E.3 Radioactive decay – IB Physics exam questions with mark scheme

1StandardHL only5 marks

The activity of a radioactive sample falls from 3.2 × 10⁵ Bq to 4.0 × 10⁴ Bq in 36 hours. Background radiation can be ignored.

(a) Determine the half-life of the sample. [2]

(b) Calculate the decay constant in s⁻¹. [1]

(a) Activity has fallen by a factor of 8 = 2³, so three half-lives passed(1)
(a) T½ = 36 ÷ 3 = 12 h(1)
(b) λ = ln2 ÷ (12 × 3600 s) = 1.6 × 10⁻⁵ s⁻¹ (1.60 × 10⁻⁵)(1)
(c) A = λN, so N₀ = A₀/λ = 3.2 × 10⁵ ÷ 1.6 × 10⁻⁵(1)
(c) N₀ = 2.0 × 10¹⁰ (allow 1.99 × 10¹⁰ from unrounded λ)(1)
Answer: (a) 12 h; (b) 1.6 × 10⁻⁵ s⁻¹; (c) 2.0 × 10¹⁰
Examiner tip: Half-life in hours must be converted to seconds before finding λ in s⁻¹.

2Standard4 marks

Radium-226 (²²⁶₈₈Ra) decays by alpha emission to radon (Rn).

(a) Complete the decay equation: ²²⁶₈₈Ra → Rn + α. Give the proton and nucleon numbers of both products. [2]

(b) Atomic masses: Ra-226 = 226.0254 u; Rn-222 = 222.0176 u; He-4 = 4.0026 u. Calculate the total energy released in the decay in MeV. (1 u = 931.5 MeV c⁻²) [2]

(a) Radon product written as ²²²₈₆Rn(1)
(a) Alpha particle written as ⁴₂He (or ⁴₂α)(1)
(b) Mass difference = 226.0254 − 222.0176 − 4.0026 = 0.0052 u (electron masses balance when atomic masses are used)(1)
(b) Energy = 0.0052 × 931.5 = 4.84 MeV (≈ 4.8 MeV)(1)
Answer: (a) ²²²₈₆Rn + ⁴₂He; (b) 4.8 MeV
Examiner tip: Both A and Z must balance (226 = 222 + 4; 88 = 86 + 2); then multiply the mass defect by 931.5.

3Standard5 marks

Masses: hydrogen-1 atom = 1.007825 u; neutron = 1.008665 u; helium-4 atom = 4.002603 u. (1 u = 931.5 MeV c⁻²)

(a) Define the binding energy of a nucleus. [1]

(b) Calculate the mass defect of helium-4. [2]

(d) Explain why fusing light nuclei to form helium releases energy. [1]

(a) The minimum energy required to separate a nucleus completely into its individual nucleons (equivalently the energy released when it forms from free nucleons)(1)
(b) Δm = 2(1.007825) + 2(1.008665) − 4.002603(1)
(b) Δm = 0.030377 u (accept 0.0304 u, or 5.04–5.05 × 10⁻²⁹ kg, or 28.3 MeV c⁻²)(1)
(c) BE = 0.030377 × 931.5 = 28.3 MeV, so BE per nucleon = 28.3 ÷ 4 = 7.07 MeV (accept answers in the range 7.07–7.08 MeV from rounding BE to 28.3 MeV first; allow 7.1 MeV; allow ECF from (b))(1)
(d) The product (helium) has a higher binding energy per nucleon than the reactant nuclei, so the total mass decreases and the mass difference is released as energy (E = Δmc²) (accept: total binding energy of the products is greater than that of the reactants; accept: total mass of the products is less than the total mass of the reactants, so energy is released; OWTTE)(1)
Answer: (a) minimum energy to separate a nucleus into its free nucleons; (b) 0.030377 u; (c) 7.07 MeV per nucleon; (d) helium has a higher binding energy per nucleon, so mass is lost and released as energy
Examiner tip: Use atomic (H-1) masses consistently so electron masses cancel, and divide by A for 'per nucleon'.

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4Core4 marks

Phosphorus-32 ($^{32}_{15}\text{P}$) is a $\beta^-$ emitter used in medical tracers. Fluorine-18 ($^{18}_{9}\text{F}$) is a $\beta^+$ emitter used in PET scans.

(a) Complete the equation for the decay of phosphorus-32, giving the proton number and nucleon number of the sulfur (S) nucleus and identifying the other particles emitted. [2]

$$^{32}_{15}\text{P}\ \rightarrow\ \text{S}\ +\ \ldots\ +\ \ldots$$

(b) State the change that occurs to a nucleon inside the fluorine-18 nucleus when it decays. [1]

(a) $^{32}_{16}\text{S}$ (both numbers correct)(1)
(a) $^{0}_{-1}\text{e}$ (or $\beta^-$) and $\bar{\nu}_e$ (electron antineutrino); do not accept neutrino for $\beta^-$ decay; ignore missing subscript $e$(1)
(b) a proton changes into a neutron (accept $p\rightarrow n + e^+ + \nu_e$); do not accept 'a proton is emitted'(1)
(c) a nucleon changes type but the total number of nucleons (protons + neutrons) stays the same / the emitted leptons are not nucleons (accept: electron/positron has nucleon number 0)(1)
Answer: (a) $^{32}_{15}\text{P}\rightarrow{}^{32}_{16}\text{S}+{}^{0}_{-1}\text{e}+\bar{\nu}_e$; (b) proton → neutron; (c) one nucleon changes type, total unchanged
Examiner tip: The most common slip is pairing a β⁻ electron with a neutrino instead of an antineutrino; lepton number must balance.

5Standard4 marks

A student investigates the radiation emitted by a sealed source using a Geiger–Müller tube placed 3 cm from the source. Each count rate below is the mean of three 1-minute counts.

Absorber between source and tube	Count rate / counts min⁻¹
none	412
thin paper	405
5 mm aluminium	98
50 mm lead	27

With the source removed, the count rate is 25 counts min⁻¹.

(a) Outline why the student takes a reading with the source removed. [1]

(b) Deduce which type or types of radiation the source emits. Refer to the data in your answer. [3]

(a) to measure the background count rate, which must be subtracted from every reading (allow 'to correct for background radiation')(1)
(b) no (significant) $\alpha$: paper produces only a small drop (412 → 405), within random variation(1)
(b) $\beta$ present: 5 mm aluminium causes a large drop (≈ 307 counts min⁻¹ absorbed)(1)
(b) $\gamma$ present: corrected count rate of ≈ 73 counts min⁻¹ passes through aluminium and is (almost) stopped only by thick lead (27 ≈ background); must use background-corrected value or comparison with 25 for this mark(1)
Answer: (a) background measurement; (b) β and γ, no α
Examiner tip: Students forget to compare the lead reading with the background: 27 is essentially background, so 98 − 25 = 73 counts per minute is the γ contribution.

6Standard4 marks

The graph shows the measured count rate from a sample of a radioactive nuclide against time. The measured values include background radiation, which is constant at 20 counts min⁻¹.

(a) Determine the half-life of the nuclide. [3]

(b) Predict the measured count rate at $t = 75$ min. [1]

(a) background subtracted from initial reading: corrected initial count rate = 140 − 20 = 120 counts min⁻¹(1)
(a) half of corrected value = 60, so measured value to look for = 60 + 20 = 80 counts min⁻¹ (allow any equivalent pair of corrected values, e.g. 60 → 30 shown on graph as 80 → 50)(1)
(a) half-life = 15 min (accept 14–16 min); an answer of about 19 min, from halving the raw 140 to 70 without subtracting background, scores [0] (ECF does not apply)(1)
(b) 75 min = 5 half-lives: corrected $120/2^5 = 3.75$, measured = 3.75 + 20 ≈ 24 counts min⁻¹ (accept 23–25; ECF from (a) half-life only if a whole number of half-lives results)(1)
Answer: (a) 15 min; (b) 24 counts min⁻¹ (23.75)
Examiner tip: Halving the raw reading (140 → 70) gives about 19 min; the background must come off before halving and go back on when reading the graph.

7HardHL only6 marks

Spectrum 1 shows the kinetic energies of the α particles emitted by nuclide X. Nuclide X also emits γ photons, all of energy 0.20 MeV. Spectrum 2 shows the kinetic energies of the β⁻ particles emitted by a different nuclide, Y. Ignore the recoil of the daughter nuclei.

Explain how these observations provide evidence for

the existence of nuclear energy levels, and
the existence of the antineutrino.

[6]

α particles from X have only a few definite (discrete) energies, 4.80 MeV and 4.60 MeV(1)
the daughter nucleus can be left in its ground state or in an excited state, so the energy available is quantised(1)
difference 4.80 − 4.60 = 0.20 MeV equals the γ photon energy(1)
γ photon emitted when the daughter nucleus drops from the excited state to the ground state, so nuclei have discrete energy levels(1)
β⁻ particles have a continuous range of energies from 0 up to a maximum of 1.71 MeV(1)
if only two particles (electron + daughter) were produced, conservation of energy and momentum would require one definite electron energy(1)
energy released in each decay is fixed (≈ 1.71 MeV) and the missing energy is carried by a third, undetected particle: the antineutrino(1)
antineutrino is neutral and (almost) massless, so very weakly interacting/hard to detect (also needed to conserve lepton number); ignore 'neutrino' if $\bar{\nu}$ is otherwise clear(1)

Level	Marks	What it looks like
Level 3	5–6	Links both features of the α/γ data to discrete nuclear energy levels with a correct energy-level argument (including the 0.20 MeV match) AND explains the continuous β spectrum in terms of energy shared with a third particle, concluding that an (anti)neutrino exists. Reasoning is logical and uses correct terminology throughout.
Level 2	3–4	Explains one of the two pieces of evidence fully, or both partially; e.g. states that discrete α energies imply energy levels and that the β spectrum is continuous, but the conservation argument or the link to the γ energy is missing or unclear.
Level 1	1–2	Relevant statements made (e.g. α energies are discrete, β energies vary) but with little or no explanation of what they show.
Level 0	0	Nothing worthy of credit.

Answer: Discrete α energies differing by exactly the γ energy (0.20 MeV) show discrete nuclear energy levels; the continuous β spectrum up to 1.71 MeV shows a fixed decay energy shared with an undetected third particle, the antineutrino.
Examiner tip: Top-band answers quote the numbers: 4.80 − 4.60 = 0.20 MeV matching the γ energy is the decisive link that weaker answers never state.

8HardHL only4 marks

Potassium-40 has a half-life so long that the activity of a sample does not change measurably during an experiment. A sample containing 2.50 mg of potassium-40 is found to have an activity of 661 Bq. The molar mass of potassium-40 is 40.0 g mol⁻¹.

(a) Determine the half-life of potassium-40 in years. [3]

(b) Explain why the half-life of potassium-40 cannot be found by measuring the activity of the sample at intervals and plotting a decay curve. [1]

(a) $N = \dfrac{2.50\times10^{-3}}{40.0}\times6.02\times10^{23} = 3.76\times10^{19}$(1)
(a) $\lambda = \dfrac{A}{N} = \dfrac{661}{3.76\times10^{19}} = 1.76\times10^{-17}\ \text{s}^{-1}$(1)
(a) $T_{1/2} = \dfrac{\ln 2}{\lambda} = 3.95\times10^{16}\ \text{s} = 1.25\times10^{9}$ years (accept $1.2$–$1.3\times10^{9}$ y; must be converted to years for this mark)(1)
(b) the activity changes by a negligible / unmeasurable fraction over any practical time (smaller than random fluctuations in count), so no decrease can be detected (accept: the decay curve would be a horizontal line)(1)
Answer: (a) $1.25\times10^{9}$ years; (b) activity effectively constant over the experiment
Examiner tip: Candidates often use the mass in grams with the molar mass in kg (or vice versa) and are out by 10³; keep units consistent before finding N.

9StretchHL only5 marks

Rubidium-87 decays by β⁻ emission to the stable nuclide strontium-87. The half-life of rubidium-87 is $4.9\times10^{10}$ years. A mineral crystal contained no strontium-87 when it formed. Today the ratio

$$\frac{\text{number of strontium-87 nuclei}}{\text{number of rubidium-87 nuclei}} = 0.0600$$

(a) Show that this ratio $R$ is related to the age $t$ of the crystal by $R = e^{\lambda t} - 1$, where $\lambda$ is the decay constant of rubidium-87. [2]

(b) Determine the age of the crystal. [2]

(a) number of Rb remaining $N = N_0e^{-\lambda t}$ and number of Sr formed $= N_0 - N = N_0(1-e^{-\lambda t})$ (each Sr nucleus comes from one decayed Rb nucleus)(1)
(a) $R = \dfrac{N_0(1-e^{-\lambda t})}{N_0e^{-\lambda t}} = e^{\lambda t}-1$ (clear algebra to the given result)(1)
(b) $\lambda = \dfrac{\ln 2}{4.9\times10^{10}} = 1.41\times10^{-11}\ \text{y}^{-1}$ and $\lambda t = \ln(1.0600)$ seen(1)
(b) $t = 4.1\times10^{9}$ years (accept 4.12 × 10⁹; allow answer in s ≈ 1.3 × 10¹⁷ s)(1)
(c) after billions of years / ~700 000 half-lives essentially no C-14 would remain (accept: crystal is not organic/never exchanged carbon with the atmosphere)(1)
Answer: (b) $4.1\times10^{9}$ years
Examiner tip: Many candidates set the Sr/Rb ratio equal to the fraction decayed; the ratio compares daughter with remaining parent, not with the original amount.

10StandardHL only5 marks

To measure the half-life of a short-lived nuclide, a student records the count rate $C$ (corrected for background) every 40 s and plots $\ln(C/\text{s}^{-1})$ against time $t$. The graph and best-fit line are shown.

(a) Explain how the graph shows that the nuclide decays exponentially. [1]

(b) Determine the decay constant of the nuclide. [2]

(d) Calculate the time taken for the corrected count rate to fall to 5.0 % of its value at $t = 0$. [1]

(a) $\ln C = \ln C_0 - \lambda t$ so a straight line with negative gradient (constant gradient) shows $C = C_0e^{-\lambda t}$(1)
(b) gradient from large triangle on the best-fit line, e.g. $\dfrac{4.00-6.40}{200-0} = -0.012\ \text{s}^{-1}$(1)
(b) $\lambda = 1.2\times10^{-2}\ \text{s}^{-1}$ (accept 0.0115–0.0125 s⁻¹; sign must be positive for λ; units required)(1)
(c) $T_{1/2} = \dfrac{\ln 2}{0.012} = 58$ s (accept 55–60 s; ECF from (b))(1)
(d) $t = \dfrac{\ln 20}{\lambda} = 250$ s (accept 240–260 s; ECF from (b))(1)
Answer: (b) $0.012\ \text{s}^{-1}$; (c) 58 s; (d) 250 s
Examiner tip: Use the best-fit line, not two data points, for the gradient, and quote λ as a positive value: the minus sign belongs to the gradient.

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