DP Physics HL E.3 Radioactive Decay

DP Physics HL E.3 Radioactive Decay - Full notes

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DP Physics HL | E.3 Radioactive Decay

DP PHYSICS · THEME E: NUCLEAR AND QUANTUM PHYSICS

E.3

Radioactive Decay

Full notes for HL · IB-style

Contents

1	Nuclear stability and binding energy
2	Types of radioactive decay
3	Properties and detection of radiation
4	Half-life
5	Uses and safety
6	The decay constant
7	Exam technique and further practice
8	Common misconceptions and links
9	Exam-style question with markscheme
10	Check your understanding
11	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
describe the strong nuclear force and nuclear stability
calculate mass defect and binding energy and interpret the binding energy curve
write equations for alpha, beta and gamma decay
compare properties of alpha, beta and gamma radiation
use half-life, including graphs and dating
describe uses and safety of radioactive sources
use A = λN and N = N₀e−λt
describe methods for measuring long and short half-lives

Sections marked (HL) are Additional Higher Level content: decay constant and exponential decay equations. Every other section is common to SL and HL.

Guiding questions

Why are some nuclei stable while others are unstable? How can the random nature of decay lead to predictable behaviour? How do we use radioactivity safely?

How this topic is examined

E.3 appears in Papers 1 and 2. Expect decay equations, binding energy calculations, half-life graphs and dating, and properties of radiation; HL adds the decay constant and exponential equations.

Balance A and Z in decay equations and use 931.5 MeV per u for energy conversions.

Command term	What the examiner wants
State	a brief answer with no explanation
Outline	a brief account or summary
Describe	a detailed account without reasons
Explain	a detailed account including causes, reasons or mechanisms
Calculate	a numerical answer with working and correct units
Determine	obtain the only possible answer, often from a graph or data
Show that	obtain the given result, showing every step (give one more significant figure)
Estimate	an approximate value using reasonable assumptions
Sketch	a graph or diagram showing the general shape and key features
Deduce / Suggest	reach a conclusion from given information; propose a reasonable explanation

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the physics shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Nuclear stability and binding energy

What holds nuclei together, and how strongly.

Think about it

Why are some nuclei stable for billions of years while others decay in a fraction of a second?

The nucleus is held together by the strong nuclear force against the electrical repulsion of its protons. The balance between these forces decides which nuclei are stable, which decay, and how much energy nuclear reactions release.

1.1 The strong nuclear force

The strong nuclear force acts between all nucleons. It is strongly attractive at separations of about 1–3 fm, becomes repulsive below about 0.5 fm and is negligible beyond a few femtometres. Because it is so short-ranged, each nucleon interacts only with its nearest neighbours, whereas electric repulsion acts between every pair of protons.

Quick check

1. Why does electric repulsion become more important in large nuclei?

Answers: 1. it acts between all pairs of protons, whereas the strong force acts only between neighbours

1.2 The stability band

Light stable nuclei have N ≈ Z. Heavier stable nuclei need extra neutrons (N > Z) to add strong-force attraction without adding repulsion. No nucleus with Z > 82 is completely stable. Nuclei above the band (too many neutrons) tend to decay by β⁻; below it (too many protons) by β⁺; very heavy nuclei by α.

Figure 1. Stable nuclei lie in a band that curves above N = Z.

Quick check

1. Which decay is likely for a nucleus with too many neutrons?

Answers: 1. beta-minus

1.3 Mass defect

The mass of a nucleus is less than the total mass of its separate protons and neutrons. This mass defect Δm corresponds to the binding energy via E = Δmc². Using atomic mass units, 1 u corresponds to 931.5 MeV.

Figure 2. The helium-4 nucleus has less mass than its separate nucleons.

WORKED EXAMPLE 1 Helium-4 (exam level)

Calculate the mass defect and binding energy of ⁴He (nuclear mass 4.001506 u; mp = 1.007276 u; mn = 1.008665 u).

Solution

Δm = 2(1.007276) + 2(1.008665) − 4.001506 = 0.030376 uM1 A1

BE = 0.030376 × 931.5 = 28.3 MeVA1

WORKED EXAMPLE 2 In joules (standard)

Convert 0.030376 u into a binding energy in joules using E = mc².

Solution

m = 0.030376 × 1.661 × 10⁻²⁷ = 5.05 × 10⁻²⁹ kgM1

E = mc² = 4.5 × 10⁻¹² JA1

1.4 Binding energy

The binding energy of a nucleus is the energy needed to separate it completely into its individual nucleons (or the energy released when it forms). It is not energy stored inside the nucleus waiting to be released.

Common mistake

Calling binding energy 'energy stored in the nucleus'. A nucleus with more binding energy is more tightly bound and has less mass.

Quick check

1. Calculate the binding energy per nucleon of ⁴He.

Answers: 1. 28.3/4 = 7.07 MeV

1.5 The binding energy per nucleon curve

Binding energy per nucleon measures how tightly each nucleon is held. It rises steeply for light nuclei, peaks near iron-56 (about 8.8 MeV) and falls slowly for heavy nuclei. Energy is released whenever nuclei change towards the peak: by fusion of light nuclei or fission of heavy nuclei.

Figure 3. Binding energy per nucleon against nucleon number.

WORKED EXAMPLE 3 Reading the curve (exam level)

Use the curve to estimate the total binding energy of uranium-238 and of iron-56.

Solution

U-238: 7.57 × 238 ≈ 1800 MeVA1

Fe-56: 8.79 × 56 ≈ 492 MeVA1

1.6 Energy released in decay

When a nucleus decays, the total mass of the products is less than the mass of the parent. The mass difference becomes the kinetic energy of the products (and any gamma photon): Q = Δm × 931.5 MeV u⁻¹.

Figure 4. Mass before and after the alpha decay of radium-226.

WORKED EXAMPLE 4 Alpha Q-value (exam level)

Use the atomic masses Ra-226 = 226.025410 u, Rn-222 = 222.017578 u, He-4 = 4.002603 u to find the energy released.

Solution

Δm = 226.025410 − (222.017578 + 4.002603) = 0.005229 uM1

Q = 0.005229 × 931.5 = 4.87 MeVA1

1.7 Nuclear and chemical energies compared

Nuclear binding energies are millions of electronvolts per nucleon, while chemical bond energies are a few electronvolts per atom. This is why nuclear fuels release roughly a million times more energy per kilogram than chemical fuels.

Process	Energy per event
breaking a chemical bond	about 4 eV
burning one carbon atom	about 4 eV
alpha decay of Ra-226	4.87 MeV
fission of one U-235 nucleus	about 200 MeV

Quick check

1. By roughly what factor does fission exceed combustion per event?

Answers: 1. about 5 × 10⁷

2 Types of radioactive decay

Alpha, beta and gamma.

2.1 Radioactive decay

Unstable nuclei decay to become more stable, emitting radiation. Decay is random (it is impossible to predict which nucleus will decay next or when) and spontaneous (it is unaffected by temperature, pressure or chemical state).

Quick check

1. What evidence shows that decay is random?

Answers: 1. fluctuations in the count rate from a Geiger counter

2.2 Alpha decay

An alpha particle is a helium nucleus (⁴₂He). Alpha decay reduces A by 4 and Z by 2. Alpha particles are emitted with discrete energies, which reveals nuclear energy levels.

Figure 5. Alpha particles from radium-226 have discrete energies.

WORKED EXAMPLE 5 Equation (standard)

Write the equation for the alpha decay of radium-226 (Z = 88).

Solution

²²⁶₈₈Ra → ²²²₈₆Rn + ⁴₂HeA2

WORKED EXAMPLE 6 Energy shared (exam level)

In alpha decay of radium-226 at rest, 4.87 MeV is released. Using momentum conservation, find the alpha particle's kinetic energy.

Solution

Equal and opposite momenta: E ∝ 1/m; alpha gets 222/226 of the totalM1

Eα = 4.78 MeVA1

2.3 Beta-minus decay

In β⁻ decay a neutron becomes a proton, emitting an electron and an electron antineutrino: n → p + e⁻ + ν̅. A is unchanged and Z increases by 1. The energy released is shared between the electron and the antineutrino, so beta particles have a continuous range of energies.

Figure 6. The continuous energy spectrum of beta particles.

WORKED EXAMPLE 7 Carbon-14 (standard)

Write the equation for the beta-minus decay of carbon-14 (Z = 6).

Solution

¹⁴₆C → ¹⁴₇N + e⁻ + ν̅A2

Quick check

1. Why was the neutrino proposed?

Answers: 1. the continuous beta spectrum seemed to violate conservation of energy

2.4 Beta-plus decay

In β⁺ decay a proton becomes a neutron, emitting a positron (the electron's antiparticle) and an electron neutrino: p → n + e⁺ + ν. Z decreases by 1; A is unchanged. Positron emitters are used in PET scanning.

WORKED EXAMPLE 8 Fluorine-18 (standard)

Write the equation for the β⁺ decay of fluorine-18 (Z = 9).

Solution

¹⁸₉F → ¹⁸₈O + e⁺ + νA2

2.5 Positron annihilation

A positron emitted in β⁺ decay quickly meets an electron. They annihilate, converting their mass into two gamma photons of 0.511 MeV each travelling in opposite directions. PET scanners detect these pairs to locate the source in the body.

Figure 7. Electron–positron annihilation produces two gamma photons.

WORKED EXAMPLE 9 Photon energy (exam level)

Show that each annihilation photon has energy 0.511 MeV (me = 9.11 × 10⁻³¹ kg).

Solution

E = mec² = 9.11 × 10⁻³¹ × 9.0 × 10¹⁶ = 8.2 × 10⁻¹⁴ JM1

= 0.511 MeVAG

2.6 Electron capture

Some proton-rich nuclei capture one of their own inner electrons: p + e⁻ → n + ν. Like β⁺ decay it reduces Z by 1 with A unchanged; X-rays are emitted as other electrons fill the vacancy.

Quick check

1. Which other decay has the same effect on A and Z as electron capture?

Answers: 1. beta-plus decay

2.7 Gamma emission

After alpha or beta decay, the daughter nucleus is often left in an excited state and emits a gamma photon to reach its ground state. A and Z are unchanged. Gamma energies are discrete, showing that nuclei, like atoms, have energy levels.

Quick check

1. Why are gamma-ray energies discrete?

Answers: 1. nuclei have discrete energy levels

2.8 Decays on an N–Z chart

Each type of decay moves a nucleus in a fixed direction on a chart of N against Z: alpha moves 2 down and 2 left; β⁻ moves 1 down and 1 right; β⁺ moves 1 up and 1 left; gamma does not move it.

Figure 8. How alpha, beta-minus and beta-plus decays move a nucleus on an N–Z chart.

Quick check

1. Which decay moves a nucleus towards the stability band from below?

Answers: 1. beta-plus (or electron capture)

2.9 Conservation in nuclear equations

In every nuclear equation, nucleon number, charge and lepton number are conserved. Check that the top numbers add up and the bottom numbers add up on both sides.

WORKED EXAMPLE 10 Missing particle (standard)

Identify X: ²³⁹₉₃Np → ²³⁹₉₄Pu + X + ν̅.

Solution

A unchanged, Z increases by 1M1

X = electron (β⁻)A1

3 Properties and detection of radiation

Ionisation, penetration and fields.

3.1 Ionising and penetrating power

Alpha particles are heavily ionising and have a short range (a few cm of air, stopped by paper). Beta particles are moderately ionising and are stopped by a few millimetres of aluminium. Gamma rays are weakly ionising and very penetrating; their intensity is reduced by several centimetres of lead.

Radiation	Nature	Charge	Ionising ability	Stopped by
alpha	helium nucleus	+2e	very strong	paper / few cm of air
beta-minus	electron	−e	moderate	a few mm of aluminium
gamma	photon	0	weak	several cm of lead (reduced)

Figure 9. Penetrating power of alpha, beta and gamma radiation.

WORKED EXAMPLE 11 Identifying (exam level)

Radiation passes through paper but is stopped by 3 mm of aluminium. Identify it.

Solution

Not alpha (passes paper); not gamma (stopped by Al)M1

BetaA1

3.2 Deflection in fields

In magnetic or electric fields, alpha and beta particles are deflected in opposite directions because they have opposite charges. Beta particles are deflected much more because their mass is so much smaller. Gamma rays are undeflected.

Figure 10. Deflection of alpha, beta and gamma radiation in a magnetic field (not to scale).

Quick check

1. Why is beta deflected more than alpha in a magnetic field?

Answers: 1. much greater charge-to-mass ratio

3.3 Tracks in a cloud chamber

In a cloud chamber, ionising particles leave trails of condensed droplets. Alpha particles leave short, thick, straight tracks of similar length (discrete energies); beta particles leave thin, wiggly, longer tracks; gamma rays leave very few tracks.

Figure 11. Typical alpha and beta tracks in a cloud chamber (schematic).

Quick check

1. Why are alpha tracks all about the same length?

Answers: 1. alphas from one source have the same (discrete) energy

3.4 Absorption of beta radiation

Placing increasing thicknesses of aluminium between a beta source and a detector reduces the count rate until it reaches the background level. This is the principle of beta thickness gauges.

Figure 12. Count rate against aluminium thickness for a beta source.

Try this

Use a Geiger counter and a beta source (under supervision) to measure count rate for different aluminium thicknesses. Subtract background and plot the results.

Quick check

1. Why is beta, not alpha or gamma, used to monitor paper thickness?

Answers: 1. alpha would be completely stopped; gamma would hardly be affected

3.5 Random fluctuations in counts

Repeated counts from the same source over equal times vary randomly about a mean. For a mean count N, the typical fluctuation is about √N, so the percentage uncertainty falls as counts increase. Counting for longer gives more reliable rates.

Figure 13. Thirty repeated 10 s counts from the same source.

WORKED EXAMPLE 12 Uncertainty (exam level)

A 10 s count gives 40 counts. Estimate the percentage uncertainty and how long to count for 5% uncertainty.

Solution

√40/40 ≈ 16%A1

Need N = 400 counts: about 100 sM1 A1

3.6 Detecting radiation

A Geiger–Müller tube counts ionising events; cloud chambers show the tracks of particles; scintillation counters and photographic film are also used. Measured count rates must be corrected by subtracting the background count rate.

WORKED EXAMPLE 13 Corrected count (standard)

A source gives 520 counts per minute; background is 30 counts per minute. Find the corrected count rate.

Solution

520 − 30M1

= 490 counts per minuteA1

3.7 Background radiation

Background radiation comes from radon gas, rocks and buildings, food and drink, cosmic rays, and medical and industrial uses. A typical annual dose is a few millisieverts and varies with location.

Figure 14. Typical sources of background radiation dose (UK averages, approximate).

Quick check

1. Name the largest natural source of background radiation in many countries.

Answers: 1. radon gas

3.8 Gamma intensity and distance

Gamma rays from a point source spread out, so their intensity (and corrected count rate) falls with the square of distance. A graph of corrected count rate against 1/d² is a straight line through the origin.

Figure 15. Corrected count rate against 1/d² for a gamma source.

WORKED EXAMPLE 14 Distance (exam level)

A corrected count rate is 900 counts per minute at 10 cm. Predict the rate at 30 cm.

Solution

I ∝ 1/d²: factor 1/9M1

= 100 counts per minuteA1

4 Half-life

Random decay, predictable averages.

Think about it

If radioactive decay is random, how can we predict that half of a sample will decay in exactly 5730 years?

4.1 Defining half-life

The half-life T½ is the time for half of the radioactive nuclei in a sample to decay, or for the activity to fall to half. Activity A is the number of decays per second, measured in becquerels (Bq). After n half-lives the fraction remaining is (½)n.

Figure 16. Exponential decay of a sample with half-life 8 days.

WORKED EXAMPLE 15 Fraction (standard)

A sample of iodine-131 (T½ = 8 days) has activity 800 Bq. Find its activity after 24 days.

Solution

24/8 = 3 half-livesM1

800/8 = 100 BqA1

WORKED EXAMPLE 16 Time to fall (standard)

How long does it take the activity to fall from 800 Bq to 50 Bq?

Solution

800 → 50 is 4 half-livesM1

= 32 daysA1

Quick check

1. What fraction remains after 5 half-lives?

Answers: 1. 1/32

4.2 Modelling random decay with dice

Throwing many dice and removing those showing a six models radioactive decay: each die has a fixed probability of 'decaying' on each throw. The number remaining falls exponentially, with random fluctuations about the smooth curve.

Figure 17. Simulated dice 'decay' (crosses) against the expected curve.

Try this

Throw 100 coins, remove those showing heads, and repeat. Plot the number remaining against throw number and find the 'half-life'.

Quick check

1. What is the 'half-life' of the dice model in throws?

Answers: 1. about 3.8 throws, since (5/6)^n = 0.5

4.3 Finding half-life from data

Plot activity (or corrected count rate) against time. Read the time to halve from several starting points and average. For noisy data, a best-fit curve is drawn first.

WORKED EXAMPLE 17 Data (exam level)

A corrected count rate falls from 1200 to 150 counts per minute in 45 minutes. Find the half-life.

Solution

1200 → 150 is 3 half-livesM1

T½ = 15 minA1

4.4 Measuring a short half-life

Protactinium-234m (T½ about 70 s) can be separated from a uranium solution in a sealed bottle and its count rate recorded every 10 s with a Geiger counter. After subtracting background, plotting the results gives the half-life directly.

Figure 18. Count rate from protactinium-234m (including background of 30 counts per 10 s).

WORKED EXAMPLE 18 Correcting for background (exam level)

Explain why the background must be subtracted before reading the half-life from this graph.

Solution

The raw count includes background, which does not halveR1

Reading raw values would give too long a half-lifeR1

4.5 Half-lives of common isotopes

Half-lives range from fractions of a second to billions of years.

Isotope	Decay	Half-life	Use or context
technetium-99m	γ	6.0 hours	medical imaging
iodine-131	β⁻, γ	8.0 days	thyroid treatment
cobalt-60	β⁻, γ	5.3 years	radiotherapy, sterilising
caesium-137	β⁻, γ	30 years	fission waste
carbon-14	β⁻	5730 years	dating
uranium-238	α	4.5 × 10⁹ years	dating rocks

Quick check

1. Why is technetium-99m better than cobalt-60 as a medical tracer?

Answers: 1. short half-life limits the patient's dose; gamma only

4.6 Radioactive dating

Carbon-14 (T½ = 5730 years) is used to date once-living material up to about 50 000 years old. Uranium–lead and potassium–argon dating are used for rocks billions of years old.

WORKED EXAMPLE 19 Carbon dating (exam level)

A bone has 25% of the C-14 activity of living bone. Estimate its age.

Solution

25% = 2 half-livesM1

Age = 2 × 5730 = 11 460 yearsA1

In the real world

Radiocarbon dating showed that the Dead Sea Scrolls are about 2000 years old.

4.7 Decay chains

Heavy nuclei such as uranium-238 decay through a series of alpha and beta decays until a stable nucleus (lead-206) is reached. Radon-222, a gas found in homes, is one member of this chain.

Figure 19. The uranium-238 decay series on an N–Z chart.

WORKED EXAMPLE 20 Counting decays (exam level)

How many alpha and beta-minus decays take U-238 (Z = 92) to Pb-206 (Z = 82)?

Solution

A falls by 32: 8 alpha decays (Z falls by 16)M1 A1

Z must rise by 6 to reach 82: 6 beta-minus decaysA1

5 Uses and safety

Radiation at work and how to stay safe.

5.1 Uses of radioactivity

Choices of source depend on the type of radiation and the half-life required.

Use	Radiation	Reason
smoke detectors	alpha (Am-241)	strongly ionising, short range; smoke absorbs it
paper thickness gauges	beta	partly absorbed; absorption depends on thickness
medical tracers (Tc-99m)	gamma	penetrates the body; short half-life (6 h)
radiotherapy	gamma or proton beams	kills cancer cells
sterilising equipment	gamma	penetrates packaging; kills microbes
PET scanning	beta-plus (positrons)	annihilation gamma photons are detected

Quick check

1. Why is an alpha source used in smoke detectors?

Answers: 1. strongly ionising but short range; smoke absorbs alphas, reducing the current

5.2 Biological effects and safety

Ionising radiation can damage cells and DNA, causing radiation sickness at high doses and increasing cancer risk at lower doses. Alpha sources are most dangerous inside the body. Exposure is reduced by limiting time, increasing distance and using shielding.

In the real world

A typical chest X-ray gives about 0.02 mSv, while average annual background dose is about 2–3 mSv.

Quick check

1. Why is an alpha source outside the body relatively harmless?

Answers: 1. alpha particles are stopped by the dead outer layer of skin

5.3 Distance as protection

For a gamma point source, dose rate falls with the square of distance. Using long-handled tongs to handle sources, rather than fingers, reduces the dose enormously.

Figure 20. Dose rate against distance from a gamma source.

WORKED EXAMPLE 21 Tongs (exam level)

A source gives 120 μSv h⁻¹ at 1.0 m. Estimate the dose rate at 2.0 cm (fingers) and at 30 cm (tongs).

Solution

At 2.0 cm: 120 × (1/0.02)² = 3.0 × 10⁵ μSv h⁻¹A1

At 30 cm: 120 × (1/0.3)² = 1300 μSv h⁻¹: about 200 times lessA1

5.4 Nuclear waste

Spent fuel from nuclear reactors contains fission products with half-lives of decades and actinides with half-lives of thousands of years. High-level waste is cooled, vitrified (set in glass) and must be stored securely for very long times, ideally in deep geological repositories.

Quick check

1. Why must some waste be stored for tens of thousands of years?

Answers: 1. it contains isotopes with very long half-lives that remain radioactive

5.5 Choosing a medical tracer

A tracer should emit gamma (detectable outside the body), have a half-life long enough for the procedure but short enough to limit dose, and be chemically suitable for the organ being studied.

WORKED EXAMPLE 22 Iodine-131 (exam level)

Explain why iodine-131 (gamma and beta emitter, T½ = 8 days) is used to study and treat the thyroid.

Solution

Iodine is taken up by the thyroidR1

Gamma allows imaging; beta damages overactive tissue; half-life suits treatmentR1

6 The decay constant

Exponential decay mathematics.

6.1 Decay constant and activity (HL)

The decay constant λ is the probability per unit time that a given nucleus decays. Activity A = λN. The number of undecayed nuclei falls exponentially: N = N₀e−λt, and A = A₀e−λt. Half-life and decay constant are linked by T½ = ln 2/λ.

WORKED EXAMPLE 23 Decay constant (exam level)

Find λ for cobalt-60 (T½ = 5.27 years) in s⁻¹.

Solution

λ = 0.693/(5.27 × 3.16 × 10⁷)M1

= 4.2 × 10⁻⁹ s⁻¹A1

6.2 Non-integer numbers of half-lives (HL)

For times that are not whole numbers of half-lives, use N = N₀e−λt and logarithms.

WORKED EXAMPLE 24 Remaining fraction (exam level)

What fraction of a sample with T½ = 6.0 h remains after 10 h?

Solution

λ = 0.1155 h⁻¹; e−1.155M1

= 0.315A1

WORKED EXAMPLE 25 Activity from mass (exam level)

Find the activity of 1.0 μg of radium-226 (T½ = 1600 years).

Solution

N = 1.0 × 10⁻⁹/0.226 × 6.02 × 10²³ = 2.66 × 10¹⁵M1

A = λN = 1.37 × 10⁻¹¹ × 2.66 × 10¹⁵ = 3.7 × 10⁴ BqA1

6.3 Logarithmic graphs (HL)

Taking logs of A = A₀e−λt gives ln A = ln A₀ − λt. A graph of ln A against t is a straight line with gradient −λ, which is the best way to find λ from experimental data.

Figure 21. ln(activity) against time: gradient = −λ.

WORKED EXAMPLE 26 Gradient (exam level)

The gradient of a ln A–t graph is −0.0866 day⁻¹. Find the half-life.

Solution

λ = 0.0866 day⁻¹M1

T½ = 0.693/0.0866 = 8.0 daysA1

6.4 Measuring long and short half-lives (HL)

Short half-lives (seconds to days): measure activity over time and plot ln A against t. Long half-lives (thousands of years or more): the activity hardly changes, so measure the activity of a known number of nuclei (from the mass) and use λ = A/N.

Quick check

1. Why can't the half-life of uranium-238 be found by watching its activity fall?

Answers: 1. 4.5 × 10⁹ years: activity changes negligibly over a lifetime

7 Exam technique and further practice

Applying the ideas.

7.1 Method checklist

•Balance A and Z in every nuclear equation; include neutrinos in beta decay.
•Subtract background before using count rates.
•Use whole half-lives where possible: (½)n.
•Convert u to MeV with 931.5 MeV per u.
•Explain binding energy as energy needed to separate nucleons.

7.2 Multi-step problem: iodine-131

WORKED EXAMPLE 27 Medical source (exam level)

Iodine-131 (Z = 53) decays by β⁻ to xenon with T½ = 8.0 days. A sample has activity 800 Bq. Find (a) the decay equation; (b) the activity after 32 days; (c) the time for the activity to fall below 30 Bq.

Solution

(a) ¹³¹₅₃I → ¹³¹₅₄Xe + e⁻ + ν̅A1

(b) 4 half-lives: 50 BqA1

7.3 Graded practice

WORKED EXAMPLE 28 Iron binding energy (standard)

Calculate the binding energy per nucleon of ⁵⁶Fe (mass defect 0.5285 u).

Solution

0.5285 × 931.5/56M1

= 8.79 MeVA1

WORKED EXAMPLE 29 Thorium chain (exam level)

Thorium-232 (Z = 90) decays by alpha, then by two beta-minus decays. Find A and Z of the final nucleus.

Solution

Alpha: A = 228, Z = 88A1

Two β⁻: A = 228, Z = 90 (thorium-228)A1

WORKED EXAMPLE 30 Activity ratio (standard)

Sample X has twice as many nuclei as sample Y but a half-life four times longer. Compare their activities.

Solution

A = λN = (ln 2/T)NM1

AX/AY = 2/4 = 0.5A1

7.4 Graded practice: decay and dating

WORKED EXAMPLE 31 Rock dating (exam level)

A rock contains 3 atoms of lead-206 for every atom of uranium-238 (T½ = 4.5 × 10⁹ years). Assuming all lead came from uranium, estimate the rock's age.

Solution

Uranium fraction remaining = 1/4: two half-livesM1

Age = 9.0 × 10⁹ yearsA1

WORKED EXAMPLE 32 Activity after time (standard)

A 2000 Bq source has a half-life of 3.0 h. Find its activity after 15 h.

Solution

5 half-livesM1

2000/32 = 62.5 BqA1

WORKED EXAMPLE 33 Identify the radiation (standard)

A source's count rate drops sharply with a sheet of paper and does not change further with aluminium. What does it emit?

Solution

Alpha stopped by paperA1

Remaining count is background, so alpha onlyA1

7.5 Data analysis: finding a half-life

A student records the count rate from a short-lived source. Background is 24 counts per minute.

t / min	0	10	20	30	40	50
count rate / min⁻¹	824	590	424	310	224	168
corrected / min⁻¹	800	566	400	286	200	144

Figure 22. Corrected count rate against time.

WORKED EXAMPLE 34 Half-life (exam level)

Use the corrected data to find the half-life.

Solution

800 → 400 in 20 min; 400 → 200 in 20 minM1

T½ = 20 minA1

7.6 Multi-step problem: carbon dating

WORKED EXAMPLE 35 Wooden tool (exam level)

Living wood gives 15.3 decays per minute per gram of carbon. A wooden tool gives 3.8 decays per minute per gram. Find (a) the fraction of C-14 remaining; (b) the age of the tool; (c) one assumption made.

Solution

(a) 3.8/15.3 = 0.25A1

(b) 2 half-lives = 11 460 yearsA1

7.7 Graded practice: nuclear equations

WORKED EXAMPLE 36 Polonium (standard)

Polonium-210 (Z = 84) emits an alpha particle. Write the equation.

Solution

²¹⁰₈₄Po → ²⁰⁶₈₂Pb + ⁴₂HeA2

WORKED EXAMPLE 37 Strontium (standard)

Strontium-90 (Z = 38) is a beta-minus emitter. Write the equation.

Solution

⁹⁰₃₈Sr → ⁹⁰₃₉Y + e⁻ + ν̅A2

WORKED EXAMPLE 38 Sodium-22 (exam level)

Sodium-22 (Z = 11) decays by β⁺. Identify the daughter and explain why this decay occurs.

Solution

²²₁₁Na → ²²₁₀Ne + e⁺ + νA1

Na-22 has too few neutrons (11 p, 11 n is proton-rich for sodium); decay moves it towards stabilityR1

7.8 Graded practice: binding energy

WORKED EXAMPLE 39 Deuterium (standard)

The mass defect of deuterium is 0.00239 u. Find its binding energy and binding energy per nucleon.

Solution

BE = 0.00239 × 931.5 = 2.23 MeVA1

Per nucleon = 1.11 MeVA1

WORKED EXAMPLE 40 Uranium mass defect (exam level)

Uranium-238 has binding energy per nucleon 7.57 MeV. Find its mass defect in u and kg.

Solution

BE = 1802 MeV: Δm = 1802/931.5 = 1.93 uM1 A1

= 3.2 × 10⁻²⁷ kgA1

WORKED EXAMPLE 41 Which is more stable? (standard)

Oxygen-16 has 7.98 MeV per nucleon and lithium-7 has 5.61 MeV per nucleon. Which is more tightly bound?

Solution

Higher BE per nucleonM1

Oxygen-16A1

7.9 Graded practice: explanations

WORKED EXAMPLE 42 Random but predictable (exam level)

Explain how radioactive decay can be random yet have a constant half-life.

Solution

Each nucleus has a fixed probability of decaying per unit timeR1

With very large numbers, the fraction decaying in a given time is predictableR1

WORKED EXAMPLE 43 Neutron-rich nuclei (standard)

Explain why a neutron-rich nucleus tends to decay by beta-minus.

Solution

A neutron changes into a protonR1

This moves the nucleus towards the stability band (N falls, Z rises)R1

WORKED EXAMPLE 44 Alpha inside the body (standard)

Explain why an alpha source is dangerous if swallowed.

Solution

Alpha is highly ionisingR1

All its energy is deposited in nearby living cellsR1

7.10 Self-test 2

Term	Meaning
mass defect	difference between nucleon masses and nuclear mass
binding energy	energy needed to separate a nucleus into nucleons
activity	decays per second (Bq)
half-life	time for half the nuclei (or activity) to decay
random	cannot predict which nucleus decays next
spontaneous	unaffected by external conditions
decay constant (HL)	probability of decay per unit time

Students often think…	Actually…
half-life is half the time for all nuclei to decay	it is the time for half of the remaining nuclei to decay
radioactive decay can be sped up by heating	it is spontaneous and unaffected
gamma radiation changes A or Z	it changes neither
binding energy is energy stored in the nucleus	it is the energy needed to separate it into nucleons
beta particles come from electron shells	they come from the nucleus

Topic	Connection
E.1 Structure of the atom	nuclear notation and isotopes
E.4 Fission	binding energy and energy release
E.5 Fusion and stars	binding energy curve
D.3 Motion in EM fields	deflection of radiation

Idea	Remember
Stability	strong force vs electric repulsion
Binding energy	Δmc²; peak at Fe-56
Decay types	α, β⁻, β⁺, γ
Properties	ionisation vs penetration
Half-life	(½)ⁿ; dating
Safety	time, distance, shielding

Part	Answer	Marks
(a)	¹³¹₅₃I → ¹³¹₅₄Xe + e⁻	A1
	+ antineutrino	A1
(b)	time for 800 → 400 Bq (and others)	M1
	8 days	A1
(c)	800/16 = 50 Bq	A1
(d)	half-life long enough to detect, short enough to limit dose	R1
	iodine absorbed by thyroid; radiation detectable outside body	R1
(e)	energy shared between electron and antineutrino	R1
	in varying proportions	R1
(f)	λ = 0.693/(8 × 86 400) = 1.0 × 10⁻⁶ s⁻¹	A1
	N = A/λ	M1
	8.0 × 10⁸	A1