Sodium has body centred packing. Distance between two nearest atoms is $3.7 Å$. The lattice parameter is
Sodium has body centred packing. Distance between two nearest atoms is $3.7 Å$. The lattice parameter is
$6.8 Å$
$4.3 Å$
$3.0 Å$
$8.6 Å$
Solution
Neighbour distance of a body centred cubic cell $\mathrm{d}=\frac{\sqrt{3}}{2} \mathrm{a}$, where $a$ is the lattice parameter.
$\begin{array}{ll}
\Rightarrow & 3.7=\frac{\sqrt{3} \mathrm{a}}{2} \\
\text { or } & \mathrm{a}=\frac{2 \times 3.7}{\sqrt{3}}=4.3 Å
\end{array}$