The

form of energy is heat. All life on earth depends on the energy from the sun. The sun’s

temperature is much higher 6000o K than the earth’s, and is referred to as radiation.

The rate at which an object radiates energy has been found to be proportional to the

fourth power of the Kelvin temperature, T. The rate of energy transfers every area and time

unit, empirically, is shown in equation 2.1.

2.4

Here is a universal constant called the Stefan-Boltzman which has the value

. The factor e is called the emissivity, and is a is number

between 0 and 1 that is characteristic of the material. Very black surfaces have emissivity

close to 1, whereas shiny surfaces have e close to zero. Based on this fact, it can be concluded

that all object emit radiation whose total intensity is proportional to the fourth power of the

Kelvin temperature ( . At normal temperature, we are not aware of this electromagnetic

radiation because of its low intensity.

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Problem 2.1

An ideal blackbody radiates 548 J at room temperature (200C) in one day. How much energy

would it emit in a day at 10000C?

Solution:

The Stefan-Boltzmant law tells us that energy is proportional to T4. Since T is the absolute

temperature, the given T’s are (20+273) K=293 K, and (1000+273)K= 1273 K. Then

, and

So

The 6000o K curve in figure 2.1, corresponding to the temperature of the sun, peaks in

the visible part of the spectrum. For lower temperatures the total radiation drops considerably

and peak occurs at higher wavelengths. Hence, the blue of the visible spectrum and the UV is

relatively weaker.

Figure 2.1 Spectrum of frequencies emitted by a blackbody

at two different temperatures.

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It is found that the wavelength at the peak of the spectrum, , is related to the Kelvin

temperatures T by:

2.5

This is known as Wien’s displacement law. For the sun’s temperature it gives

, which is in the visible spectrum.

A mayor problem facing scientist in the 1980s was to explain blackbody radiation.

Maxwell’s electromagnetic theory had predicted that oscillating electric charges produce

electromagnetic waves, and the radiation emitted by a hot object could be due to the

oscillation of electric charges in the molecules of material. Although this theory could explain

where the radiation came from, however, it is incorrectly to predict the spectrum of emitted

light.

Problem 2.2

At what wavelength is the greatest amount of energy emitted at both room temperature and

10000C for the ideal blackbody?

Solution

T=293 K (room temperature)

T=1273 (10000C)

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3. Raleigh-Jeans Theory

Lord Rayleigh-Jeans tried to derive the expression of the characteristic of spectrum of

frequencies emitted by a blackbody. Suppose that the electromagnetic waves are in cubic

cavity of side L, in which no energy transfers for standing waves. The number of possible

oscillation between and +d per volume equals (82/c3) d. Classically, the energy for

each kind of oscillation equals kT, where k is Boltzman’s constant, that is, k= 1.3807x10 -23

J/K=8.61x10-5 eV/K. Therefore the energy per volume between and +d is

2.6

Based on Rayleigh-Jeans theory, the relation (2.3) only fits well at the very low frequency

(higher wave length), however in shorter wave length, the energy transfer is very high without

limit well-known as “the ultraviolet catastrophe”. It is very dangerous for our nature.

4. Max-Planck Theory

The break came in late 1900 when Max Planck proposed his two postulates to explain

spectrum of frequencies emitted by a blackbody. Planck’s Postulate

1. Energy of electromagnetic oscillators were quantized and could

have only allowed energies of , .; is the

oscillators energy. It is known Plank’s quantum postulate, and h is

Plank’s constancy valued h=6.623x10-34 J.s.

2. The average energy of an oscillator is

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Planck, explicitly, declared that energy of the molecular oscillators is thus quantized, then

when light is emitted by a molecular oscillator, its energy of nh must decrease by at least an

amount h to another integer times h. The to conserve energy, the light ought to be emitted

in packet or quanta, each with an energy E=h. If the number of possible oscillation between

and +d per volume equals (82/c3) d, so the energy of a blackbody per volume

between and +d is

2.7

The revolutionary thought of Planck on his postulate gave a final equation that fit the

experimental results. Another word, Max Planck’s quantization assumption did produce an

equation that agreed with the experimental result, as shown in figure 2.2. Consequently, it

would change our perception that light (wave) doesn’t merely show continuous phenomenon

in the point of view of classical physics, however, it is quantized. It exists only in discrete

amounts. The smallest amount light (wave) energy is called the quantum of energy (modern

physics). Max Planck’s quantization is regard as milestone of the quantum physics.

Figure 2.2 Comparison of the Rayleigh-Jeans and theories to that of empiric.

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Problem 2.3

Find the amount of energy per volume in the cavity at 2000K for light within the frequency

range of from 5.00x1014Hz to 5.04x1014Hz.

Solution

Since the frequency range is small compared to the actual frequency, we can replace by

and v by and still

arrive at a very good approximation. So, the equation 2.4 becomes:

=1.84x10-6J/m3

With h=6.625x10-34 J.s

c=3.00x108m/s

k=1.38x10-23J/K

T=2000K

TASK-3

1. If all object radiate energy, why can’t we see them in the dark?

2. Consider a point of source light vary with distance from the source according to (a) wave

theory, (b) particle (photon) theory? Would this help to distinguish the two theories?

3. Why do we say that light has wave properties? Why do we say that light has particle

properties?

4. Why do we say that electrons have wave properties? Why do we say that electrons have

particle properties?

47

5. (a) At what temperature will the peak of a blackbody spectrum be at 1.0 nm? (b) What is

the wavelength at the peak of a blackbody spectrum if the body is at temperature of 800K?

COURSE 4

I. Basic Competency:

Students are able to understand the theory of quantum and its implication in physics.

II. Indicators:

1. Able to explain the quantum property of wave.

2. Able to derive the Photoelectric effect of Einstein.

3. Able to derive the Compton Effect.

4. Able to derive pair production

5. Able to explain De Broglie Hypothesis to the wave of particle.

6. Able to explain the uncertainty of Heisenberg.

7. Able to derive Schrodinger equation to the wave-particle duality

III. Subject Matter

1. The Photoelectric Effect

In 1864, Maxwell predicted that oscillating electric currents would produce

electromagnetic waves that would move through a vacuum at the speed of light. In 1887,

Heinrich Hertz was attempting to confirm this prediction when he notices that a spark could

more easily be induced to jump a gap when the gap was illuminated. Continuing this research,

other workers showed that when light is incident on a metal surface, electrons are emitted

from the surface. This phenomenon is called the photoelectric effect.

48

Figure 2.3 Experimental arrangements for studying the photoelectric effect

As shown in figure 2.3, monochromatic light of one frequency of variable intensity

from source S shines on a clean metal surface in vacuum tube. Electrons, called

photoelectrons, are emitted from that electrode with some kinetic energy and travel to the top

electrode. This action produces a small current, which can be measured by a very sensitive

galvanometer, G. The battery is connected across a voltage divider circuit so that the potential

difference is measured by the voltmeter, V.

When the voltage is zero, the photoelectrons can easily travel to the collecting

electrode. However, giving the collecting electrode a negative charge will set up an electric

field to oppose the motion of the photoelectron. In moving against this field, the

photoelectrons do work. The total work done is equal to the charge, -e, times potential

difference. This work comes from the kinetic energy of the photoelectrons. As the retarding

potential difference is increased, fewer and fewer electrons will have sufficient kinetic energy

to reach the top electrode. Therefore the current decreases as the retarding potential difference

increase. The current will drop to zero when the work done, (-e)(-Vo), equals the maximum

kinetic energy of the photoelectrons, Ekmaxs. Therefore

2.8

49

Where Vo is the absolute value of the potential difference measured by the voltmeter when

the current drops to zero (if both electrodes are made of the same material) and is called the

stopping potential.

Classical physics runs into many difficulties when trying to explain various result of

photoelectric experiment, such as:

1. For a constant light frequency,, Vo is independent of the light intensity, I, as shown

in figure 2.4. Classical physics would suggest that light of greater intensity (energy per

area per time) should give electrons greater kinetic energy, so that Ekmax and therefore

Vo increase with I.

Figure 2.4 For a constant light frequency, the stopping potential is

independence of the intensity.

2. Even at extremely low intensity, photoemission occurs with essentially no delay as

soon as the light source is turned on. Classical physics would say that energy is spread

over the entire wave-front. Thus a particular electron would have to gather energy in

from a certain rather small surrounding area until the electron finally had enough

energy to break free from the surface. For extremely weal light, this occurrence could

take days in the classical point of view

3. For a given frequency,, and retarding potential, V, the photocurrent measured by G is

directly proportional to the light intensity. Since the power output is proportional to

50

the photocurrent and the power input is proportional to the intensity, classical physics

will agree with this result.

4. The maximum kinetic energy of photoelectrons is linearly related to the frequency of

the light. Below a certain threshold frequency, no photoelectrons will be emitted.

Classical physics has no explanation for this behavior, as shown in figure 2.5.

Figure 2.5. The stopping potential and the maximum kinetic energy of the

photoelectrons increase linearly with the light frequency

In 1905 , Einstein made a bold extension of the quantum idea by proposing a new

theory of light. Plank’s work suggested that the vibrational energy molecules in a radiating

object is quantized with energy E=nh. Einstein reasoned that if the energy of the molecular

51

oscillators is thus quantized, then the light is emitted by a molecular oscillator, its energy must

decrease by at least an amount h.

2.9

Since all light ultimately comes from a radiating source, this suggests that light is transmitted

as tiny particles, or photons, rather than as wave, as shown in figure 2.6.

Figure 2.6 For an electromagnetic harmonic oscillator, (a) classical physics said that can have a continuous range

of possible values; (b) Planck said that E=nh, so that E can have only certain allowed values, and

(c) Einstein said that photons with E=h are emitted in transition between adjacent energy levels.

Problem 2.4

Calculate the energy of a photon of blue light, .

Solution

Since , we have

Einstein suggested that the photoelectric effect occurred when a photon gave up all its

energy, h, to an electron near the surface of the metal. The electron would then have to do

52

some work in overcoming the forces binding it to the surface. Any remaining energy would

be the electron’s kinetic energy when free of the surface. The maximum kinetic energy of a

photoelectron thus becomes

2.10

The term is called the work function. It is the minimum amount of work that must be done

to free an electron from the surface. This quantum idea of Einstein then explains the result of

the photoelectric effect experiments:

1. At constant , Ekmax and therefore Vo don’t depend on the intensity of the light. That

is, they don’t depend on how many photons per time per area are arriving.

2. As soon as a photon reaches the surface, it can be absorbed and an electron emitted.

3. The intensity is directly proportional to the number of photons per time, as it the

number of electrons emitted per time. Therefore the photocurrent is directly

proportional to the intensity.

Problem 2.5

What is the maximum kinetic energy (Ek) and speed of an electron ejected from a sodium

surface whose work function is W0 = 2.28 eV when illuminated by light of wavelength (a) 410

nm; (b) 550 nm?

Solution:

(a) For , or 3.03 eV

Ek max=3.03 eV-2.28 eV=0.75 eV or 1.2 x 10-19 J. Since Ek max=1/2 mv2 where

m=9.1 x10-31 kg,

=5.1x105 m/s

(b) For h=3.60 x10-19J=2.25 eV. Since this photon energy is less than the work

function, no electron are rejected.

53

2. The Compton Effect

A number of other experiments were carried out in the early twentieth century which

also supported the photon theory. One of these was the Compton effect (1923). Compton

scattered short-wavelength light (X-ray) from various of material. He found the scattered light

had slightly lower frequency that did the accident light, indicating a loss energy. This

experience could only be explained on the basis of the photon theory as incident photons

colliding with electrons of the material. He applied the laws of conservation of energy and

momentum to such collision and found that the predicted energies of scattered photons were

in accord with experimental results.

Figure 2. Compton effect

In order to analyze the Compton Effect, firstly, consider that the photons is truly a relativistic

particle due to its speed. Thus, we must use relativistic formulas for dealing with its mass,

energy, and momentum. The mass m of any particle is given by . Since v=c

for a photon, the denominator is zero. So the rest mass, mo, of a photon must also be zero, or

54

its energy E=mc2 would be infinite. Of course a photon is never at rest. The momentum of a

photon with mo=o, is

2.11

Since , the momentum of a photon is related to its wavelength by

2.12

Problem 2.6

An x-ray photons has a wavelength of 100 pm. Calculate its momentum in SI and eV/c units.

Solution

We can then convert the result directly into eV/c units. Another way to

solve the problem would be to first find the frequency:

Then

If the incoming photon in figure 2.7 has wavelength , then its total energy and momentum

are

and

After the collision, the photon scattered at the angle has a wavelength which we call ’. Its

energy and momentum are

55

and

The electron, assumed at rest before the collision, is scattered at angle as shown. Its kinetic

energy is

Where mo is the rest mass of the electron and v is speed. The electron’s momentum is

We apply conservation of energy to the collision:

And we apply conservation of momentum to the x and y components of momentum:

(The x component of momentum)

(The y component of momentum)

We can combine these three equation to eliminate v and . Thus, the result for the wavelength

of the scattered photon in terms of its scattering angle :

2.13

For =0, the wavelength is unchanged (there is no collision for this case of the photons

passing straight through). At any other angle, ’ is longer than . The difference in

wavelength is

56

2.14

Equation 2.10 is called the Compton shift (The quantity , which has the dimensions of

length, is called the Compton wavelength of particle whose rest mass is mo). For free electron,

. The wave theory of light predicts no such shift: an

incoming EM wave of frequency should set electrons into oscillation at frequency , and

such oscillating electrons should reemit EM waves of the same frequency .

Problem 2.7

For 100.0-pm x-rays are scattered from a metal. What are the changes in the wavelength and

scattered wavelengths for scattering angles of 10o, 900, and 1800?

Solution

is the change in the wavelength.

At 100:

At 900:

At 1800:

3. Pair Production

When a photon passes through matter, it interacts with the atoms and their electrons.

There are four important types of interactions that a photon can undergo. First, the photon can

be scattered of an electron (or nucleus) and in the process lose some energy; this is the

Compton effect. But notice that the photon is not slowed down. It still travels with speed c, but

its frequency will be less. A second type of interaction is the photoelectric effect; a photon

may knock an electron out of an atom and in the process itself disappear. The third process is

similar: the photon may knock an atomic electron to a higher-energy state in the atom if its

57

energy is not sufficient to knock it out altogether. In this process the photon also disappears,

and all its energy is given to the atom. Such an atom is then said to be in an exited state.

Finally, a photon can actually create matter. The most common process is the production of an

electron and a positron. (A positron has the same mass as an electron, but the opposite charge,

+e and is called the antiparticle to the electron). This is called pair production and the photon

disappears in the process, as shown in figure 3.

Figure 3. Pair production: a photon disappears and produces an electron

and a positron

Problem 2.8

What is the minimum energy of a photon, and its wavelength, that can produce an electron-

positron pair?

Solution

Because , the photon must have energy E=2(9.1x10-31kg)(3.0x108m/s)2=1.64x10-23J or

1.02 MeV. A photon with less energy cannot undergo pair production. Since ,

the wavelength of a 1.02 MeV photon is

58

Which is 0.0012 nm. Thus the wavelength must be very short. Such photon are in the gamma-

ray ( or very short X-ray) region of the electromagnetic spectrum.

The photoelectric effect, the Compton Effect, and pair production have placed the

particle theory of light on a firm experimental basis. The other side, the classical experiments

of Young and others on interference and diffraction which showed that the wave theory of

light also rest on affirm experimental basis. Some experiments indicate that light behaves like

a wave, and others indicate that it behaves like a stream particles. These two theories seem to

be incompatible but both have been shown to have validity. Finally, physicist come to the

conclusion that light referred to as the wave-particle duality. The wave description and the

particle description of light are complimentary, called the principle of complementarity as

proposed by Niels Bohr in 1928. It is used to explain the nature phenomenon of light

separately.

4. De Broglie’s Wave-Nature of Matter Hypothesis

In 1923h , Louis de Broglie extended the idea of the wave-particle duality. He sensed

deeply the symmetry in nature and argued that if light sometimes behaves like a wave and

sometimes like a particle, then perhaps those things in nature thought to be particles, such as

electrons and other material objects, might also have wave properties. De Broglie proposed

that the wavelength of a particle would be related to its momentum in the same way as for a

photon. That is, for particle of mass m traveling with speed v, the wavelength is given by

2.15

59

The equation 2.11 is sometimes called the de Broglie wavelength of a particle.

Problem 2.9

Determine the wavelength of an electron that has been accelerated through a potential

difference of 100 V.

Solution

We assume that the speed of the electron will be much less than c so we use non-relativistic

mechanics.

Then

In order to verify the de Broglie’s wave nature of matter hypothesis, C.J. Davisson and

L.H. Germer performed the crucial experiment; they scattered electrons from the surface of a

metal crystal and, observed that the electrons came off in regular peaks. When they

interpreted these peaks as a diffraction pattern, the wavelength of the diffracted electron wave

was found to be just that predicted by de Broglie.

60

Figure 2.9 Diffraction pattern of electron scattered from aluminum foil, as

recorded on film.

5. Heisenberg Uncertainty Principle

Whenever a measurement is made, some uncertainty or error is always involved. For

example, we can not make an absolute exact measurement of the length of table, how good

the measuring device. But, at the point view of the duality of wave-particle, the uncertainty

appeal naturally due to the unavoidable interaction between the system observed and the

observing instrument. If we use light of wavelength , the position can be measured at best to

an accuracy of about . That is, the uncertainty in the position measurement, x is

approximately

Supposed that the object can be detected by a single photon. The photon has a momentum

, and when it strikes our object it will give some or all of this momentum to the object.

Therefore, the final momentum of our object will be uncertainty in the amount

Since we can’t tell beforehand how much momentum will be transferred? The product of

these uncertainty

Of course, the uncertainty could be worse than this, depending on the apparatus and the

number of photons needed for detection. In Heisenberg’s more careful calculation, he found

that at the very best

2.16

61

This is a mathematical statement of Heisenberg’s uncertainty principle. It tell us that we can

not simultaneously measure both the position and momentum of an object precisely. The

more accurately we try to measure the position, so that x is small, the greater will be the

uncertainty in momentum, P. And vice versa, if we try to measure the momentum very

precisely, then the uncertainty in the position becomes large.

Problem 2.10

An electron moves in a straight line with a constant speed=1.10x106 m/s which has been

measured to a precision of 0.1 percent. What is the maximum precision with which its

position could be simultaneously measure?

Solution

The momentum of the electron is

. The uncertainty in the momentum is 0.1 percent of this, or

From the uncertainty principle, the best simultaneous position measurement will have an

uncertainty of

=

or about 100 nm. This about 1000 times the diameter of an atom.

Another useful form of uncertainty principle relates energy and time. The object to be

detected has an uncertainty in position . Now the photon used to detect it travels with

62

speed c, and it takes a time to pass through the distance of uncertainty. Hence, the

measured time when our object is at given position is uncertain by about

Since the photon can transfer some or all of its energy ( ) to our object, the

uncertainty in energy of our object as a result is

The product of these two uncertainties is

Heisenberg’s more careful calculation gives

2.17

This form of the uncertainty principle tell us that the energy of an object can be uncertain, or

may even be non-conserved, by an amount for a time .

Problem 2.11

The lifetime of a typical atomic state is 10-8s. What is the uncertainty in the energy of such a

state?

Solution

Therefore the spread in energy of the state is at least 4x10-7eV.

6. The Schrodinger’s Equation

63

As we know that the important properties of any wave are its wavelength (),

frequency(), and amplitude (A) or displacement (y) at any point. The other side, the

important properties of any particle is its position (x), momentum (P) and mass (m).

Generally, we can analyze the dynamic of wave from its displacement by using Maxwell

theory, and the dynamic of particle from its position by using Newton laws. Because of the

wave-particle duality, early, there is no theory can derive the dynamic of system whose has

duality property, neither Newton Laws nor Maxwell theory.

It is de Broglie dared to reveal idea that particle characterized by momentum also

brought wave property characterized by wavelength, .

Although, de Broglie could able to predict the wavelength () of particle, but he cannot derive

the wave of particle dynamically. Physicist, then, expressed that the state of wave-particle

duality, such as electron, can be represented by the wave function ().

The wave function () depends on time (t) and position (x), and the

has physics meaningless, however, in order that it be physically meaningful, Max Born stated

that at certain point in space and time represent the probability of finding the

particle within volume dV about the given position at that time. Thus, is often

referred to as probability density or probability distribution, since it is the probability of

finding the particle per unit volume. Then, the sum of the probabilities over all space-that is,

the probability of finding the particle at one point or another-becomes

2.18

This is called the normalization condition, and the integral is taken over whatever region of

space in which the particle has a chance of being found, which is often all of space, from

to .

64

De Broglie hypothesa gave a basic affect in deal with microscopic physis system. For

a particle of mass m and velocity v, the de Broglie wavelength is , where P=mv is the

particle’s momentum. Hence

2.19

Generally, in one dimension, the equation of wave can be formulated

2.20

The solution of the wave equation above is

2.21

Where and 2.22

Equation 2.18 substitute to 2.17 yields

2.23

The dynamic of particle can be analyzed from its conservation of energy, it’s e expressed by

If both side multiplied by , we get

65

2.24

It can be proved that:

and 2.25

Substituted 2.21 to 2.20, it yields

2.26

It is Time Dependence non-relativistic Schrodinger’s equation, regarded as wave equation

for particle, in which figures the quantum state of particle from where all

information about the particle can be accessed from the quantum state. The wave function

is a complex function like , . may have both real and imaginary

part, so is not an observable quantity.

Problem 2.12

If , show that the probability of finding the particle in an infinitesimal

volume is independent of time.

Solution

The probability desired is

66

Since , , and are independent of time, the probability desired independent of time.

In many physics problem, generally the function of U (potential energy) is only depend on

position, . Hence, it is possible to write the wave function as a product of separate

function of space and time.

2.27

Substituting 2.23 into the time-independent Schrodinger equation,

We divide both sides of this equation by and obtain an equation that involves only x

on one side and only t on the other:

This separation of variable is very convenient. Since the left side is a function only of x, and

the right side is a function only of t, the quality can be valid for all values of x and all values

of t only if each side is equal to a constant, which we call E, so

Then,

2.28

The other side:

67

2.29

Thus, the total wave function is

2.30

Problem 2.13

Show that is a correct wave function for a free particle (V=0) in one dimension,

where .

Solution

Substituting the given into yields . The expression

is correct because , leaving

Now so

68

Since , so that

Recall that and , then we have . But , so

, which is the correct non-relativistic relation.

TASK-4

1. Emission of electrons from a given surface is 380 nm. What will be the maximum kinetic

energy of ejested electrons when the wavelength is changed to (a) 480 nm, (b) 280 nm?

2. X-rays of wavelength nm are scattered from a carbon block. What is the Compton

wavelength shift for photons detected at angles (relative to the incident beam) of (a) 450,

(b) 900, and (c) 1800?

3. In what ways is Newtonian mechanics contradicted be quantum mechanic?

4. If an electron position can be measured to an accurate of 1.6 x 10-8m, how accurately can

its velocity be known?

5. An electron with Ek=100 eV in free space passes over a potential well 50 eV deep that

stretches from x=0 to x= 5.0 nm. What is the electron’s wavelength (a) in free space, (b)

when over the well? (c) Draw a diagram showing the potential energy and total energy as

a function of x, and on the diagram sketch a possible wave function!

COURSE 5

I. Basic competency:

Able to understand the theory models of atom and its implication in Physics

II. Indicators:

69

1. Able to explain the J.J. Thomson models of the atom

2. Able to explain atomic spectra of the atom

3. Able to explain the Rutherford models of the atom

III. Subject matter

1. Early Models of the Atom

The idea that matter is made up atoms was accepted by most scientist by 1900. J. J.

Thomson visualized the atom as a homogeneous sphere of positive charge inside of which

there were the negatively charged electrons, a little like plums in a pudding, soon after his

discovery of the electron in 1897, argued that the electron in this model should be moving.

The Thomson model of atom is shown in figure 3.1.

Figure 3.1 Thomson’s model of the atom

Ernest Rutherford (1937) performed experiment whose result contradicted Thomson’s

model of the atom. In these experiments a beam of positively charge “alpha () particle” was

directed at a thin sheet of metal foil such as gold. It was expected from Thomson’s model that

alpha particle would not be deflected significantly since electron are so much lighter than

alpha particle. The experimental result completely contradicted these predictions. It was found

that most of the alpha particles passed through the foil unaffected, as if the foil were mostly

70

empty space. A few were deflected at very large angles, some evening nearly the direction

from which they had come.

Figure 3.2 Rutherford’s Experiments

Rutherford reasoned, only if the positive charged alpha particles were being repelled

by a massive positive charge concentrated in a very small region of space. Then, Rutherford

theories that the atom must consist of a tiny but massive positively charged nucleus,

containing over 99.9% of the mass of atom, surrounded by electrons some distance away. The

electrons would be moving in orbits about the nucleus, much like the planets move around the

sun, because if they were at rest they would fall into the nucleus due to electrical attraction.

The motion of electron encircle the nucleus can be derived from

Newton’s law. The attractive Coulomb force (Fe) between electron and nucleus works as

centripetal force (Fc) that keeps electron on the track.

3.1

and 3.2

The total energy of electron can be derived as below:

71

= +

= 3.3

where

The Rutherford’s model of the atom fails, theoretically, to explain the stability of atom.

According to Maxwell theory, electron orbits the nucleus with accelerating, consequently,

electron radiate energy continuously. Consequently, because of losing energy, electron will be

predicted to be captured by nucleus. However, it is never happen. The other side, empirically,

The Rutherford’s model can not explain the line of atomic spectra.

Figure 3.3 Rutherford’s Model of the Atom

2. Atomic Spectra

Empirically, if gas heated, it emits light. The radiation is assumed to be due to

oscillation of atoms. The radiation from exited gases had been observed, and it was found that

the spectrum was not continuous, but discrete in the form of line spectrum. The line spectrum

72

can be used to study the structure of the atom. Any theory of atomic structure should able to

explain why atom emits light only of discrete wavelength.

Figure 3.4 Gas discharge tube

Hydrogen is the simplest atom worthy to be studied. It has only one electron orbiting its

nucleus. J.J. Balmer (1885) shown that the four visible lines in the hydrogen spectrum that

would fit the following formula

…. 3.1

R is called the Rydberg constant whose value R=1.097 x 107 m-1. Later it was found that this

Balmer series of line extended into the UV region, as shown figure 3.5

73

Figure 3.5 Balmer series of lines for hydrogen

Later experiment on hydrogen showed that there were other series of lines in the UV and IR

and these had a pattern just like the Balmer series, but at different wavelengths, namely

1. Lyman series: n=2, 3, ………

2. Balmer series: n=3, 4,……….

3. Paschen series: n=4, 5, ……….

4. Bracket series : n=5,……….

5. Pfund series : n=

Problem 2.1

What is the energy of the ultraviolet photon absorbed by the atom for which the values of n i

and nf are 1 and 5, respectively?

Solution

COURSE 6

I. Basic competency:

Able to understand the theory models of atom and its implication in Physics

74

II. Indicators:

1. Able to explain the Bohr model of hydrogen atom

2. Able to derive de Broglie’s Hypothesis of hydrogen atom

3. Able to analyze hydrogen atom in the view of quantum mechanic

III. Subject matter1

1. The Bohr Model

Bohr had studied deeply the Rutherford’s model, and believe the planetary model

atom of Rutherford is still valid. But in order to make it work, Bohr proposed two postulate to

mend the Rutherford’s model of the atom, namely

1. Electrons move about the nucleus in circular orbit, but that only certain

orbits are allowed called stationery orbit without radiating energy. Light is emitted or

absorbed when an electron jumps from one stationer state to another. The amount of

radiating/absorbing energy is

3.2

Where refers to the energy of the upper state, and El the energy of the lower state.

2. The angular momentum of electron moving in circular r with speed v is

quantized by . n is called the quantum number

Mathematically,

3.3

75

Figure 3.6 Electric force (Coulomb law) keeps electron in orbit

Here n is an integer and is the radius of the nth possible orbit.

From Newton’s second law, , and , then Coulomb’s law for F can be derived:

the other side

We solve for and find:

3.4

The smallest orbit has n=1 and for Hydrogen (Z=1) it has the value

r1 is sometime called the Bohr radius.

Problem 3.2

Calculate the attractive force and the electron’s speed in the smallest orbit of the Bohr

hydrogen atom.

Solution

is the smallest orbit of Bohr hydrogen.

Since Z=1 for hydrogen, r=52.9 pm, then

The speed is

76

The total energy En for an electron in the nth orbit of radius rn is the sum of its kinetic and

potential energies:

n=1, 2, 3, … 3.5

For hydrogen (Z=1) the lowest energy level (n=1), is:

The lowest energy level or energy state, E1 is called the ground state, and the higher states E2,

E3,….. is called excited state.

Figure 3.7 Possible orbits in the Bohr model of hydrogen

Problem 3.3

Find the principal quantum number, the total energy, the binding energy, and the excitation

energy of the fifth excited state of hydrogen.

77

Solution

The fifth excited state is the fifth state above n=1, so its principal quantum number is 1+5=6.

The total energy to be . Then the binding energy equals

. It is needed minimum 0.38 eV to separated completely the

electron and proton. Since the ground state energy of hydrogen is -13.6 eV, the excitation

energy is the difference between the excited state energy and the ground state energy: -0.38

eV-(-13.60 eV)=13.22 eV. In other words, it is needed 13.22 eV to a hydrogen atom in its

ground state to excite it to its fifth excited state.

At room temperature , nearly all H atom will be in ground state. At higher

temperatures, or during an electric discharge when there are many collisions between free

electrons and atoms, many electrons can be in excited states. Once in an excited state, an

electron can jump down to a lower state, and give off a photon (light) in the process, as shown

in figure 3.6. The wavelength of the spectral lines emitted in the transition process is:

or

3.6

Where n refers to the upper state and n’ to the lower state. This theoretical formula has the

same form as the experimental Balmer formula. The strength of Bohr’s model is (1) it gives a

model for why atoms emit line spectra, and 2) it ensures the stability of the atom.

78

Figure 3.8. Energy-level diagram for the hydrogen atom

Problem 3.4

Find the energy, frequency, and wavelength of the photon emitted when the electron

transitions from the second excited state in hydrogen to the first excited state!

Solution

Using Z=1, ni=3, and nf=2 in

and

Therefore . The energy of the emitted

photon is , its frequency is

Then the wavelength is

2. De Broglie’s Hypothesis

79

Bohr’s theory was largely of an hoc nature, and could give no reason why the orbits

were quantized. Louis de Broglie proposed that electron have a wave nature. According to de

Broglie, a particle of mass m moving with speed v would have a wavelength of

De Broglie argued that each electron orbit in atom, actually, is a standing wave. Electron

wave must be a circular standing wave that closes on itself. If the wavelengths of a wave does

not close on itself, destructive interference takes place as the wave travels around the loop and

it quickly dies out. Thus, the only waves that persist are those for which the circumference of

the circular orbit contains a whole number of wavelengths. The circumference of a Bohr orbit

of radius rn is , so

n=1, 2, 3, ….. 3.7

Substitute , we get

This is just the quantum condition proposed by Bohr on ad hoc basis, but can be explained

elegantly by the wave-particle duality of de Broglie. Bohr’s theory works well for hydrogen

and for one-electron ions. It did not prove as successful for multi-electron atoms. It is needed

a new theory to give a whole comprehension for multi-electron atoms.

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Figure 3.8. De Broglie wavelength

3. Hydrogen Atom: in the view of Quantum Mechanic

Quantum mechanic is a new theory that can unified the wave-particle duality into a

single consistent theory. Quantum mechanic, supported by Schrodinger’s equation, retains

certain aspects of the Bohr Theory. The hydrogen atom is the simplest of all atoms, consisting

of a single electron of charge –e moving around a central nucleus of charge +e. For the

hydrogen, the potential energy is due to the Colomb force between electron and proton.

The time-independence Schrodinger equation, which must now be written in three

dimensions, is then

3.8

In the three-dimensional problem of the H atom, the solutions of the Schrodinger equation are

characterized by three quantum numbers. However, four different quantum numbers are

actually needed to specify each state in the H atom, the fourth coming from a relativistic

treatment.

Quantum mechanics predicts exactly the same energy levels for the H atom as does the

Bohr Theory. That is

n=1,2,3…… 3.9

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Where, n is an integer called the principal quantum number. The total energy of a state in

the H atom depends on n.

The orbital quantum number, l, is related to the orbital angular momentum of the

electron. l can take on integer values from 0 to (n-1). For the ground state, n=1, l can only be

zero. But for =3, l can be 0,1, or 2. The actual magnitude of the orbital angular momentum L

is related to the quantum number l by

3.10

For l=0 s state (sharp); l=1 p state (principal); l=2 d state (diffuse); l=3 f state (fundamental).

The value of l does not affect the total energy of the atom H, only n does to any appreciable

extent. But in atoms with two or more electrons, the energy does depend on l as well as n.

The magnetic quantum number, ml, is related to the direction of the electron’s angular

momentum, and it can take on integer values ranging from –l to +l. For example, if l=2, then

ml can be -2,-1,0,+1,+2. Since angular momentum is a vector, it is not surprising that both its

magnitude and its direction would be quantized. For l=2, the five directions allowed can be

presented by the figure 3.8

Figure 3.8. Quantization of angular momentum direction for l=2

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This limitation on the direction of L is often called space quantization. In quantum mechanics

the direction of the angular momentum is usually specified by giving its component along the

z axis. The Lz is related to ml by the equation

3.11

It was found that when a gas discharge tube was placed in a magnetic field, the spectral lines

were split into several very closely spaced lines, a phenomenon known as the Zeeman effect,

as shown in figure 3.9. This implies that energy levels must be slit, and thus that the energy

of a state depends not only on n but also on ml when magnetic field is applied.

Figure 3.9 . Transition can occur between levels n=3;l=2 to levels n=2;l=1, with photons of

several slightly different frequency being given off (the Zeeman effect)

Finally, there is the spin quantum number, ms, which can have only two values,

or . The existence of this quantum number did not come out of

Schrodinger’s origin theory, as did n, l, and ml. Instead, it came out of a subsequent

relativistic treatment due to P.A.M Diract. A careful study of the spectral lines of hydrogen

showed that each actually consisted of two (or more) very closely spaced lines even in the

absence of an external magnetic field. It was at first hypothesized that this tiny splitting of

energy levels, called fine structure, might be due to angular momentum associated with a

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spinning of the electron. The interaction between the tiny current of the spinning electron and

the magnetic due to the orbiting charge could the cause the small observed splitting of energy

levels. The electron is said to have a spin s=1/2, which produces a spin angular momentum S

given by

3.12

This spin can have two different directions, mz=1/2 or mz=-1/2, which are often said to be spin

“up” and “spin down”, as shown in figure 3.10.

Figure 3.10. The spin angular momentum S can take on only two directions,

ms=1/2 (spin up) and ms=-1/2 (spin down).

Problem 3.5

Determine (a) the energy and (b) the orbital angular momentum for each of the states n=3

Solution

(a) The energy of a state depends only on n except for the very small corrections from spin

magnetic of electron. Since n=3 for all these states have the same energy,

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(b) For l=0,

For l=1,

For l=2,

TASK-6

1. For very large values of n, show that when an electron jumps from the level n to the

leveln-1, the frequency of the light emitted is equal to

2. Calculate the orbital angular momentum, the allowed z components of the orbital angular

momentum, and the allowed angles with the z axis for an electron in a d state (l=2)!

3. The magnitude of the orbital magnetic dipole moment of an electron in a hydrogen like

atom is 1.3115x10-23 A. m2. What state is the electron in?

4. Show that for a particle of rest mass m0, if then it cannot be true that

where E is (a) kinetic energy, or (b) Ek plus rest mass energy, and is the speed of the

particle.

COURSE 7:

I. Basic competency:

Able to understand nuclear physics, radioactivity and its application

II. Indicators:

1. Able to explain structure and properties of the nucleus

2. Able to explain alpha, beta, gamma decay

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3. Able to explain half-life and rate of decay

III. Subject matter

4.1 Structure and Properties of the Nucleus

An important question to physicist in the early nineteen century was whether the

nucleus had a structure, and what structure might be. In 1930 a model of nucleus had been

developed. According to the model, a nucleus is considered as an aggravated of two types of

particles: protons and neutrons. A proton is the number of the simplest atom, hydrogen. It has

a positive charge of +e =+1.60x10-19 C and a mass mp=1.6726x10-27 kg. The neutron, whose

existence was ascertained only in 1932 by Chadwick, is electrically neutral (q=0). Its mass,

which is almost identical to that of the proton, is mn=1.6749x10-27 kg. These two constituent

of a nucleus, neutron and proton, are referred to the collectively as nucleus.

The number of protons in a nucleus is called the atomic number and is designed by

the symbol Z. The total number of nucleons, neutron plus protons, is designed by the symbol

A and is called the atomic mass number. The neutron number N is N=A-N. To specify a

given nuclide, we need only A and Z. A special symbol is commonly used which takes the

form

Where, X is the chemical symbol for the element. A is the atomic mass number, and Z is the

atomic number.

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Figure 4.1 Number of neutron and proton for stable nuclides

For a particular type of atom (say carbon), nuclei are found to contain different

number of neutrons, although they all have the same number of protons. Nuclei that contain

the same number of protons but different numbers of neutrons are call isotopes. Thus ,

, , , , and are all isotope of carbon. The approximate size of nuclei was

determined originally by Rutherford from the scattering of charge particles. It is found that

nuclei have a roughly spherical shape, with a radius that increase with A according to the

approximate formula

4.1

Since the volume of a sphere is , so that the volume is proportional to the number of

nucleons,

Nuclear masses are specified in unified atomic mass units (u). A neuron has a

measured mass of 1.008665 u, a proton 1.007276 u, and a neutral hydrogen atom,

1.007825 u. Beside that, masses are often specified using the electron-volt energy unit. This

can be done because mass and energy are related, and the precise relationship is given by

Einstein’s equation .

The result is

4.2

Just as an electron has an intrinsic spin and angular momentum, so too do nuclei and

their constituents, the proton and neutron. Both the proton and the neutron are spin-1/2

particles. A nucleus, made up of protons and neutrons, has a nuclear spin, I, that can be either

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integer or half integer, depending on whether it is made up of an even or an odd number of

nucleons. The nuclear angular momentum of a nucleus is given by . Nuclear

magnetic moments are measured in terms of the nuclear magneton.

4.3

Then,

4.2 Binding Energy and Nucleus Forces

The total of a nucleus is always less than the sum of the masses of its constituent

protons and neutrons. As we know that the mass of a neutral is 4.002603. The mass of

two neutrons and two protons is

-------------------------

Thus, the mass of is measured to be 4.032980 u – 4.002603 u =0.030377 u less than the

masses of its constituents. How can this be? Where has this mass gone? The loss mass has

gone into energy of another kind (binding energy). The mass difference in the case of , is

(0.030377 u)(931.5 MeV/u)=28.30 MeV. This difference is referred to as the total binding

energy of the nucleus. The binding energy represents the amount of energy that must be put

into a nucleus in order to break it apart into its constituent protons and neutrons. The average

binding energy per nucleon is defined as the total binding energy of a nucleus divided by A,

the total number of nucleon. For it is 28.3 MeV/4=7.1 MeV.

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Nuclei can hold together in nucleus is not only merely explained from the point of

view of energy binding, but also from the point of view of the forces that hold them together.

It is impossible a collection of protons and neutrons to come together spontaneously, since

protons are all positively charged and thus exert repulsive forces on each other. Indeed, the

question arises as to how a nucleus stays together at all in view of the fact that electrostatic

repulsion between protons would tend to break a part. Since stable nuclei do stay together, it

is clear that another force must be acting, and it is stronger than the electric force. It is called

the strong nuclear force. The strong nuclear force is an attractive force that act between all

nucleons-protons and neutron alike. The strong nuclear force is a short range force. Compare

this to electric and gravitational forces, which can act over great distances and are therefore

called long-range forces.

What we mean by a stable nucleus is one that stays together indefinitely. What then is

unstable nucleus? It is one that comes apart, and this result in radioactive decay. The subject

of radioactivity can not be separated with the weak nuclear force.

4.3 Alpha, Beta, and Gamma Decay

In related to radioactivity, Becquerel made an important discovery, in his studies of

phosphorescence, he found that a certain mineral (which happened to contain uranium) would

darken a photographic plate was wrapped to exclude light. It was clear that the mineral

emitted some new kind of radiation which, unlike X rays, occurred without any external

stimulus. This new phenomena came to be called radioactivity. Then deeply studied was done

by Marie Curie and her husband, isolated polonium and radium, other radioactive elements

were soon discovered as well. The source of radioactivity must be deep within the atom that it

must emanate from the nucleus. The radioactivity is the result of the disintegration or decay,

with the emission of some type of radiation. Rutherford began studying the nature of the rays

89

emitted in radioactivity. He concluded that there are three types of radiation named alpha (),

beta (), and gamma (), as shown in figure 4.1

When a nucleus emits an particle ( ), alpha decay, it is clear that the remaining

nucleus will be different from the original, for it has lost two protons and two neutrons. When

alpha decay occurs, a new element is formed, for example:

The daughter nucleus ( ) is different from the parent nucleus ( ). This

changing of one element into another is called transmutation of the element.

Figure 4.2 Alpha, beta, and gamma decay

(a) Alpha Decay

Alpha decay occurs because the strong nuclear force is unable to hold very large

together. Because the nuclear force is a short-range force, it acts only between neighboring

nucleons. But the electric force can act clear across the nucleus. For very large nuclei, the

large Z means the repulsive becomes very large. And, it acts between all protons. The

90

instability nuclei can be stated in energy (or mass). The mass of the parent nucleus is greater

than the mass of the daughter nucleus plus the mass of the particle. The mass difference as

kinetic energy carried away mainly by the particle. Te total energy is called the

disintegration energy (Q) or the Q-value defined:

4.4

Where , , are the masses of the parent nucleus, daughter nucleus, and particle. If

the parents have less that the daughter plus the particle the decay could not occur, for the

conservation of energy law would be violated.

Problem 4.1

Calculate the disintegration energy when a nucleus (mass=232.03714 u) decays to

(228.02873 u) with the emission of an particle.

Solution

Since the mass of a nucleus 4.002603 u, the total mass in the final state is 228.02873 u +

4.002603 u=232.03133 u. The mass lost when the nucleus decays is 232.03714-

232.03133 u. This mass appears as kinetic energy. Since 1 u=931.5 MeV, the Ek released (Q)

is (0.00581 u)(931.5 MeV/u) 5.4 MeV

We can understand decay using a model of a nucleus inside of which there is an alpha

particle bouncing around. The potential energy “ seen” by the particle would have a shape

something like that shown in figure 4.2.

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Figure 4.3 Model of Alpha decay

The approximately square well between r=0 and r=R represent the short-range attractive

nuclear force. Beyond the nuclear radius, Ro, the Coulomb repulsion dominates. Since the

potential energy just beyond r=Ro is greater than the energy of the particle, so the particle

could not escape the nucleus according to classical physics. But quantum mechanic allows

that there is a certain probability that the particle can tunnel through the Coulomb barrier,

from point A and point B.

(b) Beta decay

Transmutation of elements also occurs when a nucleus decays by decay-that is, with

the emission of an electron or particle. The nucleus , for example, emits an electron

when it decays:

No nucleons are lost when an electron is emitted, and the total number of nucleons, A, is the

same in the daughter has in the parent. But because an electron has been emitted, the charge

of the daughter is different from the parent. The electron decay is not an orbital electron.

Instead, the electron is created within the nucleus itself. Indeed, free neutrons actually do

decay in this fashion:

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Problem 4.2

How much energy is released when decays to by emission.

Solution

Assume the parent nucleus has six orbiting electrons so it is neutral and its mass is 14.003242

u. The daughter in this decay is not neutral, however, since it has the same six electrons

circling it, but the nucleus has a charge of +7e. However, the mass of this daughter with its

six electrons, plus the mass of the emitted electron is jus the mass of a neutral nitrogen atom.

That is, the mass in final state is

(mass of nucleus + 6 electrons)+ (mass of 1 electron)

= mass of neutral (includes 7 electrons)

=14.003074 u

Hence the mass before decay is 14.003242 and after decay is 14.003074 u, so the mass

difference is 0.000168 u, which corresponds to 0.156 MeV or 156 keV.

Although the energy conserves in decay, however, careful experiment also indicated

that, linier momentum and angular momentum too did not seen to be conserved. The trouble

can be resolve after Wolfgang Pauli (1930) proposed an alternative solution: perhaps a new

particle that was very difficult to detect was emitted during decay in addition to the electron.

The particle hypothesized could be carrying off the energy, momentum, and angular

momentum required maintaining the conservation laws. This new particle was named the

Neutrino, meaning “ little neutral one” by the great Italian physicist Enrico Fermi (1934). The

correct way of writing the decay of is then

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The bar (--) over the neutrino symbol is to indicate that it is an “antineutrino”.

(c) Gamma Decay

Gamma rays are photons having very high energy. Like an atom, nucleus can be in an

excited state because of a violent collision with another particle, more commonly, the nucleus

remaining after a previous radioactive decay. The nucleus in an excited state is said to be in a

metastable state, and is called an isomer. When, it jumps down to a lower energy state, or to

the ground state, it emits a photon, for example:

An exited nucleus can sometimes return to the ground state by another process, known

as internal conversion. In this process, the excited nucleus interacts with one of the orbital

electrons and ejects this electron from the atom with the same Ek that an emitted ray would

have had. There is difference between a ray and X ray. X ray if the photon is produced by

an electron-atom interaction and ray if the photon is produced in a nuclear process.

4.4 Half-life and Rate of Decay

A macroscopic sample of any radioactive isotope consists of a vast number of

radioactive nuclei. These nuclei do not all decay at one time. Rather, they decay one by one

over period of time. This is a random process; we can’t predict exactly when a given nucleus

will decay. But we can determine, on a probability basis, approximately how many nuclei in a

sample will decay over a given period. The number of decays that occur in a very short

94

time interval is found to be proportional to t and to the total number N of radioactive

present:

4.5

is a constant of proportionality called the decay constant, which is different for different

isotopes.

If we take the limit , N will be small compared to N, and we can write the

equation 4.5 in infinitesimal form as

4.6

Then

And then integrating from t=0 to t=t:

Where No is the number of parent nuclei present at t=0, and N is the number remaining at

time t. The integration give

Or

4.7

Equation 4.7 is called the radioactive decay law. The number of radioactive nuclei in a given

sample decreases exponentially in time, as shown in figure 4.3

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Figure 4.4 The number of decays per second decreases exponentially

The rate of decay, or number of decays per second, in a pure sample is , which is

called the activity of a given sample.

At t=0, the activity is

Hence 4.8

The rate of decay of any isotope is often specified by giving its half-time rather than the decay

constant . The half-life of an isotope is defined as the time it takes for half the original

amount of isotope in a given sample to decay. The precise relation is:

4.9

Problem 4.3

The isotope has a half-life of 5730 year. If at some time a sample contains 1.0x1022

carbon 14 nuclei, what is the activity of the sample

Solution

First we calculate the decay constant and obtain

96

Since there are (60)(60)(3651/4)=3.156 x 107s in a year. The activity or rate of decay is:

TASK-7

1. What do different isotopes of a given element have in common? How are they different?

2. What are the similarities and the differences between the strong nuclear force and the

electric force?

3. Use the uncertainty principle to argue why electrons are unlikely to be found in the

nucleus!

4. in an exited state emits a 1.33 MeV ray as it jumps to the ground state. What is the

mass of the excited cobalt atom?

5. The activity of a sample of (T1/2=7.5x106s) is 5.2 x 106 decays per second. What is the

mass of sample present?

COURSE 9: MID TERM TEST

COURSE 10

Able to understand nuclear physics, radioactivity and its application

II. Indicators:

1. Able to explain nuclear reaction

2. Able to derive nuclear Fission

3. Able to analyze nuclear Fusion

97

III. Subject matter

1. Nuclear Reaction

The transformation of one element into another, called transmutation, also occur by

means of nuclear reactions. A nuclear reaction is said to occur when a given nucleus is struck

by another nucleus, or by a simpler particle such as a ray or neutron, so that an interaction

takes place. Ernest Rutherford was the first to report seeing a nuclear reaction that some of the

particles passing through nitrogen gas were absorbed and protons emitted.

Where is an particle and is a proton. Nuclear reaction are sometimes written in a

shortened form, . In any nuclear reaction, both electric charge and nucleon

number are conserved.

Problem 4.4

A neutron is observed to strike an nucleus and a deuteron is given off. Deuteron is the

isotope of hydrogen containing one proton and one neutron ( )

Solution

We have the reaction . The total number of nucleons initially 16+1=17, and

the total charge is 8+0=8, the same totals to apply to the right side of the reaction. Hence the

product nucleus must have Z=7 and A=15. From the periodic table, we find that it is nitrogen

The reaction can be written

, where d represent deuterium, .

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Energy and momentum is conserved in nuclear reaction, and this can be used to

determine whether a given reaction can occur or not. If the total mass of products is less than

the total mass of the initial particles, then energy will released by the reaction-it will appear as

kinetic energy of the outgoing particle. But if the total mass of the products is greater than the

total mass of the initial reactants, the reaction requires energy. Consider a nuclear reaction of

the general form

Where a is a projectile particle or small nucleus that strikes nucleus X, producing nucleus Y

and particle b typically p, n, . We define the reaction energy or Q value, in terms of the

masses involved, as

4.10

Since energy is conserved, Q is equals to the change in kinetic energy

4.11

For Q>0, the reaction is said to be exothermic or exoergic: energy is released in the reaction,

so the total Ek is greater after the reaction than before. If Q<0, the reaction is said to be

endothermic or endoergic: energy is required to make the reaction happen.

Problem 4.5

The nuclear reaction is observed to occur even when very slowly moving

neutrons (Mn=1.0087 u) strike a boron atom at rest. For a particular reaction in which

Ekn0, the helium (MHe=4.0026 u) is observed to have a speed of 9.30x106 m/s. Determine (a)

the Ek of the Lithium (MLi=7.0160 u), and (b) the Q value of the reaction.

99

Solution

(a) Since the neutron and boron are both essentially at rest, the total momentum before the

reaction is zero, and afterward is also zero. Therefore,

Hence

Where we have used 1u=1.66x10-27 kg and 1 MeV=1.60x10-13J

(b) We set Eka=Ekx=0; so , where

Hence

2. Nuclear Fission

Otto Hahn and Fritz Strassmann (1938) made an amazing discovery, because of their

founding that uranium bombarded by neutrons sometimes produced smaller nuclei which

were roughly half the size of the original uranium nucleus. The uranium nucleus, after

absorbing a neutron, had actually split into two roughly equal pieces. This phenomenon was

named nuclear fission because of its resemble to cell division. A typical fission reaction is

A tremendous amount of energy is released in a fission reaction because the mass of is

considerably greater than that of the fission fragments. The neutrons released in fission could

be used to create a chain reaction, as shown in figure 4.4

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Figure 4.5 Chain reaction

3. Nuclear Fusion

The mass of every stable nucleus is less than the sum of the masses of its constituent

protons and neutrons. Thus, if two protons and two neutrons were to come together to form a

helium nucleus there would be a loss of mass. This mass loss is manifested in the release of a

large amount of energy. The process of building up nuclei by bringing together individual

protons and neutrons, or building larger nuclei by combining small nuclei, is called nuclear

fusion. Nuclear fusion is continually taking place within the stars, including our sun,

producing the prodigious amounts of radiant energy they emit. The energy output of our sun

is believed to be due principally to the following sequence of fusion reactions:

(0.42 MeV)

(5.49 MeV)

(12.86 MeV)

where, the Q values of each reaction are given in parentheses. The net effect of this sequence,

which is called the proton-proton cycle, is for four protons to combine to form one

nucleus plus two positron, two neutrinos, and two gamma rays:

101

4.12

Problem 4.6

One of the simplest fusion reaction involves the production of deuterium, , from a neutron

and a proton: . How much energy is released in this reaction?

Solution

The initial rest mass is 1.007825 u + 1.008665 u = 2.016490 u and after the reaction the

mass is that of the , namely 2.014102 u. The energy released is thus (0.002388 u)(931.5

MeV/u)=2.22 MeV, and is carried off by the nucleus and the ray.

TASK-10

1. A proton strikes a nucleus, and an particle is emitted. What is the residual

nucleus? Write down the reaction equation!

2. (a) Can the reaction occur if the incident proton has Ek=1500 keV?(b) If

so, what is the total kinetic energy is released?

3. Calculate the energy released in the fission reaction

Assume the initial Ek of neutron is very small.

4. What is the average kinetic energy of protons at the centre of a star where the

temperature is 107K?

5. Show that the energy released in the fusion reaction

is 17.59 MeV!

COURSE 11

I. Basic competency:

102

Able to understand elementary physics, and its reaction

II. Indicators:

1. Able to explain high-energy particles and particle accelerator

2. Able to explain The Yukawa particle

3. Able to explain particles and antiparticle

4. Able to explain particle interactions and conservation laws

.III. Subject matter

In the years after World War II, it was found that if the incoming particle in a nuclear

reaction has sufficient energy, new types of particles can be produced. In order to produce

high-energy particles, various of particle accelerator have been constructed. These high-

energy accelerators have been used to probe the nucleus more deeply, to produce and study

new particles.

5.2 High-Energy Particles and Particle Accelerator

Particles accelerated to high speeds are projectile which can probe the interior of nuclei

and nucleon that they strike. One of particle accelerator is Cyclotron developed by E.O.

Lawrence (1930). The prototype of cyclotron is shown in figure 5.1. It uses a magnetic field

to maintain the charge of particles, such as proton or heavy ion. The protons move within two

D-shave cavities. Each time they pass into the gap between the metal ”dees”, a voltage is

applied that accelerates them. The voltage applied the dees to produce the acceleration must

be alternating. The increases their speed and also increases the radius of curvature of their

path. After many revolutions, the protons acquire high kinetic energy and reach the outer edge

of the cyclotron. Then, they either strike a target placed inside the cyclotron, or leave the

cyclotron with the help of carefully placed “bending magnet” and are directed to an external

target. The frequency of the applied voltage must be equal to that of the circulating protons;

103

5.1

Where, q and m are the charge and mass of the particles moving in the magnetic field B.

Figure 5.1 The Cyclotron

An interesting aspect of the cyclotron in the frequency of the applied voltage, equation 5.1,

does not depend on the radius As the mass increases, the frequency of the applied voltage

must be reduced. To achieve large energies, complex electronic is needed to decrease the

frequency as a packet of protons increase in speed and reaches larger orbit.

Another way to deal with the increase in mass with speed is to increase the magnetic

field B as the particles speed up. Such a device is called a synchrotron. Synchrotron can

accelerate protons to energies of 500 GeV. The synchrotron do not use enormous magnets 1

km in radius. Instead, a narrow ring of magnet is used which are all place at the same radius

from the center of the circle. The magnets are injected, then must then move in a circle of

constant radius. This is accomplished by giving them considerable energy initially in a much

smaller accelerator, and then slowly increasing the magnetic field as they speed up in the

large synchrotron. One problem of any accelerator is that accelerating electric charge radiate

electromagnetic energy.

104

Problem 5.1

A small cyclotron of maximum radius R= 0.25 m accelerate protons in a 1.7 T magnetic field.

Calculate (a) what frequency is needed for the applied alternating voltage, and (b) the kinetic

energy of protons when they leave the cyclotron.

Solution

(a)

(b) The protons leave the cyclotron at r=R=0.25 m then, since

5.3 The Yukawa particle

At first, we recognized that all atoms can be considered to be made up of neutron,

protons, and electrons. The basic constituent of the universe were no longer considered to be

atoms but rather the proton, neutron, and electron. Beside these three elementary particles,

several others were also known the positron, the neutrino, and the particle, for a total of six

elementary particles. But then, in the decades that followed, hundreds of other sub-nuclear

particles were discovered.

105

Figure 5.2 Feynman Diagram, showing how a photon acts as carrier of

electromagnetic force between two electrons.

Hideki Yukawa (1935) had predicted the existence of a new particle that would in

some way mediate the strong nuclear force. To understand Yukawa idea’s, first look at the

electromagnetic force. Firstly, we know that electric force acts over a distance, without

contact, but how a force can act over a distance. Faraday introduced the idea of field. The

force that one charged particle exerts on a second can be said to be due to the electric field (E)

set up by the first. Similarly, the magnetic field (B) can be said to carry the magnetic force.

Later, we know that electromagnetic field can travel through space as waves. But, then,

electromagnetic radiation can be considered as either a wave or a collection of particles called

photons. Because of the wave-particle duality, it is possible to imagine that the

electromagnetic force between charged particle is due (1) to the EM fields set up by one and

felt by the other, or (2) to an exchange of photons or -particle between them. At the end, It

can be stated that the electromagnetic force between two charged particles, it is photon that

are exchanged between the two particles that give rise to the force, as shown in figure 5.2, the

Feynman diagram in his quantum electro-dynamic (QED).

In analogy to photon exchange to mediate the electromagnetic force, Yukawa argued

that there ought to be a particle that mediates the strong nuclear force, the force that holds

nucleons together in the nucleus. Just as the photon is called the quantum of the

electromagnetic force, so the Yukawa particle would represent the quantum of the strong

nuclear force. Yukawa predicted that this new particle have a mass intermediate between that

106

of electron and proton. Hence it was called a meson, meaning “ in the middle”, as shown in

figure 5.3, a Feynman diagram of meson exchange.

Figure 5.3 Meson exchange when proton and neutron interact via

strong nuclear force

The mass of the meson, approximately, can be determined as follows. Suppose the

proton is at rest. For it emit a meson would require energy (to make the mass) which would

have to come from nowhere, such a process would violate conservation of energy. But the

uncertainty principle allows non-conservation of energy by an amount E if it occurs only for

a time t given by . Let set E equal to the energy needed to create the mass

m of the meson: . Now conservation energy is violated only as long as the meson

exist, which is the time t required for the meson to pass from one nucleon to the other.

Assume the meson travels at relativistic speed, close to the speed of light c, then t will be at

most about t=d/c, where d is the maximum distance that can separate the interacting

nucleon, thus

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or 5.2

The range of the strong nuclear force is about m, so

The mass of the predicted meson that carries the strong force is thus very roughly 130 MeV/c 2

or about 250 times the electron mass of 0.51 meV/c2.

Just as proton can be observed as free particle, as well as acting in an exchange, so it

was expected that meson might be observed directly. The particle predicted by Yukawa

(1947). It is called the “” or pi meson, or simply the pion. It comes in three charge states, +,

-, or 0. The and have mass of 139.6 MeV/c2 and the a mass of 135.0 MeV/c2. All

three interact strongly with matter. Soon after pion discovery in cosmic rays, they produced in

the laboratory using a particle accelerator. Reaction observed included

5.3

A number of other meson were discovered in subsequent years which were also considered to

mediate the strong nucleus force. However, the recent theory of quantum chromo-dynamics,

involving quarks, has replaced mesons with gluon as the basic carries of the strong force.

As we have known that, there are four types of force or interaction in nature, namely

(1) the electromagnetic force, (2) the strong nuclear force, (3) the weak nuclear force, and (4)

the gravity force. The particle presumed to transmit the weak force is referred to as the W+,

W-, and Z0, founded by Carlo Rubbia (1983). The quantum of the gravitational force called

the graviton, however, has not yet been identified. The comparison of the four forces is

shown as follows.

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Table 5.1

The four forces in nature:

Type Relative strength Field particle

Strong nuclear 1 Meson/gluon

Electromagnetic 10-2 Photon

Weak nuclear 10-13 W+, W-, and Z0

Gravitational 10-40 Graviton

5.4 Particles and Antiparticle

The positron is basically a positive electron. Its properties are the same as for the

electron, such as mass, but it has the opposite charge. The positron is said to be the

antiparticle to the electron. The antiparticle of proton was found, the antiproton ( ). Many

other particles also have antiparticles. But the photons, the 0, and a few other particles do not

have distinct antiparticles.

5.5 Particle Interactions and Conservation Laws

The laws of conservation of energy, of momentum, of angular momentum, and of

electric charge are found to hold precisely an all particle interaction. A study of particle

interaction has revealed a number of new conservation laws, namely the conservation of

baryon number. Baryon number is the same as nucleon number. All nucleon have baryon

number B=+1, all antinucleon (antiprotons, antineutrons) have B=-1. For example, the

following reaction does conserve B and does occur if the incoming proton has sufficient

energy.

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Other useful number laws are associated with weak interactions, including decays. In

ordinary decay an electron or positron is emitted along with a neutrino or antineutrino, as

follow

To explain the nuclear reaction above, the concept of electron lepton number was proposed. If

the electron (e-) and the electron neutrino ( ) are given Le=+1, and e+ andn are given Le=-1,

whereas other particles have Le=0.

Problem 5.2

Which of the following decay schemes is possible for muon decay:

(a)

(b)

(c)

Solution

A has L=+1, and Le=0. This is the initial state, and the final state must also have L =+1,

Le=0. In (a), the final state has L=0+0=0, and Le=+1-1=0; L would not be conserved and

indeed this decay is not observed to occur. The final state of (b) has L=0+0+1=+1 and

Le=+1-1+0=0, so both L and Le are conserved. This is in fact the most common decay mode

of the -. Finally (c) does not occur because Le=+2, so it is not conserved.

TASK-10

1. What limits the maximum energy attainable by protons in an ordinary cyclotron? How is

this limitation overcome in a synchrotron?

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2. If a proton is moving at very high speed, so that its Ek is much greater that its rest energy,

can it the decay via ?

3. Why is it that a neutron decays via the weak interaction even though the neutron and one

of its decay products (proton) are strongly interacting?

4. What is the total energy of a proton whose kinetic energy is 15.0 GeV?

5. If particles are accelerated by the cyclotron, what must be the frequency of voltage

applied to the dees?

COURSE 11:

I. Basic competency:

Students are able to understand nuclear physics, radioactivity and its application

II. Indicators:

1. Able to derive particle classification

2. Able to analyze strange particle

3. Able to explain Quarks and Charm

4. Able to explain The “ Standard Model”: Quantum Chromo dynamics (QCD) and the

Electroweak Theory

III. Subject Matter

1. Particle Classification

A great many other sub-nuclear particles were discovered after 1940 year. Then, an

important aid to understanding is to arrange the particles in categories according to their

properties, and their interaction. Since not all particles interact by means of all four of the

forces known in nature, this is used as a classification scheme. At the first categories is the

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gauge bosons, which include the photons and the W and Z particles that carry the

electromagnetic and weak interaction, respectively.

The second category is the lepton, which are particles that do not interact via the

strong force but do interact via the weak nuclear force ( as well as the much weaker

gravitational force); those that carry electric charge also interact via the electromagnetic force.

The lepton include the electron, the muon, and the tau plus three types of neutrino.

The third category of particle is the hadron. Hadrons are those particles that can

interact via the strong nuclear force. Hence they are said to be strongly interacting particles.

They also interact via the other forces, but the strong force predominant at short distance. The

hadrons include nucleons, pions, and a large number of other particles. They are divided into

to subgroups: The baryon, which are those particles that have baryon number +1 and -1 in the

case of their particle; and the meson. Which have baryon number=0. Notice that, the baryon

, all decay to lighter-mass baryons, and eventually to a proton or neutron. All

these process conserve baryon number.

Table 5.2 Elementary Particle

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2. Quarks and Charm

All observed particles fall into two categories: fermion (which obey the Paulli

exclusion principle) and bosons (which don’t obey the Paulli exclusion principle). The

fermions can be further subdivided into two groups: leptons and hadrons. The principal

difference between these two groups is that the hadrons interact via the strong interaction

whereas lepton do not. The lepton are considered to be truly elementary particles since they

do not seem to break down into smaller entities, do not show any internal structure, and have

no measurable size.

The hadrons, on the other hand, are more complex. Experiment indicate they do have

an internal structure. Gell-Nman and Zweig (1963) proposed that the hadrons are made up of

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combinations of three, more fundamental, pointlike entities called quarks. Quarks, then,

would be considered truly elementary particles, like leptons. The three quarks were labeled u,

d, s and given the names up, down, strange. Each of quarks has antiquark. Meson would

consist of a quark and antiquark. A meson is considered a pair. Netron is ,

whereas antiproton is . After deeply studied, it was found a fourth quark, based on the

expectation that there exists a deep symmetry in nature, including a connection between quark

and lepton. The fourth quark was said to be charmed. The new charmed quark would have

charm C=+1 and its antiquark C=-1.

And then, theoretical physicist postulated the existence of a fifth and sixth quark. These were

named top (t) and bottom (b) quarks, since they resemble the “up” and “down” quarks.

Although, some physicist prefer that, the names truth (t) and beauty (b) quarks for top and

bottom quark.

3. Quantum Chromo dynamic (QSD) and The Electroweak Theory

Not long after the quark was proposed, it was suggested that quarks have another

proverty (or quality) called color. The distinction between the six quarks (u,d,s,c,b,t) was

referred to as flavor. According to theory, each of the flavors of quark can have three colors,

usually designed red, green, and blue. The antiquark are colored antired, antigreen, and

antiblue. Baryon are made up of three quarks, one of each color. Meson consist of a quark-

antiquark pair a articular color and its anticolor. Thus baryon and mesons are white or

colorless. The idea of color soon became, in addition, a central feature of the theory as

determining the force binding quarks together in a hadron. Each quark is assumed to carry a

color charge, analogous to electric charge, and the strong force between quarks is often

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referred to as the color force. This new theory of the strong force is called quantum

chromodynamics (QCD), to indicate that the force acts between color charges. The strong

force between hadrons is considered to be a force between the quarks that make them up. The

particles that transmit the force are called gluon. There are eight gluons, according to the

QCD Theory, all mass less, and six of them have color charge. Thus gluon have replace

mesons as the particles carrying the strong (color) force.

The weak force is thought to be mediated by the W+, W-, and Z0 particles. It acts

between the weak charge that each particle has. Each elementary particle thus has electric

charge, weak charge, color charge, and gravitational mass, although one or more of these

could be zero. The latest theory consider the truly elementary particles to be the leptons, the

quarks, and the gauge bosons. One important aspect of new theoretical work is the attempt to

find a unified basis for the different forces in nature. Weinberg, Glashow, and Salam (1960)

proposed “gauge” theory that unifies the weak and electromagnetic interaction. In this

electroweak theory, the weak and electromagnetic forces as two different manifestation of a

single, more fundamental, electroweak interaction. So, the electroweak theory and QCD for

the strong interaction are often referred to today as the standard model.

With the success of unified electroweak theory, attempts have recently been made also

to incorporate it and OCD for the strong force into a so-called Grand Unified Theory (GUT).

One type of such a grand unified theory of the electromagnetic, weak, and strong forces has

been worked ot in which there is only one class of particle: leptons and quarks belong to the

same family and are bale to change freely from one type to the other, and the three forces are

different aspect of a single underlying force.

TASK-11

115

1. What magnetic field intensity is needed at the 1.0 km radius synchrotron for 400 GeV

protons? Use the relativistic mass.

2. Show that the energy of a particle (charge e) in a synchrotron, in the relativistic limit

(vc), is given by E (in eV)=Brc, where B is magnetic field strength and r the radius of

the orbit.

3. How much energy is released when an electron and a positron annihilate each other?

How much energy is released when a proton and an antiproton annihilate each other?

4. Which of the following decays are possible? For those that are forbidden, explain which

laws are violated.

(a)

(b)

(c)

5. (a) What are the quark combinations that can form (a) a neutron, (b) an

antineutron, (c) a , (d) a

(b) What particles do the following quark combinations produce? (a) uud, (b)

, (b) , (c) , (d) , (e) ?

COURSE 12:

I. Basic competency:

Students are able to understand experimental, and applied nuclear physics

II. Indicators:

1. Able to explain any kind of experimental nuclear physics

2. Able to explain applied fission nuclear physics

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3. Able to explain applied fusions nuclear physics

4. Able to understand nuclear reactors

.III. Subject Matter

1. Overview

The most important terrestrial applications of nuclear physics and some of the

experimental methods used to detect nuclear particles and to study nuclear reactions. Beside

that, the basic physics of nuclear fission and the nuclear reactions at present has the most

prominent applications. Fission provides the energy in nuclear reactors and some of the

energy in atomic weapons. Nuclear reactors, which apply the fission reaction to the

production of usefull power and of new nuclides. The energetic charge particles and photos

emerging from nuclear reactions interact strongly with matter.

2. Nuclear Fission

Nuclear fission is the splitting of a nucleus into two nearly equal parts, illustrated

schematically in figure 6.1. The nucleus at (a) is nearly spherical in shape, but if this nucleus

is excited (e.g., by particle bombardment) it can excute violent vibrations. It should be

recalled that the list of stable nuclides terminates at the high Z end because of the electrostatic

repulsion of the protons. In a fissionable nucleus this repulsion is very powerful when a

configuration like (b) or (c) is reached. The two parts are restrained from flying apart,

however, bby the surface energy, since the surface to volume ration of the shape (b) and (c) is

than that of the sphere (a). The surface energythereby provides a barrier, and it is this barrier

that is responsible for the existence in nature of nuclide with A>~110.

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Figure 6.1 Nuclear fission

Nuclides with A of the order of 250 have such a small barrier that the spontaneous

oscillation of the nucleus can surmount it with a short half-life, but this spontaneous fission

process becomes unmeasurably fission, fission excited by an external particle. Suppose, for

example, that a thermal neutrons is incident upon . It adds its kinetic energy(negligible)

and its binding energy (6.3 MeV) to the nucleus, and the compond nucleus then has

enough energy to surmount the barrier. Once past the configuration (c), the surface energy

rapidly decrease by ”necking” as in the break-up of a large water drop. The nearly spherical

fragment in (d) fly apart, acclerated by the coulomb repulsion. Two or three neutrons are

typically ejected, presumbly during the (c)- (d) stage of the process.

Many nuclides are fissionable by fast particles, which add their kinetic energies as well

as their binding energies to nuclei, but ot the naturally occuring

nuclides, only undergoes fission by slow neutrons. This is the only natural nuclide with

a barrier low enough to be overcome by the neutron binding energy alone. Two other nuclides

in which fissions can be induced by slow neutrons are and , which can be

produced in nuclear reactors by neutrons capture in and , respectively, and

subsequent decay.

It should be clear from the foregoing sketch of the fission process that there is not a

unique reaction equation for a given fissionable nuclide and a given incoming particle. The

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fission fragments vary in A and Z, with a considerable preference for somewhat unequal

distribution of mass between them.

3. Detectors of Radioactivity

Many different types of devices are used to detect charged particles and uncharged

particles (including photons). Modern detectors now rely heavily on electronics. In many

types of these devices the detection process begins when particle first creates directly or

directly some kind of an electrical signal. This electrical signals is then detected and may tell

the energy of the particle, as well as indicating its presence. Other types of detectors utilize a

track created by the particle to gain information.

Scintilation detectors use the light emitted when particles strike a sensitive material

called a phosphor. The electrons of the phosphor atoms are knocked into excited states by the

collision. As they drop back toward their ground states, they emit photons, as shon in figure

6.2. You are familiar with this effect when the particles are electrons and the phosphore coats

the screen of a cathode-ray tube in a television set, an oscilloscope, or computer terminal. In

flourescent lihts, the incident particles include ultraviolet photons.

Figure 6.2 Scintillation detectors

A photomultiflier tube is used to increase the sensitivity of a scintilation detector well

past that of the human eye. In a photomultiflier tube, as shown in figure 6.3, an incident

photon is absorbed at the cathode, causing the emission of photoelectron. This photoelectron

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is then accelerated through a potential difference of about 100V to strike the first dynode. At

the first dynode, the accelerated photoelectrons is then accelerated through 100 V to strike a

second dynode. Each again knocks out several more electrons and the process continues. With

ten dynodes, a greatly increased pulse of current is produced. This current pulse is nearly

proportional to the energy of the particle detected. Therefore the particle energy can be

determined with proper calibration.Other advantages of the scntillation detector include the

ability to detect at high counting rates and high efficiency. If the light from the scintillation is

adequate, photodiodies may be used in place of the photomultiplier tube.

Figure 6.3 A photomultiflier tube

EXAMPLE 6.1

What two methods of removing electrons from a material are used in the photomultiplier

tube?

Solution: The photoelectric effect at the cathode and secondary emission at the dynodes.

Gas-filled detectors include ionization chambers, proportional counters,and Geiger-Muller

(G-M) counters. In these devices, charged filled chamber. The particles or photons then knock

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electrons of gas atoms, creating ion pair is created for each 30 or so eV of energy deposited in

the chamber. The actual average ionization energy depends on the gas in the chamber.The

electrons of the ion pairs are then attracted to a positive electrode and the ions are attracted to

a negative electrode. This motion of charges gives a current pulse through a resistor, R,

external to the detector, as shown in Fig.6.4. The resulting voltage pulse across R is detected

by electronic circuit.

Figure 6.4 Gas-filled detector

In the low-voltage region 1 in Fig.6.5, many of the ion pairs simply recombine,

decreasing the charge collected. In region 2 ( typically les than 300 V), the electric field is

sufficien to prevent recombination of the ion

Figure 6.5 The charge collected at the electrodes

The charge collected at the electrodes ( on a log scale) for alpha particles and electrons

as a fuction of the voltage applied to a gas filled detecor. Recall that one ion pair is created on

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the average for every 30 or so eV of energy deposited in the chamber. Therefore the charge

collected in an ionization chamber is ameasure of the energy deposited in it.

EXAMPLE 6.2

An average of 35 eV is required to create an ion pair in air. How much charge will be

collected in an air-filled ionization chamber if a 0.60-MeV electron loses all its energy in the

chamber and negligible recombination occurs?

Solution

Treating the units as algebraic quantities, we get

(0.60 x 106 eV)/(35 eV/ion pair)=1.71 x 104 ion pairs.

Since negligible recombination occurs, all 1.71 x 104 ion pairs are colected at the electrodes.

Each ion and electron has [q]=e=1.60 x10-19 C, so (1.71x104 ion pairs)/(1.60x10-19C/ion

pair)=2.7x10-15C is collected.

In region 3 (typicaly between 300V and 900V) the stringer electric field will accelerate

the electrons of the ion pairs to the energy that enables them to create more ion pairs. Then the

electrons from these ion pairs can create even more ion pairs, and so on. This multiplication

of ion pairs in region 3 may give up to million times the charge collected for the same

incident particle in region 2.

A gas filled detector operating in region 3 is called a proportional counter. Examplel 6.2

gave about 10-15 C for an ionization chamber, but the charge collected could be more like 10 -9

C for a proportional counter. The word proportional is part of the name because the carge

collected is region 3. However, the voltage applied must be held quite constant. Otherwise,

the charge collected will vary widely fir the same energy deposited because of the steepness

of the curve of figure 6.5.

Example 6.3

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A 1.33 MeV gammab ray gives 2.04 x 10-9 C of charge collectedin a proportional counter.

Another gamma ray gives 1.79 x 10-9C. What is the energy of the second ray?

Solution:

The carge collected is directly proportional to the energy deposited, so q1/q2=E1/E2, or

E2=E1q2/q1= 1.33 MeV(1.79 x10-9C)/(2.04x10-9)=1.17 MeV.

The assumtions we made in example 34.3 are that the potential difference constant, that all the

gamma-ray energy is deposited in the chamber, and the counter is being operated in the

strictly proportional part of region 3.

At higher voltage (typically in the region of 1000 V), there is a plateau region. If the

chamber is being operated in the plateau region, any ionizing radiation will trigger a large

momentary breakdown in the gas that start near the central electrode (where the electric field

is the greates). The gas-filled detector acts as aGeiger-Muller (G-M) counter in this voltage

region. For G-M counter, the charge collected does no depend on the energy of the particle.

Therefore this type of counter is used as a relatively simple methode of counting photons or

the particle of certain types of ionizing radiation.

Example 6.4.

A Geiger-Muller doesn’t require a well-regulated power supply. Why not.

Solution:

Figure 6.4 shows that changing V along the plateau region result in little change of the charge

collected. The charge collected doesn’t give the energy of the particle, any way.

If the voltage is increased even farther nto region 5 in figure 34.4. the electric field a the

central electrode will exceed the dielectric strength \of the gas. Therefore the gas in the tube

will break down and conduct even without ionizing radiation. The result is a continous

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discharge in the detector. Semiconductor detectors act mush like solar cell. Ionizing radiation

creates electron-hole paors in the semiconductor. If these pairs are sufficiently energetic, they

may lose energy in collisions that create other electron-hole pars. The process continues until

the average energy deposites per pair is a few times the width of the energy gap of the

semiconductor.

In the junction region, the built inelectric field and any applied field sweep these pairs

apart so that the charge collected is proportional to the particle energy deposited. Diffusing Li

into the junction increases its width. The Si(Li) and Ge(Li) detectors are lithiumdrifted Si and

Gedetectors. (they are humorously called “silly” and “ Jelly” detectors). Their resolution is

much better than that of proportional counters. Better resolution means that they might

distinguish two different particle energies that are very close to one another, while the

proportional counter would see just one comparatively smeared-out energy. The Si(Li) and

Ge(li) detectors also have a short response time.

An airplane moving through air faster that the speed of sound creates a shock wave of

air, which we hear as a “sonic boom” Similarly, a charge particle moving through a

transparent medium faster that the speed of light in that medium creates an electromagnetic

shock wave in that medium. The energy in this electromagnetic shock wave is called

Cerenkov radiation. A detector having a photomultiflier tube or other device to detect a

particle by means of this radiation is called a Cerenkov detector.

Examle 6.5.

The speed of light in water is 2.25 x 108 m/s. Within what speed range will electrons create

Cerekov radiation while moving through water?

Solution:

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The electron, a charged particle, must move faster that the speed of ligth in the water to create

Cerenkov radiation. However, the electron cannot move at or greater that c, the speed of light

in a vacumm. Therfore the range of the speed v is 3.00 x 108 m/s >v>2.25 x108 m/

4. Particle Track Recorders

The particle of nuclear an subnuclear physics are much too small for us to see, even

though the finest microscope. However, we can use a number of different devices to display

the tracks of submicroscopic charged particles. These tracks often pass through a known

applied magnetic field, . We can then apply qvB=mv2/r to the measured track, obtaining

information about the particle’s charge and momentum.

A nuclear emulsion is simply thicker and more sensitive that the photographic emulsion

used in ordinary camera film. The passage of charged particle expose the grains of the

emulsion along its path. The emulsion is later developed and examined by microscope.

Because of its light weight and simplicity, a nuclear emulsion is often used for high altitude

rocket or ballon studies of cosmic rays.

Conceptually, plastic track detectors are similar to nuclear emulsions. Some plastic are

easily damaged by the passage of energetic charged particles, especially heavy ions. Chemical

etches can the be used to enlarge the damaged areas, making them clearly visible. The size of

each resulting etch pit depends on the charge and the energy, so stacked sheets of these

plastics can be used as track detectors.

The cloud chamber, first developed by C.T.T Wilson in 1911, is basically a countainer

filled with s supersaturated vapor. When a charge paticle moves through the vapor, it creates

ions along its path. The ions then act as condensation centers droplets of liquids. Wee see this

trail of droplets in Fig. 6.6, not the particle itself. (This effect is mucj like seeing a high-flying

planes’s costrail but not being able to see the plane itself). Wilson originally began work on

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the chamber in 1895 to reproduce in his lab some beautiful effects of sunlight on mist that

he’d noticed.

Donald Glaser developed the bumble chamber in 1952 for studying high-energy

interactions. It is basically a container filled with a superheated liquid. The liquid is three

orders of magnitude denser that the vapor in the cloud chamber. High-energy particles are

thus much more likely to interact within the chamber. When a charged particle moves through

the superheated liquid, it creates ions along its path. These local additions of energy cause

small volumes of boiling. That, is, a track of vapor bubbles appears in the superheated liquid,

as shon in figure 6.6. Despite the well-known-but aporcryphal-story, Glaser didn’t get the idea

for the device while watching the bubbles rise from the bottom of a glass of beer. However,

he did study the properties of that energy of that beverage as a superheated liquid during his

development of the bubble chamber.

Example 6.6.

The bubble chamber has been called the “ inverse of the cloud chmber”. Explain why?

Solution:

In both types of chambers, ionizing particles leave a track through a metastable medium. In

the cloud chamber, the track is of liquid in a vapor. In a bubble chamber, the track is of vapor

in a liquid.

126

Figure 6.6. A bubble chamber photograph.

In a spark chamber and a streamer chamber a high electric field is set up in a gas so that

the gas is almost ready to break down and conduct.Passing a charged particle through creates

ion pairs in the gas. These ion pairs then multifly, creating a spark along the track of the

particle. The positions of the sparks can be detected photographically or electronically and

used to study the track.

Figure 6.7. Streamer Chamber Photograph

A drift chamber, represented in figure 6.7, contains an array of closely spaced ( a few

mm apart) wires and a low-pressure gas. Charged particles ionoze the gas and ions drift to the

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wires, depositing their charge. The charge collected and time information from each wire is

the processed on a computer to reconstruct the trajectories.

Beside those we have discussed, there are various other types of particle track recorders.

The type used for a particular experiment depends on such variable as energy, charge, time

between events, massm lifetime, and, of course budget.

TASK-12

1. If five electrons are knocked of the first dynode, and these five each knock five move

off the second dynode, and this process continues until the tent dynode, how many

electrons will be knocked of the tenth?

2. A beam of 0.50 Mev protons is lossing all its energy in an ionizaton chamber filled

with a 32 eV/ion pair gas. A current of 4.7µA is collected. Assuming negligible

recombination, calculate the beam current!

3. In a proportional counter, a 0.975 MeV gamma gives 1.236x10-9C of charge collected.

How much charge is collected for 0.585 MeV and 0.390 MeV rays?

4. A 0.438 MeV electron gives 3.7x10-10C of charge collected when stopped in a

proportional counter filled with 35 eV/ion pair gas. a) How much charge would a

0.278 MeV ray give? b) By what factor does the applied voltage cause the charge

collected to increase?

5. In an attempt to make Geiger-Muller counter more sensitive , a uranium prospector

decides to double the normal voltage. What will happen?

6. The element Si has an energy gap of 1.12 eV and Ge a gap of 0.66 eV. Will a Si or a

Ge semiconductor detector collect more charge from a given particle, all else being

equal? Explain!

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7. Approximately how much charger (ignoring recombination)could be collected by a

Si(Li) semiconductor detector (energy gap of 1.12 eV) for a 0.35 MeV - and a 0.15

MeV if one-third of the energy is used to create electron-hole pairs?

8. If 1.00 TeV protons are to move in 1.0 km radius circle, what magnetic field and

frequency are required?

9. A cyclotron is designed to deliver 10-MeV protons at a 0.75 m radius.

(a) Calculate the frequency at which the cyclotron will operate.

(b) How long does it take the proton to complete one revolution?

(c) What is the maximum proton speed?

(d) Is this a relativistic problem? Explain, using 10 MeV and also by using the answer

to part (c).

10. (a) Why does a particle getting “out of phase” with a cyclotron create a problem?

(b) Explain how cyclic accelerators compensate for the relativistic mass change of the

particle.

COURSE 13

I. Basic competency:

II. Indicators:

Able to write a paragraph in English with various kinds of writing versions.

III. Subject Matter

1. Linier Accelerators

Particle accelerators are used in nuclear physics to obtain the threshold or resonant

energy needed for some nuclear reactions and to obtain the high momentum ( and

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corresponding small de Broglie wavelength) needed to “see” small structure. Accelerators are

also used for ion implantation for selective doping of semiconductors, for alloying with

minute quantities of rare metals, for surveying for hydrocarbons surrounding well shafts, for

the production of medical isotopes and for selective radiation, for changing the properties of

plastics, for radioactive dating, and for many other purposes. In almost all practice

accelerators, the charged particles move in an excellent vacuum, as low as 10-13 atmosphere.

Otherwise, they would lose energy, be scattered in different directions, or even be absorbed in

collisions before reaching their intented target. Linier accelerators speed up charge particles in

a straight line. The acceleration results from the force applied by an electric field. In a

Cockroft- Walton accelerator, the electric field is provided by a large potential difference. A

particle of change q moving through a potential drop V can gain a kinetic energy:

KE = qV 6.1

Example 6.7

The high voltage ( up to about 2 MV) may be obtained by charging capacitors with a voltage

protons. What was the potential drop?

Solution:

The kinetic energy, KE, is 800 keV. For a proton, q = e. Then KE/q = V= 800 keV/e =800kV.

Using KE = qV is almost trivial in eV units, but recall tha Q = +2e for alpha particles.

In 1929, Robert J. Van de Graaf invented an electrostatic generator, which is now called

the Van de Graaff accelerator. It uses a moving belt to carry charge to the inside of a hollow

conductor. Recall that all excess charge resides on the surface of an electrostatic conductor

(unless that charge is insulated from the surface). Therefore the charge carried to the inside by

the belt is rapidly conducted to the outside surface of the conductor. Then the excess charge,

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external electric field, and potential difference build up and are limited only by the insulating

properties of the surroundings. Potential differences of 30 MV are even possible.

In the ingenious tandem Van de Graff accelerator, ions are first negatively charge and

accelerated toward a positively charged conductor, as shown in Fig. 6.8. Inside the conductor

electrons are stripped off to make the moving ions positive. Acceleraction of the ions then

continues, but away rom the positively charged conductor.

Figure 6.8 Van de Graff accelerator

EXAMPLE 6.8

Describe ho 30-MeV protons could be obtained from a tandem Van de Graaf accelerator

operating at 15 MV above ground. Start with H-1 atoms.

Solution:

First, an extra electron is added to each hydrogen atom, giving H ions. The H-ions are then

accelrated from ground to +15MV. Therefore +15MV is the potential rise, so the potential

drop is -15MV. For H-q = -e. Equation34.1 then tell us that the H- ions gain a kinetic energy

of qV = (-e)(-15MV) = + 15MV.

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Then both electrons are stripped off each moving H-ion, leaving bare protons. The charge is

now +e. The protons then continue from +15MV to ground, dropping 15MV. Therefore they

gain an additional qV = (+e)(+15MV) = 15 MeV of kinetic energy for a total KE grain of 30

MeV.

A driff-tube linier accelerator (or linac) is composed of a series of coaxial conducting

cylinders( driff tubes), as illustrated in Fig.6.9. The cylinders are alternately connected to an

oscillating voltage source . The electric field inside the conductors will be almost zero, but

between the conductors it will always be in the direction needed to accelerate the Charge

particle.

Figure 6.9 A driff-tube linier accelerator

For example, suppose that a proton leaves the ion source.Drift tube 1 is made negative,

attracting the proton into the cylinder. While the proton moves ( drifts) through the tube, the

charge on 1 is made positive and drift tube 2 becomes negative, so that the proton is further

accelerated between 1 and 2. Then 2 is made positive and 3 negative, again accelerating the

proton as it passes between 2 and 3. This procces continues for the length of the linac. The

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drift tubes are designed with increasing lengths to compensate for the speed increase, but their

lengths soon approach a constant value as V approaches c.

Low-mass charged particles, such as electrons and positrons, are usually accelerated to

high energies in a travelling-wave linier accelerator. This type of linier accelerator is basiclly

a long, hollow, evacuated conductor, or wave guide, designed so that electromagnetic waves

traveling down the tube have an electric-field component parallel to its axis. This electric field

accelerates the charge particles. The waveguide is also designed so that the speed of the

electromagnetic wave (which is less than the free vacuum value of c ) matches the increasing

speed of the particles. The Stanford Linier Accelerator Center (SLAC) has a traveling-wave

linier accelerator over three kilometers ( two miles) long. Essentially a 2856 MHz waveguide,

it accelerates electrons and positrons to an energy that has exceeded 50 GeV. Like

surfboarders riding a wave, the electron is this accelerator ride an electromagnetic wave down

the waveguide.

EXAMPLE 6.9

Can electrons and positrons be accelerated in the same bunch in a linier accelerator? ( hint: F

=qE)

Solution:

Even if you could figure out a way to keep them annihilating one another, the answer is still

“no.” In all types of accelerators, an electric field provides the accelerating force. The relation

F = qE tells you that the positive positons are accelerated in the direction of the electric field

but that the negative electrons are accelerated in the opposite direction.

In Example 6.9 the phrases “ in the same bunch” provides the main restriction. In drift-

tube and travelling-wave linier accelerators, the electric field reverses direction every half

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cycle. Therefore a bunch of positrons can be accelerated and then, a half cycle later, a bunch

of electrons can follow.

2. Cyclic Accelerators

In cyclic acceleratos, magnetic fields bend charged particles around closed paths and

electric fields speed them up. When a charge particle of mass m and charge q moves in a

circular path of radius r perpendicular to a magnetic field B,F = ma becomes qvB=mv2/r.

Solving for the momentum of the charged particle, we obtain

p = mv = qBr 7.2

Example 6.10

Bubble chamber photographs taken with a 0.40-T field show a particle with q=e moving in a

15-cm radius circle at one point. Calculate the particle’s momentum at that point if v is

perpendicular to B.

Solution:

We know that q=e=1.60x10-19C, B=0.40 T, and r=15x10-2 m. Therefore

P=qBr=(1.6x10-19A.s)(0.40 N/A.m)(15x10-2m)

=9.6x10-21N.s (1 MeV/c)/(5.344x10-22N.s)=18 MeV/c

The tracks in particle track recorders aren’t simple circles, even if v is perpendicular to B.

Rather, the tracks are decreasing radius spirals, as shown previously in fig. 6.10. Why?

Because the particle loses energy and momentum in creating its track. The expression p=qBr

shows that a decreasing momentum gives a decreasing radius.

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Figure 6.10 Particle Track Recorders

The time required for a single revolution in acircular cyclic accelerator equals the

circumference divided by the speed, or 2r/. Therefore the frequency, , is the reciprocal,

/2r, of that time. Solving mv=qBr for and substituting into =v/2r gives

7.3

This frequency is called the cyclotron frequency, after the cyclic acceleration shown in figure

34.12.

The cyclotron is composed of two D-shped, hollow conductors (called dess) placed

between the poles of a magnet. The magnetic field is constant and is perpendicular to the

velocity of the charged particles. This configuration makes the charged particles move in

semicircular paths within each dee.

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Figure 6.11 A Cyclotron

The dees are connected to a source of alternating voltage. The source is synchronized so that

whenever a charged particle is moving from dee 1 to dee 2, the electric field between the

oppsitely charged dees is at a maximum in the direction that will accelerate the particle. One-

half cycle later, the charged particle is moving from dee 2 to dee 1 and the electric field has

been reversed to again give the particle a maximum increase in speed. When inside the metal

dees, the charged particle is within a conductor where the electric field is zero.

As the particle gains energy and momentum from its trip between the dees, the radius

will increase, according to the expression p=qBr. If the maximum radius is R anf if KEmax is

much less the rest energy, Eo, then KEmax=p2max/2mo, with pmax=qBR, gives:

7.4

Example 6.11.

Designed and built by Ernest Lawrence and M. Stanley Livingston in 1952, the first cyclotron

used a 1.3T magnetic field and had an 11 cm radius

(a) For protons, at what frequency was it operated?

(b) Show that the protons were nonrelativistic

Solution:

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(a) For protons, q=1.6x10-19C, and mo=1.67x10-27kg. We know that B=1.3T and R=0.11

m, then

(b) The maximum kinetic energy was

=

For a proton, Eo=938 Mev, so KEmax <<Eo. Therefore the protons were not relativistic

to the precision of the given data.

If the particle’s kinetic energy begins to become an appreciable fraction of the rest

energy, the mass begin to increase significantly. As a result, the cyclotron frequency, qB/2m,

will decerease when the particle is sped up at a constant B. In a cyclotron, the particle would

get “out of phase” with the accelesating electric field.

Electron have a small rest energy, so this relativistic mass change occurs at relatively low

energies. Therefore electrons are not accelerated to high energies by cyclotrons. With a

constant B, the frequency could be changed as m changes to keep the particle anf the electric

field in phase. Alternatively, B may be varied and may or may not be varied to keep the

particles in phase with the electric field in the class of cyclic accelerators called the

synchrotrons.

Figure 6.12 shows another kind of cyclic accelerator, the betatron., It utilizes an

increasing magnetic filed, which keeps the particle moving in a circular path. Then the change

in the magnetic fieds induces an electric field( by Faraday’s law of induction), which speeds

up the charged particle. A free charged particle radiates electromagnetic energy whenever it is

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accelerated. Examples include bremmstrahlung anf transimission from radio and TV

antennas. In s cyclic accelerator, this radiation is often called synchrotron here being used as a

well-cntrolled radiation source. To produce it in a more usable form, groups of magnets called

“wigglers” and “undulators” are used to accelarate beams of electrons from side to side.

Figure 6.12 A Betatron Accelerator

If r is small, the acceleration v2/r is large. The point could be reachd where every bit of

energy given a charged particle in a cyclic accelerator is radiated right back out. This

possibility is one reason why the large electron-positron (LEP) storage ring anf collider at

what is now called the European Laboratory for Particle Physics (CERN) has a radius of 4.3

km fot its eventual goals of over 100 GeV. Well before excessive radiation loss occurs for

particles more massive than electrons, another consideration requires the huge radii of modern

high energy accelerators. The expression p=qBr shows us that increasing the momentum

proportionally increases the radius for a given magnetic field. Therefore high momentum,

high energy cyclic accelerators, such as the 1-TeV (1000 GeV) Fermilab Tevatron accelerator

in Illinois shown in figure 6.13 require radii of about a kilometer. The proposed 20 TeV

Superconductor Super Collide will require a radius of 13.5 km.

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Figure 6.13 Fermilab Tevatron Accelerator

Obviously, no lab can afford a magnet having square kilometers of pole face area. Therefore

huge-radius accelerator keep the path of the charged particles constant and utilize magnets

placed at intervals along that path. Since , B must increase as both m and v increase.

The actual radii of these huge accelerators are larger than that calculated from pmax=qBmaxr,

because dipole bending magnets alternate with quadrupole and other focusing magnets to

keep the particle beams from spreading.

3. Storage Rings and Colliding Beams

In order to collect an increased number of accelerated particles, each output pulse from

an accelerator can be placed in a circular ring where the net energy is not increased further.

This ring, called a storage ring, contains magnetic fields to keep the particles moving in a

circle. Large number of particles per bunch increases the probability that the bunch will cause

some event to occur.

Colliding beam of accelerated particles can yield more energy (for instance, to create

new particles) than collisions in which one particle is at rest. To understand this phenomenon,

recall the classical, completely n elastic collision of a particle of mass m moving at speed v

with an identical particle at rest. Conservation of linier momentum requires that the two

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particles move together after the collision at a speed v/2. Therefore the kinetic energy

converted to other forms of energy in the collision is 1/2mv2-1/22m)(v/2)2=1/2(1/2mv2). Just

one-half the initial kinetic energy has been converted to some other kind of energy.

On the other hand, if two particles with equal and opposite momentum hit and stick

together, they are motionless after the collision and all the nitial kinetic energy is converted

into other kinds of energy. So if you want to convert 1 MeV of KE into some other form of

energy, you could (classically) hit a rest particle with a 2-MeV equal-mass particle. Or you

could do it with two ½-MeV equal-mass particles in a head-on collision. Instead of requiring a

2-MeV accelerator, you could use ½ MeV accelerator and some cleverness. Since ½ MeV/ 2

MeV=1/4, the particle in the colliding beams need only one-fourth of the kinetic energy of the

particles bombarding a fixed target in the classical case. In the relativstic case, this ratio

decreases rapidly below one-fourth as the required energy increases. Therefore existing

accelerators have been altered, and new ones designed and built, to provide head-on collision

and the energy that would other-wise require an accelerator with a much higher energy for

fixed-target collisions.

To show this difference, we can prove that :

7.5

Recall that KEth is the minimum kinetic energy needed to make a reaction occur when

is negative and the target particle is at rest. The mass of the target particle is m X.

The total rest mass of all the particles before the reaction is , and the total rest mass of all

the particle after is .

Example 6.12.

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Consider a hypothetical reaction of the form , where the proton is at rest and particle

P has a rest mass of 2000 TeV/c2. The rest mass of a proton or antiproton is 938 MeV/c2.

Calculate the minimum kinetic energy of the antiproton.

Solution:

The metric prefix M represents 106, and T represents 1012, so

MX=938x10-6TeV/c2, m=2(938x10-6 TeV/c2),

And

m’=2.000 TeV/c2,

Then,

and

So, we get:

Based on the fact that physicist would have to provide a 2130 TeV antiproton to obtain

the 1.998 TeV energy increase. Such antiproton would require a circular accelerator

approximately the size of the moon. Or a 0.999 TeV antiproton could collide head on with

and annihilate a 0.999 TeV proton, as in done at Fermilab. From that viewpoint positive

protons are accelerated in a clockwise direction, and negative antiproton are accelerated in a

counterclock wise direction, but the beams are kept separate except a a designated collision

point.

Relying on colliding beams has a major disadvantage: The probability is rather small that

a tiny particle from one low-density beam will actually collide head-on with a tiny particle

from the other, equality tenuous beam. An important parameter of a beam, therefore, is the

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number of reactions that will occur per nit time per unit cross section. This quotient of

reaction rate and cross section is called the luminosity. The design goal for the

Superconducting Super Collider (SSC) is a luminosity of 1037s-1m-2.

The huge size of the proposed SC indicates that even higher energy will probably require

quite different designs. Accelerator designs that use high-power laser beams are being

considered, perhaps in conjuction with plasmas. Another line of research is exploring the use

of the electric field from a high-current, low-energy beam of charged particles to accelerated a

second low current beam to high energies, an idea reminiscent of the step-up transformer

concept.

Although we couldn’t discuss all the types of detectors and variations and refinements of

accelerators, we did give a reasonably complete introduction to those topics. The detectors we

covered obtain information about radiation by means of an electromagnetic signal, path

visualization, or both. We discussed both linier and cyclic accelerators to provide the final

energy, or linier accelerators in conjunction with storage rings and/or colliding beams.

TAsK-13

1. (a) Why do high energy accelerators need such a large radius?

(b) Express B for a constant radius synchrotron in terms of the particle speed and

constant.

(c) Why is the actual r larger that pmax/qBmax for large synchrotrons?

2. The SLAC electron-positron collider is able to create groups of particles with a total

rest energy of 100 GeV. Calculate the threshold energy for the production o such a

group of particles by a positron beam incident on a metal terget.

3. The luminescence process in solids or liquids is characterized by A Stokes shift, a

different between the wavelength of the emited light and the wave-length of

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stimulating light. It might be thought that this fact does not concern us in the

scintillation counter application, where stimulation does not involve visible light, but

the Stokes shifts is vital to the functioning of the counter. Explain!

4. Either scintillation or semiconductor detectors can be used in the following way to

determined the mass spectrum of a beam of heavy chred particles. The particles first

transver a thin detector in which they lose only a small fraction of their energies. The

current pulse for each particle is proportional to –dK/dx, which is in turn proportional

to 1/v2, where v is the velocity of the particle. The particles then pass into a thick

detector and lose all their energy; the current pulse for each particle is proportional to

K. Explain how to combine these data to obtain the particle masses.

5. A much older device than bubble chamber is the cloud chamber, in which the ions

produced by charged particles serve as nuclei for condensation in a supersaturated

vapor. Although this device gives the same type of track information as the bubble

chamber, the letter is much more popular in high energy physics. Why?

6. How would you construct an electron detector that would count only electrons with

energies greater than 500 MeV?

7. Radionuclides are useful source of small amount of power in space vehicles, remote

communications stations, and similar applications. Calculate the power in wats per

kilogram of Ce144 , a - emitter with an average energy of 0.10 MeV and =285 days,

and of Po210, an emitter with an energy of 5.30 MeV and =138 days.

8. How and why does the effective length of a semiconductor nuclear particle detector

depend on bias voltage? If we wish a large efective length, why must the device be

constructed a very pure, high resistivity semiconductor?

9. Why is the use of neutrons as reactants in a fusion reactor not an atractive possibility?

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10. There are a few delayed neutrons in fission. Why are there no delayed protons? Why

are there no positron?

COURSE 14:

I. Basic competency:

Students are able to understand the general theory of relativity to models of the universe

II. Indicators:

1. Able to explain the general theory of relativity

III. Subject Matter

1. Overview

Cosmology is the study of the universe, which is done by means of models. Most

cosmological models utilize the cosmological principle, the assumption that the universe is

homogeneous and isotropic. That is, the assumption is made that the properties of the universe

are the same at all places and in all directions. For example, the cosmological principle

requires the density, , of the universe to be constant in space (but not in time). This

assumption, of course, isn’t true at small scale. Matter exist in clumps, ranging from leptons

and quarks to you and me to galaxies and to super clusters of galaxies. In fact, the farther anf

the finer that astronomers can see, the more nonuniformity they seem to find in the

distribution of mass. At the universal scale, however, the cosmological principle seems to be a

reasonable first assumption. Many early cosmological models made the earth the unmoving

center of the universe. More modern cosmologies consider the earth to have no exceptional

location in a vast universe made up of about 1011 galaxies, each containing an average of 1011

stars. However, modern cosmologists do still assume one earth-centered concept: The laws of

physics that we discover here on earth are the same as the laws everywhere else in the

universe.

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2. Cosmological Models And Expansions

The cosmological principle states that the properties of the universe are the same

everywhere. Therefore the laws that cause these properties must be the same everywhere.

Regional changes in the laws would give regional differences in measured properties. Thus if

an astronomer finds that all the spectral lines in the light from a star or group of stars have

shifted to different frequencies, he assumes that this shift is the result of known phenomena,

such as the Doppler effect. Only as a last resort would the astronomer assume that the laws of

nature are different at the source of the light.

Vesto sliher measured the first frequency shifte of spectral lines from Galaxy at the

Lowell Observatory if Flagstaff, Arizona, in 1912. It was a blueshift( a shift to higher

frequencies) and it indicated that the Andromeda galaxy has a relative velocity component of

about 300 km/s toward us. He then continued his measurement on other galaxies. Meanwhile,

Einstein applied his GTR to the whole universe, publihsing the result in 1917. He utilized the

cosmological principle, assuming a constant density and a constan positive curvature of

spacetime. He also decided that he wanted a static universe. Newton had realized that a finite

static universe was impossible. A finite universe would eventually collapse because of the

gravitational attraction of each mass for all the others. However, Newton throught

(incorrectly) that an infinite static universe was possible.

Einstein similarly found that his field equation didn’t yield a static universe solution,

even for an infinite universe. To arrive at a static equalibrium, he introduced a term called the

cosmological constant, , into his equations. (This step was equivalent to inventing a

repulsive fifth force of nature to oppose gravity). However, because of later astronomical data

and proof that the static equilibrium was unstable. Einstein’s model fell from favor. Slipher’s

continued measurements showed that the Andromeda blueshift was unusual. By 1923, of

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galaxies studied, 36 had redshifts and only five had blue shifts. This result seemed to relate to

another cosmological model published in 1917, the decidedly odd model of the Dutch

astronomer Willem de Sitter.

Attempting to arrive at a theory for a static universe, de Sitter used the cosmological

constant. He also used a constant density, but did so by letting equal zero. The de Sitter

universe contained no matter at all. However, if particles of mass were added to this model,

they acted as if they were moving away from one another, just as most of the galaxies seemed

to be moving away from us. Then, in 1924 at Mt. Wilson in California, Edwin Hubble began

to improve techniques to measure distances to galaxies. With the help of Milton Humason, he

discovered what appeared to be a linier relationship between the distances of the galaxies

from us and their speeds. Hubble and Humason determined that the few galaxies moving from

us and their speeds. Hubble and Humason determined that the few galaxies moving toward us

are nearby. More-distant galaxies are all moving away from us. The universe seemed to be

expanding. This linier relationship is called the Hubble law.

vr=Hs 7.1

Where vr is the recession velocity, or the velocity at which a galaxies is being expanded away

from us; s is the distance the galaxy is away from us, and H is a quantity best labeled the

Hubble parameter. Apparently constant for all galaxies now (and therefore commonly called

the Hublle constant), H may change with time. It has no precisely determined value because

of the great uncertainties in measuring distances for galaxies. The current value of the Hubble

parameter is H=(238) km/s.Mc.yr. For most of our purposes, we’ll use the average value of

23 km/s.Mc.yr. Recall that Mc.yr is our abbreviation for 106 light years and that one light year

is the distance traveled in one year at the speed of light.

Example 7.1

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The recession velocity of a galaxy in Ursa Major is 1.5x104 km/s. How far away from us in

this galaxy?

Solution: Using vr=1.5x104 km/s and the average value of the Hubble parameter, H=23 km/s.

Mc.yr in r=Hs, we get

Or 650 million light years away from us.

In the 1920s, cosmological theory also indicated that the universe was not static, but

was expanding. The chief theoreticians were Alexander Friedman, a Rusian, and Georges

Lemaitre, a Belgian. Their models of the universe used the cosmological principle with a

nonzero dentsity. Lemaitre included a cosmological constant (but not Einstein’s value), but

Friedmann used no cosmological constant. Einstein realized that his rather arbitrary addition

of kept him from discovering the concept of the expanding universe and called it” the

biggest blunder of my life”.

When you first hear of the expanding universe, your mental picture may well be of a

bunch of stars and stuff all moving outward through space from some central point. That is

not what is happening according to the dynamic general relativistic cosmologies. Rather,

the”stars and stuff” are relatively fixed in space, and it is space itself that is expanding. Our

problem is trying to visualize what is happening is that we need five dimensions, four for

spacetime and one for curvature, to properly display expansion. However, suppose that we

again ignore one dimension and discuss the evolution of a three-dimensional representation in

time to try to get an understanding of the expanding universe. Figure 7.1 shows one model-the

raisin bread dough model. However, we’ll use the common spherical rubber-ballon model.

First we mark longitude and latitude lines on the ballon, jus like those on globe of the

earth. Then randomly here and there we glue pieces of confetti on the surface with a flexible

147

glue. Each piece of confetti the represents a galaxy (or even some larger clustering of

galaxies). All displacements are restricted to the surface of the ballon. The flexible glue

enables us to move wach piece of confetti a bit from the latitude and longitude coordinates

peculiar to that piece. This type of motion gives rise to what is called the peculiar velocity of a

galaxy, that is, its motion with respect to its coordinates in space.

Figure 7.1 (a) The raisin-bread-dough model of the expanding universe

Figure 7.1 (b) The spherical ruber-ballon model of theexpanding universe

Now let’s blow up the ballon slowly. What happens to the pieces of confetti? Ignoring

their peculiar velocities, we see that their coordinates don’t change, but the distances between

them ( as measured along great circles on the surface) do. Every piece of confetti moves away

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from every other piece of confetti. The distance between pieces is given by s=r, where r is

the radius of the ballon and is the angle between two radii from the centre to two pieces of

confetti. The latitude and longitude angles don’t change, so is also constant. The recession

velocity of the pieces is vr=ds/dt=(dr/dt) . Therefore vr at any time is directly proportional to

how far apart the pieces are that time.

Example 7.2

Prove that vr at any time is directly proportional to how far apart the pieces are at that time.

Solution:

First, s=r gives =s/r. We then substitute =s/r into vr=(dr/dt) to arrive at vr =[(dr/dt)/r]s.

At any time, (dr/dt)/r is the same for all points on the balloon, showing that v r is directly

proportional to s at that time.

The Andromeda galaxy is about teo million light years away from us and has a

velocity component of 300 km/s toward us. We might be tempted to try to analyze its motion

using vr=Hs, but we would be wrong to do so. The cosmological models undergirding the

Hubble law asume a uniform density of matter at any given time and therfore a uniform

curvature of spacetime. On the scale of only a few million light years, matter is manifestly not

uniformly distributed. The 300 km/s is mainly a component of a peculiar velocity, not a

recession velocity.

You should remember that the expansion of the universe involves the expansion of

space itself in most cosmological models. Therefore we can write the distance s between any

two distant objects as the productof a scale factor R times the change in the space coordinates

between the objects. This approach is similar to converting a distance on a map to an actual

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distance by multiflying by the map’s scale factor; for example, 2.6 (map cm)x50 km/(map

cm)=130 km. Solving vr=ds/dt=Hs with s proportional to R gives

7.2

Example 7.3.

Assume that R follows a simple power law, R=Ctn, where C is a constant. Find the time-

dependence of the Humbble parameter!

Solution:

Differentiating, we have

Therefore

Because the scale factor is changing with time, distance between points are also changing

with time, including the distance between adjacent crests in waves. This distance is the

wavelength of the wave. Consider a wave emitted at time to with wavelength o and

frequency o when the scale factor was Ro. We detect the wave at time t with wavelength

and frequency when the scale factor is R. Since is directly proportional to R and when

the scale factor is R. Since is directly proportional to R and is inversely proportional to ,

we can write

7.3

Example 7.4:

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Is the redshift of distant galaxies a special relativistic Doppler shift as given by equation :

Solution:

The answer is no. Equation 7.5 shows that this redshift (lower frequencies and lower

wavelengths) is a direct result of the expansion of the universe, because expansion means that

R(now) is greater than R0 (when the galxies emitted the light.

Most cosmological models predict that the expansion of the universe is showing down

or decelerating because of the attraction of the masses of the universe to one another. If there

is a deceleration, dR/dt decreases with time, making d2R/d2t negative. The deceleration

parameter, q, is a measure of this decrease:

7.4

The deceleration parameter has no units and is positive for a decelerating universe.

Example 7.5

Assume that R follows a simple power law, R=Ctn, where C is a constant. Calculate the

deceleration parameter.

Solution:

Then

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Therefore

An area of great current interest is the determination of the average density of the

universe. Below a certain critical density of about 10-26 kg/m3, the universe will eventually

stop expanding and begin to collapse, as shown in figure 7.3.

Figure 7.3 The scale factor of the universe as a function of time

for certain cosmological model

That collapse is sometimes called the “Big Crunch”. In some models, the universe will

then begin to expand again, with a new oscillation, turning the Big Crunch into “Big Bounce”.

The best current estimates of the average density of the universe range from 0.03 to 10 times

the critical density. How old is the universe? We can get a rough idea from the relation R=Ctn,

which gives H=n/t. Letting the beginning be t=0, our time now is t=T, the age of the universe.

Therefore T=n/H=n(155)x109 yr. Different cosmological models give different values for n,

with typical value of ½ to 2/3, giving T=(94)x109 yr. However, we can’t really answer the

question with our current level of understanding.

If we mentally travel backward in time, our expanding universe becomes a contracting

universe, with the scale factor approaching zero as t approaches zero in most cosmological

models. That means that the univese started out with a very small scale factor and

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corresponding very small distances between everything and then expanded. This expansion is

called the Big Bang. George Lemaitre is called “ the farther of the Big Bang”, not because he

was a catholic priest, but because he first investigated the idea (but gave it a different name).

Despite its name, the Big Bang is not an explosion in space, but a rapid expansion of space.

TASK-14

1. The steady-state model of the universe assumes a “perfect cosmological principle,”

one that says the properties are the same at all times, at all places, and in all directions.

Explain why this model then requires that matter be created constantly throughout an

expanding universe.

2. Explain why the universe has no edge and no center if it can be modeled by the

expanding ballon.

3. At what distances do recession velocities and peculiar velocities have about the same

magnitude?

4. An astronomer determines that a galaxy in Virgo is recending from us at 12 million

m/s and that it is 800 million ligth years away from us. Are these values reasonable?

COURSE 15:

I. Basic competency:

Students are able to understand experimental, and applied nuclear physics

II. Indicators:

1. Able to explain the history of the universe

III. Subject Matter

1. The History of Universe

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If we try to travel too far back in time, we run into a problem. The scale factor

becomes so small that the corresponding separations and densities require a unification of

quantum mechanics and general relativity to explain what is happening. This condition

corresponds to a time of about 10-45s (called Planck time) after the beginning of the Big Bang.

However, we simply don’t have the unified theory needed to define just what time is before

10-45 s, and so cannot truly talk about the “when” of the beginning of the Big Bang. Figure 7.4

presents an overview of the history of the universe, with all the question marks in the uper left

of the figure emphasizing our lack of understanding before the time 10-43s.

Figure 7.4 The History of The Universe

In the standard model of the Big Bang, all four forces of nature were unified before the

time 10-43s. At that time, the equacalence of the four forces broke down and gravity began to

act separately. At 10-43s, the temperature of the universe in the standard model was about 1032

K. Since Boltzman’s constant, 8,62x10-5 eV/K, is approximately 10-13 GeV/K, the average

energy per particle, approximately kT, was about 1019 GeV. By 10-35 s, the decrease in density

and temperature ( to about 1027 K) of the expanding universe brought about the separation of

the strong nuclear force from the electroweak force. Therefore the 10 -43s to 10-35 s interval is

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the GUTs era, or the time for which the grand unified theories explain what is happening. One

prediction of GUTs is that the baryon number need not be exactly conserved, so it is in this

era that the GUTs predict a very slight excess of quarks over antiquarks.

In inflationary models, this separation of the strong nuclear force corresponded to a

phase change (like the phase change between liquid and gas phases of a substance, with its

corresponding latent heat of vaporization). As a result, one inflationary model predicts that the

scale factor, R, increased by 1050 in 10-32 s. This model, with a relatively small nnumber of

assumptions, claims to explain the overall homodeneity and isotropy of the universe is

apparently so close to the critical density, the apparent rarity of magnetic monopoles, and the

small scale fluctuations that eventually led to the formationof galaxies and other features of

the universe.

The universe-a mixture of quarks, leptons, and gauge bosons-continued to expand and

cool from the inflationary period to 10-6 s when the quarks began to bind together to form

baryons and their antiparticles. Before about 10-2s, most radiation had enough energy that pair

production by two photons balanced baryon-antibaryon pair annihilation. After that time,

most of the radiation fell below the threshold energy for pair production. Then, because of the

slight excess of baryons over antibaryons, almost all the antibaryons and most of the baryons

annihilated one another. After similar processes for the electron-positron pairs ( at about t=14

s), the universe was left with its current great preponderance of matter over antimatter.

Example 7.6

Why were most of the positron completely annihilated so much later than most of the

antibaryon?

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Solution

This complete annihilation couldn’t occur until most of radiation had energy below the

electron-positron pair production energy. Since electrons have much less rest mass-energy

than baryons, this event wouldn’t occur until the universe had expanded and cooled well past

the time when baryon-antibaryon pair production ceased.

At about t=1 s, most electrons no longer had enough energy for the reaction

. The neutrinos also had a lower average energy and matter of the universe was

more spread out, so equilibrium reactions involving the absorption of neutrinos began to

decrease rapidly. Thus at this time the flux of neutrinos anf antineutrinos throughout the

universe”decoupled” from the rest of the universe. Because of the extraordinarily low cross

section for neutrino absorption, this flux is still present, although cooled by expansion to an

average temperature of about 2 K (according to the standard model, bu no one has been able

to figure out an experiment to test this prediction).

Also at about t=1 s, the ratio of protons to neutrons was , where E is the rest

energy difference between the two nucleons, (mn-mp)c2. With a temperature of roughly 1010K,

this relation gives about none protons for every two neutrons. However, neutrons are

radioactive with a half-life of 630 s, so the proton-neutron ratio continued to increase until

t=225 s. At that time, T was below 109 K ad very few photons had energies greater than the

2.22 MeV binding energy of the deuteron. Therefore a proton anf a neutron could combine

and remain combined asa deuteron,H-2. This combination halted the decay of the free

neutrons at a ratio of about seven protons to every one neutron. The nuclide H-2 can absorb a

neutron to form H-3 or a proton to form He-3, and the H-3 and He-3 can absorb a proton and

a neutron respectivelyto both form He-4, but there the building of nuclei almost completely

stops. The reason is that no stable or long-lived nuclides have a mass number of five, and the

cross section for further reactions is very small.

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Example 7.7

Almost all the protons and neutrons in the seven-to-one ratio either made He-4 or remained as

H-1. Therefore, by mass, what percent of H and He should there have been?

Solution:

The nuclide He-4 contains 2 protons and 2 neutrons, so we need at least 2 neutrons in the 7:1

(proton-to-neutron) ratio. The ratio then is 14 protons to 2 neutrons. The 2 neutrons and 2 of

the protons make up 1 He-4 nucleus. The remaining 12 protons have no neutrons to combine

with and so become H-1 nuclei. Since the masses of H-1 and He-4 are about 1u and 4 u,

respectively, there were 12 u of H-1 for every 4 u of he-4, and a total of 12 u+ 4 u= 16 u.

therefore H was 12/16=75 percent and He was 4/16 percent by mass.

The average energy per particle at that time was still much too highh to allow

electrons to start binding to nuclei to form atoms. Atomic binding didn’t occur until after 1013

s, or nearly 700,000 years later, when the temperature had dropped to about 3000K. After that

very few free electrons were left to scatter electromagnetic radiation and so the flux of

electromagnetic radiation throughout the universe decoupled from the matter of the universe.

The radiation has continued to expand its wave lengths to form the 2.7 K blackbody radiation

that we receive (spatially uniform to 0.01 percent) from space. This radiation is one of the

main experimental pillars of the standard model of the Big Bang.

Example 7.8

By approximately what ratio has the universe expanded since t=700.000 years?

Solution

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For blackbody radiation known that λp=constant/T. Therefore we can write:

or

At t=700,000 year, T was about 3000K and now it is 2.7 K (or about 3K), giving

R/Ro=3000K/3K=1000. Universal distances are proportional to the scale factor R, so the

R/Ro ratio is also the distance ratio. Thus the universe has expanded by approxomately a

factor of 1000.

The energy radiated from stars is produced in fusion reactions in which H-1 becomes

He-4. When the hydrogen is about used up, the inward gravitational pressure exceeds the

outward radiation and gas pressure, and the star’s core begins to contract. As it does so, its

gravitational potential energy decreases, and the kinetik energy of its atoms increases. For

stars of sufficient mass, there is both enough energy anf the necessary density to begin helium

fusion.

Two He-4s first fuse to form Be-8. The Be-8 has an extremely short half-life (10 -8 s)

but is compensated for by an unsually high resonance cross section for He-4 absorption at

these energies. Therefore a reasonable fraction of the Be-8 fuses`with more He-4 to give

stable C-12. (Three He-4’s have fused to form one C-12 in what is called the triple-alpha

process). Then succesive fussion with He-4 give O-16, Ne-20, and Mg-24. All these reactions

release energy to heat the star so that C-12 and O-16 can fuse to form elements having higher

and higher atomic numbers.

The binding energy per nucleon curve peaks out at Fe, so exoergic fusion reactions

stop with Fe, but succesive neutron captures can continue the synthesis of heavy elements. A

step in the evolution of some massive stars in an explosion of the star ( a supernova). Such

explosions release into space the heavy elements that ware produces by the earlier processes.

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In space, the debris and other interstellar matter gravitationally bunch together to form stars

and planets. Thus it is more than poetic to say that the Earth is formed of stardust

TASK-15

5. Intensity measurements indicate that a galaxy in Bootes is 1.7 x 109 c.year from us.

What would you expect its recession velocity to be?

6. The Hubble length is defined as the distance at which the recession velocity equals c.

(a) Find the limiting values of the Hubble length

(b) What is the Hubble length for a galaxy 5 Gc.yr from us?

7. A galaxy can’t move through space at velocities greater than c, but both its recession

velocity greater than c, but both its recession velocity and

COURSE 16: FINAL TEST

Singaraja, Dec 15th 2011

Drs. IBP. Mardana, M.Si

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