Question

The shortest distance between $I^{st}$ and $V^{th}$ layer of $HCP$ arrangement is?
(where r is the radius of the spherical atom)

• A. $8\sqrt{\frac{2}{3}}r$
• B. $4\sqrt{\frac{3}{2}}r$
• C. $16\frac{\sqrt{2}}{3}r$
• D. $8\sqrt{\frac{3}{2}}r$

Answer 1

The correct option is A $8\sqrt{\frac{2}{3}}r$

solution:

in the lattice structure let the radius be r

length of each side of tetrahedron will be 2r

2r sin 600 =$\sqrt{3}$r

distnace between centroid and mid point of any side of tetrahedron is

$\frac{1}{3}\sqrt{3}r$

height of tetrahedron is 2 x r

2 x $\sqrt{(\sqrt{3}r{)}^2-(\frac{r}{\sqrt{3}}{)}^2}$

2x$\sqrt{\frac{2r^2}{3}}$

height between 1st and 5th LAYER of H.C.P is

4x h = 4 x( 2x$\sqrt{\frac{2r^2}{3}}$)

$\to$ 8$\sqrt{\frac{2}{3}}r$