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Q. The height of a $HCP$ unit cell is $5.715$ $\text{Å}.$ What is the volume of the unit cell in $\text{Å}^{3}$ ?

NTA AbhyasNTA Abhyas 2022

Solution:

The height of the $HCP$ unit cell $=4\sqrt{\frac{2}{3}}.r=5.715$
$\text{r} = \frac{5.715 \times \sqrt{3}}{\sqrt{2} \times 4} = 1.75 \, \text{Å}$
The base area of the unit cell $=6\sqrt{3}r^{2}$
The volume of the unit cell $=base \, \, area \, \, \times $ $\text{height} = 24 \sqrt{2} \text{r}^{3} = 181.8$ $\text{Å}^{3}$