Question

If the closest distance of approach for an $\alpha$ particle when it is radiated upon an element of atomic no.$40$, is ${10}^{-14}m$, then find the initial $KE$ of the $\alpha$ particles.

Answer 1

The closest distance is defined as distance at which the $potential$ energy of the particle and nucleus system becomes equal to $kinetic$ energy of the particle initially.

So required kinetic energy is just the potential energy at $r={10}^{-14}m$

Using formula $P.E.=\left(1/4\pi {ϵ}_0\right)\frac{2Z{e}^2}{r}$

Put $Z=40$ and $e=1.6\times {10}^{-19}$ with $1/4\pi {ϵ}_0=9\times {10}^9$

we get $P.E.=1.8432\times {10}^{-12}Joule=\frac{1.8432\times {10}^{-12}}{1.6\times {10}^{-19}}ev=11.52\times {10}^{6}electron-volt$