Question

If an element (at. wt. $=50$ ) crystallizes in $fcc$ lattice, with a $=0.50nm$. What is the density of unit cell if it contains $0.25\%$ schottky defects (use $N_A=6\times 10^{23}$ ). Give your answer as multiple of $10^{-24}$.

Answer 1

Answer: 2.66

${\rho }_{\text{theoretical }}=\frac{N\times \text{ mol. wt }}{a^3\times N_a}$
$=\frac{4\times 50}{{\left(0.50\times 10^{-7}\right)}^{3}c{m}^3\times 6\times 10^{23}}$
$=\frac{200}{0.75\times 10^{-21+23}}=2.66g/c{m}^3$
$\%$ missing $=\frac{{\rho }_{\text{theo }}-{\rho }_{\exp }}{{\rho }_{\exp }}\times 100$
$0.25=\frac{2.66-{\rho }_{\exp }}{2.66}\times 100$
$0.00665=2.66-{\rho }_{\exp }$
${\rho }_{\exp }=2.66665g/c{m}^3$