Question

In a hexagonal close packing, the edge length of base of a unit cell is $100pm$. Find the volume of the hcp unit cell if the height of the cell is $160pm$.

• A. $416\times {10}^{-26}c{m}^3$
• B. $286\times {10}^{-26}c{m}^3$
• C. $543\times {10}^{-26}c{m}^3$
• D. $750\times {10}^{-26}c{m}^3$

Answer 1

The correct option is A $416\times {10}^{-26}c{m}^3$

In hcp, the structure of unit cell is hexagon.
Thus,
$\text{Volume of unit cell (Hexagon)}=\text{Base area}\times \text{Height of hexagon}$

$\text{Height of hexagon, C}=160pm=160\times {10}^{-10}cm$


We know, the base of the hexagon is made of 6 equilateral triangle.
$\text{Area of an equilateral triangle}=\frac{\sqrt{3}}{4}{a}^2$
where, a is the edge length of hexagon.
$a=100pm=100\times {10}^{-10}cm$
$\Rightarrow a={10}^{-8}cm$

$\text{Volume of unit cell}=6\times \frac{\sqrt{3}}{4}({10}^{-8}{)}^2\times 160\times {10}^{-10}$
$\text{Volume of unit cell}=416\times {10}^{-26}c{m}^3$