A nucleus with energy frac{1}{2}mv^2 is fired | vedclass

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Question

(B) At the distance of closest approach $(r)$,the initial kinetic energy of the projectile is completely converted into electrostatic potential energy

Vedclass · Accepted Answer

$1/m$

Vedclass · Answer

(B) At the distance of closest approach $(r)$,the initial kinetic energy of the projectile is completely converted into electrostatic potential energy.The initial kinetic energy is $K = \frac{1}{2}mv^2$.The electrostatic potential energy at distance $r$ is $U = \frac{k(Ze)(2e)}{r}$,where $2e$ is the charge of the alpha particle (assuming it is an alpha particle as is standard in Rutherford scattering problems).Equating the two:$\frac{1}{2}mv^2 = \frac{k(Ze)(2e)}{r}$Rearranging for $r$:$r = \frac{4kZe^2}{mv^2}$From this expression,we can see that $r \propto \frac{1}{m}$ and $r \propto \frac{1}{v^2}$.Comparing this with the given options,the correct proportionality is $r \propto 1/m$.

$A$ nucleus with energy $\frac{1}{2}mv^2$ is fired at a target nucleus of charge $Ze$. The distance of closest approach to the $Ze$ nucleus is proportional to which of the following?

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