Question

If the de Broglie wavelength of an electron is equal to ${10}^3$ times the wavelength of a photon of frequency $6\times {10}^{14}$ Hz, then the speed of the electron is equal to (Speed of light = $3\times {10}^{8}m/s$, Planck's constant = $6.63\times {10}^{34}$. Mass of electron = $9.{110}^{31}$ kg)

• A. $1.8\times {10}^{6}m/s$
• B. $1.45\times {10}^{6}m/s$
• C. $1.1\times {10}^{6}m/s$
• D. $1.7\times {10}^{6}m/s$

Answer 1

The correct option is B

$1.45\times {10}^{6}m/s$

Step 1: Given data

Speed of light = $3\times {10}^{8}m/s$

Planck's constant = $6.63\times {10}^{34}J$

Mass of electron = $9.{110}^{31}kg$

Step 2: Formula used

The wavelength of the given photon is given as,

${\lambda }_{\rho }=\frac{C}{V_{\rho }}$

Step 3: Calculation

Putting the values in the formula, we get,

${\lambda }_{\rho }=\frac{C}{V_{\rho }}$

$=\frac{3\times {10}^8}{6\times {10}^{14}}m$

$=5\times {10}^{-7}m$ ……….(1)

As it is given that, a wavelength of the electron is

${\lambda }_e={10}^{-3}\times {\lambda }_{\rho }$ [Use equation (1)]

$=5\times {10}^{-10}m$

Also, the wavelength of an electron is given as,

${\lambda }_e=\frac{h}{\rho }=\frac{h}{m{v}_e}$

$v_e=\frac{h}{{\lambda }_e{m}_e}$

Substituting the given value, we get

$=\frac{6.63\times {10}^{-34}}{5\times {10}^{-10}\times 9.1\times {10}^{-31}}m/s$

$=1.45\times {10}^{6}m/s$

The speed of the electron is equal to $1.45\times {10}^{6}m/s$.