Physics With Kate
Empowering Girls in Physics
Spec-aligned revision resources, group courses, and 1:1 tutoring for GCSE and A Level Physics — built by an experienced teacher and examiner.
Edexcel A Level Physics · Topic 11
Nuclear Radiation
FREE Online Physics A Level Course90minutes
76marks
9questions
How it works
- Answer each question. Type into the boxes or pick an option, and use paper for working and drawings.
- Press Check my answers at the end of the question. Tick-box and number answers are marked for you.
- For written answers, compare yours with the mark scheme and tick each point you made. We highlight the key words we spot, but you decide, just like an examiner.
You will need: paper, a pen and a calculator. Use 1 MeV = 1.60 × 10⁻¹³ J. Your answers save automatically in this browser.
1
Question 1
Multiple choice10 marksTen quick questions from across the topic. Each part is worth 1 mark.
(a)
Which condition, in a sample of hydrogen, is not necessary for nuclear fusion to occur?
(1 mark)
Mark scheme
High temperature gives the nuclei enough kinetic energy to overcome the electrostatic repulsion; high density (pressure) makes collisions frequent. The total mass of the sample doesn’t matter.
Answer: B very high mass
(b)
The fuel in a nuclear fission reactor is uranium. Which of the following is required for fission to proceed?
(1 mark)
Mark scheme
Fission is induced by a nucleus absorbing a (slow) neutron. It’s fusion that needs very high temperatures.
Answer: C The uranium nuclei must absorb neutrons.
(c)
A radioactive sample is a beta emitter. Beta particles are able to pass through
(1 mark)
Mark scheme
Beta passes through paper and card but is stopped by a few mm of aluminium.
Answer: B thick card.
(d)
A thin aluminium foil between a source and a detector makes no difference to the count rate, but a 0.5 cm aluminium plate brings it down to background. The source must be emitting
(1 mark)
Mark scheme
No drop with thin foil: no alpha. Back to background with 0.5 cm aluminium: no gamma (gamma would still get through).
Answer: C beta radiation only.
(e)
Which row correctly compares β particles with α particles of the same energy?
(1 mark)
Mark scheme
β is less ionising (fewer ion pairs per cm), so it loses its energy more slowly and travels further.
Answer: C Range: greater · Ion pairs per cm: smaller
(f)
A treatment for brain tumours directs many narrow beams of radiation from different directions onto the cancerous cells. Which is the most suitable radiation?
(1 mark)
Mark scheme
The beams must get through the skull to reach the tumour. Alpha wouldn’t get through the skin.
Answer: C gamma radiation, because it is very penetrating
(g)
In an experiment to measure the activity of a source, the measured activity should always be corrected by
(1 mark)
Mark scheme
Activity is a rate, so subtract the background count rate.
Answer: D subtracting the background count rate.
(h)
A student uses a Geiger counter on a weak source. To find the background count rate, the best procedure is to measure the count for
(1 mark)
Mark scheme
Decay is random, so a longer count is more reliable. With the source in place you’d be measuring the source too.
Answer: C 10 minutes before the source is in position.
(i)
Radioactive decay is described as spontaneous. In this context, spontaneous means
(1 mark)
Mark scheme
Spontaneous: the decay is not affected by external factors (such as temperature, pressure or force). No external trigger is needed. “Random” (B) is a different idea.
Answer: A nothing can influence the decay.
(j)
A radioactive isotope decays to a stable isotope. Which graph shows how the number of daughter nuclei varies with time?
Sketch graphs
(1 mark)
Mark scheme
Key equationNdaughter = N0(1 − e−λt)starts at zero, grows, levels offC is the number of parent nuclei.
Answer: D
2
Question 2
2 marksState what is meant by binding energy.
(2 marks)
Mark scheme
Key equationΔE = c²ΔmModel answer: Binding energy is the energy required to completely separate a nucleus into its individual protons and neutrons. Learn the definitions word for word: Nuclear Radiation definitions.
Tick each marking point your answer makes:
3
Question 3
9 marksIn 1939 Hans Bethe described the fusion processes in stars. In the proton–proton cycle, hydrogen is converted to helium in stages. One stage is:
73Li + X → 2 × 42He
(a)(i)
Complete the equation: give the nucleon number and proton number of X, then name X.
(2 marks)
Mark scheme
7 + 1 = 2 × 4 on top; 3 + 1 = 2 × 2 on the bottom.
X: nucleon number (top): 1
X: proton number (bottom): 1
X is: a proton (accept hydrogen / H)
💡 Deuterium would have a nucleon number of 2. Balance the top: 7 + ? = 8.
(a)(ii)
Calculate, in joules, the energy released in this stage of the cycle.
| Mass / MeV/c² | |
|---|---|
| proton | 938.3 |
| neutron | 939.6 |
| helium nucleus | 3727.4 |
| lithium nucleus | 6533.8 |
J
(3 marks)
Mark scheme
Key equationΔE = c²Δmin MeV/c², the mass difference gives the energy in MeV directly
Energy released: 2.77 × 10⁻¹² J (6533.8 + 938.3 − 2 × 3727.4 = 17.3 MeV; × 1.60 × 10⁻¹³)
💡 That’s the energy in MeV. Convert: × 1.60 × 10⁻¹³ J.
💡 1 MeV is 1.60 × 10⁻¹³ J, not 1.60 × 10⁻¹⁹ J (that’s 1 eV).
💡 Use a proton (938.3), not a neutron, as X.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
(b)
Explain why sustainable fusion has not yet been achieved for the generation of electrical power.
(4 marks)
Mark scheme
Max 4. Tick what you wrote (max 4):
⚠️ Examiners want high density (so collisions happen more frequently), not high pressure.
⚠️ It’s nuclei that fuse, not atoms or particles.
4
Question 4
4 marksIn 2012, building began on ITER in France: an experimental fusion reactor.
(a)
Describe the process of nuclear fusion.
(2 marks)
Mark scheme
Model answer: The joining of two small mass nuclei to form a large mass nucleus, releasing energy. Learn the definitions word for word: Nuclear Radiation definitions.
Tick each marking point your answer makes:
⚠️ Say nuclei, not atoms.
(b)
Explain why the fusion of hydrogen nuclei should release energy.
(2 marks)
Mark scheme
Key equationΔE = c²Δm
Tick each marking point your answer makes:
5
Question 5
7 marksSome smoke detectors contain americium-241, whose nuclei decay by emitting an alpha particle. The americium makes the air between two metal plates conduct.
Part of a smoke detector
(a)(i)
Explain why a stream of alpha particles causes charge to flow between the plates.
(2 marks)
Mark scheme
Tick each marking point your answer makes:
(a)(ii)
Suggest how smoke particles between the plates cause the current to decrease.
(1 mark)
Mark scheme
Tick each marking point your answer makes:
(b)(i)
The decay of americium-241 is random and spontaneous. State what is meant by random and by spontaneous.
(2 marks)
Mark scheme
Model answers: Random: we cannot predict which nucleus will decay next or when a particular nucleus will decay. Spontaneous: the decay is not affected by external factors (such as temperature, pressure or force). No external trigger is needed. Learn the definitions word for word: Nuclear Radiation definitions.
Tick each marking point your answer makes:
⚠️ Examiners only give the mark for nucleus: decay is a nuclear process, not an atomic one.
(b)(ii)
Complete the equation for the alpha decay of americium-241.
95Am → 237 Np + 42α
(2 marks)
Mark scheme
Am: nucleon number (top): 241
💡 The top line must balance: ? = 237 + 4.
Np: proton number (bottom): 93
💡 Alpha decay removes 2 protons: 95 − 2.
💡 An alpha particle has 2 protons: 95 − 2 = 93.
6
Question 6
13 marksPositron emission tomography (PET) detects the pairs of gamma rays made when positrons from a radioisotope annihilate with electrons. Carbon-11 is used: it has a half-life of 1220 s and decays by positron emission to stable boron-11.
(a)
Explain what is meant by a radioactive atom.
(2 marks)
Mark scheme
Tick each marking point your answer makes:
(b)
Complete the equation for the decay of carbon-11.
116C → B + e⁺ + 00νe
(2 marks)
Mark scheme
Top line: 11 = 11 + 0 + 0. Bottom line: 6 = 5 + 1 + 0.
B: nucleon number: 11
e⁺: nucleon number: 0
B: proton number: 5
💡 A positron carries away +1 charge, so the proton number goes down: 6 − 1.
e⁺: proton (charge) number: 1
💡 That’s an electron. A positron has charge +1.
(c)
Calculate the energy, in joules, released in the positron decay of one carbon-11 nucleus.
| Mass / MeV/c² | |
|---|---|
| positron | 0.511 |
| carbon-11 nucleus | 10 253.6 |
| boron-11 nucleus | 10 252.2 |
J
(3 marks)
Mark scheme
Key equationΔE = c²Δm
Energy released: 1.42 × 10⁻¹³ J (10 253.6 − 10 252.2 − 0.511 = 0.889 MeV)
💡 That’s in MeV. Convert: × 1.6 × 10⁻¹³ J.
💡 Don’t forget the positron’s mass: subtract 0.511 too.
💡 1 MeV is 1.6 × 10⁻¹³ J, not 1.6 × 10⁻¹⁹ J.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
(d)
Explain why carbon-11 is a relatively safe radioisotope to use within the body.
(2 marks)
Mark scheme
Tick each marking point your answer makes:
(e)
A patient was injected with carbon-11 of activity 1.58 × 10⁶ Bq. The sample had been prepared 3600 s earlier. Calculate its activity when it was prepared.
Bq
(4 marks)
Mark scheme
Key equationλt½ = ln 2Key equationA = A0 e−λt
Activity when prepared: 1.2 × 10⁷ Bq (λ = 0.693 ÷ 1220 = 5.68 × 10⁻⁴ s⁻¹; 1.58 × 10⁶ × e2.045)
💡 That’s the activity 3600 s after injection. The question wants the activity before: the earlier value is bigger.
💡 You used 3600 ÷ 1220 half-lives but multiplied by it. Use A0 = A × 2t/t½.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 3):
7
Question 7
6 marksTechnetium-99m decays by emitting γ radiation. Its half-life is 2.16 × 10⁴ s.
(a)
State what is meant by half-life.
(1 mark)
Mark scheme
Model answer: The average time taken for the number of undecayed nuclei of a sample to halve. Learn the definitions word for word: Nuclear Radiation definitions.
Tick each marking point your answer makes:
⚠️ It’s the number of undecayed nuclei that halves, not the mass.
(b)(i)
A sample contains 7.30 × 10¹⁹ nuclei of technetium-99m. Show that its activity is about 2.3 × 10¹⁵ Bq.
Bq
(3 marks)
Mark scheme
Key equationA = λN
Activity: 2.34 × 10¹⁵ Bq (λ = 3.21 × 10⁻⁵ s⁻¹; × 7.30 × 10¹⁹). A “show that” needs 3 significant figures.
💡 For a “show that”, give one more figure than the question: 2.34 × 10¹⁵.
💡 λ = ln 2 ÷ t½ (= 3.21 × 10⁻⁵ s⁻¹), not t½ ÷ ln 2.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
(b)(ii)
Calculate the activity 1 day after the sample was prepared. 1 day = 86 400 s.
Bq
(2 marks)
Mark scheme
Key equationA = A0 e−λt
Activity after 1 day: 1.5 × 10¹⁴ Bq (1 day = 4 half-lives: 2.34 × 10¹⁵ ÷ 16 = 1.46 × 10¹⁴)
💡 That’s 2 half-lives. 86 400 ÷ 21 600 = 4 half-lives.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 1):
8
Question 8
9 marksAfter the 2011 Fukushima accident, seaweed near Long Beach, California was contaminated with iodine-131, a β emitter. A Geiger counter measured a sample of seaweed for 10 minutes; the corrected readings are shown.
| Corrected count 1 | Corrected count 2 | Corrected count 3 | Corrected count rate / Bq |
|---|---|---|---|
| 3820 | 3830 | 3825 | 6.38 |
(a)
State what is meant by the activity of a radioactive source.
(1 mark)
Mark scheme
Model answer: The number of nuclei that decay per second. Learn the definitions word for word: Nuclear Radiation definitions.
Tick each marking point your answer makes:
⚠️ Say what decays: nuclei.
(b)(i)
State why the Geiger counter readings have to be corrected.
(1 mark)
Mark scheme
Tick each marking point your answer makes:
(b)(ii)
Explain why the count is repeated.
(2 marks)
Mark scheme
Tick each marking point your answer makes:
(b)(iii)
The same sample is measured again 30 days later. Calculate the new corrected count rate. Half-life of iodine-131 = 8.0 days.
Bq
(3 marks)
Mark scheme
Key equationA = A0 e−λt
New corrected count rate: 0.47 Bq (λ = 0.0866 day⁻¹; 6.38 × e−0.0866 × 30)
💡 That started from 2.5 Bq per gram. Start from the measured count rate, 6.38 Bq (2.5 Bq still earns the method marks).
💡 You divided by 3.75. 30 days is 3.75 half-lives: multiply by (½)3.75.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
(b)(iv)
There is a moderate risk to the public from iodine-131 building up in seaweed. Explain why.
(2 marks)
Mark scheme
Tick each marking point your answer makes:
9
Question 9
16 marksCarbon-14 is radioactive and decays by emitting β⁻ particles. It forms in the upper atmosphere when a nitrogen nucleus absorbs a neutron.
n + 147N → 146C + X
(a)
Complete the nuclear equation and identify X.
(3 marks)
Mark scheme
Top: 1 + 14 = 14 + 1. Bottom: 0 + 7 = 6 + 1.
n: nucleon number: 1
X: nucleon number: 1
n: proton number: 0
X: proton number: 1
X is: a proton
(b)
After an organism dies, the ratio of carbon-14 to carbon-12 in it falls. Use the activity–time graph to show that the half-life of carbon-14 is about 6000 years.
Activity of carbon-14
years
(2 marks)
Mark scheme
Key equationλt½ = ln 2Using one half-life only (240 → 120 Bq) is less reliable: the scheme wants at least two.
Half-life from your readings: about 5800–6000 years (e.g. 240 Bq → 30 Bq takes 17 500 years = 3 half-lives: 5833 years)
💡 That’s two half-lives (240 → 60 Bq). Divide by 2.
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 1):
(c)(i)
A sample from an ancient wooden object gives a corrected count rate of 10.9 Bq; the same mass of living wood gives 29.6 Bq. State why the count rates have to be corrected.
(1 mark)
Mark scheme
Tick each marking point your answer makes:
(c)(ii)
State one procedure that would make these count rates more accurate.
(1 mark)
Mark scheme
Tick each marking point your answer makes:
(c)(iii)
Calculate the age of the object. Assume the half-life of carbon-14 is 6000 years.
years
(3 marks)
Mark scheme
Key equationA = A0 e−λt
Age: 8650 years (λ = 1.16 × 10⁻⁴ y⁻¹; t = ln(29.6 ÷ 10.9) ÷ λ)
💡 Right, in seconds. The box wants years: keep λ in year⁻¹ and you never need seconds.
💡 You used log₁₀. Use ln (natural log).
Final answer not right? You can still earn method marks. Tick each step you wrote down (max 2):
(d)
A news report claims a dinosaur skull is 68 million years old. Explain why carbon-14 dating could not have been used to find its age.
(2 marks)
Mark scheme
Tick each marking point your answer makes:
*(e)
A website includes this statement about heavy nuclei:
Particles in the nucleus are held together by a force scientists call nuclear binding energy. It is possible to overcome the binding energy in some large atoms, such as uranium atoms, causing the atoms to undergo fission.
Comment on this statement.
Technical terms must be spelt correctly and your answer organised logically.
(4 marks)
Mark scheme
Model definitions: Binding energy is the energy required to completely separate a nucleus into its individual protons and neutrons. Fission is the splitting of a large mass nucleus into two smaller mass nuclei, releasing energy and two or three neutrons. Learn the definitions word for word: Nuclear Radiation definitions.
Tick each marking point your answer makes:
Your results
0 / 76
| Question | Q1 | Q2 | Q3 | Q4 | Q5 | Q6 | Q7 | Q8 | Q9 | Total |
|---|---|---|---|---|---|---|---|---|---|---|
| Out of | 10 | 2 | 9 | 4 | 7 | 13 | 6 | 9 | 16 | 76 |
| Your mark | – | – | – | – | – | – | – | – | – | 0 |
Physics With Kate · FREE Online Physics A Level Course · physicswithkate.com
Proud member · Accredited · Trusted partners
© 2026 Physics With Kate · All rights reserved