Question

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

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

Answer 1

The correct option is D $\frac{1}{\sqrt{2}}$

$\lambda =\frac{h}{\sqrt{2mE}}\therefore \lambda \propto \frac{1}{\sqrt{E}}\Rightarrow \frac{{\lambda }_2}{{\lambda }_1}=\sqrt{\frac{E_1}{E_2}}=\sqrt{\frac{E}{2E}}=\frac{1}{\sqrt{2}}{\lambda }_2=\frac{1}{\sqrt{2}}\lambda$