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Question
An electron is accelerated from rest through a potential V. Obtain the expression for the de-Broglie wavelength associated with it ?
Solution
Energy of electron, E = eV
Momentum,
\[p = \sqrt{2mE} = \sqrt{2meV}\]
de-Broglie wavelength,
\[\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}\]
∴\[\lambda = \frac{h}{\sqrt{2meV}}\]
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