Question

Calculate de Broglie wavelength associated with an electron, accelerated through a potential difference of $400V$.

Answer 1

The de-Broglie wavelength of electron accelerated through potential difference $V$ is

$\lambda =\frac{h}{\sqrt{2mqV}}$

$=\frac{6.626\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 1.6\times {10}^{-19}\times 400}}m$

$=6.14\times {10}^{-11}m$