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 $K=\frac{1}{2}mv^{2}=\frac{\left(mv\right)^{2}}{2\,m}$
$\Rightarrow mv=\sqrt{2\,mK}$
Thus, $\lambda =\frac{h}{\sqrt{2\,mK}}$
$\Rightarrow \lambda \,\propto\, \frac{1}{\sqrt{K}}$
$\therefore \frac{\lambda_{2}}{\lambda_{1}}=\frac{\sqrt{K_{1}}}{\sqrt{K_{2}}}=\frac{\sqrt{K_{1}}}{\sqrt{2\,K_{1}}}\left(\because K_{2}=2K_{1}\right)$
$\Rightarrow \frac{\lambda _{2}}{\lambda _{1}}=\frac{1}{\sqrt{2}}$
$\therefore \lambda_{2}=\frac{\lambda_{1}}{\sqrt{2}}$