Question

The de-Brogile wavelength of an electron moving with a velocity $1.5\times {10}^8{\text{ms}}^{-1}$ is equal to that of a photon. The ratio of the kinetic energy of the electron to that of the energy of photon is (apply non-relativistic for electron)

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

Answer 1

The correct option is D $\frac{1}{4}$

Given, $v=1.5\times {10}^8{\text{ms}}^{-1}$

The energy of a photon is,

$E=\frac{hc}{\lambda }........\left(1\right)$

The de-Broglie wavelength of the electron is,

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

$\Rightarrow p=\frac{h}{\lambda }$

$\because KE=\frac{1}{2}m{v}^2=\frac{1}{2}pv=\frac{hv}{2\lambda }........\left(2\right)$

From (1) and (2) we get,

$\frac{KE}{E}=\frac{\frac{hv}{2\lambda }}{\frac{hc}{\lambda }}=\frac{v}{2c}$

$=\frac{1.5\times {10}^8}{6\times {10}^8}=\frac{1}{4}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.