Using the classical equations for momentum and kinetic energy, show that an electron’s de Broglie wavelength in nanometres can be written as$\mathit{\lambda }\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{226}\mathbf{/}\sqrt{\mathbf{K}}$, in which Kis the electron’s kinetic energy in electron-volts.

The classical equation of momentum is,$p=mv$

The kinetic energy is,$K=\frac{1}{2}m{v}^2$

The expression to calculate the kinetic energy of the electron is given as follows.

$K=\frac{1}{2}m{v}^2$ …(i)

Here, m is the mass of the electron and v is the velocity of electron.

The expression to calculate the de Broglie wavelength is given as follows.

$\lambda =\frac{h}{p}$ …(ii)

Here, h is the plank’s constant and$p$ is the momentum.

Calculate the kinetic energy of the electron,

Substitute$\rho /m$ for v into equation (i).

$\begin{array}{rcl}K& =& \frac{1}{2}m{\left(\frac{p}{m}\right)}^2\\ K& =& \frac{1}{2}\frac{p^2}{m}\\ p^2& =& 2Km\\ p& =& \sqrt{2Km}\end{array}$

Calculate the de Broglie wavelength.

Substitute$\sqrt{2Km}$ for$p$ into equation (ii).

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

Substitute $6.62\times {10}^{-34}\text{Js}$ for h and $9.11\times {10}^{-31}\text{kg}$for m into above equation.

$\begin{array}{rcl}\lambda & =& \frac{6.62\times {10}^{-34}\mathrm{Js}}{\sqrt{2\times \left(9.11\times {10}^{-31}\mathrm{kg}\right)\times \left(1.602\times {10}^{-19}\mathrm{J}/\mathrm{eV}\right)\times K}}\\ & =& \frac{6.62\times {10}^{-34}\mathrm{Js}}{\sqrt{29.1884\times {10}^{-50}\mathrm{kgJ}/\mathrm{eV}}}\frac{1}{\sqrt{K}}\\ & =& \frac{1.226\times {10}^{-9}{\mathrm{meV}}^{{\displaystyle \frac{1}{2}}}}{\sqrt{K}}\end{array}$