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Sodium has body centred packing. Distance between two nearest atoms is 3.7 Å. The lattice parameter is
A
8.6 Å
B
6.8 Å
C
4.3 Å
D
3.0 Å
Detailed Solution
In a body-centred cubic (bcc) lattice the nearest neighbours are a corner atom and the body-centre atom.
The body diagonal has length $\sqrt{3}a$ and the body-centre atom lies at its midpoint, so the nearest-neighbour distance is $d = \frac{\sqrt{3}a}{2}$
$a = \frac{2d}{\sqrt{3}} = \frac{2\times3.7}{1.732}$
a = 4.27 Å $\approx$ 4.3 Å
The body diagonal has length $\sqrt{3}a$ and the body-centre atom lies at its midpoint, so the nearest-neighbour distance is $d = \frac{\sqrt{3}a}{2}$
$a = \frac{2d}{\sqrt{3}} = \frac{2\times3.7}{1.732}$
a = 4.27 Å $\approx$ 4.3 Å