Nuclear DecayEdexcel International A Level Physics: Topic test
20 questions, 54 marks
Edexcel International A Level Physics
Nuclear Decay topic test
Total 54 marks
Name
Class
Date
- 1A smoke detector contains a small sealed source of americium-241 (proton number 95), an alpha emitter with a half-life of 432 years. Take 1 year = 3.16 × 10⁷ s.(a)Which nuclide is formed when americium-241 decays by alpha emission?[1 mark]
- A
- B
- C
- D
(b)What is the decay constant of americium-241?[1 mark]- A5.1 × 10⁻¹¹ s⁻¹
- B7.3 × 10⁻¹¹ s⁻¹
- C1.6 × 10⁻³ s⁻¹
- D1.4 × 10¹⁰ s⁻¹
(c)Explain why an alpha emitter is used in a smoke detector rather than a source of gamma radiation.[2 marks]Total for question 1: 4 marks
- 2An oxygen-16 nucleus has a mass of 15.990526 u. The mass of a proton is 1.007276 u and the mass of a neutron is 1.008665 u. Use 1 u = 931.5 MeV/c² = 1.66 × 10⁻²⁷ kg and c = 3.00 × 10⁸ m s⁻¹.(a)What is the mass deficit of the oxygen-16 nucleus?[1 mark]
- A16.1275 u
- B0.0137 u
- C0.2740 u
- D0.1370 u
(b)What is the binding energy per nucleon of oxygen-16?[1 mark]- A127.6 MeV
- B7.98 MeV
- C63.8 MeV
- D15.9 MeV
(c)Calculate, in joules, the energy equivalent of the mass deficit of the oxygen-16 nucleus.[2 marks]Total for question 2: 4 marks
- 3A sealed cobalt-60 source (proton number 27) is used to irradiate packaged food. It has an initial activity of 3.7 × 10¹⁴ Bq and a half-life of 5.27 years. Cobalt-60 decays by beta-minus emission to nickel-60 (proton number 28), and the nickel-60 nucleus then emits gamma radiation.(a)Write a balanced nuclear equation for the beta-minus decay of cobalt-60, and explain why the gamma radiation rather than the beta particles is used to irradiate the packaged food.[3 marks](b)The source is replaced when its activity has fallen to 20% of its initial value. Calculate the time, in years, after which it is replaced.[4 marks]
Total for question 3: 7 marks
- 4A paper mill monitors the thickness of a moving sheet of paper. A strontium-90 source (proton number 38) is placed on one side of the sheet and a Geiger–Müller tube is placed on the other side. Strontium-90 is a beta-minus emitter with a half-life of 29.0 years. The count rate from the tube is used to control the rollers. Take 1 year = 3.16 × 10⁷ s.(a)Explain why a beta source is suitable for this application but alpha and gamma sources are not, and why the long half-life of the source is an advantage.[6 marks](b)The source has an activity of 3.0 × 10⁸ Bq. Calculate the number of strontium-90 nuclei in the source and the activity after 12.0 years.[6 marks]
Total for question 4: 12 marks
- 5In a nuclear reactor, a uranium-235 nucleus absorbs a slow neutron and splits into two fission fragments and several neutrons. Each fission releases about 200 MeV of energy. Use 1 MeV = 1.60 × 10⁻¹³ J, 1 u = 1.66 × 10⁻²⁷ kg and c = 3.00 × 10⁸ m s⁻¹.(a)Why does the fission of a uranium-235 nucleus release energy?[1 mark]
- AThe fission fragments have a lower binding energy per nucleon than uranium-235.
- BThe fission fragments have a higher binding energy per nucleon than uranium-235.
- CThe fission fragments contain fewer nucleons than the uranium-235 nucleus.
- DThe neutrons released have a greater total mass than the absorbed neutron.
(b)Which statement about the masses in one fission is correct?[1 mark]- AThe mass of the fragments and neutrons is greater than the mass of uranium-235 and the neutron.
- BThe mass of the fragments and neutrons is equal to the mass of uranium-235 and the neutron.
- CMass is conserved but the number of nucleons falls.
- DThe mass of the fragments and neutrons is less than the mass of uranium-235 and the neutron.
(c)Calculate the mass converted to energy in one fission, in u.[2 marks]Total for question 5: 4 marks
- 6A technician investigates a sealed source using a Geiger–Müller tube placed 2.0 cm from the source. With the source removed, the tube records a background count rate of 28 counts per minute. With the source in place the following count rates are recorded: no absorber, 412 counts per minute; a sheet of paper, 408 counts per minute; a 3 mm aluminium sheet, 145 counts per minute; 10 cm of lead, 30 counts per minute.(a)What is the corrected count rate from the source with no absorber?[1 mark]
- A384 counts per minute
- B412 counts per minute
- C440 counts per minute
- D145 counts per minute
(b)Which radiations does the source emit?[1 mark]- Aalpha and beta only
- Balpha and gamma only
- Cbeta and gamma only
- Dgamma only
(c)Explain how the data show that the source does not emit alpha particles.[2 marks]Total for question 6: 4 marks
- 7A researcher is studying a radioactive sample that initially contains 2.4 × 10¹² undecayed nuclei of an isotope whose half-life is 40 minutes.(a)A student states that all the nuclei will have decayed after 80 minutes, because this is two half-lives. Explain why the student is wrong, and calculate the number of undecayed nuclei after 80 minutes.[3 marks](b)Calculate the initial activity of the sample, and the time taken for its activity to fall to 1.0 × 10⁸ Bq.[4 marks]
Total for question 7: 7 marks
- 8A space probe is powered by a generator containing plutonium-238 (proton number 94), an alpha emitter with a half-life of 87.7 years. Each decay releases 5.59 MeV, almost all of which is converted to thermal energy within the generator. At launch the activity of the plutonium-238 is 3.6 × 10¹⁴ Bq. Use 1 MeV = 1.60 × 10⁻¹³ J.(a)Explain, with reference to the nature of alpha radiation and to nuclear binding energy, why plutonium-238 is a suitable source of heat for the generator. Include a balanced nuclear equation for the decay.[6 marks](b)Calculate the thermal power of the generator at launch, the power after 20.0 years, and the time at which the power has fallen to 200 W. Assume the power is proportional to the activity.[6 marks]
Total for question 8: 12 marks
End of questions
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).