Solid State - Result Question 2
2. An element has a face-centred cubic (fcc) structure with a cell edge of $a$. The distance between the centres of two nearest tetrahedral voids in the lattice is
(2019 Main, 12 April I)
(a) $\sqrt{2} a$
(b) $a$
(c) $\dfrac{a}{2}$
(d) $\dfrac{3}{2} a$
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Answer:
Correct Answer: 2. (c)
Solution:
- In fcc unit cell, two tetrahedral voids are formed on each of the four non-parallel body diagonals of the cube at a distance of $\sqrt{3} a / 4$ from every corner along the body diagonal.
The angle between body diagonal and an edge is $\cos ^{-1}(1 / \sqrt{3})$. So, the projection of the line on an edge is $a / 4$. Similarly, other tetrahedral void also will be $a / 4$ away. So, the distance between these two is $\left[a-\dfrac{a}{4}\right]-\dfrac{a}{4}=\dfrac{a}{2}$.
BETA
