Question

How is the de-Broglie wavelength associated with an electron accelerated through a potential difference of $100$ volts?

Answer 1

Using energy conservation, $KE=PE$ or $\frac{p^2}{2m}=eV$ or $p=\sqrt{2meV}$ where $p=$ momentum of electron, $m=$ mass of electron, $e=$ charge of electron and $V=$ potential difference.

Now, de-Broglie wavelength $\lambda =\frac{h}{p}=\frac{h}{\sqrt{2meV}}=\frac{6.6\times {10}^{-34}}{\sqrt{2(9.1\times {10}^{-31})(1.6\times {10}^{-19})\left(100\right)}}=1.22\times {10}^{-10}m$