An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of closest approach is of the order of :

10^-12 cm

From conservation of mechanical energy decrease in kinetic energy = increase in potential energy 

or $\frac{1}{4 \pi \varepsilon_0} \frac{(Z e)(2 e)}{r_{\min}}$ = 5 MeV

$=5 \times 1.6 \times 10^{-13} J$

∴ $r_{\min }=\frac{1}{4 \pi \in_0} \frac{2 Z e^2}{5 \times 1.6 \times 10^{-13}}$

$=\frac{\left(9 \times 10^9\right)(2)(92)\left(1.6 \times 10^{-19}\right)^2}{5 \times 1.6 \times 10^{-13}}$

$=5.3 \times 10^{-14} m$

$=5.3 \times 10^{-12} cm$

i.e., $r_{\min }$ is of the order of $10^{-12} cm.

∴ correct option is (c)