Silver: states per unit volume vs atoms

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.