---
title: "Density of states in a 10 eV band"
subject: "Electrical Engineering Material"
chapter: 1
key: "semester_3/electrical_engineering_material:density-of-states-band-numerical"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/density-of-states-band-numerical
site: BCT Engineering
---

# Density of states in a 10 eV band

Electrical Engineering Material · Chapter 1: Theory of Metals

- Asked in: 2071 Shrawan, 4 marks
- Video: [Density of States Derivation Part 1](https://www.youtube.com/watch?v=z7YGS67GETo) — Jordan Edmunds Chetty
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/density-of-states-band-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-1

## Question

The width of energy band is typically 10ev calculate: i) The density of states at the center of the band ii) The number of states per unit volume within a small energy range KT above the center.

## Short answer

(i) Using g(E) = (8π√2 m^{3/2}/h³) E^{1/2} at E=5 eV gives g = 1.52×10^{28} eV^{-1} m^{-3}. (ii) With kT ≈ 0.0259 eV at 300 K, the number of states per unit volume within kT above the center is ΔN ≈ g·kT = 3.94×10^{26} m^{-3}.

## Answer

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).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How does g(E) vary with E for a free-electron band?
   A. Proportional to E^2
   B. Proportional to sqrt(E)
   C. Constant
   D. Proportional to 1/E
   Answer: B. g(E) = 8 pi sqrt(2) m^1.5 E^0.5 / h^3. (Key point: g(E) formula)

2. For a 10 eV band measured from its bottom, at what E is g evaluated for the centre?
   A. 10 eV
   B. 5 eV
   C. 2.5 eV
   D. 0 eV
   Answer: B. Centre is at half the width, 5 eV. (Key point: Centre of band)

3. How is the number of states within kT above the centre found?
   A. Delta N = g(E) x kT, treating g as nearly constant
   B. Delta N = g(E)/kT
   C. Delta N = g(E) x 10 eV
   D. Delta N = g(E)^2 kT
   Answer: A. Since kT (0.0259 eV) is much smaller than 5 eV, Delta N = g Delta E = 1.52e28 x 0.0259 = 3.94e26 per m^3. (Key point: Number of states in kT)

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