An alpha nucleus of energy frac{1}{2}mv^2 bom | vedclass

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(D) The distance of closest approach $(r)$ is the distance at which the initial kinetic energy of the alpha particle is completely converted into elec

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

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(D) The distance of closest approach $(r)$ is the distance at which the initial kinetic energy of the alpha particle is completely converted into electrostatic potential energy.The kinetic energy of the alpha particle is $K.E. = \frac{1}{2}mv^2$.The charge of the alpha particle is $q = 2e$ and the charge of the target nucleus is $Q = Ze$.At the distance of closest approach $(r)$,the electrostatic potential energy is given by $U = \frac{1}{4\pi\epsilon_0} \cdot \frac{qQ}{r} = K \cdot \frac{(2e)(Ze)}{r}$,where $K = \frac{1}{4\pi\epsilon_0}$.Equating kinetic energy to potential energy:$\frac{1}{2}mv^2 = \frac{K(2e)(Ze)}{r}$Solving for $r$:$r = \frac{4KZe^2}{mv^2}$From this expression,it is clear that $r \propto \frac{1}{v^2}$ and $r \propto \frac{1}{m}$.Therefore,the distance of closest approach is proportional to $\frac{1}{v^2}$.

An alpha nucleus of energy $\frac{1}{2}mv^2$ 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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