Question

A proton is fired from very far away towards a nucleus with charge $Q=120e$, where $Q$ is the electric charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is $(\text{proton -mass}=\frac{5}{3}\times {10}^{-27}\text{kg},\frac{h}{e}=4.2\times {10}^{-15}Js{C}^{-1})$

Answer 1

$\frac{p^2}{2m}=\frac{KZ{e}^2}{r}\Rightarrow p=\sqrt{\frac{KZ{e}^2.2m}{r}}$
$\Rightarrow \lambda =\frac{h}{p}=\frac{\frac{h}{e}}{\sqrt{2K\frac{Zm}{r}}}=\frac{4.2\times {10}^{-15}}{\sqrt{\frac{2\times 9\times {10}^9\times 120\times \frac{5}{3}\times {10}^{-27}}{10\times {10}^{-15}}}}=7\text{fm}$