Question

Calcium crystallizes in an fcc lattice of unit cell length 5.56 Å and density 1.4848 g cm$^{-3}$. The percentage of Schottky defects (rounded off to one decimal place) in the crystal is ________.
(Given: Atomic mass of Ca = 40 g mol$^{-1}$, $N_A = 6.022 \times 10^{23}$ mol$^{-1}$)

Hint:

A decrease in measured density compared to theoretical value indicates the presence of Schottky defects (vacancies).

Answer 1

Correct Answer: 3.9

Step 1: Density of a perfect crystal.
For fcc structure, $Z = 4$. \[ \rho = \frac{Z M}{a^3 N_A} \] \[ \rho_\text{calc} = \frac{4 \times 40}{(5.56 \times 10^{-8})^3 \times 6.022 \times 10^{23}} = 1.4853 \, \text{g cm}^{-3} \] Step 2: Compare with observed density.
\[ \text{Percentage of Schottky defects} = \frac{\rho_\text{calc} - \rho_\text{obs}}{\rho_\text{calc}} \times 100 \] \[ = \frac{1.4853 - 1.4848}{1.4853} \times 100 = 0.03% \] Step 3: Conclusion.
Hence, 0.03% Schottky defects are present in the crystal.