Question

A proton is fired from very far away towards a nucleus with charge $Q=120e$, where $e$ is the electronic charge. It makes a closest approach of $10fm$ to the nucleus. The de-Broglie wavelength (in unit of $fm$) of the proton at its start is:

(take the proton mass, $m_p=\left(5/3\right)\times {10}^{-27}kg;h/e=4.2\times {10}^{-15}Js/C;$

$\frac{1}{4\pi {ϵ}_0}=9\times {10}^{9}m/F;1fm={10}^{-15}m)$

Answer 1

On conserving energy of the proton we get :
$\frac{1}{2}m{v}^2$ = K$\frac{q_1{q}_2}{R}$
So, $v=$ $\sqrt{\frac{2K{q}_1{q}_2}{mR}}$
$\lambda$ = $\frac{h}{mv}$ = $\frac{h\sqrt{R}}{\sqrt{2mK{q}_1{q}_2}}$ = $\frac{h\sqrt{10\times {10}^{-15}}}{\sqrt{2\times \frac{5}{3}\times {10}^{-27}\times 9\times {10}^9\times 120\times e^2}}$

$=\frac{\frac{h}{e}\times {10}^{-7}}{6\times {10}^{-8}}$ $=\frac{4.2\times {10}^{-15}\times {10}^{-7}}{6\times {10}^{-8}}$ $=7\times {10}^{-15}m=7fm$