18 IONISING RADIATION AND RISKQuantifying risk

• Review properties of alpha, beta and gamma radiation

• Quantify risk associated with exposure to ionising radiation

Starter: On the paper, record a fact about alpha, beta or gamma radiation. Pass it on around the group. Starter: Define the term risk.

Radiation dose measurements and units

activity in becquerel =disintegrations per second

Radiation quality factors

alphabetagammax-rays

neutrons

20 1 1 110

20 Sv 1 Sv 1 Sv 1 Sv

10 Sv

radiation factor dose equivalentof 1 gray

dose in gray = energy deposited per kg

dose equivalent in sievert = dose in gray quality factor

energy in different typesof radiation

radiationsource

dose ingray

Typical whole body dose equivalent per year from various sources (Europe)

Artificial sources530 µSv22% of total

Natural sources1860 µSv78% of totalwork 9 µSv

air travel 8 µSvnuclear power 3 µSvnuclear weapons fallout 10 µSv

cosmic and solar radiation310 µSv

gamma radiation fromground and buildings380 µSv

radon gas from groundand buildings 800 µSvfood and drink 370 µSv

medical procedures500 µSv

QuestionThe activity of an alpha source is 2 x 106 Bq. An alpha particle carries 5 MeV of kinetic energy.

(a) Calculate the absorbed dose in Grays (J kg-1) for a 60 kg person exposed to this source for 30 minutes.

(b) What is the dose equivalent in Sievert? (Quality factor for alpha radiation = 20)

(c) How does your answer to (b) compare with the annual dose equivalent of 2000 μSv?

(d) The best estimate of the probability of developing a radiation-induced cancer is about 5 % per Sievert. What is the risk, expressed as both a percentage and as a fraction, of developing a cancer due to this exposure?

Shielding from x-rays and gamma rays

fraction absorbed extra thickness

dI/I = –µ dxintensity decreases exponentially

= exp (–µx)II0

half-thickness x1/2 = ln 2µ

Ways photons are absorbed or lose energy

1. ionise atom or put atom in higher energy level

2. scattering from electrons

3. electron–positron pair production (important at high energies)

thickness x

half-thicknessof absorber

intensity Ipenetratingthickness x

exponential decreaseof intensity withthickness

Half-thicknessesphoton energy

0.5 mm

200 keVx-ray

25 mm

concrete

50 mm

1 MeV ray 10 mm

lead

Absorption by successive thicknesses

intensity reduced to one half by each block

The half thickness of aluminiumUse these data to determine the half thickness of aluminium towards beta radiation :

Thickness (mm) Corrected count rate (Bq)0.1 13040.2 11500.3 10110.7 5060.8 4130.9 4011.0 3052.0 10

Modelling radioactive decay• Model the process of

radioactive decay, including decay series

• Explain practical consequences of radioactive equilibrium

Starter:A sample containing 0.5 g of uranium 238 has an activity of 6200 Bq. Use this information to show that the half life of uranium 238 is about 4.5 x 109 years.(Molar mass of U-238 = 0.238 kg mol-1; Avogadro’s number = 6 x 1023 mol-1)

Алекс ндр В льтеа́ а́ рович Литвин нкое́

Alexander Litvinenko: former KGB agent who criticised the Russian government.

In November 2006 he was admitted to hospital in London with suspected heavy metal poisoning. He died three weeks later from radiation poisoning.

It turned out that an ex-colleague had spiked his drink with a lethal dose of polonium-210.

Polonium-210 has a half life of only 138 days, so it cannot be stored for long periods.

It can only be made in nuclear reactors, none of which is in the hands of terrorists or criminals.

You don’t need to be Sherlock Holmes to work out who had him killed.

Nuclear binding energy

• Explain how the stability of a nucleus arises from the Einstein mass-energy relation

• Calculate stability of specific nuclides

Starter: Why doesn’t a nucleus just blow apartdue to repulsion between positively-charged protons?

Stable and unstable nuclei: balance of numbers of protons and neutrons

150

100

50

0

0 20 40 60 80 100

proton number Z

stable

alpha decay

beta decay

electron capture

positron emission

fission

N=Z

Binding energy of carbon-12 nucleus

Mass of carbon-12 atom1 atomic mass unit u= 1/12 of mass of C-12 atom1 u = 1.66056 10–27 kg

mass of C-12 atom = 12.0 u

Mass of 6 electrons

= 9.1095 10–31 kg= 0.000549 u

mass of 6 electrons = 0.0033 u

mass of electron6 electrons

Calculate mass of carbon-12 nucleus

mass of carbon-12 nucleus= mass of carbon-12 atom– mass of 6 electrons

Calculate mass of all theprotons and neutrons

mass of proton= 1.67265 10–27 kg= 1.00728 u

mass of neutron= 1.67495 10–27 kg= 1.00866 u

mass of carbon-12 nucleus= (12.000 – 0.0033) u= 11.9967 u

mass of 6 protons and 6 neutrons= 6 (1.00728 + 1.00866) u= 12.0956 u

Difference in mass= mass of carbon-12 nucleus – mass of protons and neutrons= (11.9967 – 12.0956) u= – 0.0989 u

Binding energyin mass units:= – 0.0989 u= – 1.643 10–28 kg

Erest = mc2 in energy units:= – 1.477 10–11 J= – 92.16 MeV

Binding energy per nucleon

–92.16 MeV for 12 nucleons= – 7.7 MeV per nucleon

Binding energy of a nucleus is the difference between its mass and the sum of themasses of its neutrons and protons

Energy well for carbon-12 is7.7 MeV per nucleon deep

7.7 MeV

6 protons6 neutrons

Binding energy of carbon-12 nucleus

Mass of carbon-12 atom1 atomic mass unit u= 1/12 of mass of C-12 atom1 u = 1.66056 10–27 kg

mass of C-12 atom = 12.0 u

Mass of 6 electrons

= 9.1095 10–31 kg= 0.000549 u

mass of 6 electrons = 0.0033 u

mass of electron6 electrons

Calculate mass of carbon-12 nucleus

mass of carbon-12 nucleus= mass of carbon-12 atom– mass of 6 electrons

Calculate mass of all theprotons and neutrons

mass of proton= 1.67265 10–27 kg= 1.00728 u

mass of neutron= 1.67495 10–27 kg= 1.00866 u

mass of carbon-12 nucleus= (12.000 – 0.0033) u= 11.9967 u

mass of 6 protons and 6 neutrons= 6 (1.00728 + 1.00866) u= 12.0956 u

Difference in mass= mass of carbon-12 nucleus – mass of protons and neutrons= (11.9967 – 12.0956) u= – 0.0989 u

Binding energyin mass units:= – 0.0989 u= – 1.643 10–28 kg

Erest = mc2 in energy units:= – 1.477 10–11 J= – 92.16 MeV

Binding energy per nucleon

–92.16 MeV for 12 nucleons= – 7.7 MeV per nucleon

Binding energy of a nucleus is the difference between its mass and the sum of themasses of its neutrons and protons

Energy well for carbon-12 is7.7 MeV per nucleon deep

7.7 MeV

6 protons6 neutrons

The stability curve for nucleons

• Analyse data to plot a graph of binding energy per nucleon against nucleon number

• Interpret fission, fusion and radioactive decay in the context of the graph

• Calculate which nuclear processes are spontaneous

Starter:Do Q4-6 on p235 of textbook.Note: Nuclear, not atomic, masses are given so you don’t need to subtract electron masses.

Nuclear landscape: The nuclear valley of stability

Stable nuclei lie along a narrow band of values of numbers of protonsand neutrons. The more negative the binding energy, the more stablethe nucleus.

These nucleigo down hillby alphadecay

0 40

Bindingenergy pernucleon/MeV

neutron number NZN

Z–2N–2

‘Iron lake’Nuclei near iron in the PeriodicTable are the most strongly bound,and lie in the bottom of the nuclearvalley. Other nuclei tend to ‘rundownhill’ to become like iron.

Fusion hillLarger numbers ofneutrons and protons arebound more strongly thansmaller numbers, by thestrong nuclearinteraction.

These nuclei go down hillby nuclear fusion.

100

0

–2.5

–5.0

–7.5

0

25

50

75

80 120 160

proton number Z

Alpha decay

Z–1N+1

ZN

+

ZN

Z+1N–1

–

proton number Z proton number Z

Positron decay Electron decay

Free particle plains

Binding energy taken as zero forfree protons and neutrons.

Bound nuclei have negativebinding energies, down to about–8 MeV per nucleon

Coulomb slopeMore neutrons thanprotons are neededto overcomeelectrical repulsionof many protons.

Pauli cliffsThe sides of the valley rise steeply. The Pauli exclusion principle keeps numbers ofprotons and neutrons approximately equal. Binding is less strong if either are in excess.

protonnumber Z

A nuclear decay can occur spontaneously if energy is released in the process.

Q1. Write down the equation for alpha emission from silicon 28, using the subscript-superscript notation, identifying the daughter nucleus formed.

Q2. Use the following nuclear mass data to determine if alpha emission from silicon 28 can occur.

Nucleus Mass / uSilicon 28 27.97693Daughter 23.98505Helium 4 (α) 4.00260

Decay sequences• Investigate radioactive decay

chains• Explain significance of radioactive

equilibrium in the context of decay chains

Starter: Q1. What is the most likely mode of decay for the nucleus oxygen-15? Explain your choice and write a balanced nuclear equation for the decay (Hint: the stable isotope of oxygen is oxygen-16).

Q2. What is the nuclide formed when plutonium 241 undergoes the following sequence of decays: βαα βαα

The thorium series The thorium series starts from the nuclide Th-232, which decays by alpha emission.

The decay sequence from thorium is α, β, β , α, α, α, α, β, β, α. Draw a chain which show all the nuclides created in this decay series.

The nuclide Bi-212 in this series can also decay by alpha emission followed by beta emission. Add these decays to the series.

Fission

• Explain how self-sustaining fission is achieved and controlled in a nuclear reactor

• Calculate the energy released by a single fission event and the rate of fission needed for practical power generation

Starter: Write out and balance the following reactions involved in fission:

1. Uranium 235 absorbs a neutron to give uranium 236

2. Uranium 236 fissions to give barium 140, krypton 93 and other particles

3. Use the equations to explain why fission can rapidly run out of control.

Fission questions

Q1. Explain the meaning of the term chain reaction. What is the purpose of the control rods in a reactor? What is the purpose of the moderator?

Q2. Calculate the energy released per fission event (in J and MeV) by doing Q1-5 fromQs 250S.

Q3. What rate of fission is needed to give a reactor thermal power of 1200 MW? (see P239)

Q4. How much U 235 would be consumed in running the reactor for a year?

Q5. If the fuel contains 4 % U 235, how much would be needed each year?

Fission questions

Q1. Explain the meaning of the term chain reaction. What is the purpose of the control rods in a reactor? What is the purpose of the moderator?

Q2. Calculate the energy released per fission event (in J and MeV) by doing Q1-5 fromQs 250S. Answers to Q2-5 (numerical)2. 236.052 588 u3. 235.859 799 u4. –0.192 789 u; energy lost. 5. 179.49 MeV

Q3. What rate of fission is needed to give a reactor thermal power of 1200 MW? (see P239) Answer: 4.2 x 1019 fissions s-1

Q4. How much U 235 would be consumed in running the reactor for a year?Answer: 516 kg

Q5. If the fuel contains 4 % U 235, how much would be needed each year? Answer: 12 917 kg

Critical massA fission reaction cannot become self-sustaining unless the rate at which neutrons are lost from the surface of the mass of fissile material is less than the rate at which they are produced within it.

Consider the fissile mass to be spherical.

Q1. Write down an expression for the rate of production of neutrons, if the rate of production per unit volume is P.Q2. The rate at which neutrons are lost from the fissile mass is proportional to its surface area. Write down an expression for the rate of loss of neutrons from the mass, if the loss rate per unit area is L.Q3. Using your answers to Q1 and Q2, explain why there is a minimum critical size (critical mass) for self-sustaining fission.

Fusion • Estimate temperature needed

for fusion in stars• Explain nuclear reaction

schemes taking place in the Sun

• Explain why the Sun’s lifetime is billions, not millions of years

Starter: Q1. Why are high temperatures required in order for protons to fuse together ?Q2. Write down an expression for the potential energy of a system of 2 charges q1 and q2, separated by a distance r. (Ch.16)Q3. Write down an expression for the average energy of a particle at a temperature T. (Ch.14)Q4. Write down an expression for the radius of a nucleus in terms of its nucleon number A. (Ch.17)

If two protons are to fuse, they must get close enough for the attractive nuclear strongforce to dominate electrostatic repulsion. As a first approximation, we may assume that they need to touch for this to happen.[Review: PE-KE exchanges in nuclear collisions; nuclear radii from electron diffraction; average energy per particle]

1. What is the potential energy of two protons touching? 2. How much kinetic energy will two colliding protons need to have if they are going to get close

enough to touch?3. How is the average kinetic energy of a system of particles related to the temperature? Use the

relevant equation and your answer to the previous question to estimate the temperature needed for fusion.

4. In fact, the core of the Sun is at a temperature of only 10 million K. Which assumption made in your model does this observation suggest is at fault?

5. The energy ε that protons need to fuse is about 8.4 x 10-15 J. What is the Boltzmann factor for proton fusion in the core of the Sun? Comment on the feasibility of the fusion process in light of the numerical value of the Boltzmann factor you have just calculated.

6. The Boltzmann factor may be interpreted as either the fraction of particles with energies in excess of ε at one instant, or as the fraction of time that an individual particle spends with energy in excess of ε.

If a proton in the core of the Sun makes on average 109 collisions s-1, show that it takes an average of 3 billion years before it gains enough energy to fuse with another proton. (1 year = 3.2 x 107 seconds)

7. What are the implications for this period of time for life on Earth?

Fusion in the Sun and on EarthFusion in the Sun: three-stage process

proton proton

positron

neutrino

deuterium H-2

proton deuterium H-2

helium He-3 helium He-3

two protons

Two protons fuse, converting one toa neutron, to form deuterium H-2.

The deuterium H-2 capturesanother proton, to form He-3.

Two He-3 nuclei fuse, givingHe-4 and freeing two protons.

neutron lithium Li-6

Deuterium and tritium are heated to very high temperature. Neutrons from theirfusion then fuse with lithium in a ‘blanket’ around the hot gases. Tritium is renewed.

Fusion on Earth: two-stage process

helium He-3helium He-4

tritium H-3

helium He-4 neutron helium He-4 tritium H-3

deuterium H-2

Fusion plenary

Q1. What is the kinetic energy of a proton when it has reached its distance of closest approach to another proton prior to fusion? What is the potential energy of the system at this point? (give equation)

Q2. Why are higher temperatures needed to initiate the fusion of heavier nuclei?

Q3. Giant stars with high mass are generally much shorter-lived than low-mass stars like the Sun. Explain this observation carefully

Homework: Write out the nuclear fusion reactions on p240 as balanced equations, paying attention to conservation.Revise for Ch.18 test next lesson