Question

The relation between edge length (a) and radius of atom (r) for BCC lattice is $\sqrt{3a}=4r$.
If true enter 1,else enter 0.

Answer 1

Usually, the length of the cell edge is represented by $a$.

The direction from a corner of a cube to the farthest corner is called body diagonal (say bd).

Face diagonal = fd

${bd}^2={fd}^2+a^2$

$=a^2+a^2+a^2$

$=3a^2$

Atoms along the body diagonal (say bd) touch each other. Thus, the body diagonal has a length that is four times the radius of the atom, R.

$bd=4R$

The relationship between a and R can be worked out by the Pythagorean theorem:

${\left(4R\right)}^2=3a^2$

Thus,

$4R=\sqrt{3}a$