Question

If the radius of an atom of an element which forms a body centered cubic unit cell is $173.2pm$. the volume of unit cell in $c{m}^3$ 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 $=173.2pm$.

$\because$ For bcc structure

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

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,

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

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

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

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