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When an $α$ -particles of mass ‘m’ moving with velocity ‘V’ bombards on a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on ‘m’ as

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1/m²

1/m

$\frac{1}{\sqrt{m}}$

answer is C.

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Detailed Solution

$\frac{1}{2} {mV}^{2} = \frac{1}{4 {πϵ}_{0}} \frac{(2 e) (ze)}{r_{0}}; r_{0} \propto \frac{1}{m}$

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