A Level Physics 23.2 Radioactive Decay MCQs – 30 Practice Questions with Answers (Cambridge 9702)
These 30 original practice questions cover Cambridge AS & A Level Physics (9702) Section 23.2, Radioactive Decay, an A Level (A2) topic. They practise half-life, activity, the decay constant, the exponential decay laws, and the uses of radioactive isotopes, with a full worked explanation for every answer.
Formula recap: activity A = λN, where λ is the decay constant. λ = ln 2 / t1/2 = 0.693 / t1/2. Number of nuclei remaining: N = N0 e−λt. Activity: A = A0 e−λt. 1 becquerel (Bq) = 1 decay per second. After n half-lives, the fraction remaining is (1/2)n. See the A Level Physics Formula Sheet for the full list of equations.
Radioactive Decay & Half-Life
Q1. Radioactive decay is described as a random process. What does this mean?
- (A) Every nucleus in a sample decays after exactly one half-life
- (B) The nuclei decay one after another in a regular, repeating pattern
- (C) It is impossible to predict which nucleus will decay next, or when a particular nucleus will decay
- (D) A nucleus only decays when the sample is disturbed or shaken
Show Answer
Answer: C – In a random process we know the probability of decay per unit time, but we cannot predict which individual nucleus decays or the moment at which a given nucleus will decay.
Q2. Radioactive decay is also described as spontaneous. A sample is heated strongly and placed under high pressure. What happens to its rate of decay?
- (A) The rate of decay increases
- (B) The rate of decay is unchanged
- (C) The rate of decay decreases
- (D) The decay stops completely
Show Answer
Answer: B – Decay is spontaneous: it is not affected by temperature, pressure or the chemical state of the material. The decay rate stays the same.
Q3. Which statement is the correct definition of the half-life of a radioactive nuclide?
- (A) The time taken for all the nuclei in a sample to decay
- (B) The time taken for one nucleus to decay completely
- (C) The time taken for the mass of a sample to fall to half its value because of heating
- (D) The time taken for the number of undecayed nuclei in a sample to fall to half its original value
Show Answer
Answer: D – The half-life is the time for half of the undecayed nuclei present to decay, which is also the time for the activity of the sample to halve.
Q4. A sample contains 64 000 undecayed nuclei of a nuclide with a half-life of 5.0 days. How many undecayed nuclei remain after 15 days?
- (A) 8000
- (B) 16 000
- (C) 21 000
- (D) 32 000
Show Answer
Answer: A – Number of half-lives = 15 / 5.0 = 3. Nuclei remaining = 64 000 × (1/2)3 = 64 000 / 8 = 8000.
Q5. What fraction of the original nuclei of a sample remains undecayed after 4 half-lives?
- (A) 1/8
- (B) 1/16
- (C) 1/4
- (D) 15/16
Show Answer
Answer: B – After n half-lives the fraction remaining is (1/2)n, so after 4 half-lives it is (1/2)4 = 1/16.
Q6. What fraction of the original nuclei in a sample has decayed after 3 half-lives?
- (A) 1/8
- (B) 3/8
- (C) 7/8
- (D) 1/2
Show Answer
Answer: C – After 3 half-lives the fraction remaining is (1/2)3 = 1/8, so the fraction that has decayed is 1 − 1/8 = 7/8.
Q7. The count rate from a radioactive source falls from 480 counts per second to 120 counts per second in 12 minutes. What is the half-life of the source?
- (A) 3.0 min
- (B) 4.0 min
- (C) 6.0 min
- (D) 24 min
Show Answer
Answer: C – The count rate halves twice: 480 → 240 → 120. Two half-lives take 12 min, so one half-life is 12 / 2 = 6.0 min.
Q8. On a graph of activity against time, a source has activity 800 Bq at time zero and 200 Bq after 30 minutes. What is its half-life?
- (A) 7.5 min
- (B) 15 min
- (C) 10 min
- (D) 20 min
Show Answer
Answer: B – The activity falls from 800 Bq to 200 Bq, which is two halvings (800 → 400 → 200). Two half-lives = 30 min, so the half-life is 15 min.
Q9. Two samples, X and Y, contain the same number of undecayed nuclei. X has a half-life of 2.0 hours and Y has a half-life of 2.0 days. Which sample has the greater initial activity?
- (A) X, because its nuclei decay more quickly
- (B) Y, because its nuclei last longer
- (C) They have the same initial activity
- (D) It is impossible to tell without knowing the masses of the samples
Show Answer
Answer: A – Activity A = λN. With the same N, the sample with the shorter half-life has the larger decay constant λ, so X has the greater initial activity.
Q10. Which statement about half-life is correct?
- (A) After two half-lives, all the nuclei in a sample have decayed
- (B) The half-life of a nuclide depends on how much of the sample is present
- (C) The half-life is also the time taken for the activity of a sample to fall to half its value
- (D) After one half-life, exactly half of every individual nucleus has decayed
Show Answer
Answer: C – Since activity is proportional to the number of undecayed nuclei, the activity halves in the same time as the number of nuclei halves. The half-life is a property of the nuclide only.
Activity & the Decay Constant
Q11. The activity of a radioactive source is measured in becquerels. One becquerel is equal to
- (A) one decay per minute
- (B) one decay per hour
- (C) 1000 decays per second
- (D) one decay per second
Show Answer
Answer: D – The becquerel (Bq) is the SI unit of activity: 1 Bq = 1 disintegration (decay) per second.
Q12. The decay constant λ of a nuclide is best defined as
- (A) the probability that a given nucleus will decay per unit time
- (B) the number of nuclei that decay each second
- (C) the time taken for half the nuclei to decay
- (D) the fraction of nuclei that decay in one second in every sample, whatever its size
Show Answer
Answer: A – The decay constant is the probability per unit time that a particular nucleus will decay. The activity is then A = λN.
Q13. A sample contains 5.0 × 106 undecayed nuclei of a nuclide with decay constant 0.20 s−1. What is the activity of the sample?
- (A) 2.5 × 105 Bq
- (B) 1.0 × 106 Bq
- (C) 2.5 × 107 Bq
- (D) 1.0 × 105 Bq
Show Answer
Answer: B – A = λN = 0.20 × 5.0 × 106 = 1.0 × 106 Bq.
Q14. A nuclide has a half-life of 69 s. What is its decay constant?
- (A) 0.010 s−1
- (B) 0.10 s−1
- (C) 100 s−1
- (D) 0.69 s−1
Show Answer
Answer: A – λ = 0.693 / t1/2 = 0.693 / 69 = 0.010 s−1.
Q15. The decay constant of a nuclide is 0.0347 h−1. What is its half-life?
- (A) 0.050 h
- (B) 20 h
- (C) 35 h
- (D) 200 h
Show Answer
Answer: B – t1/2 = 0.693 / λ = 0.693 / 0.0347 = 20 h.
Q16. A sample initially contains 8.0 × 1018 undecayed nuclei. After a time t for which λt = 1.386, how many nuclei remain?
- (A) 4.0 × 1018
- (B) 2.9 × 1018
- (C) 1.0 × 1018
- (D) 2.0 × 1018
Show Answer
Answer: D – N = N0 e−λt = 8.0 × 1018 × e−1.386 = 8.0 × 1018 × 0.25 = 2.0 × 1018. (λt = 1.386 corresponds to two half-lives.)
Q17. A nuclide has a decay constant of 0.050 year−1. In the equation N = N0 e−λt, the time t must be measured in
- (A) seconds
- (B) hours
- (C) years
- (D) any unit, as the units do not matter
Show Answer
Answer: C – The product λt must have no units, so t must be measured in the unit that matches λ. Here λ is per year, so t must be in years.
Q18. Nuclide P has a larger decay constant than nuclide Q. This means that, compared with Q, P has
- (A) a longer half-life
- (B) the same half-life, because the decay constant does not affect it
- (C) a larger number of nuclei in every sample
- (D) a shorter half-life, and its activity falls more rapidly
Show Answer
Answer: D – Since λ = 0.693 / t1/2, a larger decay constant means a shorter half-life, so the number of nuclei and the activity decrease more rapidly.
Q19. A source has initial activity 6.0 × 103 Bq and decay constant 0.0693 min−1. What is its activity after 20 min?
- (A) 3.0 × 103 Bq
- (B) 2.1 × 103 Bq
- (C) 1.5 × 103 Bq
- (D) 7.5 × 102 Bq
Show Answer
Answer: C – λt = 0.0693 × 20 = 1.386, so A = A0 e−1.386 = 6.0 × 103 × 0.25 = 1.5 × 103 Bq.
Q20. As a radioactive sample decays, which quantity stays constant?
- (A) The number of undecayed nuclei
- (B) The activity of the sample
- (C) The decay constant of the nuclide
- (D) The count rate measured by a detector near the sample
Show Answer
Answer: C – The decay constant is a fixed property of the nuclide. N, the activity and the count rate all fall exponentially as the sample decays.
Decay Calculations & Uses
Q21. A sample contains 1.6 × 1020 undecayed nuclei of a nuclide with half-life 8.0 days. How many nuclei remain after 24 days?
- (A) 2.0 × 1019
- (B) 4.0 × 1019
- (C) 5.3 × 1019
- (D) 8.0 × 1019
Show Answer
Answer: A – Number of half-lives = 24 / 8.0 = 3. N = 1.6 × 1020 × (1/2)3 = 1.6 × 1020 / 8 = 2.0 × 1019.
Q22. A source has an activity of 960 Bq and a half-life of 4.0 minutes. What is its activity after 12 minutes?
- (A) 240 Bq
- (B) 320 Bq
- (C) 480 Bq
- (D) 120 Bq
Show Answer
Answer: D – 12 min is 3 half-lives, so A = 960 × (1/2)3 = 960 / 8 = 120 Bq.
Q23. A sample contains 5.0 × 1018 undecayed nuclei of a nuclide with decay constant 0.0347 s−1. How many nuclei remain after 40 s?
- (A) 1.25 × 1018
- (B) 2.5 × 1018
- (C) 6.3 × 1017
- (D) 3.6 × 1018
Show Answer
Answer: A – λt = 0.0347 × 40 = 1.386. N = N0 e−λt = 5.0 × 1018 × e−1.386 = 5.0 × 1018 × 0.25 = 1.25 × 1018.
Q24. A nuclide has a half-life of 3.0 hours. How long does it take for the activity of a sample of this nuclide to fall to 1/32 of its original value?
- (A) 5.0 h
- (B) 9.0 h
- (C) 96 h
- (D) 15 h
Show Answer
Answer: D – 1/32 = (1/2)5, so 5 half-lives are needed. Time = 5 × 3.0 = 15 h.
Q25. The background count rate in a laboratory is 40 counts per minute. A source gives a measured count rate of 240 counts per minute, and 14 days later a measured count rate of 90 counts per minute. What is the half-life of the source?
- (A) 7.0 days
- (B) 3.5 days
- (C) 14 days
- (D) 28 days
Show Answer
Answer: A – First subtract the background: corrected rates are 240 − 40 = 200 and 90 − 40 = 50 counts per minute. The fall from 200 to 50 is two halvings (200 → 100 → 50) in 14 days, so the half-life is 7.0 days.
Q26. Which isotope is the most suitable to use as a medical tracer that is injected into a patient and detected outside the body?
- (A) An alpha emitter with a half-life of several years
- (B) A gamma emitter with a half-life of a few hours
- (C) A beta emitter with a half-life of several weeks
- (D) A gamma emitter with a half-life of several years
Show Answer
Answer: B – Gamma radiation can pass out of the body to be detected, and a half-life of a few hours is long enough for the investigation but short enough that the activity soon falls to a safe level.
Q27. Carbon-14 has a half-life of 5730 years. A wooden artefact has a carbon-14 activity that is 1/8 of the activity of the same mass of living wood. What is the approximate age of the artefact?
- (A) 5730 years
- (B) 11 460 years
- (C) 45 840 years
- (D) 17 190 years
Show Answer
Answer: D – A fall to 1/8 = (1/2)3 means 3 half-lives have passed. Age = 3 × 5730 = 17 190 years.
Q28. A smoke detector contains an alpha source with a half-life of several hundred years. Why is such a long half-life useful in this application?
- (A) It makes the alpha particles more penetrating
- (B) The activity of the source stays almost constant over the working life of the detector
- (C) It makes the source completely safe to handle
- (D) It increases the ionisation produced by each alpha particle
Show Answer
Answer: B – With a half-life of hundreds of years, only a tiny fraction of the nuclei decay in a few years, so the activity, and therefore the ionisation current in the detector, stays almost constant.
Q29. A radioactive source is used in a beam for external radiotherapy, in which a tumour inside the patient is treated from outside the body. The source should emit
- (A) penetrating gamma radiation, so that it can reach the tumour from outside the body
- (B) alpha radiation, because it is the most ionising
- (C) beta radiation, because it is stopped by a few millimetres of aluminium
- (D) radiation with a half-life of a few seconds, so that treatment is quick
Show Answer
Answer: A – For a beam applied outside the body, the radiation must be penetrating enough to pass through tissue to reach the tumour. Gamma radiation is used; alpha and beta radiation would be absorbed before reaching the tumour.
Q30. A source has a half-life of 20 minutes. Its activity falls from 640 Bq to 10 Bq. How long does this take?
- (A) 60 min
- (B) 100 min
- (C) 120 min
- (D) 1280 min
Show Answer
Answer: C – Halving repeatedly: 640 → 320 → 160 → 80 → 40 → 20 → 10, which is 6 half-lives. Time = 6 × 20 = 120 min.
Keep practising: try more Physics MCQs by topic, and keep the A Level Physics Formula Sheet open while you revise radioactive decay.
These are original practice questions written for revision. They are not Cambridge past-paper questions, and this page is not affiliated with or endorsed by Cambridge Assessment International Education.