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 77.07 pm
$\sqrt{3} a = 4 (2 r)$

Here , $a = 356 p m$

$∴ r = \frac{\sqrt{3} \times 356}{8} = 77.07 p m$

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Similar questions

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

Q. 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 the radius (in pm) of carbon atom is:

Q. In diamond, carbon atom occupy FCC lattice points as well as alternate tetrahedral voids. If edge length of the unit cell is

356 p m

, then the radius of carbon atom is :

Q. If in diamond, there is unit cell of carbon atoms as fcc and if carbon atom is

s p^{3}

hybridized what fractions of void occupied by carbon atom?

Q. Calculate the radius of tetrahedral void present in the unit cell of an atom X which crystallizes in fcc unit cell having an edge length

a = 50 \sqrt{2} p m

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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 :-

The correct option is A 77.07 pm√3a=4(2r)Here , a=356 pm∴r=√3×3568=77.07 pm

The correct option is A 77.07 pm
$\sqrt{3} a = 4 (2 r)$

Here , $a = 356 p m$

$∴ r = \frac{\sqrt{3} \times 356}{8} = 77.07 p m$