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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
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B
154.14 pm
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C
251.7 pm
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D
89 pm
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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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