Question

The energy of a photon is equal to the kinetic energy of a proton. The energy of photon is $E$. Let ${\lambda }_1$ be the de Broglie wavelength of the proton and ${\lambda }_2$ be the wavelength of the photon. Then, ${\lambda }_1/{\lambda }_2$ is proportional to :

• A. $E^0$
• B. $E^{1/2}$
• C. $E^{-1}$
• D. $E^{-2}$

Answer 1

The correct option is B $E^{1/2}$

The energy photon , $E=h{\nu }_2=\frac{hc}{{\lambda }_2}$ or ${\lambda }_2=\frac{hc}{E}$

As energy of photon $=$ energy of proton so $E=\frac{p^2}{2m}$ where p be the momentum of proton.

Now, ${\lambda }_1=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$

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

Hence, $\frac{{\lambda }_1}{{\lambda }_2}\propto E^{1/2}$