91Ó°ÊÓ

In this problem, we are given the relationship between the energy$E$and the quantum state$n$:

$E=b{n}^{2/3}\text{,}$

Where$b$ is constant. Our task is to determine whether the density of states $D\left(E\right)$is increasing, decreasing, or independent of $E$.

Given the quantized energy equation above, let us first express $n$in terms of $E$:

$\begin{array}{l}E=b{n}^{2/3}\\ n^{2/3}=\frac{E}{b}\\ n={\left(\frac{E}{b}\right)}^{3/2}\end{array}$

Now that we have $n$in terms of$E$, let us take the derivative of both sides:

$\begin{array}{l}d\left(n\right)=d{\left(\frac{E}{b}\right)}^{3/2}\\ dn=\frac{3}{2}\frac{E^{1/2}}{b^{3/2}}dE\\ =\frac{3}{2}\sqrt{\frac{E}{b^3}}dE\end{array}$

Dividing both sides by $dE$we obtain the expression for the density of states $D\left(E\right)$associated with the energy relation above:

$\begin{array}{l}\frac{dn}{dE}=\frac{3}{2}\sqrt{\frac{E}{b^3}}\\ \frac{dE}{dE}=\frac{3}{2}\sqrt{\frac{E}{b^3}}\\ D\left(E\right)=\frac{3}{2}\sqrt{\frac{E}{b^3}}\end{array}$

Based on our above expression for $D\left(E\right)$, the density of states is thus an increasing function of $E$. More accurately, the density of states is directly proportional to $E^{1/2}$.