The de-Broglie wavelength of an electron moving with a velocity `1.5 xx 10^(8)ms^(-1)` is equal to that of a photon. The ratio of the kinetic energy of the electron to that of the photon is:

Options

A. 2
B. 4
C. `(1)/(2)`
D. `(1)/(4)`

Answer

D

Explanation

`lambda_(d) = (h)/(m v)`
`E_(lambda)` = energy of photon `= (h c)/(lambda) = m v c`
Energy of electron `= (1)/(2) m v^(2)`
The required ratio `= ((1)/(2) mv^(2))/(m v c) = (1)/(2) (v)/(c) = (1)/(4)`