Q83 / 120 · VITEEE 2019 · Physics

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
$\sqrt{2}$
Correct answer
D
$\frac{1}{\sqrt{2}}$

Solution

de-Broglie wavelength, $$ \begin{array}{l} \lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \cdot \mathrm{m} \cdot(\mathrm{K} \cdot \mathrm{E})}} \\ \therefore \lambda \propto \frac{1}{\sqrt{\mathrm{K} \cdot \mathrm{E}}} \end{array} $$ If $\mathrm{K} . \mathrm{E}$ is doubled, then de-Broglie wavelength becomes $\frac{\lambda}{\sqrt{2}}$

Hence, the correct answer is option: 'D'.

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