Question

In diamond, carbon atom occupy FCC lattice points as well as alternate tetrahedral voids. If edge length of the unit cell is $356pm$, then the radius of carbon atom is :

• A. $77.07pm$
• B. $154.14pm$
• C. $251.7pm$
• D. $89pm$

Answer 1

The correct option is A $77.07pm$

For diamond unit cell, $\frac{\sqrt{3}}{4}a=2r$

Here a is the edge length and r is the radius of C atom.

$a=356$ pm

Hence, $\frac{\sqrt{3}}{4}\times 356=2r$

$r=77.07pm$