Question

An electron moves in an electric field with a kinetic energy of $2.5eV$. What is the associated de Broglie wavelength?

Answer 1

Kinetic energy $=\frac{1}{2}m{v}^2$ $(v=\frac{h}{m\lambda })$

$=\frac{1}{2}m{\left(\frac{h}{m\lambda }\right)}^2$

$=\frac{1}{2}\frac{h^2}{m{\lambda }^2}$

or ${\lambda }^2=\frac{1}{2}\frac{h^2}{m\times KE}$

$\lambda =\frac{h}{\sqrt{2m\times KE}}$ $\left(\begin{array}{c}m=9.108\times {10}^{-28}g\\ h=6.626\times {10}^{-27}erg-sec\\ 1eV=1.602\times {10}^{-12}erg\end{array}\right)$

$=\frac{6.626\times {10}^{-27}}{\sqrt{2\times 9.108\times {10}^{-28}\times 2.5\times 1.602\times {10}^{-12}}}$

$=7.7\times {10}^{-8}cm$.