ib hl study guide organizer-nuclear physics

7/29/2019 Ib Hl Study Guide Organizer-nuclear Physics

1/20

IB 12

1

Atomic Number 26 13 6 6

Mass Number 56 27 12 14

Neutron Number 30 14 6 8

Atomic Mass 56 u 27 u 12 u 14 u

Molar Mass 56 g 27 g 12 g 14 g

Nuclear Physics

Nuclide: a particular type of nucleus

Nucleon: a proton or a neutron

Atomic number (Z) (proton number): number of protons in nucleus

Mass number (A) (nucleon number): number of protons + neutrons

Neutron number (N): number of neutrons in nucleus (N = A Z)

Isotopes: nuclei with same number of protons but different numbers of neutrons

Unified atomic mass unit (u): 1/12th

the mass of a carbon-12 nucleus

Atomic mass A * u

1 u = 1.661 x 10-27 kg 1 u = 1 g/mol 1 u = 931.5 MeV/c2

56

26Fe12

6 C27

13 l14

6 CAZ X

How big are atomic nuclei? 10-15

m 10-14

m

How do we know this? Alpha-particle

scattering experiments (Geiger-Marsden)

Formulas:ET = ET

EK= Ee mv

2= qV

mv2

= q(kQ/r)

Note that most nuclei have approximately the same . . .density

= 2 x 1017 kg/m3 = 1014 times denser than water

How do we know that neutrons exist?

By the existence of isotopes

How do we know that isotopes exist?

By measurements in the mass spectrometer

Formulas:v = E/B

r = mv/qB


2/20

IB 12

2

What interactions exist in the nucleus?

1. Gravitational: (long range) attractive but very weak/negligible

2. Coulomb or Electromagnetic: (long range) repulsive and very strong between protons

3. Strong nuclear force: (short range) attractive and strongest between any two nucleons

4. Weak nuclear force: (short range) involved in radioactive decay

Nuclear Stability

Why are some nuclei stable while others are not?

The Coulomb force is a long-range force which means that every

proton in the nucleus repels every other proton. The strongnuclear force is an attractive force between any two nucleons

(protons and/or neutrons). This force is very strong but is shortrange (10

-15m) which means it only acts between a nucleon and

its nearest neighbors. At this range, it is stronger than the

Coulomb repulsion and is what holds the nucleus together.

Neutrons in the nucleus play a dual role in keeping it stable. Theyprovide for the strong force of attraction, through the exchange of

gluons with their nearest neighbors, and they act to separate

protons to reduce the Coulomb repulsion.

Each dot in the plot at right represents a stable nuclide and theshape is known as the band (or valley) of stability. With few

exceptions, the naturally occurring stable nuclei have a numberNof

neutrons that equals or exceeds the numberZof protons. For small

nuclei (Z < 20), number of neutrons tends to equal number ofprotons (N = Z).

As more protons are added, the Coulomb repulsion rises faster than the strong force of attraction

since the Coulomb force acts throughout the entire nucleus but the strong force only acts among

nearby nucleons. Therefore, more neutrons are needed for each extra proton to keep the nucleustogether. Thus, for large nuclei (Z > 20), there are more neutrons than protons (N > Z).

After Z = 83 (Bismuth), adding extra neutrons is no longer able to counteract the Coulomb repulsionand the nuclei become unstable and decay in various ways.

Nuclei above (to the left of) the band of stability have too many neutrons and tend to decay by alpha

or beta-minus (electron) emission, both of which reduce the number of neutrons in the nucleus.

Nuclei below (to the right of) the band of stability have too few neutrons and tend to decay by beta-

plus (positron) emission which increases the number of neutrons in the nucleus.


3/20

IB 12

3

ParticleElectric Charge

(e)

Electric Charge

(C)

Rest Mass

(kg)

Rest Mass

(u)

Rest Mass

(MeV/c2)

Proton +1 +1.60 x 10-19

1.673 x 10-27

1.007276 938

Neutron 0 0 1.675 x 10-27

1.008665 940

Electron -1 -1.60 x 10-19

9.110 x 10-31

0.000549 0.511

Binding Energy

The total mass of a nucleus is always less than the sum of the masses its nucleons. Because mass is another

manifestation of energy, another way of saying this is the total energy of the nucleus is less than the combined

energy of the separated nucleons.

Mass defect (mass deficit) (m)

Difference between the mass of the nucleus and the sum of the masses ofits individual nucleons

Nuclear binding energy (E)

1. energy released when a nuclide is assembled from its individual

components

2. energy required when nucleus is separated into its individual

components

Formulas:

mnucleus + m = mnucleons

E = ( m)c2

Different nuclei have different total binding energies. As a general trend, as the atomic numberincreases . . . the total binding energy for the nucleus increases.

1. The most abundant isotope of helium has a 24He nucleus whose mass is 6.6447 10-27 kg. For this nucleus, find the

mass defect and the binding energy.

helium: 0.050404 x 10-27

kg

4.53636 x 10-12

J

2. Calculate the binding energy and mass defect for816O whose measured mass is 15.994915 u.

Oxygen: 0.132613 u

123.5 MeV


4/20

IB 12

4

Binding Energy per Nucleon

To see how the nuclear binding energy varies from nucleus to nucleus, it is useful to compare the binding

energy for each nucleus on a per-nucleon basis, as shown in the graph below.

Your Turn

a) This graph is used to compare the energy states of different nuclides and to determine what nuclear reactions are energeticallyfeasible. As binding energy per nucleon increases so does the stability of the nucleus. Higher binding energies representlower energy states since more energy was released when the nucleus was assembled.

b) Binding energy per nucleon increases up to a peak at 2656Fe then decreases, so 2656Fe is the most stable nuclide. Most nuclideshave a binding energy per nucleon of about 8 MeV. Lighter nuclei are held less tightly than heavier nuclei.

c) Nuclear reactions, both natural (radioactive decay) and artificial/induced (fission, fusion, bombardments) occur if theyincrease the binding energy per nucleon ratio. Fusion occurs for light nuclei (below 26

56Fe) and fission occurs for heavy nuclei

(above 2656Fe).

d) For both natural and induced nuclear reactions, the total rest mass of the products is less than the total rest mass of thereactants since energy is released in the reaction. Also, the products are in a lower energy state since energy was released inthe reaction and so the products have a greater binding energy per nucleon than the reactants.

Types of Nuclear Reactions

1) Artificial (Induced) Transmutation: A nucleus is bombarded with a nucleon, an alpha particle or another small nucleus,

resulting in a nuclide with a different proton number (a different element).

2) Nuclear Fusion: Two light nuclei combine to form a more massive nucleus with the release of energy.

3) Nuclear Fission: A heavynucleus splits into two smaller nuclei of roughly equal mass with the release of energy.

4) Natural Radioactivity: When an unstable (radioactive) nucleus disintegrates spontaneously, the nucleus emits a particle

of small mass and/or a photon.

Binding energy per nucleon:greater for product nuclei than for original nuclei since energy is released.

Release of energy in nuclear reactions: m = m + m

Energy is usually released in the form of kinetic energy for the products.

1. Use the graph above to estimate the total binding energy of an oxygen-16 nucleus. 8 MeV x 16 = 128 MeV


5/20

IB 12

5

Radioactive Decay Reactions

Alpha Decay

Alpha particle: helium nucleus, , 24He

Example reaction: 226 222 4

88 86 2a Rn He energy

4 4

2 2

A A

Z ZX Y He energy

1. A radium nucleus, initially at rest, decays by the emission of an alpha particle into radon in the reaction described above.

The mass of88226Ra is 226.025402 u and the mass of86222Rn is 222.017571 u and the mass of the alpha particle is4.002602 u.

a) Calculate the energy released in this decay.

c) If the kinetic energy of the alpha particle is

4.77 MeV, calculate its speed.

m = 0.005229 u

4.87 MeV

b) Compare the momenta, speeds, and kinetic energies of the two particles produced by this reaction.

d) Calculate the recoil speed of the radon nucleus.

Use momentum no need to change units

Result: nucleus is in a more stable state with higher binding energy and higher BE per nucleon

since it released energy

Where does the kinetic energy come from? Rest mass of nucleons


6/20

IB 12

6

Gamma Decay

Gamma particle: high energy photon,

Example reaction: 12 * 12

6 6C C energy

*A A

Z ZX X energy General equation:

Before Decay

After DecayWhere does the photon (energy) come from? Rest mass of nucleons

Beta Decay

Beta-minus particle: electron, -, -10e

Example reaction:

1

14 14 0 0

6 7 0C N e energy

1

0 0

1 0

A A

Z Z Y e energy

General equation:

Consider the following two mysterious results of beta decay:

a)

Observe the before and after picture of beta decay.Whats wrong?

b) Inspect the graph of kinetic energy carried away bythe beta particles. Why are so few beta particles

leaving with the majority of the kinetic energy?

Where did this missing kinetic energy go?

Conclusion: there is third particle involved with beta decay that carries away some KE

and momentum virtually undetectable

Beta-plus particle: positron, +, +10e

Neutrino and anti-neutrino: fundamental particles - no charge very small mass

Beta-minus decay Beta-plus decay

Example reaction:

How does this happen? Weak force

1

1 1 0 0

0 1 0n p e energy

1

12 12 0 0

7 6 0N C e energy

1

0 0

1 0

A A

Z ZX Y e energ

General equation:

How does this happen? Weak force

1

1 1 0 0

1 0 0p n e energy


7/20

IB 12

7

Ionizing Radiation

Particle helium nucleusElectron or

positronhigh-energy photon

Penetration ability low medium high

Material needed to

absorb it

Sheet of paper; a

few centimeters ofair

1 mm of

aluminum10 cm of lead

Path length in air a few cm less than 1 meter effectively infinite

Ionizing Radiation As this radiation passes through materials, it knocks off electrons from neutral atoms thercreating an ion pair: free electrons and a positive ion. This ionizing property allows the radiation to be detected b

is also dangerous since it can lead to mutations in biologically important molecules in cells, such as DNA.

Detection of Radiation: the Geiger-Muller tube (Geiger counter)

The Geiger counter consists of a gas-filled metal cylinder. The , , or rays enter the cylinder through a thwindow at one end. Gamma rays can also penetrate directly through the metal. A wire electrode runs along

center of the tube and is kept at a high positive voltage (1000-3000 V) relative to the outer cylinder.

When a high-energy particle or photon enters the cylinder, it collides with and ionizes a gas molecule. Theelectron produced from the gas molecule accelerates toward the positive wire, ionizing other molecules in it

path. Additional electrons are formed, and an avalanche of electrons rushes toward the wire, leading to a pu

current through the resistorR. This pulse can be counted or made to produce a "click" in a loudspeaker. Th

number of counts or clicks is related to the number of disintegrations that produced the particles or photons.

Biological Effects of Ionizing Radiation

Alpha and beta particles have energies typically measured in MeV. To ionize an atom requires about 10 eV so each particle can

potentially ionize 105 atoms before they run out of energy. When radiation ionizes atoms that are part of a living cell, it can affectthe ability of the cell to carry out its function or even cause the cell wall to rupture. In minor cases, the effect is similar to a burn. If

a large number of cells that are part of a vital organ are affected then this can lead to death. Alternatively, instead of causing the cell

to die, the damage done by ionizing radiation might just prevent cells from dividing and reproducing. Or, it could be the cause of thetransformation of the cell into a malignant form. If these malignant cells continue to grow then this is called cancer.

The amount of harm that radiation can cause is dependent on the number and energy of the particles. When a gamma photon is

absorbed, the whole photon is absorbed so one photon can ionize only one atom. However, the emitted electron has so much energythat it can ionize further atoms, leading to damage similar to that caused by alpha and beta particles.

On a positive note, rapidly diving cancer cells are very susceptible to the effects of radiation and are more easily killed than normal

cells. The controlled use of the radiations associated with radioactivity is of great benefit in the treatment of cancerous tumors.

Energy Spectra of Radiation

The nucleus itself, like the atom as a whole, is a quantum system

with allowed states and discrete energy levels. The nucleus can be

in any one of a number of discrete allowed excited states or in its

lowest energy relaxed state. When it transitions between a higherenergy level and a lower one, it emits energy in the form of alpha,

beta, or gamma radiation. When an alpha particle or a gamma

photon is emitted from the nucleus, only discrete energies are

observed. These discrete energy spectra give evidence that a

nucleus has energy levels. (However, the spectrum of energiesemitted as beta articles is continuous due to its sharing the energy

with a neutrino or antineutrino in any proportion.)

Alpha spectra Beta spectraGamma

spectra

Discrete continuous discrete

Importance: discrete energy spectra give evidence

for nuclear energy levels


8/20

IB 12

8

Mathematical Description of Radioactive Decay

Radioactive decay:

1) Random process: It cannot be predicted when a particular nucleus will decay, only theprobability that it will decay.

2) Spontaneous process: It is not affected by external conditions. For example, changing thepressure or temperature of a sample will not affect the decay process.

3) Rate of decay decreases exponentially with time: Any amount of radioactive nuclei will reduceto half its initial amount in a constant time, independent of the initial amount.

Half-life (T1/2)

- the time taken for of the radioactive nuclides in a sample to decay

- the time taken for the activity of a sample to decrease to of its initial value

Units:

time

= s or hr or

d or yr

N0 = number of radioactive nuclei originally present

N = number of radioactive nuclei present at any one time

Radioactive X stable Y + particle

Show how amount of daughter Y mirrors X

Your Turn

Radioactive tritium has a half-life of about 12

years. Complete the graph below.

A nuclide X has a half-life of 10 s. On decay the stable nuclide Y is formed. Initially a sample contains only atoms of X. After

what time will 87.5% of the atoms in the sample have decayed into nuclide Y? 30 s


9/20

IB 12

9

Activity

N

A t

Activity (A) the number of radioactive disintegrations per unit time (decay rate)

Formula:

Units:

Standardunits:

decays/time = s-1

or hr-1

or d-1

or yr-1

Becquerel (Bq)

1 Bq = 1 decay per second

3. The isotope Francium-224 has a half-life of 20 minutes. A sample of the isotope has an initial activity of 800

disintegrations per second. What is the approximate activity of the sample after 1 hour?

N = 1/8 N0

So A = 1/8 A0

A = 100 disintegrations s-1

2. Samples of two nuclides X and Y initially contain the same number of radioactive nuclei, but the half-life of nuclide

X is greater than the half-life of nuclide Y. Compare the initial activities of the two samples.

Activity of sample Y is greater

1. A sample originally contains 8.0 x 1012 radioactive nuclei and has a half-life of 5.0 seconds. Calculate the activity of

the sample and its half-life after:

c) 15 secondsb) 10. secondsa) 5.0 seconds

8.0 x 1011

Bq

5.0 s

6.0 x 1011

Bq

5.0 s

4.7 x 1011

Bq

5.0 s

The Radioactive Decay Law: The rate at which radioactive nuclei in a

sample decay (the activity) is proportional to the number of radioactive

nuclei present in the sample at any one time.

[As the number of radioactive nuclei decreases, so does the average

rate of decay (the activity).]

N

N

Activity

Initial Activity

0 0NThe initial activity (A

0) is directly related to the number of radioactive

nuclei originally present (N0) in the sample.


10/20

IB 12

10

2. A sample of a radioactive isotope X has the same initial activity as a sample of the isotope Y. The sample of X

contains twice the number of atoms as the sample of Y. If the half-life of X is TX then the half-life of Y is 0.5 TX

Decay constant ()constant of proportionality between the decay rate (activity) and the

number of radioactive nuclei present.

- probability of decay of a particular nuclei per unit time.

Units:inverse time

= s-1

or hr-1

or d-1

or yr-1

Deriving the Radioactive Decay Law Relating the Decay Constant and Half-life

1/ 2 0

0

1/ 20 0

1

2

1

2

At then

The decay equation becomes

t

t T N N

N N e

TN N e

1/ 2

1/ 2

1/ 2

1/ 2

1/ 2 1/ 2

1/ 2

11

2

2

ln 2

ln2

ln 2 0.693

ln 2 0.693

T

T

T

e

e

e

T

T T

T

1. The half-life of a certain radioactive isotope is 2.0 minutes. A particular nucleus of this isotope has not decayed within

a time interval of 2.0 minutes. What is the probability of it decaying in:

a) the next two minutes b) the next one minute c) the next second

0

o 0

A solution to this equation

is an exponential function

of the form

alsoA = A

t

t t

NA N

t

dNN

dt

N N e

e N e


11/20

IB 12

11

4. The half-life of a radioactive substance is 10 days. Initially, there are 2.00 x 1026 radioactive nuclei present.

a) What is the probability of any one particular nucleus decaying?

b) What is the initial activity?

c) How many radioactive nuclei are left after 25 days?

d) What is the activity of the sample after 25 days?

e) How long will it take for the activity to fall to 1.0 x 1024 dy-1?

3. The half-life of a radioactive isotope is 10 days. Calculate the fraction of the sample that will be left after 15 days. 35%


12/20

IB 12

12

Graphs of Radioactive Decay

Activity of sample vs. time Straightening by natural log

ln(N0)

Slope = -

Straightening by natural logRadioactive nuclei vs. time

0

0

0

0

ln ln

ln ln

ln ln

t

t

N N e

N N e

N N t

N t N

y mx b

T1/2 = ln 2/(-slope)

Methods of Determining Half-life

If the half-life is short, then readings can be taken of activity versus time using a Geiger counter, for example. Then,either

1. A graph of activity versus time would give the exponential shape and several values for the half-life could beread from the graph and averaged.

OR

2. A graph of ln (activity) versus time would be linear and the decay constant can be calculated from the slope.

If the half-life is long, then the activity will be effectively constant over a period of time. If a way could be found to

calculate the number of nuclei present chemically, perhaps using the mass of the sample and Avogadros number, then theactivity relation or the decay equation could be used to calculate half-life.


13/20

IB 12

13

3. The radioactive isotope potassium-40 undergoes beta decay to form the isotope calcium-40 with a half-life

of 1.3 x 109 yr. A sample of rock contains 10 mg of potassium-40 and 42 mg of calcium-40.

a) Determine the age of the rock sample. 3.1 x 109 yr

b) What are some assumptions made in this determination of age?

No calcium-40 originally present = all potassium-40

No loss of either isotope in intervening years

1. Cesium-138 decays into an isotope of barium. Measurements of

the activity of a particular sample of cesium-138 were taken and

graphed at right.

b) Use the graph to estimate the half-life of cesium-138.

35 minutes

a) Suggest how the data for this graph could have been obtained.

Geiger counter

c) Use the graph to estimate the half-life of the barium isotope.

Wait until 250/300 minutes when very little cesium left

90 minutes

2. A 2.0 mg sample of carbon-14 is measured to have an activity of 6.5 x 1010 Bq.

a) Use this information to determine the half-life of carbon-14 in years. 5700 years

b) A student suggests that the half-life can be determined by taking repeated measurements of the activity andanalyzing the data graphically. Use your answer to part (a) to comment on this method of determining the half-life.


14/20

IB 12

14

Nuclear Fission

Nuclear Fission: A heavynucleus splits into two smaller nuclei of roughly equal mass with the release of energy.

Nuclear Fusion: Two light nuclei combine to form a more massive nucleus with the release of energy.

Release of energy in nuclear reactions:

m = m + m

Energy is usually released in the form of . . .

kinetic energy for the products.

Binding energy per nucleon:greater for

product nuclei than for original nuclei sinceenergy is released.

One Common Fission Reaction

235 1 236 *

92 0 92 neutronsU n U X Y

There are about 90 different daughter nuclei (X and Y) that can be formed.

Here is a typical example:

235 1 141 92 1

92 0 56 36 0

3U n Ba Kr n

1. Estimate the amount of energy released when a uranium nucleus fissions.

Use plot above

Original: 7.5 MeV per nucleon

Daughters: 8.5 MeVEnergy difference: 1.0 MeV per nucleon x 235 nucleons

= 235 MeV as KE of daughters and neutrons

Masses:

94239Pu = 239.052157 u

3896Sr = 95.921750 u

56140Ba = 139.910581 u

01n = 1.008665 u

m = m + m

0.19383 u = 180 MeV

2. A neutron collides with a nucleus of plutonium and the following fission reaction occurs. Determine the number of neutronsproduced and calculate the amount of energy released.

239 1 140 96

94 0 56 38Pu n Ba Sr


15/20

IB 12

15

Naturally Occurring Isotopes of Uranium:

1) Uranium-238: most abundant, 99.3%, very small probability of fissioning when it captures aneutron, not used for fuel, more likely to capture high energy neutron than low energy one

2) Uranium-235: 0.3%, 500 times greater probability of fissioning when captures a neutron but mustbe a low-energy (thermal) neutron, used for fuel

Fuel Enrichment: process of increasing proportion of uranium-235 in a sample of uranium

1) formation of gaseous uranium (uranium hexafluoride) from uranium ores

2) Separated in gas centrifuges by spinning heavier U-238 moves to outside3) increases proportion of U-235 to about 3% to be used as fuel in nuclear reactors

Chain Reaction neutrons released from one fission reaction go on to initiate further reactions

Chain Reactions

Uncontrolled Chain Reaction

Controlled Chain Reaction

Uncontrolled nuclear fission: nuclear weapons

Controlled nuclear fission: nuclear power production

1) some material absorbs excess neutrons before striking nucleus

2) leaving only one neutron from each reaction to produceanother reaction

Thermal Neutron: low-energy neutron (1eV) that favors fission reactions energy comparable to

gas particles at normal temperatures

Critical Mass: if mass of uranium is too small, too many

neutrons escape without causing further fission in uranium so

the reaction cannot be sustained

Advantage: more uranium is available for fission and reaction can be sustained

Disadvantage: enriched fuel can be used in the manufacture of nuclear weapons

threat to world peace 85% = weapons grade


16/20

IB 12

16

Nuclear Reactors

Most nuclear reactors: thermal fission reactor using uranium-235 as fuel

Fuel Rods: enriched solid uranium

Moderator: material (water, graphite) used to slow down high-energy neutrons emitted from fissionreactions to thermal levels for use in further fission reactions to sustain the chain reaction - slow

neutrons by collisions

Control Rods: inserted between fuel rods made of neutron-absorbing cadmium or boron -used to control reactor temperature to prevent overheating lowered if too many

neutrons/reactions and excess thermal neutrons are absorbed

Heat Exchanger: hot fluid circulating around fuel rods (primary loop) is fed into tank of water

heat is transferred to water and makes steam steam expands adiabatically against fan blades of

turbines and turns a magnet is a coil of wire to generate electricity


17/20

IB 12

17

Neutron Capture and Plutonium-239

239 1 140 96

94 0

1

056 38 3Pu n Ba Sr n

Uranium-238 is a non-fissionable isotope but is considered fertile

0

1

238 1 239 239

92 0 92 93U n U eNp Neutron capture and

Beta-minus decay

239 239

93 94

0

1

Np Pu e

Beta-minus decay

Fission

reaction

Advantage:plutonium-239used as fuel in breeder reactors

Disadvantage:plutonium-239used in nuclear weapons

Safety Issues and Risks in the Production of Nuclear Power

Thermal Meltdown: overheating and melting of fuel rods may be caused by malfunction in cooling

system or pressure vessel overheating may cause pressure vessel to burst sending radioactive material

and steam into atmosphere (as in Chernobyl, Ukraine 1986) hot material may melt through floor China syndrome as in Three Mile Island limited by containment vessel and containment building

Uranium Mining:

open-cast mining: environmental damage, radioactive waste rock (tailings)

underground mining: release of radon gas (need ventilation), radioactive rock dangerous for workers,radioactive waste rock (tailings)

leaching: solvents pumped underground to dissolve uranium and then pumped back out

contamination of groundwater

Nuclear Waste:

Low-level waste: radioactive material from mining, enrichment and operation of plant must be

disposed of left untouched or encased in concrete

High-level waste: disposal of spent fuel rods- some isotopes have lives of thousands of years

plutonium 240,000 years

1)

stored under water at reactor site for several years to cool of then sealed in steel cylinders,buried underground

2) reprocessed to remove any plutonium and useful uranium, remaining isotopes have shorter lives and lon -term stora e need is reduced

Nuclear Weapons Manufacture:Enrichment technology could be used to make weapons grade uranium (85%) rather than fuel

grade (3%)

Plutonium is most used isotope in nuclear weapons and can be gotten from reprocessing spent

fuel rods


18/20

IB 12

18

1. Suppose the average power consumption for a household is 500 W per day. Estimate the amount of uranium-235 that would have

to undergo fission to supply the household with electrical energy for a year. Assume that for each fission, 200 MeV is released.

2. A fission reaction taking place in a nuclear power station might be

Estimate the initial amount of uranium-235 needed to operate a 600

MW reactor for one year assuming 40% efficiency and 200 MeV

released for each fission reaction.

235 1 141 92 1

92 0 56 36 03U n Ba Kr n


19/20

IB 12

19

Nuclear Fusion

Nuclear Fusion: Two light nuclei combine to form a more massive nucleus with the release of energy.

12H (deuterium, 2.0141 u)

13H (tritium, 3.0161 u)

24He (4.0026 u)

neutron (1.0087 u)

1. Write the reaction equation for the fusion reaction shown at right.

2 3 4 1

1 1 2 0H H He n 2. Calculate how much energy is released in this fusion reaction.

0.0189 u = 17.6 MeV

m = m + m

Important occurrence of fusion: main source of Suns energy fusion of hydrogen to helium

3. Calculate the energy released per nucleon and compare this with a fission reaction.

5 nucleons react = 17.6 MeV / 5 = 3.5 MeV

Compared with 1.0 MeV for fission

Suggested Mechanism: proton-proton cycle

1 1 2 0

1 1 1 1

1 2 3

1 1 2

H H H e

H H He

Then either: Or:

1 3 4 0

1 2 2 1

H He He e 3 3 4 1 1

2 2 2 1 1He He He H H


20/20

IB 12

20

Fusion Reactor

Plasma: fuel for reactor high energy ionized gas (electrons and nuclei are

separate) if energy is high enough (hot enough), nuclei can collide fast enough to

overcome Coulomb repulsion and fuse together

Magnetic confinement: charged particles are contained via magnetic fields travel in a circle in a doughnut shaped ring (tokamak)

Heating Plasma: accelerate nuclei by means of magnetic fields and forces = high

temperatures (high kinetic energies)

Problems with current fusion technology: maintaining and confining very high-density and

high-temperature plasmas very difficult to do uses more energy input than output notcommercially efficient

Artificial (Induced) Transmutation

Artificial (Induced) Transmutation: A nucleus is bombarded with a nucleon, an alpha particle or another small nucleus,

resulting in a nuclide with a different proton number (a different element).

1. In 1919, Ernest Rutherford discovered that when nitrogen gas is bombarded with alpha particles, oxygen and protons are

produced. Complete the equation for this reaction.

Requirement: the bombarding particle must have sufficient kinetic energy to overcome the Coulomb

repulsion

14 4 1

12

7 1

87 N He O H

NOTE: isotopes produced

Importance: artificial isotopes produced are used in medical tests and therapies

2. Neutron bombardment of lithium can produce the radioactive isotope of hydrogen known as tritium. Complete the reaction.

6 1 3

3 0

4

1 2Li n H He

ib hl study guide organizer-nuclear physics

Documents