Radioactive decay, nuclear radiation and nuclear equations – GCSE Physics exam questions with mark scheme

1Grade 4–54 marks

(a) Polonium-210 decays by emitting an alpha particle. Complete the nuclear equation.

$$^{210}_{84}\text{Po} \rightarrow {}^{\,\square}_{\,\square}\text{Pb} + {}^{4}_{2}\text{He}$$ [2]

(b) Carbon-14 ($^{14}_{6}\text{C}$) decays by emitting a beta particle and becomes an isotope of nitrogen (N).

Write the nuclear equation for this decay. [2]

(a) mass number 206(1)
(a) atomic number 82(1)
(b) $^{14}_{7}\text{N}$ (both numbers correct)(1)
(b) beta particle shown as $^{\;0}_{-1}\text{e}$ (allow $^{\;0}_{-1}\beta$)(1)
Answer: (a) $^{206}_{82}\text{Pb}$ (b) $^{14}_{6}\text{C} \rightarrow {}^{14}_{7}\text{N} + {}^{\;0}_{-1}\text{e}$
Examiner tip: In beta decay the mass number stays the same and the atomic number goes UP by 1 — a neutron turns into a proton.

2Grade 4–55 marks

A student uses a Geiger–Müller tube connected to a counter to measure the radiation from a radioactive source.

She records the number of counts in five separate 10-second intervals. Her results are shown.

Interval	1	2	3	4	5
Counts	52	47	58	44	49

(a) The source and the tube were not moved. Explain why the number of counts is different in each interval. [2]

(b) Calculate the mean count rate in counts per second. [2]

(a) radioactive decay is random(1)
(a) it is not possible to predict which nucleus will decay or when, so the number of nuclei decaying in each 10 s varies allow the number of decays in each interval changes by chance(1)
(b) mean = 250 ÷ 5 = 50 (counts in 10 s) allow total 250 counts in 50 s(1)
(b) 5 (counts per second) allow 5.0; correct answer with no working gains 2 marks(1)
(c) to reduce the effect of the random variation / so a mean can be calculated / to give a more accurate value ignore "to make it fair"; ignore "more reliable" on its own(1)
Answer: (a) Decay is random, so the number of nuclei decaying in each interval varies. (b) 5 counts/s. (c) To reduce the effect of random variation (more accurate mean).
Examiner tip: Many students divide the mean by 5 instead of by the interval length (10 s): the count rate is counts ÷ time.

3Grade 4–55 marks

(a) Americium-241 decays to neptunium-237 by emitting a particle X.

$$^{241}_{95}\text{Am} \rightarrow {}^{237}_{93}\text{Np} + {}^{\,a}_{\,b}\text{X}$$

Determine the values of $a$ and $b$ and name particle X. [2]

(b) A nucleus of beryllium-13 is very unstable. It decays by emitting a neutron.

$$^{13}_{4}\text{Be} \rightarrow {}^{\,\square}_{\,\square}\text{Be} + {}^{1}_{0}\text{n}$$

Complete the nuclear equation. [1]

Explain why emitting gamma radiation does not change the mass number or the atomic number of a nucleus. [2]

(a) $a = 4$ and $b = 2$ (both needed)(1)
(a) alpha particle allow helium nucleus; allow ecf: if a and b are wrong, name must match their values(1)
(b) 12 and 4 (both needed) — $^{12}_{4}\text{Be}$(1)
(c) gamma radiation is an electromagnetic wave / has no mass and no charge(1)
(c) so the number of protons and the number of neutrons in the nucleus do not change allow the nucleus just loses energy(1)
Answer: (a) $a = 4$, $b = 2$; X is an alpha particle. (b) $^{12}_{4}\text{Be}$. (c) Gamma is EM radiation with no mass or charge, so the numbers of protons and neutrons are unchanged.
Examiner tip: In neutron emission the atomic number stays the same: the neutron carries mass number 1 but no charge.

4Grade 8–9Higher4 marks

Uranium-238 ($^{238}_{92}\text{U}$) decays through a long series of alpha and beta decays. The series ends with the stable isotope lead-206 ($^{206}_{82}\text{Pb}$).

Determine how many alpha decays and how many beta decays take place in the series.

Show how you work out your answer. [4]

mass number falls by 238 − 206 = 32(1)
only alpha decay changes the mass number (by 4 each time), so 32 ÷ 4 = 8 alpha decays(1)
8 alpha decays lower the atomic number by 16 / to 92 − 16 = 76(1)
each beta decay raises the atomic number by 1, so 82 − 76 = 6 beta decays allow ecf from their number of alpha decays(1)
Answer: 8 alpha decays and 6 beta decays.
Examiner tip: Do the alpha decays first: only they change the mass number, then use the atomic number to find how many beta decays are needed.

5Grade 6–76 marks

In a paper factory, wet pulp is pressed between rollers to make a long sheet of paper.

The thickness of the paper is monitored using a radioactive source and a detector, as shown in the diagram. A control unit uses the reading from the detector to change the pressure of the rollers on the paper.

Explain which type of radiation, alpha, beta or gamma, the source should emit, and how the system keeps the thickness of the paper constant. [6]

beta radiation should be used(1)
alpha would be completely stopped / absorbed by the paper, so the detector reading would not change with thickness(1)
gamma would pass through the paper almost unaffected, so small changes in thickness would cause no measurable change in count rate(1)
beta is partly absorbed by the paper, and the amount absorbed depends on the thickness(1)
if the paper becomes too thick, more beta is absorbed so the count rate falls(1)
the control unit then increases the pressure of the rollers (moves them closer together) to make the paper thinner(1)
if the paper becomes too thin, the count rate rises and the roller pressure is reduced(1)
the source should have a long half-life so its activity does not fall noticeably, so changes in count rate are caused only by changes in thickness(1)
decay is random, so the count rate should be averaged over a short time before the rollers are adjusted(1)

Level	Marks	What it looks like
Level 3	5–6	Beta is chosen and the choice is justified by explaining why both alpha and gamma are unsuitable in terms of absorption by paper. A clear, logical description of the feedback is given, linking a change in thickness to a change in count rate and to the correct adjustment of the rollers.
Level 2	3–4	Beta is chosen with a reason, and at least one of alpha or gamma is ruled out. There is some description of how the count rate changes with thickness or how the rollers are adjusted, but the link between them is incomplete or partly unclear.
Level 1	1–2	Simple statements, e.g. a correct choice of beta with little or no reason, or a relevant comment about penetration or that the detector reading is used to adjust the rollers, with no linked explanation.
No relevant content	0	Nothing worthy of credit.

Answer:
Beta: alpha is fully absorbed and gamma barely absorbed by paper, so only beta gives a count rate that changes with thickness. Thicker paper → lower count rate → roller pressure increased; thinner paper → higher count rate → pressure reduced.
Examiner tip: Say why alpha and gamma do not work in terms of what happens to the count rate as thickness changes, not just "alpha is too weak".

6Grade 1–3Foundation5 marks

An unstable nucleus can give out nuclear radiation when it decays.

(a) Complete the table to show what each type of nuclear radiation is.

Type of radiation	What it is
alpha ($\alpha$)
beta ($\beta$)
gamma ($\gamma$)

[3]

(b) A fourth type of nuclear radiation is a particle that has no charge. It is not electromagnetic radiation. Name this particle. [1]

(a) alpha: two protons and two neutrons allow a helium nucleus; do not accept helium atom(1)
(a) beta: a (high-speed) electron (emitted from the nucleus) do not accept an electron from the shell / orbit(1)
(a) gamma: electromagnetic radiation / an electromagnetic wave allow EM wave; do not accept particle(1)
(b) neutron(1)
(c) becquerel / Bq do not accept counts per second or count rate(1)
Answer: (a) alpha: 2 protons + 2 neutrons (helium nucleus); beta: high-speed electron; gamma: electromagnetic wave. (b) neutron. (c) becquerel (Bq).
Examiner tip: Writing "helium atom" for alpha loses the mark: an alpha particle has no electrons, so it is a helium nucleus.

7Grade 6–76 marks

A student investigates which types of radiation are emitted by a radioactive source.

She places the source 3 cm from a Geiger–Müller tube. She puts different absorbers between the source and the tube and measures the count rate.

With the source removed, the background count rate is 25 counts per minute.

Absorber	Count rate in counts per minute
none	395
sheet of paper	394
5 mm of aluminium	140
5 cm of lead	26

(a) Explain what the results show about the types of radiation emitted by the source. [4]

(b) Determine the count rate due to beta radiation from the source. [1]

(a) no alpha (is emitted) because paper does not (significantly) reduce the count rate allow 395 → 394 is only random variation(1)
(a) beta is emitted(1)
(a) because the aluminium greatly reduces the count rate allow reference to 395 falling to 140(1)
(a) gamma is emitted because the count rate with aluminium (140) is still well above background (25) / only lead reduces it to (about) background(1)
(b) 255 (counts per minute) allow 254(1)
(c) alpha particles only travel a few cm in air, so at 30 cm alpha would not reach the tube (even if it were emitted) allow the range of alpha in air is about 5 cm; allow at 30 cm the count rate would be much lower / too close to background to show the changes clearly(1)
Answer: (a) No alpha (paper makes no difference); beta present (aluminium causes a large fall); gamma present (count with aluminium still well above background, reduced to background only by lead). (b) 255 counts/min. (c) Alpha has a range of only a few cm in air.
Examiner tip: A conclusion with no evidence from the table scores little: quote the change in count rate that shows each radiation is present or absent.

8Grade 4–55 marks

Potassium-40 is a radioactive isotope found in rocks and in the human body.

A nucleus of potassium-40 contains 19 protons and 21 neutrons. It decays by emitting a beta particle.

(a) Describe what happens inside the nucleus when a beta particle is emitted. [2]

(b) Give the number of protons and the number of neutrons in the nucleus formed. [2]

(a) a neutron changes into a proton allow neutron → proton + electron(1)
(a) and a (high-speed) electron is emitted from the nucleus do not accept an electron from the shell / orbit(1)
(b) 20 protons(1)
(b) 20 neutrons allow ecf: 40 − their number of protons(1)
(c) the total number of protons and neutrons stays the same (40) / one neutron is replaced by one proton allow the electron has (almost) no mass(1)
Answer: (a) A neutron turns into a proton and a high-speed electron is emitted. (b) 20 protons, 20 neutrons. (c) The total number of nucleons stays at 40.
Examiner tip: Beta particles come from the nucleus, not from the electron shells: say a neutron changes into a proton.

9Grade 1–3Foundation5 marks

The diagram shows three types of radiation, A, B and C, from radioactive sources. It shows how far each type of radiation travels through different materials.

(a) Identify radiations A, B and C. [3]

(b) Which radiation, A, B or C, is the most strongly ionising? [1]

(a) A: alpha allow α(1)
(a) B: beta allow β(1)
(a) C: gamma allow γ(1)
(b) A allow alpha / α(1)
(c) C allow gamma(1)
Answer: (a) A alpha, B beta, C gamma. (b) A. (c) C.
Examiner tip: Link the two ideas: the most ionising radiation (alpha) is the least penetrating, because it loses its energy quickly by ionising the atoms it passes.

10Grade 6–7Higher7 marks

The diagram shows part of a radioactive decay series. Each letter represents a nucleus. Each arrow shows one nucleus decaying to form the next nucleus.

(a) State the type of decay when P decays to Q. Give a reason for your answer using the diagram. [2]

(b) State the type of decay when S decays to T. [1]

(d) Nucleus U decays by emitting an alpha particle to form nucleus V.

Give the mass number and the atomic number of V. [2]

(a) alpha (decay)(1)
(a) mass number decreases by 4 and atomic number decreases by 2 allow 226 → 222 and 88 → 86(1)
(b) beta (decay) allow β; a description of the change in numbers without naming beta does not score(1)
(c) R and U(1)
(c) same atomic number / same number of protons (84) but different mass numbers / different numbers of neutrons(1)
(d) mass number 210(1)
(d) atomic number 82(1)
Answer: (a) alpha: A falls by 4, Z falls by 2. (b) beta. (c) R and U: both have atomic number 84 but different mass numbers. (d) mass number 210, atomic number 82.
Examiner tip: Read both axes for each step: alpha moves 2 left and 4 down, beta moves 1 right with no change in mass number.

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