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A proton of mass $m$ and charge $e$ is projected from a very large distance towards an $\alpha$-particle with velocity $v$. Initially $\alpha$-particle is at rest, but it is free to move. If gravity is neglected, then the minimum separation along the straight line of their motion will be
$e^2 / 4 \pi \varepsilon_0 m v^2$
$5 e^2 / 4 \pi \varepsilon_0 m v^2$
$2 e^2 / 4 \pi \varepsilon_0 m v^2$
$4 e^2 / 4 \pi \varepsilon_0 m v^2$
Solution
(b)
As $\alpha$-particle is free to move, initial kinetic energy of system will be
$k_i=\frac{1}{2} \mu v^2$
where, $\mu=$ reduced mass of system
$=\frac{m \cdot 4 m}{m+4 m} \text {. }$
Now, by energy conservation, we have Initial kinetic energy = Potential energy at minimum separation $r$
$\frac{1}{2} (\frac{m .4 m}{m+4 m})v^2=\frac{1}{4 \pi \varepsilon_0} \cdot \frac{2 e^2}{r}$
$\Rightarrow r=\frac{5 e^2}{4 \pi \varepsilon_0 m v^2}$
