Question

If the kinetic energy of a free electron doubles, it’s de-Broglie wavelength changes by the factor

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

Answer 1

Answer: (D)

de-Broglie wavelength,
$\lambda =\frac{h}{p}=\frac{h}{\sqrt{2.m.\left(K.E\right)}}$
$\therefore \lambda \propto \frac{1}{\sqrt{K.E}}$
If K.E is doubled, then de-Broglie wavelength becomes $\frac{\lambda }{\sqrt{2}}$