Question:SSM Show that the angular wave number $\mathbf{k}$for a nonrelativistic free particle of mass $\mathbf{m}$can be written as $\mathbf{k}\mathbf{=}\frac{\mathbf{2}\mathbf{\pi }\sqrt{\mathbf{2}\mathbf{mK}}}{\mathbf{h}}$ in which $\mathbf{K}$is the particle’s kinetic energy

The angular wave number of the particle is $k$.

The mass of the particle is m.

The kinetic energy of the particle is K.

Wave number

The spatial frequency of a wave is known as the wavenumber in the physical sciences. It is measured in cycles per unit distance (common wavenumber) or radians per unit distance (angular wavenumber)

We know that the angular wave number $k$is related to the wavelength

$k=\frac{2\mathrm{\pi }}{\lambda }$

Suppose the particle momentum is $p$.

The wavelength $\lambda$of the particle according to de-Broglie is

$\lambda =\frac{h}{p}$

So, wave number

$\begin{array}{l}k=\frac{2\pi }{h/p}\\ k=\frac{2\mathrm{\pi p}}{h}\end{array}$

Where $h$is plank’s constant

The particle kinetic energy in terms of momentum and mass can be given as

$\begin{array}{l}K=\frac{p^2}{2m}\\ p=\sqrt{2mK}\end{array}$

Substituting the value of $p=\sqrt{2mK}$into the wave number formula

Therefore, the wave number of the particles

$k=\frac{2\mathrm{\pi }\sqrt{2\mathrm{mK}}}{h}$

Hence it is shown that the angular wave number k for a nonrelativistic free particle can be written as $k=\frac{2\mathrm{\pi }\sqrt{2\mathrm{mK}}}{h}$.