Question: 37. A photon and an electron (mass m) have the same energy E. The ratio ( \lambda_{\text{photon}} / \lambda_{\text{electron}} ) of their de Broglie wavelengths is ( c is the speed of light)
(1) \sqrt{E / 2 m}
(2) c \sqrt{2 m E}
(3) c \sqrt{\frac{2 m}{E}}
(4) \frac{1}{c} \sqrt{E / 2 m}
Answer: Option (3)
Explanation:
The de Broglie wavelength of a particle is given by \lambda = \dfrac{h}{p},
where h is Planck’s constant and p is momentum.
For a photon, the energy–momentum relation is E = pc.
Hence, momentum of the photon is p_{\text{photon}} = \dfrac{E}{c}.
Therefore, de Broglie wavelength of the photon is:
\lambda_{\text{photon}} = \dfrac{h}{p_{\text{photon}}} = \dfrac{hc}{E}For an electron having kinetic energy E, its momentum is given by:
E = \dfrac{p_{\text{electron}}^2}{2m} p_{\text{electron}} = \sqrt{2mE}Thus, the de Broglie wavelength of the electron is:
\lambda_{\text{electron}} = \dfrac{h}{\sqrt{2mE}}Now, take the ratio of the two wavelengths:
\dfrac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} = \dfrac{\dfrac{hc}{E}}{\dfrac{h}{\sqrt{2mE}}}Simplifying:
\dfrac{\lambda_{\text{photon}}}{\lambda_{\text{electron}}} = c \dfrac{\sqrt{2mE}}{E} = c \sqrt{\dfrac{2m}{E}}Hence, the correct answer is Option (3).