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 Brogle wavelength (in units of fm) of the proton at its start is : (take the proton mass, $\mathrm{m}_{\mathrm{p}}=(5 / 3) \times 10^{-27} \mathrm{~kg}, \mathrm{~h} / \mathrm{e}=4.2 \times 10^{-15} \mathrm{~J} . \mathrm{s} / \mathrm{C} ; \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \mathrm{~m} / \mathrm{F} ; 1 \mathrm{fm}=10^{-15} \mathrm{~m}$ )