Question
de Broglie wavelength associated with an electron accelerated through a potential difference V is $\lambda$ What will be the de Broglie wavelength when the accelerating potential is increased to 4V?
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Answer
$\frac{\gamma}{2}$Reason: de Broglie wavelength associated with electron is,
$\lambda=\frac{\text{h}}{\sqrt{2\text{meV}}}\Rightarrow\ \lambda\propto\frac{1}{\sqrt{\text{V}}}$
Obviously when accelerating potential becomes 4V, the de-Broglie wavelength reduces to half.
$\lambda=\frac{\text{h}}{\sqrt{2\text{meV}}}\Rightarrow\ \lambda\propto\frac{1}{\sqrt{\text{V}}}$
Obviously when accelerating potential becomes 4V, the de-Broglie wavelength reduces to half.
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