Question

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

A
77.07 pm
B
154.14 pm
C
251.7 pm
D
89 pm

Solution

## The correct option is A 77.07 pm$$\sqrt{3}a = 4(2r)$$Here , $$a = 356 \ pm$$$$\therefore r = \frac{\sqrt{3} \times 356}{8} = 77.07 \ pm$$Chemistry

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