Question

An electron is accelerated from rest through a potential V. Obtain the expression for the de-Broglie wavelength associated with it ?

Answer 1

Energy of electron, E = eV
Momentum,

\[p = \sqrt{2mE} = \sqrt{2meV}\]

de-Broglie wavelength,

\[\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}\]

∴\[\lambda = \frac{h}{\sqrt{2meV}}\]