Question

A proton is fired from very far away source towards a nucleus with charge q = 120e, where 'e' is the electronic 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___? (Take the proton mass, mp = (5/3) $\times {10}^{-27}kg;h/e=4.2\times {10}^{-15}J.s/C;\frac{1}{4\pi {\epsilon }_0}=9\times {10}^{9}m/F;1fm={10}^{-15}m)$

Answer 1

The potential Energy of the final orientation has to be equal to the initial kinetic energy.
$\frac{(9\times {10}^9)\left(120e\right)\left(e\right)}{10\times {10}^{-15}}=\frac{p^2}{2m}$
$\lambda =\frac{h}{p}$
${\lambda }^2=\frac{h^2}{p^2}=\frac{h^2\times 10\times {10}^{-15}}{2m\times 9\times {10}^{-9}\times 120e^2}$
$=\frac{4.2\times 4.2\times {10}^{-30}\times 10\times {10}^{-15}}{2\left(5/3\right)\times {10}^{-27}\times 9\times {10}^{-9}\times 120}$
$=\frac{42\times 42}{36}\times {10}^{-30}$
$\Rightarrow \lambda =7\times {10}^{-15}m=7fm$