Physics - Modern Physics Question with Solution | TestHub

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 $10\mathrm{fm}$ to the nucleus. The de Broglie wavelength (in units of $\mathrm{fm}$ ) of the proton at its start is: (take the proton mass, $m_p=(5/3)\times 10^{-27}\mathrm{kg};h/e=4.2\times$ $10^{-15}\mathrm{J}.\mathrm{S}/\mathrm{C}$; $\frac{1}{4\pi {\epsilon }_0}=9\times 10^9\mathrm{m}/\mathrm{F};1\mathrm{fm}=10^{-15}\mathrm{m}$