Aluminium crystallizes in cubic system with unit cell edge length equal to 4.0 Å. If its density is

Question

Aluminium crystallizes in cubic system with unit cell edge length equal to 4.0 Å. If its density is $\frac{45}{16}$g/cm3, then the atomic radius of Al-atom is (NA = 6 × 1023, Al = 27)

Accepted Answer

$d = \frac{Z.M}{N_{A}.V} \Rightarrow \frac{45}{16} = \frac{Z \times 27}{\left( 6 \times 10^{23} \right) \times \left( 4 \times 10^{- 8} \right)^{3}}$

⇒ Z = 4

Hence, Al crystal is FCC.

For FCC, $\sqrt{2}a = 4r \Rightarrow r = \frac{\sqrt{2} \times 4.0}{4} = 1.414\overset{0}{A}$

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