Question

For 3-D hexagonal closed packing structure (with similar particles), the correct relation between radius of atoms (r) and height of unit cell (h) is :

• A. $h=\sqrt{\frac{2}{3}}\times 2r$
• B. $h=\sqrt{\frac{2}{3}}\times 4r$
• C. $h=\sqrt{\frac{3}{2}}\times 2r$
• D. $h=\sqrt{\frac{3}{2}}\times 4r$

Answer 1

The correct option is B $h=\sqrt{\frac{2}{3}}\times 4r$


In the above figure, distance between B layer and A layer is equal to the distance between the centre of B layer and centre of sphere at the face centre in A layer , as shown :


In fig (ii) ,

1. A is the centre of the sphere at face centre of A layer.
2. C is the centre of B layer .
3. B is the centre of sphere of B layer touching the sphere at the face centre of A layer

Distance AC is to be found out.
Now,

$AB=a=2r$

To find BC, look at the triangle formed by the three spheres of B layer as shown in (iii) :


Let
$B_{1}C=B_{2}C=x$
$\text{From}△B_1{B}_{2}C$
$\cos {120}^{\circ }=\frac{x^2+x^2-a^2}{2x^2}\because \text{In any}△ABC,\cos C=\frac{a^2+b^2-c^2}{2ab}$
$\therefore -\frac{1}{2}=\frac{2x^2-a^2}{2x^2}\Rightarrow 3x^2=a^{2}x=\frac{a}{\sqrt{3}}$
Applying pythagoras theorem fig (ii)
$\text{In}△ABCA{C}^2=A{B}^2-B{C}^2{\left(\frac{h}{2}\right)}^2=a^2-x^2{\left(\frac{h}{2}\right)}^2=a^2-{\left(\frac{a}{\sqrt{3}}\right)}^2$
${\left(\frac{h}{2}\right)}^2=\frac{2a^2}{3}h=2\times \sqrt{\frac{2}{3}}\times a=4\times r\times \sqrt{\frac{2}{3}}$
Option (b) is correct