Question 38. The de-Broglie wavelength of an electron is the same as that of a photon. The velocity of the electron is 25% of the velocity of light. The ratio of kinetic energy of the electron to the energy of the photon is :
(1) 1:1
(2) 1:8
(3) 8:1
(4) 1:4
Answer: (2)
Explanation:
For a photon, the energy is given by E_{p}=\frac{hc}{\lambda_{p}}
So, \lambda_{p}=\frac{hc}{E_{p}}
For an electron, the de-Broglie wavelength is
\lambda_{e}=\frac{h}{m_{e}v_{e}}The kinetic energy of the electron is K_{e}=\frac{1}{2}m_{e}v_{e}^{2}
So, \lambda_{e}=\frac{hv_{e}}{2K_{e}}
Given, the velocity of the electron is v_{e}=0.25c
Substitute this value,
\lambda_{e}=\frac{h\times0.25c}{2K_{e}} \lambda_{e}=\frac{hc}{8K_{e}}Since the de-Broglie wavelength of the electron is equal to the wavelength of the photon,
\lambda_{e}=\lambda_{p}So, \frac{hc}{E_{p}}=\frac{hc}{8K_{e}}
Cancel common terms, \frac{K_{e}}{E_{p}}=\frac{1}{8}
Hence, the required ratio is 1:8.