Question

If calcium crystallizes in bcc arrangement and the radius of Ca atom is $96$ pm, then the volume (in m${}^3$) of the unit cell of Ca is:

• A. $10.9\times {10}^{-36}$
• B. $10.9\times {10}^{-30}$
• C. $21.8\times {10}^{-30}$
• D. $21.8\times {10}^{-36}$

Answer 1

Answer: (B)

$10.9\times {10}^{-30}$

For bcc lattice,

$\sqrt{3}a=4r\Rightarrow a=\frac{4\times 96}{\sqrt{3}}$ pm $=221.7$ pm
Here, $r$ is the radius of Ca atom.

Volume of unit cell $=a^3=(221.7\times {10}^{-12}{)}^3$ m${}^3$ $=10.9\times {10}^{-30}$ m${}^3$

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