Question

The kinetic energy of an electron get tripled then the de-Broglie wavelength associated with electron changes by a factor of

• A. $\frac{1}{3}$
• B. $\sqrt{3}$
• C. $\frac{1}{\sqrt{3}}$
• D. $3$

Answer 1

Answer: (C)

$\frac{1}{\sqrt{3}}$

de-Broglie wavelength of an electron $\lambda =\frac{h}{\sqrt{2mK}}$

Or $\lambda \propto \frac{1}{\sqrt{K}}$

where $K$ is the kinetic energy of the electron and $m$ is the mass.
$\therefore \frac{{\lambda }^{\prime }}{\lambda }=\frac{1}{\sqrt{3K}}.\frac{\sqrt{K}}{1}=\frac{1}{\sqrt{3}}$
Or ${\lambda }^{\prime }=\frac{\lambda }{\sqrt{3}}$
i.e., de-Broglie wavelength will decrease by a factor of $\frac{1}{\sqrt{3}}$.