Question

Calcium crystallizes in face-centred cubic unit cell with $a=0.556nm$. Calculate density
(i) if contained $0.1$% Frenkel defect
(ii) it contained $0.1$% Schottky defect.

Answer 1

Theoretical density of the crystal is $\rho =\frac{nM}{N_o{a}^3}g/c{m}^3$

Now, since in Frenkel Defect, there is no change in the theoretical density of the crystal because, in this defect, the only atom switches its position from lattice site to interstitial site.

So, $\rho$ (with $0.1$% Frenkel defect) $=\frac{nM}{N_o{a}^3}g/c{m}^3$

$a=0.056\times {10}^{-7}cm;M=40g;n=4$ (For FCC)

$\rho =\frac{40\times 4}{6.022\times {10}^{23}\times (0.056{)}^3\times ({10}^{-7}{)}^3}=1.5455g/c{m}^3$

Now, in case of Schottky defect, the density of the crystal changes because the atom leaves the crystal. So, in general for FCC, $n=4$

But, because o $0.1$% Schottky defect, the value of $n$ will change.

So, $n^{\prime }=4–4\times \frac{0.1}{100}=3.996$

${\rho }_{new}=\frac{n^{\prime }M}{N_o{a}^3}=\frac{3.996\times 40}{6.022\times {10}^{23}\times (0.556{)}^3\times ({10}^{-7}{)}^3}=1.5440g/c{m}^3$