A proton and an anti-proton come close to each other in vacuum such that the distance between them is $10 \,cm$. Consider the potential energy to be zero at infinity. The velocity at this distance will be ........... $\,m / s$
KVPY 2020, Diffcult
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$(a)$
The situation can be shown as
From energy conservation,
$( PE )_{\text {initial }}+( KE )_{\text {initial }}=( PE )_{\text {final }}+( KE )_{\text {final }}$
$0+0=\frac{K e^{2}}{r}+\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}$
$\Rightarrow v=\sqrt{\frac{K e^{2}}{m r}}=\sqrt{\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{1.67 \times 10^{27} \times 10 \times 10^{-2}}}$
$=1.17 \,ms ^{-1}$
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