Physics - Modern Physics Question with Solution | TestHub

Energy of the photon is given by $E=\frac{hc}{{\lambda }_2}$

${\lambda }_2=\frac{hc}{E}$

Kinetic energy of the proton is $E=\frac{p^2}{2m}$

Momentum of the proton is $p=\sqrt{2mE}$

de-Broglie wavelength of the proton is ${\lambda }_1=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$

Taking the ratio of the wavelengths:

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{\frac{h}{\sqrt{2mE}}}{\frac{hc}{E}}=\frac{1}{c\sqrt{2m}}\sqrt{E}$

Therefore, $\frac{{\lambda }_1}{{\lambda }_2}\propto \sqrt{E}$

Answer: $\sqrt{E}$