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If the kinetic energy of a free electron doubles, it's de-Broglie wavelength changes by the factor
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2233 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{\sqrt{2}}$
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 $\frac{\lambda}{\sqrt{2}}$
$$
\begin{array}{l}
\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \cdot \mathrm{m} \cdot(\mathrm{K} \cdot \mathrm{E})}} \\
\therefore \lambda \propto \frac{1}{\sqrt{\mathrm{K} \cdot \mathrm{E}}}
\end{array}
$$
If $\mathrm{K} . \mathrm{E}$ is doubled, then de-Broglie wavelength becomes $\frac{\lambda}{\sqrt{2}}$
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