Density of states in a 10 eV band

Treat the band like a free-electron band, with energy measured from the bottom of the band. The density of states per unit volume per unit energy is

$$g(E) = \frac{8\pi\sqrt{2}\, m^{3/2}}{h^3}\, E^{1/2}$$

Given: band width $= 10\ \text{eV}$, so the centre is at $E = 5\ \text{eV} = 8.01 \times 10^{-19}\ \text{J}$. Take $T = 300\ \text{K}$.

i) Density of states at the centre of the band

$$\begin{aligned} g(5\ \text{eV}) &= \frac{8\pi\sqrt{2}\,(9.109 \times 10^{-31})^{3/2}}{(6.626 \times 10^{-34})^3} \times (8.01 \times 10^{-19})^{1/2} \\ &= 9.51 \times 10^{46}\ \text{J}^{-1}\,\text{m}^{-3} \\ &= 9.51 \times 10^{46} \times 1.602 \times 10^{-19} = 1.52 \times 10^{28}\ \text{eV}^{-1}\,\text{m}^{-3} \end{aligned}$$

ii) Number of states per unit volume in a range kT above the centre

At 300 K, $kT = \frac{1.381 \times 10^{-23} \times 300}{1.602 \times 10^{-19}} = 0.0259\ \text{eV}$. Since $kT \ll 5\ \text{eV}$, $g(E)$ is almost constant over this small range, so

$$\Delta N \approx g(E)\,\Delta E = 1.52 \times 10^{28} \times 0.0259 = 3.94 \times 10^{26}\ \text{m}^{-3}$$

Answer: (i) $g = 9.51 \times 10^{46}\ \text{J}^{-1}\text{m}^{-3} = 1.52 \times 10^{28}\ \text{eV}^{-1}\text{m}^{-3}$; (ii) about $3.94 \times 10^{26}$ states per m³ (at 300 K).