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If the potential difference used to accelerate electrons is doubled, by what factor does the de-Broglie wavelength associated with the electrons change?
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The correct answer is:
Wavelength is decreased to $\frac{1}{\sqrt{2}}$ times.
The de Broglie wavelength is given by
$\lambda(\text { in } \mathrm{nm})=\frac{1.23}{\sqrt{V}}$
Hence it becomes $\frac{1}{\sqrt{2}}$ times
$\lambda(\text { in } \mathrm{nm})=\frac{1.23}{\sqrt{V}}$
Hence it becomes $\frac{1}{\sqrt{2}}$ times
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