(a) In the Geiger-Marsden experiment, calculate the distance of closest approach for an alpha particle with energy 2.56 x 10^–12 J. Consider that the particle approaches the gold nucleus (Z=79) in head-on position.
OR
(b) If the above experiment is repeated with a proton of the same energy, then what will be the value of the distance of closest approach?
 

Let d be the distance of closest approach then by the conservation of energy.
Initial kinetic energy of incoming α-particle K = Final electric potential energy U of the system
As $$\begin{gathered} \mathrm{K}=\frac{1}{4\mathrm{\pi \epsilon }0}\mathrm{x}\frac{\left(\mathrm{ze}\right)\left(2e\right)}{\mathrm{d}} \\ \therefore \mathrm{d}=\frac{1}{4\mathrm{\pi \epsilon }0}\mathrm{x}\frac{2\left(\mathrm{ze}\right)\left(2e\right)}{\mathrm{K}} .....\left(\mathrm{i}\right) \end{gathered}$$
Here, 
$\begin{array}{l}\frac{1}{4\mathrm{\pi \epsilon }0}=9\times {10}^9{\mathrm{Nm}}^2{\mathrm{C}}^{-2}\\ \mathrm{Z}=79,\mathrm{e}=1.6\times {10}^{-19}\mathrm{C}\\ \mathrm{K}=2.56\times {10}^{-12}\mathrm{J}\end{array}$
Substituting these values in (i)
$\begin{array}{l}\mathrm{d}=9\times {10}^9\times \frac{2\times {\left(1.6\times {10}^{-19}\right)}^2\times 79}{2.56\times {10}^{-12}}\\ \mathrm{d}=1.42\times {10}^{-14}\mathrm{m}\\ \mathrm{d}=14.2\mathrm{fm}\left(\because 1\mathrm{fm}={10}^{-15}\mathrm{m}\right)\end{array}$
OR

Since, the d is directly proportional to square of charge and directly propotional to Z, Hence distance of closest approach becomes 1/8 times