If $a$ be the edge length of the unit cell and $r$ be the radius of an atom then for fcc arrangement, the correct relation is

• A

  $4 a=\sqrt{3} r$
• B

  $4 r=\sqrt{3} a$
• Correct answer
  C

  $4 r=\sqrt{2} a$
• D

  $4 r=\frac{a}{\sqrt{2}}$

Solution

Given, the edge length of the unit cell $=a$ and, radius of the sphere $=r$ Consider $\triangle A B C$, $\begin{aligned} A C^2 & =A B^2+A C^2 \\ & =a^2+a^2 \\ & =2 a^2 \\ \therefore \quad A C & =\sqrt{2} \cdot a \end{aligned}$ But, from the figure, $A C=4 r$ $\therefore \quad 4 r=\sqrt{2} \cdot a$

Hence, the correct answer is option: 'C'.