OCR A Level Physics A: Radioactive decay, activity, half-life and dating (H556) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 6.4.3 (Radioactivity), statements (a) to (h), of the OCR Advanced GCE in Physics A (H556) specification, Version 3.0 (March 2026), for first assessment in 2017. Use Avogadro constant = 6.02 × 10²³ mol⁻¹ and 1 year = 3.16 × 10⁷ s. Start with the study guide or the revision notes. The course hub is OCR A Level Physics, the printable list is the checklist, and the free 10-minute diagnostics show where to focus.

Questions

1. State what is meant by saying that radioactive decay is (i) spontaneous and (ii) random. [2]

2. (a) State what an α-particle is and name a material that stops it. [2] (b) State how the penetration of β-radiation differs from that of γ-radiation. [2]

3. Complete each decay equation by giving the missing particle or nuclide. (a) ²¹⁰₈₄Po → ²⁰⁶₈₂Pb + ? [1] (b) ³²₁₅P → ? + ⁰₋₁e + ? [2] (c) ¹¹₆C → ¹¹₅B + ? + ? [2]

4. Describe an experiment to show that a β-source is absorbed by aluminium sheets of increasing thickness. Include the measurement of background and two safety precautions. [6]

5. A sample contains 2.0 mg of an isotope of molar mass 90 g mol⁻¹. Its activity is 1.0 × 10¹⁰ Bq. (a) Determine how many nuclei the sample contains. [2] (b) Calculate the decay constant, in s⁻¹. [1] (c) Calculate the half-life, in years. [2]

6. A source has an initial activity of 3.2 × 10⁶ Bq and a half-life of 4.0 years. (a) Calculate the decay constant, in year⁻¹. [1] (b) Calculate the activity after 10 years. [2] (c) Calculate the time for the activity to fall to 4.0 × 10⁵ Bq. [2]

7. A class simulates decay with 150 dice. After each throw, every die showing a 4 is removed. (a) Calculate the expected number of dice left after four throws. [2] (b) Show that the half-life of this model is about 3.8 throws. [2] (c) Explain why a class’s results will not match the expected values exactly. [1]

8. A spreadsheet models ΔN = −λNΔt with N₀ = 8000, λ = 0.020 s⁻¹ and Δt = 5.0 s. (a) Write the formula for the value in a row in terms of the value in the row above, and calculate N at 5.0 s and 10.0 s. [3] (b) The exact value at 10.0 s from N = N₀e^(−λt) is 6550. State why the model gives a different value. [1] (c) State one change that makes the model more accurate. [1]

9. Corrected count rate C (after subtracting background) from a protactinium source is recorded.

t / s	0	30	60	90	120	150
C / s⁻¹	160	121	92	69	52	40

(a) Explain why a graph of ln C against t is plotted. [1] (b) Calculate the gradient of the line using the first and last points. [2] (c) Determine λ and the half-life. [2] (d) State one reason why the plotted points do not lie exactly on a line. [1]

10. Living wood gives 0.230 Bq per gram of carbon. A bone sample gives 0.092 Bq per gram. Take the half-life of carbon-14 as 5730 years. (a) Calculate the decay constant, in year⁻¹. [1] (b) Calculate the age of the bone. [2] (c) State two assumptions made in the calculation. [2] (d) Explain why the method is not suitable for a sample about 200 000 years old. [1]

11. The activity of a source is 4.8 × 10⁵ Bq at 09:00 and 1.5 × 10⁵ Bq at 15:00 on the same day. Background has been subtracted. (a) Calculate the decay constant, in h⁻¹, and the half-life. [3] (b) Calculate the number of undecayed nuclei at 09:00. [3] (c) Calculate the time after 09:00 at which the activity is 1.0 × 10⁴ Bq. [2] (d) A technician suggests heating the source to make it decay faster. Evaluate the suggestion. [1]

Answers

1. (i) Decay happens without any external cause or trigger, and is unaffected by conditions such as temperature [1]. (ii) It is impossible to predict when a given nucleus will decay; every undecayed nucleus has the same fixed chance of decaying in each second [1]. [2] Examiner insight: Answers tend to gain credit when each word is defined separately, so avoid giving a single sentence that blends the two.

2. (a) An α-particle is a helium nucleus (two protons and two neutrons) [1]. A sheet of paper, or a few centimetres of air, stops it [1]. (b) β is stopped by a few millimetres of aluminium [1], whereas γ is only reduced in intensity by thick lead and is never stopped completely [1]. [4] Examiner insight: Naming a material together with an approximate thickness makes a penetration comparison much stronger than “β goes further than α”.

3. (a) ⁴₂α, a helium nucleus [1]. (b) The nuclide is ³²₁₆S [1] and the other particle is an antineutrino, ν̄ [1]. (c) The particle is a positron ⁰₊₁e [1] with a neutrino ν [1]. [5] Examiner insight: Check each line by adding the nucleon numbers and the proton numbers on both sides; omitting the neutrino or antineutrino is a common way to lose credit.

4. Measure the background count with no source nearby over a long time, such as five minutes, and find the rate [1]. Place the source and GM tube a fixed distance apart [1]. Put one aluminium sheet between them, count for a fixed time such as 60 s, and record the thickness with a micrometer [1]. Add more sheets and repeat [1]. Subtract the background rate from each reading and plot corrected count rate against thickness, which should fall to the background level at a definite thickness [1]. Safety: handle the source with tongs, keep it stored when not in use and never point it at anyone [1]. [6] Examiner insight: A description is easier to credit when it names what is kept constant and what is varied; “measure the count” alone leaves the method incomplete.

5. (a) Moles = 2.0 × 10⁻³ / 90 = 2.22 × 10⁻⁵ mol [1]; N = 2.22 × 10⁻⁵ × 6.02 × 10²³ = 1.3 × 10¹⁹ [1]. (b) λ = A/N = 1.0 × 10¹⁰ / 1.34 × 10¹⁹ = 7.5 × 10⁻¹⁰ s⁻¹ [1]. (c) t½ = 0.693 / 7.48 × 10⁻¹⁰ = 9.27 × 10⁸ s [1]; dividing by 3.16 × 10⁷ gives 29 years [1]. [5] Examiner insight: Carry unrounded values between parts, and convert the mass to grams or the molar mass to kilograms consistently.

6. (a) λ = ln 2 / 4.0 = 0.173 year⁻¹ [1]. (b) A = 3.2 × 10⁶ × e^(−0.173 × 10) [1] = 5.7 × 10⁵ Bq [1]. (c) The ratio 3.2 × 10⁶ / 4.0 × 10⁵ = 8, which is three half-lives [1], so t = 12 years [1]. [5] Examiner insight: Spotting a whole number of half-lives gives a quick route and a check on the logarithm method.

7. (a) The chance a die survives a throw is 5/6 [1], so expected number = 150 × (5/6)⁴ = 72 [1]. (b) (5/6)ⁿ = e^(−λn) gives λ = ln(6/5) = 0.182 per throw [1]; t½ = ln 2 / 0.182 = 3.8 throws [1]. (c) Each die is removed at random, so real counts scatter about the expected values [1]. [5] Examiner insight: In a “show that” answer, writing out the intermediate value of λ is what lets a reader follow the route to the given result.

8. (a) N(new) = N(old) − 0.020 × N(old) × 5.0 [1]; N = 7200 at 5.0 s [1] and 6480 at 10.0 s [1]. (b) The model uses the number at the start of each step for the whole step, so it removes too many nuclei [1]. (c) Use a smaller time step [1]. [5] Examiner insight: A reason for the difference is stronger when it refers to the way the step is calculated, not to “rounding” or “error”.

9. (a) Since ln C = ln C₀ − λt, a straight line confirms exponential decay and its gradient gives −λ [1]. (b) ln 160 = 5.075 and ln 40 = 3.689 [1]; gradient = (3.689 − 5.075) / 150 = −0.0092 s⁻¹ [1]. (c) λ = 0.0092 s⁻¹ [1]; t½ = 0.693 / 0.00924 = 75 s [1]. (d) Decay is random, so the counts fluctuate [1]. [6] Examiner insight: Reading the gradient from the ends of the line, as here, reduces the effect of any one scattered point.

10. (a) λ = ln 2 / 5730 = 1.21 × 10⁻⁴ year⁻¹ [1]. (b) t = ln(0.230 / 0.092) / λ = 0.916 / 1.21 × 10⁻⁴ [1] = 7600 years [1]. (c) The ratio of carbon-14 in the atmosphere was the same in the past as now [1]; the sample has not been contaminated by other carbon [1]. (d) Too few carbon-14 nuclei remain, so the activity cannot be distinguished from background [1]. [6] Examiner insight: Give the age to the precision supported by the data, which here is two significant figures.

11. (a) λ = ln(4.8 × 10⁵ / 1.5 × 10⁵) / 6.0 [1] = 0.194 h⁻¹ [1]; t½ = 0.693 / 0.194 = 3.6 h [1]. (b) λ = 0.194 / 3600 = 5.38 × 10⁻⁵ s⁻¹ [1]; N = A/λ [1] = 4.8 × 10⁵ / 5.38 × 10⁻⁵ = 8.9 × 10⁹ [1]. (c) t = ln(4.8 × 10⁵ / 1.0 × 10⁴) / 0.194 = ln 48 / 0.194 [1] = 20 h, which is about 05:00 the next day [1]. (d) Heating would not help: decay is spontaneous, so temperature does not change the decay rate [1]. [9] Examiner insight: When a calculation moves between hours and seconds, stating the unit of λ at each step makes slips easy to spot.

Where marks are usually lost

• Failing to subtract background before using a count rate.
• Using minutes or years for t with λ in s⁻¹.
• Leaving the minus sign out of the exponent.
• Leaving the neutrino or antineutrino out of a β equation.
• Describing γ as having a fixed range.
• Giving “careful” instead of a specific safety precaution.
• Rounding λ early and carrying an error into a later part.
• Describing the dice model but not saying what is varied or counted.

Next steps

• Read the revision notes and the study guide.
• Open the OCR A Level Physics hub and the checklist.
• Take the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR Advanced GCE in Physics A (H556), specification Version 3.0 (March 2026), first assessment 2017, published by OCR: section 6.4.3, Radioactivity.