Calculate the volume of fcc unit cell if radius of a particle in it is 106.05 pm.

Calculate the volume of fcc unit cell if radius of a particle in it is 106.05 pm.
  1. $\quad 7.4 \times 10^{-23} \mathrm{~cm}^3$
  2. $9.9 \times 10^{-23} \mathrm{~cm}^3$
  3. $2.7 \times 10^{-23} \mathrm{~cm}^3$
  4. $6.4 \times 10^{-23} \mathrm{~cm}^3$

Solution

For fcc unit cell, $a=\frac{4 r}{\sqrt{2}}$ $\begin{aligned} \therefore \quad a & =\frac{4 \times 106.05}{\sqrt{2}}=\frac{424.2}{1.41} \\ & =300 \mathrm{pm}=300 \times 10^{-10} \mathrm{~cm} \end{aligned}$ $\therefore \quad$ Volume of the unit cell $\left(\mathrm{a}^3\right)=\left(300 \times 10^{-10}\right)^3$ $=2.7 \times 10^{-23} \mathrm{~cm}^3$

Asked in: MHT CET 2024 (09 May Shift 2)

Practice more Solid State questions on Aicharya