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

77.07 pm

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154.14 pm

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251.7 pm

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89 pm

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Solution

The correct option is
A

154.14 pm

As we know that for a diamond unit cell
$\frac{\sqrt{3}}{4} a = 2 r$
$W h e r e a = e d g e l e n g t h$
$r = r a d i u s o f c e n t r a l a t o m$
$S i n c e a i s g i v e n a s = 356 p m$
$S o \frac{\sqrt{3}}{4} \times 356 = 2 r$
$r = 77.07 p m .$

Final answer
- Therefore correct option is option (A)

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