Question

Compute the radius r of an impurity atom that will just fit into the body-center hole of a
simple cubic (SC) lattice in terms of the atomic radius R of the host atom (without
introducing lattice strains).
(b) Perform the same derivation for an FCC tetrahedral interstitial hole.

Answer 1

(a) In SC lattice since the atoms at the corners touch each other, the void radius will be the space between the atoms along the body diagonal.

The length of body diagonal can be caculated by applying pythagorus theorem, first to a faceto find the face diagonal length and then to plane along the body diagonal to find the length of body diagonal.

Length of edge = 2R

Length of face diagonal = =

Length of face diaonal = =

Radius of impurity = (Length of body diagonal - 2R)/2 =( - 2R)/2 = - R

(b) Using the approach used in part "a" the radius ratio for FCC body center void can be calculated.

R_void/R_atom = 0.225

R_void = 0.225*R