Question

The de Broglie wavelength of an electron moving with a velocity $2.25\times {10}^8$ $m{s}^{-1}$ is equal to that of a photon. The ratio of the kinetic energy of the electron to that of the energy of photon is :

• A. $\frac{3}{8}$
• B. $4$
• C. $\frac{1}{2}$
• D. $2-\sqrt{3}$

Answer 1

Answer: (A)

$\frac{3}{8}$

$K.E_{particle}=\frac{1}{2}m{v}^2$

Also,

$\lambda =\frac{h}{mv}$

$⟹K.E_{particle}=\frac{1}{2}\left(\frac{h}{\lambda v}\right)v^2=\frac{vh}{2\lambda }$ ............(i)

$K.E_{photon}=\frac{hc}{\lambda }$ ..............(ii)

Therefore,

$\frac{K.E_{particle}}{K.E_{photon}}=\frac{v}{2c}=\frac{2.25\times {10}^8}{2\times 3\times {10}^8}=\frac{3}{8}$