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(A) The distance of closest approach $r_0$ is the distance at which the initial kinetic energy of the $\alpha$-particle is entirely converted into ele

Vedclass · Accepted Answer

$\frac{1}{m}$

Vedclass · Answer

(A) The distance of closest approach $r_0$ is the distance at which the initial kinetic energy of the $\alpha$-particle is entirely converted into electrostatic potential energy.Using the conservation of energy:$\frac{1}{2} mv^2 = \frac{1}{4\pi \epsilon_0} \cdot \frac{(Ze)(2e)}{r_0}$Rearranging for $r_0$:$r_0 = \frac{1}{4\pi \epsilon_0} \cdot \frac{2Ze^2}{\frac{1}{2}mv^2} = \frac{Ze^2}{\pi \epsilon_0 m v^2}$From this expression,we can see that $r_0 \propto \frac{1}{m}$ and $r_0 \propto \frac{1}{v^2}$.Comparing this with the given options,the correct proportionality is $r_0 \propto \frac{1}{m}$.

When $\alpha$-particles with kinetic energy $\frac{1}{2} mv^2$ are bombarded on a heavy nucleus of charge $Ze$,the distance of closest approach is proportional to .........

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