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Al (at. wt 27) crystallizes in the cubic system with a cell edge of 4.05 $A^{o}$ .Its density is 2.7 g per cm³ .Determine the unit cell type calculate the radius of the Al atom

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$fcc, 1 .432 A^{o}$

$bcc, 2 .432 A^{o}$

$bcc, 1 .432 A^{o}$

$fcc, 2 .432 A^{o}$

answer is D.

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Detailed Solution

$ρ = \frac{Z \times M}{N_{o} \times a^{3}} 2 .7 = \frac{Z \times 27}{6 .02 \times 10^{23} \times (4 .05)^{3} \times 10^{- 24}} ∴ Z = 4$
Hence it is face centred cubic unit lattice.
Again $4 r = a \sqrt{2} = 5 .727 A^{o}$
$∴ r = 1 .432 Å$

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