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(B) The distance of the closest approach $r_{0}$ for an alpha particle of kinetic energy $K$ is given by the formula:$r_{0} = \frac{1}{4 \pi \varepsil

Vedclass · Accepted Answer

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

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(B) The distance of the closest approach $r_{0}$ for an alpha particle of kinetic energy $K$ is given by the formula:$r_{0} = \frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{2Ze^{2}}{K}$From this expression,we can see that the distance of the closest approach is inversely proportional to the kinetic energy:$r_{0} \propto \frac{1}{K}$Let $r_{01} = r_{0}$ when kinetic energy $K_{1} = K$. When the kinetic energy is increased to $K_{2} = 2K$,let the new distance be $r_{02}$.Using the proportionality $r_{01} K_{1} = r_{02} K_{2}$,we get:$r_{0} \cdot K = r_{02} \cdot (2K)$$r_{02} = \frac{r_{0} \cdot K}{2K} = \frac{r_{0}}{2}$Therefore,the new distance of the closest approach is $\frac{r_{0}}{2}$.

The distance of the closest approach of an alpha particle fired at a nucleus with kinetic energy $K$ is $r_{0}$. 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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