Solid State - Result Question 32
####32. The volume of this hep unit cell is
(a) $24 \sqrt{2} r^{3}$
(b) $16 \sqrt{2} r^{3}$
(c) $12 \sqrt{2} r^{3}$
(d) $\frac{64 r^{3}}{3 \sqrt{3}}$
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Answer:
Correct Answer: 32. (a)
Solution:
- In close packed arrangement, side of the base $=2 r$
$$ \Rightarrow \quad R S=r $$
Also MNR is equilateral triangle, $\angle P R S=30^{\circ}$
In triangle $P R S, \cos 30^{\circ}=\frac{R S}{P R}=\frac{\sqrt{3}}{2}$
$$ \Rightarrow \quad P R=\frac{2}{\sqrt{3}} R S=\frac{2}{\sqrt{3}} r $$
In right angle triangle $P Q R: P Q=\sqrt{Q R^{2}-P R^{2}}=2 \sqrt{\frac{2}{3}} r$
$\Rightarrow$ Height of hexagon $=2 P Q=4 \sqrt{\frac{2}{3}} r$
$\Rightarrow$ Volume $=$ Area of base $\times$ height $=6 \frac{\sqrt{3}}{4}(2 r)^{2} \times 4 \sqrt{\frac{2}{3}} r$
$$ =24 \sqrt{2} r^{3} $$
