JEE Main 2021 — Solid State Question with Solution

$\mathrm{HCP}$ structure : Per atom, there will be one octahedral void $\left(\mathrm{OV}\right)$ and two tetrahedral voids $\left(\mathrm{TV}\right)$.

Therefore, total three voids per atom are present in $\mathrm{HCP}$ structure.

$\to$ Therefore total no. of atoms of $\mathrm{Ga}$ will be-

$=\frac{ \mathrm{Mass} }{ \mathrm{Molar} \mathrm{Mass} }\times {\mathrm{N}}_{\mathrm{A}}=\frac{0.581 \mathrm{g}}{70 \mathrm{g}/\mathrm{mol}}\times 6.023\times {10}^{23}$

$\to$ Now, total Number of voids $=3$ $\times$ total no. of atoms 

$=3\times \frac{0.581}{70}\times 6.023\times {10}^{23}=14.99\times {10}^{21}$

$\simeq 15\times {10}^{21}$