Question

Derive an expression for de Broglie wavelength associated with an electron accelerated through a potential v.

Answer 1

Similar to the crystal diffraction patterns produced by X-rays, even the beam of electrons of appropriate momentum could produce crystal diffraction pattern. Imparting the desired momentum to the electron obtained by thermionic emission from a heated filament was done by electrostatic acceleration. As electron accelerated through a potential difference of "V" volts, acquires a kinetic energy.
According to work energy principle, work done on the electron appears as the gain in the kinetic energy of the electron
$\therefore eV=\frac{1}{2}m{v}^2$
$\Rightarrow V=\sqrt{\frac{2eV}{m}}$
De-Broglie wavelength associated with moving electron is given by
$\lambda =\frac{h}{mv}=\frac{h}{\sqrt{\frac{2eV}{m}}}$
Substituting $h=6.62\times {10}^{-34}Js,m=9.1\times {10}^{-31}Kgande=1.6\times {10}^{-19}$, we get
$\lambda =\frac{6.62\times {10}^{-34}}{\sqrt{(2\times 9.1\times {10}^{-31}\times 1.6\times {10}^{-19}\times V)}}$
$=\frac{(12.25\times {10}^{-10}}{\sqrt{V}}m$