Question

The de-Broglie wavelength of an electron moving with a velocity $\frac{c}{2}$ (where c is velocity of light in vacuum) is equal to the wavelength of a photon. The ratio of the kinetic energies of electron and photon is:

• A. $1:4$
• B. $1:2$
• C. $1:1$
• D. $2:1$

Answer 1

Answer: (A)

$1:4$

De-broglie wavelength of the electron $=\frac{h}{m_{e}v}=\frac{h}{m_e\left(c/2\right)}$

Wavelength of photon $=\lambda$(say)

Thus, $\lambda =\frac{h}{m_e\left(c/2\right)}$

Ration of kinetic energies $=\frac{\frac{1}{2}{m}_e(\frac{c}{2}{)}^2}{h\frac{c}{\lambda }}=\frac{\frac{1}{2}{m}_e(\frac{c}{2}{)}^2}{m_e\frac{c^2}{2}}=1:4$