An alpha nucleus of energy frac{1}{2}m{v^2} b | vedclass

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Question

Work done to stop the $\alpha$ particle is equal to $K.E.$

$	herefore q V=\frac{1}{2} m v^{2} \Rightarrow q 	imes \frac{K(Z e)}{r}=\frac{1}{2} m

Vedclass · Accepted Answer

$\frac{1}{{{v^2}}}$

Vedclass · Answer

Work done to stop the $\alpha$ particle is equal to $K.E.$

$	herefore q V=\frac{1}{2} m v^{2} \Rightarrow q 	imes \frac{K(Z e)}{r}=\frac{1}{2} m v^{2}$

$\Rightarrow r=\frac{2(2 e) K(Z e)}{m v^{2}}=\frac{4 K Z e^{2}}{m v^{2}}$

$\Rightarrow r \propto \frac{1}{v^{2}}$ and $r \propto \frac{1}{m}$

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

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