Question

A proton strikes another proton at rest with speed $V_0$. Assume impact parameter to be zero. Their closest distance of approach is (mass of proton is $m$)

• A. $\frac{e^2}{4\pi \epsilon m{v}_0}$
• B. $\frac{e^2}{\pi {\epsilon }_{0}m{v_0}^2}$
• C. $\frac{e^2}{m{v}_0^2}$
• D. Zero

Answer 1

The correct option is B $\frac{e^2}{\pi {\epsilon }_{0}m{v_0}^2}$

The kinetic energy of the system initially $=\frac{m{v}_0^2}{2}$
Assume initially they have infinite separation.

Hence potential energy $=0J$
At distance of closest approach, $PE$ is equal to $KE$ which is half the total energy.
By law of conservation of energy:
$P{E}_{atclosestdistance}=\frac{K{E}_{initial}}{2}$
$\frac{1}{4\pi {ϵ}_0}\frac{e^2}{d}=\frac{m{v}_0^2}{4}$
i.e. Distance for closest approach $\left(d\right)=\frac{1}{\pi {ϵ}_0}\frac{e^2}{m{v}_0^2}$