In the alpha particle scattering experiment,i | vedclass

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(D) The distance of closest approach $r_0$ is determined by equating the initial kinetic energy of the alpha particle to the electrostatic potential e

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$\frac{d}{4}$

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(D) The distance of closest approach $r_0$ is determined by equating the initial kinetic energy of the alpha particle to the electrostatic potential energy at the point of closest approach:$\frac{1}{2} m v^2 = \frac{1}{4 \pi \varepsilon_0} \cdot \frac{(Ze)(2e)}{r_0}$From this equation,we can see that $v^2 \propto \frac{1}{r_0}$,which implies $r_0 \propto \frac{1}{v^2}$.Given that the initial distance of closest approach is $d$ for velocity $v$,let the new distance be $d'$ for velocity $2v$.$\frac{d'}{d} = \frac{v^2}{(2v)^2} = \frac{v^2}{4v^2} = \frac{1}{4}$Therefore,$d' = \frac{d}{4}$.

In the alpha particle scattering experiment,if $v$ is the initial velocity of the particle,then the distance of closest approach is $d$. If the velocity is doubled,then the distance of closest approach changes to:

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