Question

If the deBroglie wavelength of an electron is equal to ${10}^{-3}$ times the wavelength of a photon of frequency $6\times {10}^{14}\text{Hz},$ then the speed of an electron is equal to:
$\text{(Speed of light,}c=3\times {10}^8{\text{ms}}^{-1}\text{Planck's constant,}h=6.63\times {10}^{-34}\text{Js}\text{Mass of electron,}{m}_e=9.1\times {10}^{-31}\text{kg})$

• A. $1.1\times {10}^6{\text{ms}}^{-1}$
• B. $1.7\times {10}^6{\text{ms}}^{-1}$
• C. $1.8\times {10}^6{\text{ms}}^{-1}$
• D. $1.45\times {10}^6{\text{ms}}^{-1}$

Answer 1

The correct option is D $1.45\times {10}^6{\text{ms}}^{-1}$

Given that, the de-Broglie wavelength,
$\lambda ={10}^{-3}\left(\frac{3\times {10}^8}{6\times {10}^{14}}\right)[\because {\lambda }_p=\frac{c}{\nu }]$

The deBroglie wavelength is given by,

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

$\Rightarrow v=\frac{6.63\times {10}^{-34}\times 6\times {10}^{14}}{9.1\times {10}^{-31}\times 3\times {10}^5}$

$\Rightarrow v=1.45\times {10}^6{\text{ms}}^{-1}$

Hence, option $\left(D\right)$ is correct.