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In a Rutherford scattering experiment when a projectile of charge $Z_1$ and mass $M_1$ approaches a target nucleus of charge $Z_2$ and mass $M_2$, the distance of closest approach is $r_0$. The energy of the projectile is :
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The correct answer is:
Directly proportional to $Z_1 Z_2$
$$
\text { } \text { Energy }=\frac{1}{4 \pi \in_0} \frac{Z_1 Z_2}{\mathrm{r}_0}
$$
\text { } \text { Energy }=\frac{1}{4 \pi \in_0} \frac{Z_1 Z_2}{\mathrm{r}_0}
$$
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