Question

If the kinetic energy of a free electron doubles, its de-Broglie wavelength changes by the factor:

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

Answer 1

Answer: (C)

We know
$\lambda =\frac{h}{mv}$
and $K=\frac{1}{2}m{v}^2=\frac{{\left(mv\right)}^2}{2m}$
$\Rightarrow mv=\sqrt{2mK}$
Thus, $\lambda =\frac{h}{\sqrt{2mK}}$
$\Rightarrow \lambda \propto \frac{1}{\sqrt{K}}$
$\therefore \frac{{\lambda }_2}{{\lambda }_1}=\frac{\sqrt{K_1}}{\sqrt{K_2}}=\frac{\sqrt{K_1}}{\sqrt{2K_1}}\left(\because K_2=2K_1\right)$
$\Rightarrow \frac{{\lambda }_2}{{\lambda }_1}=\frac{1}{\sqrt{2}}$
$\therefore {\lambda }_2=\frac{{\lambda }_1}{\sqrt{2}}$