Question

The de Broglie wavelength of an electron moving with a velocity 0f $1.5X{10}^8$ m/s is equal to that of a photon. The ratio of kinetic energy of that electron to the photon is.

[IIT JEE 2004]

• A. 2
• B. 4
• C. 1/2
• D. 1/4

Answer 1

The correct option is D

1/4

Given ${\lambda }_e={\lambda }_p$
$\Rightarrow \frac{h}{m_{e}v}=\frac{hc}{E}$ …(1)
(∵ P = $\frac{h}{\lambda },$ for electron
E = $\frac{hc}{\lambda },$ for photon)

From $e{q}^n\left(1\right)E=m_{e}vc$ ……(II)

Now ratio of electron energy to photon = $\frac{\frac{1}{2}{m}_e{V}^2}{E}$
from $e{q}^n\left(II\right)$
$\Rightarrow \frac{\frac{1}{2}{m}_e{V}^2}{m_{e}VC}$
$\Rightarrow \frac{1}{2}\frac{V}{C}=\frac{1}{2}\frac{1.5x{10}^8}{3x{10}^8}$
$\Rightarrow \frac{1}{4}$