Consider a system of one-dimensional spinless particles in a box (see Section 5.5) somehow exchanging energy. Through steps similar to those giving equation (9-27). show that

$\mathit{D}\mathbf{\left(}\mathit{E}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{m}}^{\mathbf{1}\mathbf{/}\mathbf{2}}\mathbf{L}}{\mathbf{\hslash }\mathbf{\pi }\sqrt{\mathbf{2}}}\frac{\mathbf{1}}{{\mathbf{E}}^{\mathbf{1}\mathbf{/}\mathbf{2}}}$

The density of states$D\left(E\right)$is defined as the number of different states in the energy interval Eand$E+DE$over the range$D\left(E\right)$. Mathematically, this can be expressed as follows,

$D\left(E\right)=\frac{dn}{dE}$

Here, n is the allowed energies of a particle.

Write the expression for the allowed energies of a particle in a one-dimensional box.

$E=\frac{n^2{\pi }^2{\hslash }^2}{2m{L}^2}$

Here n is the quantum number, $\hslash$is reduced Planck's constant, m is the particle mass, and L is the width of the box.

Rearrangethe above equation.

$n=\frac{L\sqrt{2mE}}{\pi \hslash }$

Take the derivative of both sides.

$\begin{array}{rcl}dn& =& d\left(\frac{L\sqrt{2mE}}{\pi \hslash }\right)\\ & =& \frac{L}{\pi \hslash }·\frac{1}{2}·(2mE{)}^{-1/2}·2m·dE\\ & & \end{array}$

Divide both sides by dE to obtain the density of states corresponding to the particleinabox problem

$\begin{array}{rcl}D\left(E\right)& =& \frac{dn}{dE}\\ & =& \frac{L}{\pi \hslash }·\frac{1}{2}·(2mE{)}^{-1/2}·2m·dE\\ & =& \frac{mL}{\pi \hslash }·\frac{1}{{\left(2mE\right)}^{1/2}}\\ & =& \frac{L}{\hslash \pi }·\frac{1}{\sqrt{2}}·\frac{1}{E^{1/2}}·\frac{m}{m^{1/2}}\\ & =& \frac{m^{1/2}L}{\hslash \pi \sqrt{2}}\frac{1}{E^{1/2}}\\ & & \end{array}$

Thus, the density of states corresponding to the particle inabox problem is$\frac{m^{1/2}L}{\hslash \pi \sqrt{2}}\frac{1}{E^{1/2}}$