A proton is fired from very far away towards a nucleus with charge $Q=120 e$, where $e$ is the electronic charge. It makes a closest approach of $10 fm$ to the nucleus. The de Broglie wavelength (in units $fm$) of the proton at its start is: (take the proton mass, $m_p=\left(\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)\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; 1 fm={10}^{-15}m$