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Chapter 1 Nuclear physics ........................................................................................................................3

1.1 The nuclear atom and Radioactivity............................................................................................3

1.1.1 Nucleus structure of atoms Atomic nucleus..................................................................3

1.1.2 Constituents of the atom...................................................................................................41.1.3 atomic measurements .......................................................................................................5

1.1.4 Natural radiation phenomenon Decay Half-life ............................................................5

1.2 Nuclear fission and fusion...........................................................................................................7

1.2.1 Nuclear reaction ...............................................................................................................7

3.2.2 Mass defect & binding energy .........................................................................................7

1.2.3 Fission ..............................................................................................................................8

1.2.4 Nuclear power station ......................................................................................................9

1.2.5 Fusion of light nuclei .....................................................................................................10

1.3 19 Worked examples ..............................................................................................................10

Chapter 2 Fundamental particles.............................................................................................................22

2.1 Particles, antiparticles and photons ...........................................................................................22

2.2 Particle interactions...................................................................................................................22

2.2.1 Electromagnetic force and Feynman diagram................................................................22

2.2.2 The weak nuclear force and Feynman diagram..............................................................22

2.3 Properties of particles and antiparticles ....................................................................................22

2.4 Quarks and Antiquarks..............................................................................................................22

2.4.1 Quarks and antiquarks properties...................................................................................22

2.4.2 Quark combinations .......................................................................................................22

2.4.3 Quarks and beta decays..................................................................................................222.5 45 Worked examples ..............................................................................................................22

Chapter 3 Electromagnetic Radiation and Quantum Phenomena ...........................................................23

3-1 The photoelectric effect ............................................................................................................23

3-2 Collisions of electrons with atoms............................................................................................23

3.2.1 The Bohr model and Energy levels ................................................................................23

3.2.2 The emission and absorption of photons........................................................................23

3.2.3 Atomic spectrum ............................................................................................................23

3.2.4 Fluorescent tube explanation..........................................................................................23

3-3 Wave-particle duality................................................................................................................23

3-4 42 Worked examples......................................................................................................................23

Chapter 4 Operational amplifier (op amp)..............................................................................................23

4-1 differential amplifier .................................................................................................................23

4-2 Negative feedback amplifiers ...................................................................................................25

4-3 Op amp as a comparator ...........................................................................................................26

4-4 Summing amplifier ...................................................................................................................29

4-5 10 Worked examples......................................................................................................................30

Chapter 5 Relativity ................................................................................................................................38

5-1 basic assumptions of special relativity......................................................................................38

5-2 relativity of time and space.......................................................................................................385-3 Relativistic mass and mass-energy equation.............................................................................38


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II

5-4 9 Worked examples........................................................................................................................38


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Chapter 1 Nuclear physics

1.1 The nuclear atom and Radioactivity

1.1.1 Nucleus structure of atoms Atomic nucleus

1.1 The nucleus structure model of atom

In 1909-1911, the British physicist Rutherford (1871-1937) and his

assistants did experiments using particle scattering. Fig. 1-1 shows the

experimental setup. Using particle to shine on gold foil, some particle

change direction after passing through the gold foil. This is because the tiny

charged particles inside the gold atom have Coulomb interactions with the

particles. This phenomenon is called the scattering of particle.

The result of the experiment was that most particles passed through the

gold foil and almost all moved in their previous direction. But there were a

tiny number of particles which showed a large deflection.

Rutherford made precise records of the number of particles scattered in

various directions. Based on this, he put forward the nucleus structure model

of the atom. There is a tiny nucleus at the center of the atom called the atomic

nucleus. All the positive charge and nearly all the mass of the atom are

concentrated in the atomic nucleus. The electrons carrying a negative charge

move in space outside the nucleus.

According to this model, since the atomic nucleus is tiny when it passes

through the gold foil, most of the particles are far from the nuclei and feel

a very small repulsive force. Their movement is hardly influenced. Only a

very small number of particles pass close to the atomic nucleus, and are

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affected considerably by the nucleus Coulomb repulsive force and so make a

large angle deflection (Fig. 1-2)

particles

1.1.2 Constituents of the atom

Inside an atom: a positively charged nucleus composed of proton and

neutrons, Electrons that surround the nucleus.

Note: inside the atom, the number of proton = the number of electron, and if

the atom do not loss or gain electrons, it is neutral (that is no charge)

Nucleon: a proton or a neutron in the nucleus.

Inside an atom:

Charge ( /C ) Mass ( /Kg )

a proton +1.601019 1.6710-27

a neutron 0 1.6710-27

an electron 1.601019

9.1110-31

Isotopes: atoms of the same element that has the same numbers of proton but

different numbers of neutrons.

Represent an atom: AZ X

X: chemical symbol

A: total number of protons and neutrons, sometimes called nucleon number

ormass number.Z: number of protons, sometimes called atomic numbers.

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So, the number of neutrons in the nucleus = AZ

Specific charge:

Forcharged particles, specific charge is defined as its charge divided by itsmass, that is

argarg

ch especific ch e

mass

Unit: C/kg

1.1.3 atomic measurements

The unified atomic mass unit (u) is used for measuring the masses of

atomic particles. And

271 1.66 10u kg

For example

Mass of proton: 1.00728 u

Mass of neutron: 1.00867 u

Mass of electron: 0.00055 u

1.1.4 Natural radiation phenomenon Decay Half-life

1 Natural radiation phenomenon

The property of a substance emitting rays is called radioactivity. Elementswith radioactivity are called radioactive elements.

The phenomenon of elements spontaneous radiation of rays is called

natural radioactivity phenomenon.

2 Three types of radiation

(i) Alpha radiation

Alpha particle is the atom ; symbol is42He4

2 , sometimes in symbol.

An unstable nucleus of an element X emits an alpha particle; the product

nucleus is a different element Y.

Equation below:4 4

2 2

A A

Z ZX Y

Note: conservation of mass number and atomic number.

(ii) Beta decay

Beta is electron, in symbol 01 or ,

Equation below:

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0

1 1

A A

Z ZX Y

: is called antineutrino, which has no charge and no mass.

(iii) Gamma radiation

In symbol , it is emitted by a nucleus with too much energy, following

an alpha or beta emission.

3 Half-life

We can use a half-life to indicate the rate of decay of radioactive elements.

The amount of time required for half of the nuclei of a radioactive

element to decay is called the half-life of the element.

Note: different radioactive elements have different half-lives, and

sometimes the difference is very big.The measure of radioactivity is the number of radioactive disintegrations

per second, called the activity (in symbol A). The SI unit of activity is the

Becquerel (Bq), which is one disintegration per second.

And if unstable atoms are present at time zero, then the number

remaining at some time t is

0N

0tN N e , where is a proportionality

constant (called the decay constant).

Since the activity is proportional to the number of radioactive atomspresent, the activity satisfies an exponential decay law also: 0

tA A e

When the time is equal to one half-life,1

2

t , the number of atoms

remaining is 02

NN , thus the equation can be written as0

tN N e

1

2te

We can use this relation to obtain the half-life in terms of .

12

0.693t

Thus, we can get1

2

0.693

t 0

t, therefore the equation N N e can be

written as:

12

0.693

0

tt

N N e

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1.2 Nuclear fission and fusion

1.2.1 Nuclear reaction

In nuclear physics, the process of making a new nucleus when a nucleus is

bombarded by other particles is called a nuclear reaction.

For example, if a high-energy particle strikes and is absorbed by a

nucleus of nitrogen-14, the new nucleus immediately decays to form a

nucleus of oxygen-17 and a proton. This is an examples of a nuclear reaction.

It can be described by the following equation:14 4 17 17 2 8N He O 1p

Note: the mass and charge numbers are all conserved in a nuclear reaction.

There is energy change in a nuclear reaction. The energy released in a

nuclear reaction is called nuclear energy.

Einsteins relativity theory points out that energy has mass. If an object

gains energy, it gains mass. If it loses energy, it loses mass. The change of

energy is linked to the change of massE m by this equation:2E mc

191 1.60 10eV J

Where c is the speed of light: 83 10 /c m s

With nuclear particles, energy is often measured in eV:

271 1.66 10u kg 2

And with nuclear particles, mass is usually measured in u

( ). by converting 1 u into kg and applying E mc , it

is possible to show that:

3.2.2 Mass defect & binding energy

A helium-4 nucleus is made up of 4 nucleons (2 protons and 2 neutrons).

The calculation below (Fig. 1-3) shows that the nucleus has less mass than its

four nucleons would have as free particles. The nucleus has a mass defect of

0.029267u.

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p p

n n

p

p

n

n

Helium-4

nucleus

4.002603u

1.007276u

1.007276u

1.008665u

1.008665u

4.03187u

Difference in mass = 0.029267u

(mass defect = 0.029267u)

Fig. 1-3 mass defect

The binding energy of a nucleus is the energy equivalent of its mass

defect. (Binding energy is the energy holds the nucleus together). Therefore,it is the energy needed to separate the nucleus into its component parts. For

, the mass defect is42He

42

2(2 ) 0.029267e

p n Hmm m m u

0.029267 931.5 27.3

And 1 u is equivalent to 931 MeV, the binding energy is

E MeV

238 1 139 97 192 0 56 36 03U n Ba Kr n

1.2.3 Fission

In nuclear physics, a reaction in which a heavy nucleus splits into nuclei withsmaller masses and releases nuclear energy is called fission.

During nuclear fission, a heavy nucleus splits to form two nuclei of

roughly the same mass, plus several neutrons. Fission occurs when a neutron

hits and is captured by the nucleus. For example, here is a typical fission

reaction for uranium-238:

Chain reaction: the fission reaction above is started by one neutron. It

gives off neutrons which may cause further fission and so on in a chain

reaction (Fig. 1-4).

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1.2.4 Nuclear power station

Nuclear power stations generate electricity using nuclear energy. Its core

facility is a nuclear reactor. In the reactor, there is a steady release of heat as

fission of uranium-235 takes place.

Uranium 235 catches slow neutrons easily and catches fast neutrons

difficultly. The neutrons produced in fission all have high speed, and are

difficult for uranium 235 to catch to encourage fission. To slow down their

speed, decelerating materials are placed around uranium bars. Fast neutrons

collide with decelerating materials and lose energy to become slow neutrons.

Commonly used decelerating materials are graphite, heavy water or

ordinary water (sometimes called light water).

To adjust the number of neutrons and control the rate of reaction, some

control bars have to be inserted between the uranium bars. Control bars are

made ofcadmium. Cadmium strongly absorbs neutrons. If a reaction is too

fierce, place the control bars more deeply so that it can absorb more neutrons,

and the chain reaction will be slowed down. Otherwise, the control bars

should be pulled out. A computer adjusts the lift-up or push-down of thecontrol bars automatically keeping the reactor at a constant power and

operating safely.

Most of the energy released from nuclear fuel fission changes into heat,

and increases the temperature of the reaction area. Water or liquid-state metal

sodium flows in and out of the reactor cyclically and carries the heat inside

the reactor to the outside to be used to generate electricity. Meanwhile, the

reactor cools down and operates safely.

Too much radiation causes harm to human and biological objects, so

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special attention must be paid to prevent the leak of radiation rays and

radioactive materials in nuclear power station construction. This will avoid

damage of rays to humans and the radioactive pollution of radioactive

materials to water supply, air and the working sites. For this purpose, the

outside of a reactor needs very thick concrete walls to stop various rays

escaping from the fission products.

1.2.5 Fusion of light nuclei

A reaction in which light nuclei combine to form nucleus with a larger mass

and also release nuclear energy is called fusion.

For example, when a deuteron and a triton combine to form helium

(meanwhile a neutron is released), 17.6MeV energy is released. The nuclear

reaction equation is2 3 4 11 1 2 0H H H n

The advantages of a fusion reactor will be:

(i) Fuels will be readily obtainable. For example, deuterium can be extracted

fro sea-water.

(ii) The main waste product, helium, is not radioactive.

(iii) Fusion reactors have built-in safety. If the system fails, fusion stops.

1.3 19 Worked examples

1. In the reaction shown, a proton and a deuterium nucleus, , fuse together

to form a helium nucleus,

2

1H

3

2He

1 2 31 1 2p H He Q

What is the value of Q, the energy released in this reaction?

Mass of a proton = 1.00728 u

Mass of a nucleus = 2.01355 u21H

Mass of a nucleus = 3.01493 u32He

Solution:

The mass defect is given by

2.01355 1.00728 3.01493 0.0059m u u u u

0.0059 931.5 5.5Q MeV

Since 1 u is equivalent to 931.5 Mev,

The energy released is given by

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2. For a nuclear reactor in which the fission rate is constant, which one of the

following statements is correct?

A There is a critical mass of fuel in the reactor.

B For every fission event, there is, on average, one further fission event.

C A single neutron is released in every fission event.

D No neutrons escape from the reactor.

Solution:

Nuclear fission occurs when a massive nucleus splits into two less-massive

fragments. Fission can be induced by the absorption of a thermal neutron.

When a massive nucleus fissions, mass is transformed into energy when the

binding energy per nucleon is greater for the fragments than for the original

nucleus. Several neutrons are also released during nuclear fission. These

neutrons can, in turn, induce other nuclei to fission and lead to a process

known as a chain reaction.

(B) is correct.

3. (a) When a nucleus of uranium -235 fissions into barium -141 and krypton

-92, the change in mass is28

3.1 10 kg

. Calculate how many nuclei mustundergo fission in order to release 1.0 J of energy by this reaction.

Solution:

Energy released by a nucleus is given by2 28 8 2 113.1 10 (3 10 ) 2.79 10E mc J

Number of nuclei required 1011

1.03.6 10

2.79 10

J

J

(b) A nuclear power station produces an electrical output power of 600MW. If

the overall efficiency of the station is 35%, calculate the decrease in the mass

of the fuel rods, because of the release of energy, during one week of

continuous operation.

Solution:

The output power from reactor:600

17140.35

MWP MW

6 15

1714 10 7 24 60 60 1.04 10E Pt J

Thus, energy output from fuel rods in one week is given by

Therefore, decrease in the mass is given by

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15

2 8 2

1.04 1011.6

(3 10 )

Em g

c

4. The reaction shown below occurs when a proton and a deuterium

nucleus, , fuse to form a helium nucleus, .21H3

2He

1 2 31 1 2p H He Q

If the energy released, Q, is 5.49 MeV, what is the mass of the helium

nucleus?

Mass of nucleus = 2.01355 u

Mass of proton = 1.00728 u

1u is equivalent to 931.5MeV

Solution:

Since 1 u is equivalent to 931.3 Mev. The mass defect is given by

5.491 0.00589372

931.3

MeVm u

MeV u

Therefore, the mass of the helium nucleus is given by

2.01355 1.00728 0.00589372 3.01494hm u u u u

5. What materials that is commonly used for moderating, controlling andshielding respectively in a nuclear reactor?

Answers:

Graphite, cadmium, concrete

6. (a) Explain why, after a period of use, the fuel rods in a nuclear reactor

become

(i) Less effective for power production,

Solution:

The amount of uranium 235 in fuel decreases, and the fission fragments

absorb neutrons.

(ii) More dangerous.

Solution:

The fission fragments are radioactive, emitting and radiation.

(b) Describe the stages in the handling and processing of spent fuel rods after

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they have been removed from a reactor, indicating how the active wastes are

dealt with.

You may be awarded marks for the quality of written communication in your

answer.

Solution:

Moved by remote control

Placed in cooling ponds for several months

Separation of uranium from active wastes

High level waste stored

7. The output power of a nuclear reactor is provided by nuclear fuel which

decreases in mass at a rate of . What is the maximum

possible output power of the reactor?

64 10 kg per hour

Solution:

In an hour, the decreases in mass is 6 64 10 / 1 4 10m kg hour hour kg

Thus, the energy loss is given by2 6 8 2 114 10 (3 10 ) 3.6 10E mc J

Therefore the output power:

1183.6 10 10 100

3600

E JP W MW

t s

8. In a Rutherford scattering experiment, an particle approaches a gold

nucleus along the straight line joining their centers and comes momentarily to

rest at point P, as shown in Fig. 8-1.

particle

P

Gold nucleus

Fig. 8-1

The particle then returns along its previous path.

(a) The distance from the centre of the gold nucleus , to the point P is

. For the point P

197

79Au

143.0 10 m

(i) show that the strength of the electric field associated with the charge of the

nucleus is ,20 11.3 10 Vm

Solution:

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For the nucleus , its electric field strength can be written as19779

Au

2

kqE

r

19 1779 1.60 10 1.26 10q C C

And the charge of is given by19779Au

Thus17

9 2 2 20

2 14 2

1.26 10(8.99 10 / ) 1.3 10 /

(3.0 10 )

kq CE N m C V m

r m

(ii) Calculate the magnitude of the force acting on the particle,

Solution:

The charge of particle 192 1.60 10Q C

19 202 1.60 10 1.3 10 41.6F QE N

Thus

The magnitude of the force is given by

(iii) Calculate the electric potential due to the charge of the nucleus.

Solution:

The electric potential at a distance r from a point charge q is:

0

1

4

kq qV

r r

Therefore17

9 2 2 6

14

1.26 10(8.99 10 / ) 3.78 10 /

3.0 10

kq CV N m C J C

r m

(b) (i) State the energy changes of the particle during its interaction with the

gold nucleus.

Solution:

Kinetic energy electric potential energy kinetic energy

(ii) Calculate the initial kinetic energy, in J, of the particle, explaining your

reasoning.

Solution:

The particle moves to the point P, its kinetic energy changes totally to

potential energy. Thus

Initial kinetic energy = potential energy at point P = QV19 6 122 1.60 10 3.78 10 1.21 10QV J

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9. (a) You may be awarded additional marks to those shown in brackets for

the quality of written communication in your answers.

(i) Describe the physical process ofnuclear fusion.

Solution:

Nuclear fusion:

In a fusion process, two nuclei with smaller masses combine to form a single

nucleus with a larger mass. Energy is released by fusion when the binding

energy per nucleon is greater for the larger nucleus than for the smaller

nuclei.

(ii) Describe the physical process ofnuclear fission.

Solution:

Nuclear fission occurs when a massive nucleus splits into two less-massive

fragments. Fission can be induced by the absorption of a thermal neutron.

When a massive nucleus fissions, mass is transformed into energy when the

binding energy per nucleon is greater for the fragments than for the original

nucleus. Several neutrons are also released.

(iii) Explain why each of these processes releases energy.

Solution:Binding energy per nucleon increases when light nuclei combine; binding

energy per nucleon also increases when heavy nuclei split.

(b) Energy is also released by radioactive decay, such as the decay of

radon-220 as represented by the equation220 21686 84Rn Po

Calculate the energy released, in J, by the decay of one nucleus of radon-220.

Solution:

The mass defect is .219.96410 213.94899 4.00150 2.01361m u u u u

13 102.01361 2.01361 931.5 1875.677715 1875.677715 1.60 10 3.0 10

Since 1 u is equivalent to 931.5 Mev. The energy released is given by:

E u Mev J J

10. (a) In order for fusion of two nuclei to take place, they have to be brought

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together to a separation of about 2 fm.

(i) Show that the electrostatic potential energy of a system consisting of two

deuterium ( ) nuclei at a separation of 2 fm is about .21H131 10 J

Solution:The electric potential V at a given point is the electric potential energy EPE

of a small test charge situated at that point divided by the charge itself:q

EPEV

q

SI unit of electric potential: joule/coulomb (J/C) = volt (V)

Thus, we firstly must calculate the electric potential V of the deuterium. And

the electric potential at a distance r from a point charge q is:

0

1

4

kq qV

r r

191.60 10q C

Charge of deuterium nuclei: .

Therefore, the electrostatic potential energy of the system is given by:2 19 2

13

12 2 2 15

0 0

1 1 1 (1.60 10 )( ) 1.15 104 4 4 8.85 10 / ( ) 2 10

q q CEPE qV q J

r r C Nm m

(ii) Two deuterium nuclei may be brought to this separation by causing themto collide with equal and opposite velocities. Calculate the minimum speed

required by each nucleus for the system to have the potential energy

calculated in part

(a)(i).

Solution:

The two deuterium nuclei have equal but opposite velocities, thus their

kinetic energy is given by:

2122

mv , which is equal to the potential energy. And the mass of is12H

272 1.67 10m kg , so

2 1312 1.15 102

mv

Gives13

6

27

1.15 105.87 10 /

2 1.67 10v m s

(b) One reaction that can occur when deuterium nuclei undergo fusion is2 2 3 11 1 1 1H H H p

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(i) Calculate the energy released, in J, by this reaction.

Mass of 21H nucleus = 2.01355u

Mass of 31H nucleus = 3.01550u

Mass of proton = 1.00728uSolution:

The mass defect is .2 2.01355 3.01550 1.00728 0.00432m u u u u

13 130.00432 0.00432 931.5 4.02408 4.02408 1.60 10 6.4 10

Since 1 u is equivalent to 931.5 Mev. The energy released is given by:

E u Mev J J

(ii) How much energy is released, in J, from 1 kg of reactant in the above

fusion reaction?

Solution:

From what we obtain in (i), the energy per unit mass is given by13

13

27

6.4 109.58 10 /

4 1.67 10

JJ kg

kg

13 139.58 10 / 1 9.58 10E J kg kg J

2 3 8 2 13(1 10 )(3 10 / ) 9 10E mc kg m s J

Thus the energy released from 1 kg of reactant is:

(c) State two reasons why fusion reactions would be preferable to fission

reactions as an energy resource, provided the necessary conditions required

for continuous fusion could be maintained.

Solution:

The energy released per unit mass is greater. And the supply of fuel is almost

unlimited (deuterium from sea water).

11. If 1g of matter is completely transformed into energy, how much energy

is released in MeV?Solution:

Energy produced:

And13

13 26

13

9 109 10 5.6 10

1.60 10J Mev Mev

12. Which one of the following statements correctly describes the changes

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that occur when a uranium nucleus undergoes fission?

A The binding energy per nucleon decreases and one or more neutrons are

released.

B The binding energy per nucleon decreases and one or more protons are

released.

C The binding energy per nucleon increases and one or more neutrons are

released.

D The binding energy per nucleon increases and one or more protons are

released.

Solution:

14. Nuclear fission occurs when a massive nucleus splits into two

less-massive fragments. Fission can be induced by the absorption of a

thermal neutron. When a massive nucleus fissions, mass is transformed into

energy when the binding energy per nucleon is greater for the fragments than

for the original nucleus. Several neutrons are also released.

Choose (C)

13. A nucleus of absorbs a neutron and undergoes fission. Which one of

the following gives possible products of this process?

235

92U

A 4 2282 882 He Ra

B 141 92 156 36 03Ba Kr n

C 0 2361 942 e P u

D 212 4 184 2 04 8Po He n

Solution:

235 1

92 0

U n

From the definition of the fission and the conservation of mass number, we

can get the correct answer(B).

14. (a) You may be awarded additional marks to those shown in brackets for

the quality of written communication in your answer.

In the context of nuclear fission, explain what is meant by

(a) (i) a chain reaction,

(a) (ii) critical mass.

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Solution:

(i) Nuclear fission occurs when a massive nucleus splits into two

less-massive fragments. Fission can be induced by the absorption of a

thermal neutron. When a massive nucleus fissions, mass is transformed into

energy when the binding energy per nucleon is greater for the fragments than

for the original nucleus. Several neutrons are also released during nuclear

fission. These neutrons can, in turn, induce other nuclei to fission and lead to

a process known as a chain reaction.

(ii) The minimum mass of fissile material.

(b)Moderation and cooling are essential processes in the operation of a

nuclear power reactor using thermal neutrons. For each process, name a

suitable material that is used to achieve the required effect, and state why it is

suitable.

(b) (i) moderation

(b) (ii) cooling

Solution:

(i) The material that slows down the neutrons is called a moderator. One

commonly used moderator is ordinary water.

(ii) The water has high specific heat capacity which can be used to carryaway the heat by water flow.

15.Fig. 15.1 shows the apparatus in which particles are directed at a metal

foil in order to investigate the structure of the atom.

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(a) () Give two reasons why the metal foil should be thin.

Solution:

(1) To prevent too many particles from being absorbed.

(2) To make the particles be scattered only once.() Explain why the incident beam of particles should be narrow.

Solution:

(1) In order to have a small collision area to reduce the uncertainty in the

scattering area.

(2) To make the scattering angle be determined accurately.

(b) Describe and explain one feature of the distribution of the scattered

particles that suggests the nucleus contains most of the mass of an atom.

Solution:

(i) The positive charge of an atom is concentrated in the nucleus as it

provides a strong enough repulsion to cause backscattering of the

particles.

(ii) The nucleus contains most of the mass of the atom for it to be massive

enough to backscatter the particles.

(c) Fig. 15-2 shows three particles with the same constant velocity

incident on an atom in the metal foil. They all approach the nucleus close

enough to be deflected by at least 10.

Draw on Figure 2 the paths followed by the three particles whose initial

directions are shown by the arrows.

Solution:

Top ray: showing at least 10deflection upwards.

Middle ray: approaches closer than about 1 cm but not touching nucleus and

is backscattered.Bottom ray: smooth curve deflected downwards at an angle greater than the

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top ray is deflected upwards.

16. Monoenergetic particles, directed normally in a parallel beam at a

thin metal foil in a vacuum, are scattered at various angles.

You may be awarded additional marks to those shown in brackets for the

quality of written communication in your answer to parts a, b and c of this

question.

(a) (i) Explain why the incident particles should be in a narrow beam.

(a) (ii) the metal foil should be very thin to prevent too many particles

from being absorbed.

State another reason for using a very thin metal foil.

Solution:

(i) In order to have a small collision area to reduce the uncertainty in the

scattering area.

(ii) To make the particle be scattered only once.

(b) Describe the angular distribution of the scattered particles coming

from the foil.

State the scattering angles at which the number of particles will be amaximum and a minimum.

Solution:

(i) The maximum number of particles occurs at zero scattering angle.

(ii) The minimum number of particles occurs at 180 scattering angle.

(c) State and explain two deductions made from the scattering distribution

about the structure of the atoms in the foil.

Solution:

(i) The positive charge of an atom is concentrated in the nucleus as it

provides a strong enough repulsion to cause backscattering of the

particles.

(ii) The nucleus contains most of the mass of the atom for it to be massive

enough to backscatter the particles.

(d) On Fig. 16-1 complete the paths taken by the parallel monoenergetic

particles which all come close enough to the nucleus to be deflected.

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{

Solution:

Top line: largest bend down with minimum radius of curvature closest to

nucleus.

Middle line: less bending down.

Bottom line: even less bending down.

Chapter 2 Fundamental particles

2.1 Particles, antiparticles and photons

2.2 Particle interactions2.2.1 Electromagnetic force and Feynman diagram

2.2.2 The weak nuclear force and Feynman diagram

2.3 Properties of particles and antiparticles

2.4 Quarks and Antiquarks2.4.1 Quarks and antiquarks properties

2.4.2 Quark combinations

2.4.3 Quarks and beta decays

2.5 45 Worked examples

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Chapter 3 Electromagnetic Radiation and Quantum Phenomena

3-1 The photoelectric effect

3-2 Collisions of electrons with atoms

3.2.1 The Bohr model and Energy levels

3.2.2 The emission and absorption of photons

3.2.3 Atomic spectrum

3.2.4 Fluorescent tube explanation

3-3 Wave-particle duality

3-4 42 Worked examples

Chapter 4 Operational amplifier (op amp)4-1 differential amplifier

Differential amplifier has two inputs and one output. It amplifies the

difference between the two input voltages. (Fig. 4.1)

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Note:

(i) if , the output voltage0inV V V 0outV

(ii) if , the output voltage0inV V V 0outV

(iii) if , the output voltage0inV V V 0outV (iv) if only one input is in use (the other is at zero voltage), then

if the inverting input (or0V 0V ), the output voltage (or

)

0outV

0outV

if the non-inverting input (or0V 0V ), the output voltage

(or )

0outV

0outV

1.1open loop voltage gain

For the Op amp above, the relationship between input voltage and output

voltage is given by

0 0( )out inV A V V A V

Where 0out

in

VA

V is calledopen loop voltage gain.

1.2 features ofideal Op amp (open loop)

has an extremely high voltage gain.

has a very high input impedance

has a very low output impedance

1.3 Graph for an open-loop op amp

In the Op amp, a supply with typically value 9V is required.

Vout

Vin

+Vsupply

-Vsupply

Linear amplification

saturation

saturation

Fig. 4.2 graph for an open-loop op amp

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Note:

(i) For the linear amplification, the out voltages are proportional to the input

voltages.

(ii) At the saturation region, the output is distorted and no longer proportional

to the input.

(iii) The output voltage can not exceed the supply voltage.

4-2 Negative feedback amplifiers

The open Op amp has an extremely high voltage gain; the output voltage is

easy saturated. In order to reduce gain, a resistor or wire linking the output to

one input is used, which is calledclosed loop.

An amplifier circuit uses a closed loop to feed back a set fraction of the

output voltage to the inverting input is callednegative feedback.

Negative means that the signal being fed back partly cancels the input

signal.

2.1 Op amp as an inverting amplifier

The figure below shows an inverting amplifier

The closed-loop voltage gain is given by

f

f

i

RA

R

2.2 op amp as a non-inverting amplifier

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The figure below shows a non-inverting amplifier

+

Supply +

Supply -

Vin

Output voltage Vout

0 V(earth)

Op amp

A

Fig. 4.4 non-inverting amplifier

Closed

loop

Feedback resistor

Rf

R1

R2

For the non-inverting amplifier, the closed-loop voltage gain is given by

1

1ffR

AR

Note:

(i) A non-inverting amplifier has a much higher input impedance than the

inverting type.

(ii) For the above non-inverting amplifier, if or , output

voltage equals to the input voltage, and the voltage gain equals to 1:

1R 0fR

out inV V , 1fA

The Op amp becomes a voltage follower, which can be used as impedance

matching.

4-3 Op amp as a comparator

3.1 The input signals ( ) goes to the non-inverting input, the reference

voltage ( ) goes to the inverting input. The diagram below shows the Op

amp as a comparator:

inV

RV

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+

Q

VR

Vin

Vout

If , , thenRV V i nV V

in RV V , ;0( )out sat V V

in RV V , 0( )out sat V V

The graph below shows the relationship between output voltage and input

voltage:

Vin

V out

+V0(sat)

V0(sat)

VR

0

Fig. 4.6 output vs. input voltage

3.2 The input signals ( ) goes to the inverting input, the reference voltage

( ) goes to the non-inverting input. The diagram below shows the Op amp

as a comparator:

inV

RV

+

Q

VR

Vin

Vout

If , , thenRV V i nV V

in RV V , ;0( )out sat V V

in RV V , 0( )out sat V V

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The graph below shows the relationship between output voltage and input

voltage:

3.3 For example, Fig. 4.9 shows the application of comparator: the LED is

switched on and off automatically as the change of light intensity.

In figure 4.9,R1 andRLDRare connected to be potential dividers;R2 andR3 is

potential dividers. Thus

11 1

SLDRS

LDR

VRV V

LDRR

RR R

3

22 3

3

1

SSR VV V RR R

R

The LED is controlled by LDR, when the LDR is in bright light, it has a

lower resistance. Thus V , thenV 0( ) 0out sat V V , the LED can not

conduct.

When the LDR is in dark light, t has a higher resistance. Thus

V V , then 0( ) 0out sat V V , the LED conducts and lights up

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0 V

LDR

Indicator lamp

(LED)

R1

RLDR

R2

R3 R4

+

Supply + (VS)

Fig. 4.9 application of comparator

Open loop

4-4 Summing amplifier

Summing amplifier: the output signal is amplified by the sum of input signal

with certain proportion.

Fig. 4.10 shows the summing amplifier and the relationship of output and

input:

(i) For two input signals, the relationship of output and input is given by

1 2

1 2out f

V VV R

R R

1 2out V VV Note: if 1 2 fR R R , then

(ii) For three input signals, the relationship of output and input is given by

3

3

1 2

1 2

out f

V

R

V VV R

R R

1 2 3out V V VV Note: if 1 2 3 fR R R R , then

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1. The diagram below (Fig. 4.11) shows an amplifier circuit incorporating an

ideal operational amplifier (op-amp).

Fig. 4.11

The operational amplifier uses a +9 V / 0 / 9 V supply.

(a) Calculate the gain of the amplifier circuit.

Solution:

The figure above shows a non-inverting amplifier, the gain of the amplifier is

given by

1

101 6

21f

f k

k

RA R

(b) Determine the output potential Vout for values of input potential Vin equal

to

(i) - 0.9 V

6 0.9 5.4out infV A V V V

V

Solution:

(ii) + 2.0 V

6 2.0 12.0out inf

V A V V

2. (a) On the axes below draw a sketch graph to show the variation with input

voltage Vin of the output voltage Vout of a non-inverting operational

amplifier.

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(b) A temperature warning device makes use of a buzzer that sounds whenthe potential difference across it is 24 V. The circuit in the warning device is

shown in Fig. 4.12.

It is required that the buzzer should sound when the temperature of the

thermistor rises above 50C.

(i) State the voltage at point Q.

Solution:

12AV V

, , thus12BV

V 12 12 ( 12) 12AQ A Q QR

V V V V V V R R

V

Therefore, .0QV V

(ii) At a temperature of 50C the resistance of the thermistor is R. Explain

why the buzzer will sound when the temperature rises above 50C.

Solution:

When the temperature rises above 50 the resistance of the thermistor falls

below R; The output voltage then saturates at 12 V giving the required

potential difference for the buzzer to sound.

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3. (a) Fig. 4.13 shows a circuit that uses an operational amplifier as an

inverting amplifier.

Fig. 4.13

The point P is a virtual earth, that is at the same potential (0 V) as the earthline.

(i) State the two properties of the operational amplifier which make P a

virtual earth.

Solution:

Extremely high input resistance and high open loop gain.

(ii) In the circuit Rf= 100 k and Rin = 10 k. Calculate the gain of the

amplifier.

Solution:

10010

10f

f

i

k

k

RA

R

(b) Fig. 4.14 shows a circuit that uses an operational amplifier as a

non-inverting Schmitt trigger.

Fig. 4.14

In the situation shown, the potential at point X is 2.0 V and the output

potential Vout is at its minimum value of10 V. Show that for the output

potential to switch to its maximum value of +10 V

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(i) The current in the resistors R1 and R2 is 0.08 mA.

Solution:

3

2

10 20.08

150 10

out X V VI mAR

(ii) Vin must rise to 5.8 V.

Solution:

From (i), 3 31

2 0.08 10 47 10 2 5.8inV IR V

4.Fig. 4.15 shows a non-inverting amplifier circuit.

(a) Suggest why the amplifier is referred to as non-inverting.

Solution:The signal goes into the non-inverting input.

(b) The input voltage for the amplifier in (a) is V in = 2.0 mV.

Calculate

(i) The gain G of the amplifier.

Solution:


1

90

1 10101

f k

G k

R

R

(ii) The output voltage Vout .

Solution:

10 2.0 20.0out inV G V mV m V

(c) Fig. 4.16 shows a particular non-inverting amplifier.

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Fig. 4.16

Explain, in terms of the properties of an op-amp, why the gain of this

non-inverting amplifier is equal to 1.

Solution:

In diagram, the feedback resistance is equal to zero and the input is extremely

high, thus1

01 11 fG RR

(d) Fig. 4.17 shows a circuit in which the battery has an emf of 6.0 V and

negligible internal resistance. Two 2.0Mresistors are connected in series to

the battery.

Fig. 4.17

(i) State the value of the potential difference between points A and B.

Solution:

The circuit current is given by 66 6

6.01.5 10

2.0 10 2.0 10

VI A

V

, thus

6 6 62.0 10 1.5 10 2.0 10 3.0ABV I A

(ii) A voltmeter of resistance 100 kis used to measure the potential

difference across points A and B.

State why the reading on the voltmeter is not equal to the value stated in

(d)(i).

Solution:The voltmeter has resistance and be connected in parallel to the R2. Thus the

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reading is less than 3.0 V.

(iii) The circuit in Fig. 4.17 is modified to include the circuit shown in Fig.

4.18.

Fig. 4.18

Explain why the voltmeter reads the value of the potential difference as stated

in (d)(i).

Solution:

R1 and R2 are potential divider, thus 3.0AV V , that is 3.0in AV V V

3.0V

. For the

diagram 4, the voltage gain G = 1, therefore, out inV V

5. A student designs a circuit to give an output voltage, VA, of +5.0V when

the temperature exceeds a certain value. The incomplete circuit diagram is

shown in Fig. 4.19.

(i) Complete the circuit in Fig. 4.19 by adding a thermistor and a variable

resistor to allow the circuit to function correctly.

Solution:

When resistance of the thermistor changes as the temperature, the

non-inverting input changes, which can be used to control the output voltage.

(ii) State the purpose of the variable resistor.

Solution:

To set switching the resistance of the thermistor(iii) If the circuit switches at a temperature of 40 C, calculate the value of

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the variable resistor. The characteristic curve of the thermistor is shown in

Fig. 4.20.

Solution:

The circuit switches at a temperature of 40 C, thus

20

30

V

th

R k

R k

And, in the figure 4, the resistance of thermistor 4t h

R k , thus

20 204 2.7

30 30V th

k kR R k

k k

k

Fig. 4.20

6. (a) For an ideal operational amplifier, state the value of

(i) The output impedance:

(ii) The input impedance:

(iii) The open loop gain:

Solution:

Features ofideal Op amp (open loop)

has an extremely high voltage gain.

has a very high input impedance

has a very low output impedance

(b) Fig. 4.21 shows an operational amplifier being used as a non-inverting

amplifier.

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Fig. 4.21

(i) Name the type of feedback used in this circuit.

Solution:

Negative feedbackNote: An amplifier circuit uses a closed loop to feed back a set fraction of the

output voltage to the inverting input is callednegative feedback.

Negative means that the signal being fed back partly cancels the input

signal.

(ii) State one advantage of this type of feedback in an amplifier circuit.

Solution:

To reduce the distortion

(c) Calculate the value of the resistor R when

Vin = 200mV

Vout = 5.0V.

Solution:


1

1ffR

AR

And3

5.0 25200 10

outf

in

V VV V

A

, gives

1

25100

1 1ffR k

AR R

R = 4.2 k

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Chapter 5 Relativity

5-1 basic assumptions of special relativity

5-2 relativity of time and space

5-3 Relativistic mass and mass-energy equation


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