Date | May 2011 | Marks available | 6 | Reference code | 11M.3.SL.TZ1.6 |
Level | Standard level | Paper | Paper 3 | Time zone | Time zone 1 |
Command term | Calculate, Draw, and State | Question number | 6 | Adapted from | N/A |
Question
This question is about radioactive decay.
Iodine-124 (I-124) is an unstable radioisotope with proton number 53. It undergoes beta plus decay to form an isotope of tellurium (Te).
State the reaction for the decay of the I-124 nuclide.
The graph below shows how the activity of a sample of iodine-124 changes with time.
(i) State the half-life of iodine-124.
(ii) Calculate the activity of the sample at 21 days.
(iii) A sample of an unknown radioisotope has a half-life twice that of iodine-124 and the same initial activity as the sample of iodine-124. On the axes opposite, draw a graph to show how the activity of the sample would change with time. Label this graph X.
(iv) A second sample of iodine-124 has half the initial activity as the original sample of iodine-124. On the axes opposite, draw a graph to show how the activity of this sample would change with time. Label this graph Y.
Markscheme
\({}_{53}^{124}{\rm{I}} \to {}_{52}^{124}{\rm{Te + }}{}_1^0\beta ^+ \);
\({}_0^0v/v\);
Do not allow an antineutrino.
Award [1 max] for \({}_{53}^{124}{\rm{I}} \to {}_{54}^{124}{\rm{Te + }}{}_1^0\beta^- + \bar v\).
(i) 4 days;
(ii) \(\lambda = \frac{{\ln 2}}{{{T_{\frac{1}{2}}}}} = \frac{{\ln 2}}{4} = \left( {0.173{\rm{da}}{{\rm{y}}^{ - 1}}} \right)\);
\(A = {A_0}{e^{ - \lambda t}} = 16 \times {10^7} \times {e^{ - 0.173 \times 21}}\left( {{\rm{Bq}}} \right)\);
A=4.2×106Bq;
Award [2 max] for bald answer in range 4.2−4.5×106 Bq, or linear interpolation between half lives giving 4.4×106Bq.
(iii) graph passing through or near (0,16), (8,8) and (16,4) – see below;
(iv) graph passing through or near (0,8), (4,4) and (8,2) – see below;
Do not penalize if graph does not pass through (12,1) and (16,0.5).