Question

Edge length in a face centered cubic unit cell of calcium is 0.556 nm. If it contains 0.1% schottky defect, the density of the metal is

• A. $1.5463gc{m}^{-3}$
• B. $1.56gc{m}^{-3}$
• C. $1.5448gc{m}^{-3}$
• D. $3.42gc{m}^{-3}$

Answer 1

The correct option is C $1.5448gc{m}^{-3}$

$Z=\{4-\frac{4\times 0.1}{100}\}=3.996$

$d=\frac{Z\times M}{a^3\times N_a}$

$=\frac{3.996\times 40}{(0.556\times {10}^{(}-7){)}^3\times (6.022\times (10{)}^{2}3)}gc{m}^{-3}$

$\Rightarrow d=1.5448gc{m}^{-3}$