Question

Relationship between atomic radius and the edge length a of a body – centred cubic unit cell is

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

Answer 1

Answer: (C)

$r = \frac{\sqrt{3}}{4}a$

Answer 2

Distance between nearest neighbours.

$d = \frac{AD}{2}$

In right angled

$\Delta ABC,AC = AB^{2} + BC^{2}$

$AC^{2} = a^{2} + a^{2}$or $AC = \sqrt{2a}$

Now in right angled $\Delta ADC$,

$AD^{2} = AC^{2} + DC^{2}$

$AD^{2} = (\sqrt{2}a)^{2} + a^{2} = 3a^{2} \Rightarrow AD = \sqrt{3}a$

$\therefore d = \frac{\sqrt{3}a}{2}$

Radius, $r = \frac{d}{2} = \frac{\sqrt{3}}{4}a$