JEE PYQ: Solid State Question 11
Question 11 - 2020 (05 Sep Shift 2)
An element crystallises in a face-centred cubic ($fcc$) unit cell with cell edge $a$. The distance between the centres of two nearest octahedral voids in the crystal lattice is:
(1) $\dfrac{a}{\sqrt{2}}$
(2) $a$
(3) $\sqrt{2a}$
(4) $\dfrac{a}{2}$
Show Answer
Answer: (1)
Solution
Distance between two nearest octahedral voids: $x = \sqrt{\left(\dfrac{a}{2}\right)^2 + \left(\dfrac{a}{2}\right)^2} = \sqrt{\dfrac{a^2}{4} + \dfrac{a^2}{4}} = \dfrac{a}{\sqrt{2}}$.
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