Question

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

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

Answer 1

The correct option is A

$\frac{1}{\sqrt{2}}$

$\lambda =\frac{h}{\sqrt{2mE}}\Rightarrow \lambda \propto \frac{1}{\sqrt{E}}$.