In Introduction to solid states physics Eighth Edition by Kittel Page 141 Eq 20. and Eq 21.

The density of states for electron in conduction in 3D $$D(\epsilon)\equiv \frac{dN}{d\epsilon}=\frac{V}{2\pi^2}(\frac{2m}{\hbar^2})\epsilon^{1/2}=\frac{3}{2} \frac{N}{\epsilon}$$

The same argument could apply such that in 2D $$D(\epsilon)= \frac{2}{2}\frac{N}{\epsilon}$$ and in 1D $$D(\epsilon)= \frac{1}{2}\frac{N}{\epsilon}$$

i.e. with each increase of dimension, the density of states increase by $\frac{1}{2}\frac{N}{\epsilon}$.

1. Why does each dimension increase the density of states $D(\epsilon)$ by $\frac{1}{2}\frac{N}{\epsilon}$?

2. What's the order of $D(\epsilon)$ in 3D, $\epsilon^{1/2}$ or $\epsilon^{-1}$?

In Introduction to Solid State Physics, eighth edition, by Kittel, page 141, eqs. (20,21), the density of states for electron in conduction in three dimensions is $$D(\epsilon)\equiv \frac{dN}{d\epsilon}=\frac{V}{2\pi^2}\left(\frac{2m}{\hbar^2}\right)\epsilon^{1/2}=\frac{3}{2} \frac{N}{\epsilon}.$$

The same argument could apply such that in two dimensions $$D(\epsilon)= \frac{2}{2}\frac{N}{\epsilon},$$ and in one dimension $$D(\epsilon)= \frac{1}{2}\frac{N}{\epsilon}$$ —i.e. with each increase of dimension, the density of states increase by $\frac{1}{2}\frac{N}{\epsilon}$.

1. Why does each dimension increase the density of states $D(\epsilon)$ by $\frac{1}{2}\frac{N}{\epsilon}$?

2. What's the order of $D(\epsilon)$ in 3D, $\epsilon^{1/2}$ or $\epsilon^{-1}$?