| Date | May 2012 | Marks available | 6 | Reference code | 12M.3.SL.TZ1.4 |
| Level | Standard level | Paper | Paper 3 | Time zone | Time zone 1 |
| Command term | Determine | Question number | 4 | Adapted from | N/A |
Question
This question is about plutonium as a power source.
Plutonium (\({}_{94}^{238}{\rm{Pu}}\)) decays by alpha emission. The energy of the alpha particle emitted is 8.8×10–13J. The decay constant of plutonium-238 is 8.1×10–3yr–1.
Define decay constant.
Plutonium-238 is to be used as a power source in a space probe.
(i) Determine the initial activity of plutonium such that the power released by plutonium is 6.0 W.
(ii) The power source becomes useless when the power released decreases to 4.0 W. Determine the time, in years, for which the power source can be used in the space probe.
Markscheme
the probability of decay per unit time / constant of proportionality in the equation relating activity to number of nuclei;
(i) power \(P = {A_0} \times E \Rightarrow {A_0} = \frac{P}{E}\);
\({A_0} = \left( {\frac{{6.0}}{{8.8 \times {{10}^{ - 13}}}} = } \right)6.8 \times {10^{12}}{\rm{Bq}}\);
(ii) realization that power is proportional to activity / \(P = {P_0}{{\rm{e}}^{ - \lambda t}}\);
\(4.0 = 6.0{{\rm{e}}^{ - 8.1 \times {{10}^{ - 3}}t}}\);
taking logs to get \(\ln \frac{4}{6} = - 8.1 \times {10^{ - 3}}t\);
\(t = \left( {\frac{{\ln \frac{4}{6}}}{{ - 8.1 \times {{10}^{ - 3}}}} = } \right)50{\rm{yr}}\);
First marking point may be implicit in second.
Award [2 max] for using half-life (86 years) and linear fit to give 57 years.
Award [4] for correct answer by other methods.
