A proton has kinetic energy $E=100eV$  which is equal to that of a photon. The wavelength of photon is ${\lambda }_2$ and that of proton is  ${\lambda }_1$ . The ratio of ${\lambda }_2/{\lambda }_1$   is proportional to 

$$\begin{gathered} \text{For photon, }{\lambda }_{Ph}=\frac{hc}{E} \\ \begin{array}{l}\text{Here, E is energy of photon.}\\ \text{For proton, }{\lambda }_{pr}=\frac{h}{\sqrt{2mK}}\end{array} \\ \text{Here, m and K be mass and kinetic energy of proton respectively.} \\ \begin{array}{c}\frac{{\lambda }_p}{{\lambda }_{Ph}}=\frac{E}{c}\sqrt{2mK}\propto E^{1/2}\\ (\because E=K\text{ Given })\end{array} \\ \therefore \frac{{\lambda }_2}{{\lambda }_1}\propto E^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.} \end{gathered}$$