Question

The de-Broglie wavelength of a free electron with kinetic energy ${}^{\prime }{E}^{\prime }$ is $\lambda$. If the kinetic energy of the electron is doubled, the de-Broglie wavelength is:

• A. $\frac{\lambda }{\sqrt{2}}$
• B. $\sqrt{2}\lambda$
• C. $\frac{\lambda }{2}$
• D. $2\lambda$

Answer 1

The correct option is A $\frac{\lambda }{\sqrt{2}}$

Kinetic energy initially is given by:

$E=\frac{1}{2}m{v}^2=\frac{p^2}{2m}$

$⟹p=\sqrt{2mE}$

where

$m:$ Mass of the particle

$v:$ Speed of the particle

$p:$ Momentum of the particle

De-broglie wavelength is given by,

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

$\therefore \lambda =\frac{h}{\sqrt{2mE}}$

Given $E_2=2E_1$

$⟹{\lambda }_2=\frac{{\lambda }_1}{\sqrt{2}}$

Given ${\lambda }_1=\lambda ,\therefore {\lambda }_2=\frac{\lambda }{\sqrt{2}}$