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An alpha nucleus of energy $\frac{1}{2} m v^{2}$ bombards a heavy nuclear target of charge Ze. The distance of closest approach for the alpha nucleus will be proportional to

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$\frac{1}{m}$

$\frac{1}{Ze}$

$\frac{1}{v^{4}}$

$v^{2}$

answer is D.

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Detailed Solution

$A s, KE = P . E \Rightarrow \frac{1}{2} m v^{2} = k \frac{(2 e) (Z e)}{r_{0}}$

$\Rightarrow r_{0} = \frac{4 k Z e^{2}}{m v^{2}}$

$∴ r_{0} \propto \frac{1}{m}$

[ The distance of closest approach r₀, m = mass alpha particle ]

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