IB Physics HL E.3.12 Half-life changes Question Bank

Practise IB Physics SL/HL E.3.12 by applying half-life changes concepts to exam-style physics questions.

Exam points

identify the relevant physical quantity, model or diagram
apply the correct equation and units to the evidence
justify the result using conservation laws, fields or energy relationships

[Maximum number: 3]

The radioactive nuclide beryllium-10 (Be-10) undergoes beta minus ( $\beta$ -) decay to form a stable boron (B) nuclide.

The initial number of nuclei in a pure sample of beryllium-10 is $\mathrm{N}_{0}$ . The graph shows how the number of remaining beryllium nuclei in the sample varies with time.

Figure for Question E.3.12—Half-life changes question 1 — IB Physics HL

After $4.3 \times 10^{6}$ years,

\frac{\text { number of produced boron nuclei }}{\text { number of remaining beryllium nuclei }}=7 .

Show that the half-life of beryllium-10 is $1.4 \times 10^{6}$ years.

ALTERNATIVE 1

fraction of $\mathrm{Be}=\frac{1}{8}, 12.5 \%$ , or 0.125

therefore 3 half lives have elapsed
$t_{\frac{1}{2}}=\frac{4.3 \times 10^{6}}{3}=1.43 \times 10^{6} \ll \approx 1.4 \times 10^{6}>\ll y \gg$

ALTERNATIVE 2

fraction of $\mathrm{Be}=\frac{1}{8}, 12.5 \%$ , or 0.125
$\begin{aligned} & \frac{1}{8}=e^{-\lambda}\left(4.3 \times 10^{6}\right) \text { leading to } \lambda=4.836 \times 10^{-7} \ll \mathrm{y}^{-1} \gg & \frac{\ln 2}{\lambda}=1.43 \times 10^{6}<\mathrm{y} \gg \end{aligned}$

Must see at least one extra sig fig in final answer.

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E.3.12—Half-life changes question 1