Q. The energy of a photon is equal to the kinetic energy of a photon. The energy of a the photon is $E.$ Let $\lambda_{1}$ be the de-Broglie wavelength of the photon and ${\ \lambda}_{2}$ be the wavelength of the photon. The ratio $\frac{\lambda_{1}}{\lambda_{2}}$ proportional to
A) $E^{0}$
B) $E^{1/2}$
C) $E^{- 1}$
D) $E^{- 2}$
Explanation
(b) $\frac{\lambda_{1}}{\lambda_{2}} = \frac{h}{\frac{\sqrt{2mE}}{\ \ \ \ \ \frac{hc}{E}}}\ \ $ or $\ \ \ \frac{\lambda_{1}}{\lambda_{2}} \propto E^{1/2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.