Question

The de-Broglie wavelength of an electron moving with a velocity $ 1.5\times 10^{8}m/s $ is equal to that of a photon. The ratio of the kinetic energy of the electron to the energy of the photon is

• A. $ \frac{1}{4} $
• B. $ \frac{1}{2} $
• C. 2
• D. 4

Answer 1

Answer: (A)

$K_{e}=\frac{1}{2} m v^{2}$ and
$\lambda=\frac{h}{m v} K_{e}=\frac{1}{2}\left(\frac{h}{\lambda v}\right) \cdot v^{2}$ $=\frac{v h}{2 \lambda}$
$K_{p}=\frac{h c}{\lambda}$
$\therefore \frac{K_{e}}{K_{p}}=\frac{v}{2 c}=\frac{1.5 \times 10^{8}}{2 \times 3 \times 10^{8}}$
$=\frac{1}{4}$