A pure metal crystallizes in a body-centered cubic (BCC) lattice where the metal atoms have a radius R . A small interstitial guest atom of radius r perfectly fits at the centre of the edge of the unit cell without distorting the lattice. The exact ratio of r to R ( r R ) is :

Step-by-step solution

In a body-centered cubic (BCC) lattice, the atoms touch each other along the body diagonal. Let the edge length of the unit cell be a . The relationship between a and the host atom radius R is given by: 3 a = 4R a = 4R 3 The atoms at the corners of the unit cell are separated by distance a along the edge. The space occupied by the two corner atoms along the edge is 2R . The empty space available at the center of the edge is: Empty space = a - 2R Since the interstitial guest atom fits perfectly at the edge center, i