An α - particle of energy 5 MeV is scattered through 180 ° by a fixed Uranium nucleus. The distance of the closest approach is (Atomic number of uranium = 92 )

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According to the law of conservation of energy,

The kinetic energy of $α -$ particle = Potential energy of alpha $α -$ particle at a distance of closest approach.

$\Rightarrow \frac{1}{2} m v^{2} = \frac{1}{4 π ε_{0}} \frac{q_{1} q_{2}}{r}$

$\Rightarrow 5 MeV = \frac{9 \times 1 0^{9} (2 e) (92 e)}{r}$

$r = 5.3 × 1 0^{- 14} m$

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