JEE Main 2023 · 10 April · evening · Single correct · +4 / −1
The correct relationships between unit cell edge length 'a' and radius of sphere 'r' for face-centred and body-centred cubic structures respectively are :
- (a) r=2 √2 a and √3 r=4 a
- (b) r=2 √2 a and 4 r=√3 a
- (c) 2 √2 r=a and √3 r=4 a
- (d) 2 √2 r=a and 4 r=√3 a
Answer and solution
Answer (d) 2 √2 r=a and 4 r=√3 a
Solution
Chemistry ma’am · 3 steps
Find the line along which the spheres touch: the face diagonal in fcc, the body diagonal in bcc.
- Where spheres touch. In any cubic cell, the spheres do not touch along the edge. Find the diagonal where they do touch. Along that diagonal the length equals a whole number of radii.
Structure Spheres touch along Length of that line Relation Face-centred cubic face diagonal (corner, face centre, corner) √2 a = r + 2r + r = 4r a = 4r/√2 = 2√2 r Body-centred cubic body diagonal (corner, body centre, corner) √3 a = r + 2r + r = 4r 4r = √3 a - Sense check. Both relations give r smaller than a: r = 0.354 a for fcc and r = 0.433 a for bcc. A sphere cannot be bigger than the cube that holds it, so any option with r larger than a is impossible.
- So. The fcc relation is 2√2 r = a and the bcc relation is 4r = √3 a. That is option (d).
The idea to keep. Find the line where the atoms touch. In fcc it is the face diagonal, √2 a = 4r. In bcc it is the body diagonal, √3 a = 4r. Then check that r comes out smaller than a.
Answer from the NTA final answer key. Solution by PrepWiser.