Number of Atoms in Face Centred Cubic Crystal Structure

To determine the value of $N$, we can calculate the total number of unit cells present in the given mass of the crystal and then use the number of atoms per unit cell for a face-centred cubic (FCC) lattice.

1. Calculate the Volume of One Unit Cell ($V_{\text{cell}}$): The unit cell edge length is given as $a = 400\text{ pm}$. Converting this to centimeters: $a = 400 \times 10^{-10}\text{ cm} = 4 \times 10^{-8}\text{ cm}$

   The volume of a cubic unit cell is: $V_{\text{cell}} = a^3 = (4 \times 10^{-8}\text{ cm})^3 = 64 \times 10^{-24}\text{ cm}^3$

2. Calculate the Total Volume of $256\text{ g}$ of the Crystal ($V_{\text{total}}$): Given density $\rho = 8\text{ g cm}^{-3}$ and mass $m = 256\text{ g}$: $V_{\text{total}} = \frac{\text{Mass}}{\text{Density}} = \frac{256\text{ g}}{8\text{ g cm}^{-3}} = 32\text{ cm}^3$

3. Calculate the Total Number of Unit Cells ($n_{\text{cells}}$): $n_{\text{cells}} = \frac{V_{\text{total}}}{V_{\text{cell}}} = \frac{32\text{ cm}^3}{64 \times 10^{-24}\text{ cm}^3} = 0.5 \times 10^{24}$

4. Calculate the Total Number of Atoms ($N_{\text{total}}$): For a face-centred cubic (FCC) structure, the effective number of atoms per unit cell ($Z$) is $4$: $Z = 8 \times \left(\frac{1}{8}\right) + 6 \times \left(\frac{1}{2}\right) = 4$

   Therefore, the total number of atoms in the crystal is: $N_{\text{total}} = Z \times n_{\text{cells}} = 4 \times (0.5 \times 10^{24}) = 2 \times 10^{24}$

Comparing $N_{\text{total}} = 2 \times 10^{24}$ with $N \times 10^{24}$, we get: $N = 2$