Date | May 2014 | Marks available | 1 | Reference code | 14M.3.SL.TZ1.5 |
Level | Standard level | Paper | Paper 3 | Time zone | Time zone 1 |
Command term | Define | Question number | 5 | Adapted from | N/A |
Question
This question is about nuclear physics and radioactive decay.
Define decay constant.
A sample of 1.6 mol of the radioactive nuclide radon-210 \(\left( {{}_{86}^{210}{\rm{Rn}}} \right)\) decays into polonium-206 \(\left( {{}_{84}^{206}{\rm{Po}}} \right)\) with the production of one other particle.
\[{}_{86}^{210}{\rm{Rn}} \to {}_{84}^{206}{\rm{Po + X}}\]
(i) Identify particle X.
(ii) The radioactive decay constant of radon-210 is 8.0×10–5s–1. Determine the time required to produce 1.1 mol of polonium-206.
Particle X has an initial kinetic energy of 6.2MeV after the decay in (b). In a scattering experiment, particle X is aimed head-on at a stationary gold-197 \(\left( {{}_{76}^{197}{\rm{Au}}} \right)\) nucleus.
Determine the distance of closest approach of particle X to the Au nucleus.
Markscheme
the probability of decay of a nucleus per unit time;
(i) alpha particle / helium nucleus;
(ii) number of Po nuclei produced=number of Rn nuclei decayed (seen or implied);
\(0.5 = 1.6{e^{ - \lambda t}}\);
\(t = \left( { - \frac{{\ln \frac{{0.5}}{{1.6}}}}{\lambda } = } \right)\frac{{1.163}}{{8.0 \times {{10}^{ - 5}}}}\);
1.5×104(s);
initial kinetic energy=electric potential energy at closest distance;
kinetic energy \(E = \left( {6.2 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}} = } \right)9.9 \times {10^{ - 13}}\left( {\rm{J}} \right)\);
\(d = k\frac{{{q_1}{q_2}}}{E} = 8.99 \times {10^9}\frac{{2 \times 79 \times {{\left[ {1.6 \times {{10}^{ - 19}}} \right]}^2}}}{{9.9 \times {{10}^{ - 13}}}}\left( {\rm{m}} \right)\) or \( = 3.7 \times {10^{ - 14}}\left( {\rm{m}} \right)\);