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JEE Main 20205 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
5 September
Shift
Evening Shift
Subject
Chemistry
Chapter
Solid State

JEE Main 2020 · 5 September · Evening Shift · Chemistry · Solid State

An element crystallises in a face–centred cubic (fcc) unit cell...

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:

An element crystallises in a face–centred cubic (fcc) unit cell with cell edge a. The distance between the centres of

Options

  1. Aa2\dfrac{a}{\sqrt{2}}
  2. Baa
  3. C2a\sqrt{2}a
  4. Da2\dfrac{a}{2}

Correct answer

A. a2\dfrac{a}{\sqrt{2}}

Step-by-step solution

In FCC octahedral voids are present at the edge centers and body center.

An element crystallises in a face–centred cubic (fcc) unit cell with cell edge a. The distance between the centres of

Minimum distance between centers of two octahedral voids

=x=(a2)2+(a2)2= x = \sqrt{{\left(\dfrac{a}{2}\right)}^{2} + {\left(\dfrac{a}{2}\right)}^{2}}
=a24+a24=a2= \sqrt{\dfrac{{a}^{2}}{4} + \dfrac{{a}^{2}}{4}} = \dfrac{a}{\sqrt{2}}

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