Question

The length of the unit cell edge of a bcc lattice metal is $350pm$. What is the volume of atoms in one mole of the metal?

• A. $8.79c{m}^3$
• B. $6.25c{m}^3$
• C. $1.21c{m}^3$
• D. $4.00c{m}^3$

Answer 1

The correct option is A $8.79c{m}^3$

In a bcc structure, the atoms touch each other along the body diagonal.
Let,
$\text{Radius of an atom}=r$
$A{B}^2=B{C}^2+A{C}^2$
$B{C}^2=B{D}^2+C{D}^2=a^2+a^2=2a^2$
$\therefore A{B}^2=2a^2+a^2=3a^2$
$\therefore AB=\sqrt{3}a$
Also, $AB=4r$
$\therefore 4r=\sqrt{3}a\Rightarrow r=\frac{\sqrt{3}a}{4}$
$r=\sqrt{3}\times \frac{350}{4}pm$
$=\sqrt{3}\times \frac{350}{4}\times {10}^{-12}m$
$=\sqrt{3}\times \frac{350}{4}\times {10}^{-10}cm=151.6\times {10}^{-10}cm$
$\text{Volume of one atom}=\frac{4}{3}\pi r^3=\frac{4}{3}\pi (151.6\times {10}^{-10}{)}^{3}c{m}^3=1.46\times {10}^{-23}c{m}^3$
Volume of $N_0$ atoms is one mole of the metal is
$6.02\times {10}^{23}\times 1.46\times {10}^{-23}c{m}^3=8.79c{m}^3$