The distance of the closest approach of an alpha particle fired at a nucleus with kinetic energy K is r₀. The distance of the closest approach when the alpha particle is fired at the same nucleus with Kinetic Energy 2k will be

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2r₀

$\frac{r_{0}}{4}$

$\frac{r_{0}}{2}$

4r₀

answer is C.

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Detailed Solution

$\begin{array}{l} K E = \frac{1}{4 π \in_{0}} \frac{q_{1} q_{2}}{r} \\ \Rightarrow \frac{k_{1}}{k_{2}} = \frac{r_{2}}{r_{1}} \Rightarrow \frac{k}{2 k} = \frac{r_{2}}{r_{0}} \Rightarrow r_{2} = \frac{r_{0}}{2} \end{array}$

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