Date | November 2013 | Marks available | 4 | Reference code | 13N.3.SL.TZ0.5 |
Level | Standard level | Paper | Paper 3 | Time zone | Time zone 0 |
Command term | Estimate | Question number | 5 | Adapted from | N/A |
Question
This question is about radioactive decay.
Potassium-40 (K-40) is a radioactive isotope that occurs naturally in many different types of rock. A very small percentage of the isotope undergoes β+ decay to form an isotope of argon (Ar). Construct and complete the nuclear reaction equation for this decay.
Overall about 10% of a sample of K-40 will decay to argon. In a particular rock sample it is found that there are 1.6×1022 atoms of K-40 and 8.4×1021 atoms of argon. The half-life of K-40 is 1.2×109 yr. Estimate the time elapsed since the rock sample was formed.
Markscheme
\(\left( {{}_{19}^{40}{\rm{K}} \to {}_{18}^{40}{\rm{Ar}} + {}_1^0{\beta ^ + } + v} \right)\)
\({{}_{18}^{40}{\rm{Ar}}}\);
v; (do not accept \(\bar v\))
original number of K-40 atoms=(1.6×1022+[8.4×1021×10]=)1.0×1023;
decay constant=\(\frac{{\ln 2}}{{1.2 \times {{10}^9}}}\) or 5.8×10-10(yr-1);
\(1.6 \times {10^{22}} = 1.0 \times {10^{23}}{\rm{ }}{{\rm{e}}^{ - {\rm{5.8}} \times {\rm{1}}{{\rm{0}}^{ - {\rm{10}}}}t}}\);
to give t=3.2×109 (yr);
Accept any alternative method that leads to the correct answer.
Award [3 max] ECF after incorrect value for N₀ (eg. use of 2.44×1022 to give 7.3×1022 yr).
Award [2 max] for approximate answers (eg. 3.0×109 yr based on an estimate of between two and three half-lives.)