Question

The edge length of unit cell of a metal having molecular weight 75 g/mol is 5 $\mathring{A}$ which crystallises in cubic lattice. If the density is 2 g/cc then find the radius of a metal atom. ($N_A=6\times {10}^{23}$ ). Give the answer in pm.

• A. 217pm
• B. 210pm
• C. 220pm
• D. 205pm

Answer 1

The correct option is A 217pm

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

$\Rightarrow n=\frac{\rho N_0}{M}{a}^3$

$\Rightarrow n=\frac{2\times 6.022\times {10}^{23}\times (5{)}^3\times ({10}^{-8}{)}^3}{75}=2$

So, it is body centered cubic (BCC) structure. For BCC, if $a$ is the edge length of the cube and $r$ is the radius of the atom then,

$4r=\sqrt{3}a$

$\Rightarrow r=\frac{\sqrt{3}}{4}a$

$=\frac{\sqrt{3}}{4}\times 5\mathring{A}$

$\Rightarrow r=2.165\mathring{A}=2.165\times {10}^{-8}cm=216pm$