Question

Calculate the number of states per cubic metre of sodium in 3s band. The density of sodium is 1013 kgm⁻³. How many of them are empty?

Answer 1

Density of sodium, d = 1013 \[\text{ kg/ m}^3\]

Volume, V = 1m³ 

Mass of sodium, m = d × V = 1013 × 1 = 1013 kg

 

Molecular mass of sodium, M = 23
We know that 23 g sodium contains 6 atoms, so the number of atoms in 1023 kg sodium will be

\[\frac{1013 \times {10}^3 \times 6 \times {10}^{23}}{23}\] 

\[ = \left( \frac{1013 \times 6}{23} \right) \times  {10}^{26} \] 

\[ = 264 . 26 \times  {10}^{26}\]

(a) As the number of maximum possible electrons that can occupy the 3s band is 2, the total number of states in the 3s band will be

\[N = 2 \times 264 . 26 \times  {10}^{26} \] 

\[ = 528 . 52 \times  {10}^{26}  \approx 5 . 3 \times  {10}^{28}\]

(b) As the atomic number of sodium is 11, its electronic configuration is \[1 s^2 , 2 s^2 , 2 p^6 , 3 s^1\] .

This implies that the 3s band is half-filled in case of sodium, so the total number of unoccupied states is 2.65 × 10²⁸.