Question

The energy of a photons is equal to the kinetic energy of a proton. If ${\lambda }_1$ is the de-Broglie wavelength of a proton, ${\lambda }_2$ the wavelength associated with the photon, and if the energy of the photon is E, then $\left({\lambda }_1/{\lambda }_2\right)$ is proportional to:

• A. $E^4$
• B. $E^{1/2}$
• C. $E^2$
• D. $E$

Answer 1

Answer: (B)

Given : $E_{proton}=E_{photon}=E$

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

$⟹$ ${\lambda }_1=\frac{h}{\sqrt{2mE}}$ .......(1)

Wavelength associated with photon is ${\lambda }_2$

$\therefore$ $E=\frac{hc}{{\lambda }_2}$ $⟹{\lambda }_2=\frac{hc}{E}$ .........(2)

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

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