A proton and an anti-proton come close to eac | vedclass

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Question

$(a)$

The situation can be shown as

From energy conservation,

$( PE )_{	ext {initial }}+( KE )_{	ext {initial }}=( PE )_{	ext {final }}+(

Vedclass · Accepted Answer

$1.17$

Vedclass · Answer

$(a)$

The situation can be shown as

From energy conservation,

$( PE )_{	ext {initial }}+( KE )_{	ext {initial }}=( PE )_{	ext {final }}+( KE )_{	ext {final }}$

$0+0=\frac{K e^{2}}{r}+\frac{1}{2} m v^{2}+\frac{1}{2} m v^{2}$

$\Rightarrow v=\sqrt{\frac{K e^{2}}{m r}}=\sqrt{\frac{9 	imes 10^{9} 	imes\left(1.6 	imes 10^{-19}ight)^{2}}{1.67 	imes 10^{27} 	imes 10 	imes 10^{-2}}}$

$=1.17 \,ms ^{-1}$

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 ........... $\,m / s$

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