Explanation:
At the distance of closest approach \(d\), initial kinetic energy of \(\alpha \) particle
\( = \) Potential energy of \(\alpha \) particle and gold nucleus
i.e., \(K = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left( {2e} \right)\left( {Ze} \right)}}{d} = \frac{{2Z{e^2}}}{{4\pi {\varepsilon _0}d}}\)
\({\rm{or}},\;\,d = \frac{{2Z{e^2}}}{{4\pi {\varepsilon _0}K}}\)
Here, \(K = 5MeV = 5 \times 1.6 \times {10^{ - 13}}J\)
\(\left( {\because \,\,\,1\,\,MeV = 1.6 \times {{10}^{ - 13}}J} \right)\)
For gold, \(Z = 79\)
\(\therefore \;\;d = \frac{{(2)(9 \times {{10}^9}\;N{m^2}{C^{ - 2}})(79){{(1.6 \times {{10}^{ - 19}}C)}^2}}}{{(5 \times 1.6 \times {{10}^{ - 13}}J)}}\)
\( = 4.55 \times {10^{ - 14}}m = 4.55 \times {10^{ - 12}}cm\)