| Date | May 2016 | Marks available | 2 | Reference code | 16M.2.HL.TZ0.11 |
| Level | Higher level | Paper | Paper 2 | Time zone | Time zone 0 |
| Command term | Suggest | Question number | 11 | Adapted from | N/A |
Question
An alpha particle with initial kinetic energy 32 MeV is directed head-on at a nucleus of gold-197 (19779Au).
(i) Show that the distance of closest approach of the alpha particle from the centre of the nucleus is about 7×10−15m.
(ii) Estimate the density of a nucleus of 19779Au using the answer to (a)(i) as an estimate of the nuclear radius.
The nucleus of 19779Au is replaced by a nucleus of the isotope 19579Au. Suggest the change, if any, to your answers to (a)(i) and (a)(ii).
Distance of closest approach:
Estimate of nuclear density:
An alpha particle is confined within a nucleus of gold. Using the uncertainty principle, estimate the kinetic energy, in MeV, of the alpha particle.
Markscheme
(i)
32 MeV converted using 32×106×1.6×10–19«=5.12×10–12J»
d=≪kQqE=8.99×109×2×79×(1.6×10−19)232×106×1.6×10−19=≫8.99×109×2×79×1.6×10−1932×106
OR 7.102×10-15m
«d≈7×10-15m»
Must see final answer to 2+ SF unless substitution is completely correct with value for k explicit.
Do not allow an approach via r=R0A13.
(ii)
m≈197×1.661×10-27
OR
3.27×10-25kg
V=4π3×(7×10−15)3
OR
1.44×10-42m-3
ρ=≪mV=3.2722×10−251.4368×10−42=≫2.28×1017≈2×1017kgm−3
Allow working in MeV: 1.28×1047MeVc–2m–3.
Allow ECF from incorrect answers to MP1 or MP2.
Distance of closest approach: charge or number of protons or force of repulsion is the same so distance is the same
Estimate of nuclear density: «ρ∝A(A13)3 so» density the same
Δx≈7×10–15 m
Δp≈6.63×10−344π×7×10−15≪=7.54×10−21Ns≫
E≈≪Δp22m=(7.54×10−21)22×6.6×10−27=4.3×10−15J=26897eV≫≈0.027MeV
Accept Δx≈3.5×10–15m or Δx≈1.4×10–14m leading to E≈0.11MeV or 0.0067MeV.
Answer must be in MeV.
