---
title: "Silver: states per unit volume vs atoms"
subject: "Electrical Engineering Material"
chapter: 2
key: "semester_3/electrical_engineering_material:silver-states-per-volume"
source: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/silver-states-per-volume
site: BCT Engineering
---

# Silver: states per unit volume vs atoms

Electrical Engineering Material · Chapter 2: Free Electron Theory of Conduction in Metal

- Asked in: 2076 Asoj, 4 marks; 2073 Shrawan, 4 marks (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electrical_engineering_material/silver-states-per-volume
- Whole chapter: https://www.bctengineer.com/md/semester_3/electrical_engineering_material/chapter-2

## Question

For silver with EF0 = 5.5 eV and Φ = 4.5 eV, calculate the total number of states per unit volume and compare this with atomic concentration of silver. Density and atomic mass of silver are 10.5 g/cm³ and 107.9 g/mol respectively.

## Short answer

Total states per unit volume up to the vacuum level (10 eV) is S = (16 pi sqrt(2)/3) m^(3/2) E^(3/2)/h^3 = 1.44e29 m^-3. Atomic concentration n_at = rho N_A / M = 5.86e28 m^-3. S/n_at = 2.45, i.e., about 2.5 states per atom.

## Answer

Electrons can be removed from silver once they reach the vacuum level, which lies at $E = E_{F0} + \Phi$ above the bottom of the band. The total number of states per unit volume from the band bottom up to energy $E$ is found by integrating the free-electron density of states:

$$
S(E) = \int_0^{E} g(E')\,dE' = \frac{16\pi\sqrt{2}}{3}\,\frac{m_e^{3/2}}{h^3}\, E^{3/2}
$$

**Given:** $E_{F0} = 5.5\ \text{eV}$, $\Phi = 4.5\ \text{eV}$, $\rho = 10.5\ \text{g/cm}^3 = 10\,500\ \text{kg/m}^3$, $M = 107.9\ \text{g/mol}$.

#### Total states up to the vacuum level

$E = E_{F0} + \Phi = 10\ \text{eV} = 1.602 \times 10^{-18}\ \text{J}$

$$
\begin{aligned}
S &= \frac{16\pi\sqrt{2}}{3} \times \frac{(9.109 \times 10^{-31})^{3/2}}{(6.626 \times 10^{-34})^3} \times (1.602 \times 10^{-18})^{3/2} \\
&= 1.44 \times 10^{29}\ \text{m}^{-3}
\end{aligned}
$$

#### Atomic concentration of silver

$$
n_{at} = \frac{\rho N_A}{M} = \frac{10\,500 \times 6.022 \times 10^{23}}{0.1079} = 5.86 \times 10^{28}\ \text{m}^{-3}
$$

#### Comparison

$$
\frac{S}{n_{at}} = \frac{1.44 \times 10^{29}}{5.86 \times 10^{28}} = 2.45
$$

There are about 2.5 states per atom (states counted with spin) up to the vacuum level. Silver is monovalent, so it has $5.86 \times 10^{28}$ free electrons per m³. At 0 K these fill states only up to $E_{F0}$: $S(5.5\ \text{eV}) = 5.86 \times 10^{28}\ \text{m}^{-3}$, exactly equal to the electron concentration, which confirms $E_{F0} = 5.5$ eV. The remaining states between $E_{F0}$ and the vacuum level (about $8.5 \times 10^{28}\ \text{m}^{-3}$) are empty at 0 K and become partly filled only by thermal excitation, which is why emission needs high temperature.

**Answer:** total states $\approx 1.44 \times 10^{29}\ \text{m}^{-3}$, atomic concentration $= 5.86 \times 10^{28}\ \text{m}^{-3}$; the states are about 2.45 times the number of atoms.

## Check yourself

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

1. Up to what energy above the band bottom are states counted for the total number of states?
   A. E_F0 = 5.5 eV
   B. E_F0 + Φ = 10 eV
   C. Φ = 4.5 eV
   D. E_F0 - Φ = 1 eV
   Answer: B. The vacuum level where electrons can be removed lies at E_F0 + Φ above the bottom of the band. (Key point: Vacuum level energy)

2. How does the number of free-electron states per volume up to energy E vary?
   A. As E^(3/2)
   B. As E^(1/2)
   C. Linearly with E
   D. As E^2
   Answer: A. Integrating g(E) proportional to E^(1/2) gives S proportional to E^(3/2). (Key point: Dependence of S on E)

3. How is the atomic concentration of silver found?
   A. n = M N_A / ρ
   B. n = ρ M / N_A
   C. n = ρ / (M N_A)
   D. n = ρ N_A / M
   Answer: D. Mass density times Avogadro's number divided by molar mass gives atoms per unit volume: 5.86e28 m^-3. (Key point: Atomic concentration)

4. The total states up to the vacuum level compare with the atomic concentration as about
   A. Equal, one state per atom
   B. 0.4 times (fewer states than atoms)
   C. 2.45 times more states than atoms
   D. 10 times more
   Answer: C. S = 1.44e29 and n_at = 5.86e28 give a ratio of 2.45. (Key point: Comparison)

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