| Date | May 2015 | Marks available | 4 | Reference code | 15M.2.HL.TZ2.2 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | State | Question number | 2 | Adapted from | N/A |
Question
Part 2 Nuclear energy and radioactivity
The graph shows the variation of binding energy per nucleon with nucleon number. The position for uranium-235 (U-235) is shown.
U-235 \(\left( {{}_{92}^{235}{\rm{U}}} \right)\) can undergo alpha decay to form an isotope of thorium (Th).
(i) State the nuclear equation for this decay.
(ii) A sample of rock contains a mass of 5.6 mg of U-235 at the present day. The half-life of U-235 is 7.0\( \times \)108 years. Determine the initial mass of the U-235 if the rock sample was formed 3.9\( \times \)109 years ago.
Markscheme
(i) \({}_{92}^{235}{\rm{U}} \to {}_{90}^{231}{\rm{Th}} + {}_2^4\alpha \); (allow He for \(\alpha \); treat charge indications as neutral)
(ii) \(\lambda = \frac{{1{\rm{n2}}}}{{7.0 \times {{10}^8}}}\left( { = 9.9 \times {{10}^{ - 10}}{\rm{yea}}{{\rm{r}}^{ - 1}}} \right)\);
\({m_0} = 5.6{{\rm{e}}^{3.9 \times {{10}^9} \times 9.9 \times {{10}^{ - 10}}}}\left( {{\rm{mg}}} \right)\)
266 mg; } (unit must match eg: allow 266 mg or 0.226 g but not 266 g or 0.266 kg)
or
number of half-lives \( = \left( {\frac{{3.9 \times {{10}^9}}}{{7 \times {{10}^8}}} = } \right)5.57\)
initial mass = 5.6×25.57 ;
266 mg; } (unit must match eg: allow 266 mg or 0.226 g but not 266 g or 0.266 kg)
Award [3] for a bald correct answer.
