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A metal has zinc blende type unit cell. If the atomic radius of this metal is 86.6 pm, then the number of atoms in a cube of edge length of 1 cm is about

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0.62 x 10 ²³ atoms

1.25 x 10 ²³ atoms

2.50 x 10²³ atoms

5.0 x 10 ²³ atoms

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detailed solution

Correct option is B

$2 r_{M} = \sqrt{3} a / 4$ Hence, $a = 8 r_{M} / \sqrt{3} = 8 \times 86.6 \times 10^{- 10} cm / 1.732 = 400 \times 10^{- 10} cm$

$v = a^{3} = {(4.00 \times 10^{- 8} cm)}^{3} = 6.4 \times 10^{- 23} {cm}^{3}$ $N = (\frac{8 atoms}{v}) V = (\frac{8 atoms}{6.4 \times 10^{- 23} {cm}^{3}}) (1 {cm}^{3}) = 1.25 \times 10^{23} atoms$

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