Question

If the radius of an atom of an element which forms a body centered cubic unit cell is $173.2\, pm$. the volume of unit cell in $cm^3$ is

• A. $3.12 \times 10^{-23}$
• B. $6.4 \times 10^{-23}$
• C. $3.2 \times 10^{-24}$
• D. $2.13 \times 10^{-23}$

Answer 1

Answer: (B)

Given,

Radius of an atom in body centered cubic (bcc) unit cell $=173.2\, pm$.

$\because$ For bcc structure

$\sqrt{3} \cdot a=4 r$

where, $a=$ edge-length

$r=$ radius of atom

and $a^{3}=V$ (volume of cubic unit cell).

$\therefore a=\frac{4}{\sqrt{3}} \cdot r$

or, $a=\frac{4}{1.73} \times 173.2 \times 10^{-10} \,cm$

$a=400 \times 10^{-10} \,cm$

Therefore,

$a^{3} =V=\left[400 \times 10^{-10}\right]^{3} $

$=6,40,00,000 \times 10^{-30} $

$=6.4 \times 10^{-23} cm ^{3}$

Hence, $6.4 \times 10^{-23} \,cm ^{3}$ is the correct answer.