An $α$ -particle after passing through a potential difference of $V$ volt collides with a nucleus. If the atomic number of the nucleus is $Z$ then the distance of closest approach of $α$ -particle to the nucleus will be

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$7.2 (\frac{Z}{V_{0}}) \overset{\circ}{A}$

$14.4 (\frac{Z}{V_{0}}) m$

$14.4 (\frac{Z}{V_{0}}) c m$

$14.4 (\frac{Z}{V_{0}}) \overset{o}{A}$

answer is D.

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Detailed Solution

KE of $α$ -particle is $K_{α} = q V_{0} = (2 e) V_{0}$

PE of $α$ -particle is $U_{α} = \frac{1}{4 π ε_{0}} \frac{(Z e) (2 e)}{r_{0}}$

By Law of Conservation of Energy, $U_{α} = K_{α}$

$2 e V_{0} = \frac{2 Z e^{2}}{4 π ε_{0} r_{0}} \Rightarrow r_{0} = \frac{Z e}{4 π ε_{0} V_{0}} \Rightarrow r_{0} = \frac{(9 \times 10^{9}) (1.6 \times 10^{- 19}) (Z)}{V_{0}} \Rightarrow r_{0} = 14.4 \times 10^{- 10} (\frac{Z}{V_{0}}) m \Rightarrow r_{0} = 14.4 (\frac{Z}{V_{0}}) \overset{\circ}{A}$