The ratio of the nearest neighbour atomic distances in body-centered cubic (bcc) and face-centered cubic… · ABC26PH0807
Subject: Physical Chemistry · Chapter: Solid State Chemistry · Topic: Solid State – General · Exam: JAM 2001-2022 · Marks: 2 · Difficulty: Medium
The ratio of the nearest neighbour atomic distances in body-centered cubic (bcc) and face-centered cubic (fcc) crystals with the same unit cell edge length is
(a)$\sqrt{\frac{3}{2}}$
(b)$\frac{\sqrt{3}}{2}$
(c)$\frac{1}{\sqrt{2}}$
(d)$\frac{1}{2}$
Answer
Answer: A ✓ checked by 4AB · confidence high
The source book printed C; on checking, A is correct — see the explanation.
Book printed (c).
Explanation
Nearest-neighbour distances: bcc √3a/2, fcc a/√2; ratio = √(3/2) (a).