An alpha particle with initial kinetic energy 32 MeV is directed head-on at a nucleus of gold-197 \(\left( {{}_{79}^{197}{\rm{Au}}} \right)\).

(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 \({}_{79}^{197}{\rm{Au}}\) using the answer to (a)(i) as an estimate of the nuclear radius.

The nucleus of \({}_{79}^{197}{\rm{Au}}\) is replaced by a nucleus of the isotope \({}_{79}^{195}{\rm{Au}}\). 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.

(i)
32 MeV converted using 32×10⁶×1.6×10^–19«=5.12×10^–12J»

\(d =  \ll \frac{{kQq}}{E} = \frac{{8.99 \times {{10}^9} \times 2 \times 79 \times {{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}}}{{32 \times {{10}^6} \times 1.6 \times {{10}^{ - 19}}}} =  \gg \frac{{8.99 \times {{10}^9} \times 2 \times 79 \times 1.6 \times {{10}^{ - 19}}}}{{32 \times {{10}^6}}}\)

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 \(r = {R_0}{A^{\frac{{\rm{1}}}{{\rm{3}}}}}\).

(ii)
m≈197×1.661×10^-27
OR
3.27×
10^-25kg

\(V = \frac{{4\pi }}{3} \times {\left( {7 \times {{10}^{ - 15}}} \right)^3}\)

OR

1.44×10^-42m^-3

\(\rho  =  \ll \frac{m}{V} = \frac{{3.2722 \times {{10}^{ - 25}}}}{{1.4368 \times {{10}^{ - 42}}}} =  \gg 2.28 \times {10^{17}} \approx 2 \times {10^{17}}{\rm{kg}}{{\rm{m}}^{ - 3}}\)

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: «\(\rho  \propto \frac{A}{{{{\left( {{A^{\frac{1}{3}}}} \right)}^3}}}\) so» density the same

Δ
x≈7×10^–15 m

\(\Delta p \approx \frac{{6.63 \times {{10}^{ - 34}}}}{{4\pi  \times 7 \times {{10}^{ - 15}}}} \ll  = 7.54 \times {10^{ - 21}}{\rm{Ns}} \gg \)

\(E \approx  \ll \frac{{\Delta {p^2}}}{{2m}} = \frac{{{{\left( {7.54 \times {{10}^{ - 21}}} \right)}^2}}}{{2 \times 6.6 \times {{10}^{ - 27}}}} = 4.3 \times {10^{ - 15}}{\rm{J}} = 26897{\rm{eV}} \gg  \approx 0.027{\rm{MeV}}\)

Accept Δ
x≈3.5×10^–15m or Δ
x≈1.4×10^–14m leading to E≈0.11MeV or 0.0067MeV.
Answer must be in MeV.