Date | November 2011 | Marks available | 1 | Reference code | 11N.2.HL.TZ0.3 |
Level | Higher level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Identify | Question number | 3 | Adapted from | N/A |
Question
This question is about nuclear processes.
Describe what is meant by
(i) radioactive decay.
(ii) nuclear fusion.
Tritium is a radioactive nuclide with a half-life of 4500 days. It decays to an isotope of helium.
Determine the time taken for 90% of a sample of tritium to decay.
A nuclide of deuterium \(\left( {{}_1^2{\rm{H}}} \right)\) and a nuclide of tritium \(\left( {{}_1^3{\rm{H}}} \right)\) undergo nuclear fusion. The reaction equation for this process is
\[{}_1^2{\rm{H + }}{}_1^3{\rm{H}} \to {}_2^4{\rm{He + X}}\]
Identify X.
Markscheme
(i) refers to unstable nucleus/isotope / refers to spontaneous/random process;
which emits named radiation (from nucleus) / forms different nucleus/isotope;
(ii) combination of two nuclei / OWTTE; (do not allow “particles” or “atoms”)
to form new nuclide with greater mass/larger nucleus/greater number of nucleons;
\(\lambda = \frac{{\ln 2}}{{4500}}\left( { = 1.54 \times {{10}^{ - 4}}} \right)\);
\(0.1{N_0} = {N_0}{{\rm{e}}^{ - 1.54 \times {{10}^{ - 4}}t}}\);
1.5×104(d) or 1.3×109(s);
Award [2 max] if answer is time to lose 10% (680 d).
Allow answer to be expressed in any time units.
Award [3] for a bald correct answer.
or
\(\ln 0.1 = \frac{{ - 0.69t}}{{{t_{\frac{1}{2}}}}}\);
t=3.3×4500;
1.5×104(d);
Award [2 max] if answer is time to lose 10% (680 d).
Allow answer to be expressed in any time units.
Award [3] for a bald correct answer.
\({}_0^1n\)/neutron;