OxfordAQA International AS and A-level Physics: Radioactive decay modes, half-life, radiation properties and the inverse-square law (9630) – 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 3.3.3 (Radioactivity) of the OxfordAQA International AS and A-level Physics (9630) specification, version 4.4, for exams from May/June 2020 onwards. The content is shared by International AS and International A-level. Exponential decay equations belong to International A2 (section 3.9) and are not needed. Any constants or data you need are given in the question.

Links: study guide, revision notes, the OxfordAQA Physics hub, the printable checklist and the free diagnostics.

Questions

1. (a) Radium-224 has nucleon number 224 and proton number 88. It decays by alpha emission to a nuclide X. Find the nucleon number and proton number of X, and name X given that proton number 86 is radon. [3] (b) Caesium-137 (A = 137, Z = 55) decays by beta-minus emission to a barium nuclide. Write the complete decay equation using A and Z values. [3]

2. (a) Fluorine-18 (A = 18, Z = 9) decays by beta-plus emission. Write the decay equation and state what changes inside the nucleus. [3] (b) A free neutron decays. Write the equation and state why an antineutrino appears in it. [2]

3. Alpha particles from a source of one nuclide all have the same kinetic energy. Beta particles from a source of another nuclide have kinetic energies spread continuously from near zero up to a maximum. Explain how this difference led to the hypothesis that the neutrino exists. [4]

4. A detector measured the count rate near a source. With the source removed the background rate was 30 counts per minute.

t / min	0	5	10	15	20	30	40
Measured rate / counts per min	1230	846	586	408	287	149	85

(a) Describe how you would use these data to find the half-life of the source. [3] (b) Use your method to show the half-life is about 9 minutes, and find the corrected rate at t = 36 min. [3]

5. A sample of sodium-24 has an activity of 2.4 × 10⁶ Bq. Its half-life is 15 h. Find the activity after 60 h, and the least time, in whole half-lives, after which the activity is below one thousandth of its starting value. [4]

6. (a) Technetium-99m (A = 99, Z = 43) emits a gamma ray and becomes technetium-99. Write the equation. [2] (b) Explain two reasons why technetium-99m is suitable for medical diagnosis. [2] (c) A dose has an activity of 360 MBq at 07:00. Take the half-life as 6.0 h. Find its activity at 19:00. [2]

7. A student identifies the radiation from a source using a Geiger-Müller tube at a fixed distance. The background rate is 18 counts per minute.

Absorber	None	Paper	1 mm aluminium	3 mm aluminium	6 mm aluminium	10 mm aluminium	10 mm aluminium + 25 mm lead
Measured rate / counts per min	1450	1442	1019	396	236	229	92

(a) Find the corrected rates and decide which of alpha, beta and gamma are emitted. [5] (b) Give two precautions the student should take when carrying out the experiment. [2]

8. (a) Explain why an alpha emitter is considered more hazardous inside the body than outside it. [3] (b) Aluminium foil is rolled to a required thickness. Suggest a suitable type of radiation for a thickness gauge, and justify the choice of both the radiation and the half-life of the source. [4]

9. The count rate from a small gamma source was measured at several distances. A background of 0.6 counts per second was present in each measured value.

r / m	0.20	0.30	0.40	0.50	0.80
Measured rate / counts per s	90.6	40.6	23.1	15.0	6.2

(a) Show that the data are consistent with I = I₀/r². [4] (b) Give one source of error in this experiment and one safety precaution. [2] (c) Predict the corrected rate at 1.2 m. [1]

10. A hand-held detector counts for 1 minute with no source three times, giving 54, 61 and 57 counts. With a gamma source at 0.30 m it gives 1337 counts in 1 minute. (a) Find the corrected count rate at 0.30 m. [2] (b) Find the count rate that the detector would record at 0.90 m, including background. [3] (c) State one reason the measured value could be higher than predicted. [1]

Answers

1. (a) Alpha emission removes 4 from A and 2 from Z [1]. A = 220 and Z = 86 [1]. Proton number 86 is radon-220 [1]. (b) Beta-minus: Z rises by 1 so the barium nuclide has Z = 56 [1]. A stays 137 [1]. Cs-137 (137, 55) → Ba-137 (137, 56) + e⁻ (0, −1) + antineutrino (0, 0) [1]. Examiner insight: In a decay equation, each particle and each number needs to be right, so leaving out the antineutrino or writing the electron’s A as 1 loses credit even when the daughter is right.

2. (a) F-18 (18, 9) → O-18 (18, 8) + e⁺ (0, +1) + neutrino (0, 0) [2]. A proton changes into a neutron [1]. (b) n (1, 0) → p (1, 1) + e⁻ (0, −1) + antineutrino [1]. An antineutrino accompanies a beta-minus electron, as in beta-minus decay [1]. Examiner insight: Beta-plus takes the neutrino and beta-minus the antineutrino. Linking the particle to the correct decay is a typical point to state clearly.

3. Alpha decay shares a fixed energy release between two bodies, so each alpha has one energy [1]. If beta decay also had two bodies, beta particles would all have one energy [1]. A continuous range and missing energy and momentum means a third particle shares the release [1]. This particle must be neutral and very weakly interacting, which was called the neutrino [1]. Examiner insight: A strong answer states the comparison with alpha decay and refers to both energy and momentum, as the specification does; “something else is emitted” alone is too vague.

4. (a) Subtract the 30 counts per minute background from every reading [1]. Plot corrected rate against time and draw a smooth curve [1]. Read the time for the corrected rate to halve, repeating from more than one starting value, and compare them [1]. (b) Corrected values are 1200, 816, 556, 378, 257, 119 and 55 [1]. Rate falls from 1200 to 600 at about 9 min, and from 600 to 300 at about 18 min, so each halving takes about 9 min [1]. 36 min is 4 half-lives, so 1200 ÷ 16 = 75 counts per minute corrected (105 measured) [1]. Examiner insight: Working with the uncorrected readings makes the half-life come out too long, so show the subtraction explicitly and say which rate you quote.

5. 60 ÷ 15 = 4 half-lives [1]. Activity = 2.4 × 10⁶ ÷ 16 = 1.5 × 10⁵ Bq [1]. (½)⁹ = 1/512 is not below 1/1000, but (½)¹⁰ = 1/1024 is [1]. Time = 10 × 15 = 150 h [1]. Examiner insight: Show the number of half-lives and the fraction. A bare final time with no reason for taking 10 rather than 9 is hard to credit.

6. (a) Tc-99m (99, 43) → Tc-99 (99, 43) + γ [1], with A and Z unchanged and no other particle [1]. (b) It emits gamma, which leaves the body to be detected [1], and no alpha or beta, so the dose is lower; its half-life of about 6 h is long enough for a scan but short enough to limit exposure [1]. (c) 07:00 to 19:00 is 12 h = 2 half-lives [1]. 360 ÷ 4 = 90 MBq [1]. Examiner insight: “Suitable” needs a property linked to a consequence. Saying only “short half-life” without the effect on the patient does not give the reason.

7. (a) Corrected rates: 1432, 1424, 1001, 378, 218, 211, 74 counts per minute [1]. Paper changes the rate by less than 1%, so alpha is not emitted [1]. Aluminium cuts 1424 to about 215, so beta is emitted [1]. The rate hardly changes between 6 mm and 10 mm of aluminium, so the remaining radiation is not beta [1]. Lead then reduces it from 211 to 74, so gamma is emitted [1]. The source emits beta and gamma. (b) Measure the background first and subtract it [1]. Keep the distance and counting time fixed, and handle the source with tongs [1]. Examiner insight: Quote corrected values and say what each change shows. Alpha only travels a few centimetres in air, so a source far from the tube shows no paper effect because the alpha never arrives; tie the conclusion to the distance used.

8. (a) Outside the body, the outer layer of dead skin absorbs alpha [1]. Inside the body, alpha is strongly ionising [1] and deposits all its energy in a very small volume of living tissue [1]. (b) Beta [1], because a few millimetres of aluminium absorbs beta, so the count rate changes with thickness [1]. Alpha would be stopped entirely and gamma would pass through almost unchanged [1]. A long half-life keeps the source activity steady, so a change in count rate is due to thickness [1]. Examiner insight: A choice of radiation alone is not enough; give a reason for each choice and compare it with the other two.

9. (a) Subtract 0.6: corrected 90.0, 40.0, 22.5, 14.4, 5.6 [1]. Find I × r²: 3.6, 3.6, 3.6, 3.6, 3.58 [2]. These are constant within rounding, so I ∝ 1/r², consistent with I = I₀/r² [1]. (b) Error: the effective positions of source and tube are uncertain, which matters most at small r [1]. Safety: use tongs, store the source in a shielded box, keep exposure short [1]. (c) I = 3.6 ÷ 1.2² = 2.5 counts per second [1]. Examiner insight: A consistency check needs calculated values and a conclusion, not only “rate falls as distance rises”.

10. (a) Mean background = (54 + 61 + 57) ÷ 3 = 57.3 per minute [1]. Corrected = 1337 − 57.3 = 1280 per minute [1]. (b) 0.90 m is 3 times 0.30 m, so the corrected rate falls by 9 [1]. 1280 ÷ 9 = 142 per minute [1]. Add background: 142 + 57.3 = 199.5, so about 200 per minute [1]. (c) Gamma scattered from nearby walls or objects, or a higher background during the count, adds counts [1]. Examiner insight: The inverse-square law applies to the corrected rate, so subtract before dividing and add the background back only if the question asks for the recorded value.

Where marks are usually lost

• Leaving out the neutrino or antineutrino in a beta equation, or choosing the wrong one.
• Writing an electron or positron with a nucleon number other than 0.
• Not subtracting the background before finding a half-life or applying the inverse-square law.
• Taking half-life from one starting point and not checking a second.
• Ranking the hazards without stating where the source is: in the body, or beyond the skin.
• Choosing a radiation for gauging without explaining why the others do not work.
• Dividing the activity by 2 once for each hour rather than once for each half-life.
• Using 1/r instead of 1/r² when checking the law.

Next steps

• Revision notes
• Study guide
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Official syllabus

OxfordAQA International AS and A-level Physics (9630), specification version 4.4, for exams from May/June 2020 onwards, published by Oxford International AQA Examinations: section 3.3.3 Radioactivity.