| Date | May 2011 | Marks available | 2 | Reference code | 11M.2.SL.TZ2.5 |
| Level | Standard level | Paper | Paper 2 | Time zone | Time zone 2 |
| Command term | Draw | Question number | 5 | Adapted from | N/A |
Question
Part 2 Unified atomic mass unit and a nuclear reaction
Define the term unified atomic mass unit.
α+147N→178O+X
(i) Identify X.
(ii) The following data are available for the reaction.
Rest mass of α = 3.7428 GeVc–2
Rest mass of 147N = 13.0942 GeVc–2
Rest mass of 178O+X = 16.8383 GeVc–2
The initial kinetic energy of the α-particle is 7.68 MeV. Determine the sum of the kinetic energies of the oxygen nucleus and X. (Assume that the nitrogen nucleus is stationary.)
The reaction in (c) produces oxygen (O-17). Other isotopes of oxygen include O-19 which is radioactive with a half-life of 30 s.
(i) State what is meant by the term isotopes.
(i) Define the term radioactive half-life.
A nucleus of the isotope O-19 decays to a stable nucleus of fluorine. The half-life of O-19 is 30 s. At time t=0, a sample of O-19 contains a large number N0 nuclei of O-19.
On the grid below, draw a graph to show the variation with time t of the number N of O-19 nuclei remaining in the sample. You should consider a time of t=0 to t=120s.
Markscheme
1/12th mass of an atom of carbon-12/12C ;
(254.1001×931.5 =)236.7(GeVc−2 ); (only accept answer in GeV c−2 )
(i) proton / hydrogen nucleus / H+ / 11H / 11p;
(ii) ∆m=(16.8383-[3.7428+13.0942]=)0.0013(GeVc−2);
energy required for reaction = 1.3 (MeV);
178O+X=(7.68−1.3=)6.4(6.38)MeV; (allow correct answer in any valid energy unit)
(i) (nuclei of same element with) same proton number, different number of neutrons / OWTTE;
(ii) the time for the activity of a sample to reduce by half / time for the number of the radioactive nuclei to halve from original value;
scale drawn on t axis; (allow 10 grid squares ≡ 30 s or 40 s)
smooth curve passes through N02 at 30s, N04 at 60s, N08 at 90s, N016 at 120s
(to within 1 square); (points not necessary)
