Question

Height (c) of the hexagonal primitive unit cell in term of radius of atom (r) is $c=4r\sqrt{\frac{2}{3}}$.
If true enter 1, else enter 0.

Answer 1

Edge length of primitive hexagonal lattice is $a=2r$ and height of unit cell is $c$.

Height $c$ of the unit cell is twice the height of tetrahedron ABCD since height of tetrahedron with base side a is $\sqrt{\frac{2}{3}}a$

Height of unit cell is $2\times \sqrt{\frac{2}{3}}a=2\sqrt{\frac{2}{3}}a$

or $c=4r\sqrt{\frac{2}{3}}$ as $a=2r$

So, height is $4r\sqrt{\frac{2}{3}}$.