The volume of the HCP unit cell is

Options

A. `24sqrt(2)r^(3)`
B. `(16)/sqrt(2)r^(3)`
C. `(12)/sqrt(2)r^(3)`
D. `(64)/(3sqrt(3))r^(3)`

Answer

1

Explanation

Base area of regular hexagon=Area of 6 equilateral trinagles each with side 'a' `=6sqrt(3)/(4)a^(2)`
Height of the unit cell (h)=2xx distance between
adjacent layers `=2xxsqrt((2)/(3))a`
`therefore` Volume of the unit cell=Base area xx height
`==((6sqrt(3))/(4)a^(2))xx(2sqrt((2)/(3))a)=3sqrt(2)a^(3)`
Putting a=2 r, volume `=24sqrt(2)r^(3)`