An $\alpha$ particle with a kinetic energy of $1MeV$ is projected towards a stationary nucleus with a charge $|75e|$. Neglecting the motion of the nucleus, determine the distance of closest approach of the $\alpha$ particle.

The correct option is D $2.14\times {10}^{-14}m$

Given, Kinetic energy $=1\mu eV$

$={10}^6\times 1.6\times {10}^{-19}Joules$

$Q=$ charge of nucleus $=75e$

$q=$ charge of $\alpha -$particle $=2e$

We have to find the distance of closest approach, we use the formula.

Kinetic energy $=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{r}$

$r=$ distance of closest approach

$r=\frac{1}{4\pi {\epsilon }_0}\frac{Qq}{K.E}$

On substituting the values,

$=\frac{9\times {10}^9\times 75\times 2\times (1.6\times {10}^{-19}{)}^2}{1.6\times {10}^{-19}\times {10}^6}$

$=2.14\times {10}^{-14}m$