Estimate the average mass density of a sodium atom assuming its size to be about $2.5 \; \mathring{A}$. (Use the known values of Avogadro's number and the atomic mass of sodium). Compare it with the mass density of sodium in its crystalline phase: $970 \; kg \; m^{-3}$. Are the two densities of the same order of magnitude? If so,why?

(A) Diameter of sodium atom $= 2.5 \; \mathring{A} = 2.5 \times 10^{-10} \; m$.
Radius of sodium atom,$r = 1.25 \times 10^{-10} \; m$.
Volume of one sodium atom,$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.14 \times (1.25 \times 10^{-10})^3 \approx 8.18 \times 10^{-30} \; m^3$.
Mass of one sodium atom,$m = \frac{\text{Atomic mass}}{\text{Avogadro's number}} = \frac{23 \times 10^{-3} \; kg}{6.023 \times 10^{23}} \approx 3.82 \times 10^{-26} \; kg$.
Density of sodium atom,$\rho = \frac{m}{V} = \frac{3.82 \times 10^{-26}}{8.18 \times 10^{-30}} \approx 4.67 \times 10^3 \; kg \; m^{-3}$.
Given density of crystalline sodium $= 970 \; kg \; m^{-3} \approx 10^3 \; kg \; m^{-3}$.
Both densities are of the order of $10^3 \; kg \; m^{-3}$. They are of the same order of magnitude because in the solid phase,atoms are closely packed,but the inter-atomic spacing is comparable to the atomic size.