A proton is fired from a very far distance towards a nucleus with charge Q = 120 e, Where e is the electronic charge. It makes the 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,  ${\text{m}}_{\text{p}}\text{=}\left(\text{5/3}\right){\text{×10}}^{\text{-27}}{\text{\hspace{0.17em}kg;h/e=4.2×10}}^{\text{-15}}\text{\hspace{0.17em}J.s/C;}\frac{\text{1}}{{\text{4πε}}_{\text{0}}}{\text{=9×10}}^{\text{9}}{\text{m/F;1fm=10}}^{\text{-15}}\text{m}$ ).

At closest approach electrostatic potential energy is equal to kinetic energy
$\frac{K{q}_1{q}_2}{r}=\frac{p^2}{2m}$ 
 $\Rightarrow \frac{(9\times {10}^9)\left(120e\right)\left(e\right)}{10\times {10}^{-15}}=\frac{p^2}{2m}$
De-Broglie wavelength of proton at start is given as
$\lambda =\frac{h}{p}$ 
$\Rightarrow p^2=\frac{h^2}{{\lambda }^2}$ 
From equation 1, we have 
 $2(\frac{5}{3}\times {10}^{-27}){10}^{15}(9\times {10}^9)\left(12\right)e^2=\frac{h^2}{2m{\lambda }^2}$
$$\begin{gathered} \Rightarrow \left(120\right)\left(3\right){10}^{-27+18+9{\lambda }^2}={\left(4.2\right)}^2\times {10}^{-30} \\ {\lambda }^2=\frac{4.2\times 4.2\times {10}^{-30}}{360\times {10}^{-3}}=\frac{42\times 42}{360}\times {10}^{-29} \\ {\lambda }^2=7^2\times {10}^{-30} \\ \lambda =7fm \end{gathered}$$