E.3 Radioactive decay

Practice Worksheet — name: ______________________ date: ____________

Formulas for this topic

Mass–energy equivalence

E = mc^{2}

Decay law (HL)

N = N_{0}e^{-\lambda t}

Activity (HL)

A = \lambda N = A_{0}e^{-\lambda t}

Half-life and decay constant (HL)

T_{\frac{1}{2}} = \frac{\ln 2}{\lambda}

Section A — Multiple choice

A1. A sample contains $8.0\times10^{20}$ radioactive nuclei with a half-life of 6.0 hours. After one day, the number remaining is:

A $4.0\times10^{20}$
B $2.0\times10^{20}$
C $1.0\times10^{20}$
D $0.5\times10^{20}$

A2. In beta-minus decay, the emitted electrons show a continuous energy spectrum. This was evidence for:

A Discrete nuclear energy levels
B The existence of the antineutrino, sharing the decay energy
C The quantisation of charge
D The strong nuclear force

A3. The binding energy per nucleon curve peaks near iron ( $A \approx 56$ ). This explains why:

A Iron is the most abundant element
B Fusion releases energy below iron and fission above it
C Iron cannot undergo any nuclear reactions
D All nuclei decay to iron eventually

Section B — Short answer

B1. Complete and balance: $^{226}_{88}\text{Ra} \to\ ?\ + \alpha$ , and $^{14}_{6}\text{C} \to\ ?\ + \beta^- + \bar{\nu}$ . [4 marks]

B2. The deuteron $^2_1\text{H}$ has mass 2.013553 u; the proton 1.007276 u and neutron 1.008665 u. Calculate its binding energy in MeV. [3 marks]

B3. A detector reads 240 counts per minute near a source; background is 30 counts per minute. The source's half-life is 20 minutes. Predict the reading after one hour. [3 marks]

Answer key

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Section A

A1: $0.5\times10^{20}$ — One day = 24 h = 4 half-lives, so the fraction remaining is $(1/2)^4 = 1/16$ : $8.0\times10^{20}/16 = 0.5\times10^{20}$ .

A2: The existence of the antineutrino, sharing the decay energy — A two-body decay would give the electron a single fixed energy. The observed spread means a third, undetected particle shares the energy randomly — Pauli's antineutrino, later confirmed directly. Alpha spectra, by contrast, ARE discrete, revealing nuclear energy levels.

A3: Fusion releases energy below iron and fission above it — Reactions release energy when the products climb towards the peak — fusing light nuclei or splitting heavy ones both increase binding energy per nucleon. Iron itself sits at the top: no energy can be extracted from it either way, which is why fusion in stellar cores stops there.

Section B

B1: Alpha decay reduces $A$ by 4 and $Z$ by 2: $^{226}_{88}\text{Ra} \to\ ^{222}_{86}\text{Rn} + ^4_2\alpha$ . Beta-minus converts a neutron to a proton ( $Z$ up by 1, $A$ unchanged): $^{14}_{6}\text{C} \to\ ^{14}_{7}\text{N} + \beta^- + \bar{\nu}$ .

B2: Mass defect: $(1.007276 + 1.008665) - 2.013553 = 0.002388\ \text{u}$ . Binding energy: $0.002388 \times 931.5 \approx 2.22\ \text{MeV}$ — the energy needed to pull the deuteron apart.

B3: Source contribution now: $240 - 30 = 210$ cpm. One hour = 3 half-lives: $210/8 \approx 26$ cpm. Reading = source + background $= 26 + 30 \approx 56$ cpm. Subtract background before halving, add it back after.