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

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(C) At the distance of closest approach $(r_0)$,the entire initial kinetic energy of the alpha particle is converted into electrostatic potential ener

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$\frac{1}{m}$

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(C) At the distance of closest approach $(r_0)$,the entire initial kinetic energy of the alpha particle is converted into electrostatic potential energy.The kinetic energy is given by $K = \frac{1}{2}mv^2$.The potential energy at distance $r_0$ is $U = \frac{1}{4\pi\varepsilon_0} \cdot \frac{(Ze)(2e)}{r_0}$,where $2e$ is the charge of the alpha particle.Equating kinetic energy to potential energy:$\frac{1}{2}mv^2 = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2Ze^2}{r_0}$Solving for $r_0$:$r_0 = \frac{4Ze^2}{4\pi\varepsilon_0 mv^2} = \frac{Ze^2}{\pi\varepsilon_0 m v^2}$From this expression,we can see that $r_0 \propto \frac{1}{m}$.Therefore,the distance of closest approach is proportional to $\frac{1}{m}$.

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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