Calculating 3D Density of States for a Dispersion Relation | Homework Solution

[Kara386]

Homework Statement

Calculate the single-particle density of states $g(\epsilon)$ for the dispersion relation $\epsilon(k) = ak^{\frac{3}{2}}$ in 3D. Use $g(k) = \frac{Vk^2}{2\pi^2}$.

Homework Equations

The Attempt at a Solution

This question is worth lots of marks. My solution is a few lines, so it must be wrong, but I have no idea why. So rearranging the dispersion relation gives
$k = (\frac{\epsilon}{a})^{\frac{3}{2}}$, so that $dk = \frac{3}{2a} \epsilon^{0.5} d\epsilon$.

$g(k)dk = \frac{Vk^2}{2 \pi^2} dk$
Sub in the expressions for k and dk:
$g(\epsilon)d\epsilon = \frac{V\epsilon^6}{2\pi^2a^6} \frac{3}{2a}\epsilon^{\frac{1}{2}}$
$= \frac{3V\epsilon^{\frac{13}{2}}}{4\pi^2a^7} d\epsilon$
And then you can look at that and go
$g(\epsilon) = \frac{3V\epsilon^{\frac{13}{2}}}{4\pi^2a^7}$

I've clearly missed something massive or a few massive things, so any help is very much appreciated! If it looks ok let me know, but I'm slightly paranoid that I'm getting more marks for this than something that required two pages of workings. The only thing I can think is that the expressions I've been given aren't for 3D but looking on the internet they do seem to be 3D.

 

[TSny]

So rearranging the dispersion relation gives
$k = (\frac{\epsilon}{a})^{\frac{3}{2}}$

You made a mistake in solving for $k$ as a function of $\epsilon$.

Otherwise, I believe your general method is correct.