Solid StateHard
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)
Options
A.1.414 Å
B.1.732 Å
C.4.0 Å
D.2.0 Å
Solution
$d = \frac{Z.M}{N_{A}.V} \Rightarrow \frac{45}{16} = \frac{Z \times 27}{\left( 6 \times 10^{23} \right) \times \left( 4 \times 10^{- 8} \right)^{3}}$
⇒ 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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