Question

Show that de Broglie wavelength of electrons accelerated $V$ volts is very nearly given by:
$\lambda \left(in\mathring{A}\right)={\frac{150}{V}}^{1/2}$.

Answer 1

According to the expression for wavelength:

$\lambda =\frac{h}{\sqrt{2eVm}}$
$\lambda ={[\frac{h^2}{2eVm}\times {10}^{20}]}^{1/2}\mathring{A}$
$={\left[\frac{{(6.626\times {10}^{-34})}^2\times {10}^{20}}{2\times 1.6\times {10}^{-19}\times V\times 9.1\times {10}^{-31}}\right]}^{1/2}={\left[\frac{150}{V}\right]}^{1/2}$.