Looking for classes? Ksquare Career Institute, Bengaluru →
In BCC, the distance between two nearest atoms will be : - (a is the edge of the unit cell)
A
$\sqrt{3}a$
B
$\dfrac{\sqrt{3}}{2}a$
C
$\dfrac{3}{4}a$
D
$\dfrac{a}{\sqrt{2}}$
Detailed Solution
In a BCC cell, the atom at the body centre touches the atoms at the corners along the body diagonal.
Length of the body diagonal $= \sqrt{3}a$.
The nearest neighbour distance is half of the body diagonal (centre to corner): $d = \dfrac{\sqrt{3}}{2}a$.
(For BCC, the atomic radius $r = \dfrac{\sqrt{3}}{4}a$.)
Length of the body diagonal $= \sqrt{3}a$.
The nearest neighbour distance is half of the body diagonal (centre to corner): $d = \dfrac{\sqrt{3}}{2}a$.
(For BCC, the atomic radius $r = \dfrac{\sqrt{3}}{4}a$.)