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If the potential difference used to accelerate electrons is doubled, by what factor does the de-Broglie wavelength associated with electrons change?
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The correct answer is:
Wavelength in decreased to $\frac{1}{\sqrt{2}}$ times.
From $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mqV}}}$,
$\lambda \propto \frac{1}{\sqrt{V}}$
If potential difference is doubled, $\lambda \propto \frac{1}{\sqrt{2 \mathrm{~V}}}$
$\therefore \quad \lambda$ is decreased by $\frac{1}{\sqrt{2}}$ times.
$\lambda \propto \frac{1}{\sqrt{V}}$
If potential difference is doubled, $\lambda \propto \frac{1}{\sqrt{2 \mathrm{~V}}}$
$\therefore \quad \lambda$ is decreased by $\frac{1}{\sqrt{2}}$ times.
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