Question

$Al$(at.wt.$27$) crystallizes in the cubic system with a cell edge of $4.05\mathring{A}$. Its density is $2.7g/{cm}^3$. Determine the unit cell type and calculate the radius of the $Al$ atom.

• A. $fcc,2.432\mathring{A}$
• B. $bcc,2.432\mathring{A}$
• C. $bcc,1.432\mathring{A}$
• D. $fcc,1.432\mathring{A}$

Answer 1

Answer: (D)

$fcc,1.432\mathring{A}$

The density is given by the formula
$\rho =\frac{Z\times M}{N_o\times a^3}$

$2.7g/c{m}^3=\frac{Z\times \text{27 g/mol}}{6.02\times {10}^{23}\times {(4.05\times {10}^{-8}\text{cm})}^3}$

The number of atoms per unit cell $Z=4$

Hence it is face centred cubic unit lattice.

Again $4r=a\sqrt{2}=4.05\mathring{A}\times 1.414=5.727\mathring{A}$

The radius of the $Al$ atom $=1.432\mathring{A}$

Hence, the correct option is $\text{D}$