The energy of a photon is equal to the kinetic energy of a proton. The energy of the photon is $E$. Let ${\lambda }_1$ be the de- Broglie wavelength of the proton and ${\lambda }_2$ be the wavelength of the photon. The ratio $\frac{{\lambda }_1}{{\lambda }_2}$ is proportional to 

where e is the electronic charge It makes a closest approach of 10 f m to the nucleus.

The de Broglie wavelength (in units of f m) of the proton at its start is : (take the proton mass,$$\begin{gathered} {\mathit{m}}_{\mathit{p}}=\left(\frac{5}{3}\right)\times {10}^{-27}\mathit{k}\mathit{g}\frac{\mathit{h}}{\mathit{e}}=4.2\times {10}^{-15}\mathit{J}\mathit{s}\frac{\mathit{J}\mathit{s}}{\mathit{c}};\frac{1}{4\mathit{\pi }{\mathit{e}}_0}=9\times {10}^9\mathit{m}/\mathit{F};\mathit{ } \\ \mathbf{1}\mathbf{ }\mathit{f}\mathit{m}\mathbf{ }\mathbf{=}\mathbf{ }{\mathbf{10}}^{\mathbf{15}}\mathit{ }\mathit{m} \end{gathered}$$