Determine the number of states per cubic metre in the 3s band of...

The number of sodium atoms per cubic metre is given by:

$n = \dfrac{\rho N_A}{M}$

Substituting the given values, $\rho = 1013 \text{ kg m}^{-3}$, $N_A = 6.022 \times 10^{23} \text{ mol}^{-1}$, and $M = 23 \times 10^{-3} \text{ kg mol}^{-1}$:

$n = \dfrac{1013 \times 6.022 \times 10^{23}}{23 \times 10^{-3}} \approx 2.65 \times 10^{28} \text{ m}^{-3}$

Each sodium atom contributes one $3s$ orbital to the $3s$ band. Since each orbital can accommodate $2$ electrons (spin up and spin down), the total number of states per cubic metre in the $3s$ band is:

$N_{\text{states}} = 2n = 2 \times 2.65 \times 10^{28} = 5.3 \times 10^{28}$

Sodium has one valence electron per atom in the $3s$ state. Thus, the number of occupied states per cubic metre is equal to the number of atoms, $n$.

The number of empty states per cubic metre is the total number of states minus the number of occupied states:

$N_{\text{empty}} = 2n - n = n = 2.65 \times 10^{28}$