Q48 / 200 · NEET 2009 (Screening) · Chemistry
Sodium has body centred packing. Distance between two nearest atoms is $3.7 Å$. The lattice parameter is
- A$6.8 Å$
- Correct answerB$4.3 Å$
- C$3.0 Å$
- D$8.6 Å$
Solution
Neighbour distance of a body centred cubic cell $\mathrm{d}=\frac{\sqrt{3}}{2} \mathrm{a}$, where $a$ is the lattice parameter.
$$
\begin{array}{ll}
\Rightarrow & 3.7=\frac{\sqrt{3} \mathrm{a}}{2} \\
\text { or } & \mathrm{a}=\frac{2 \times 3.7}{\sqrt{3}}=4.3 Å
\end{array}
$$
Hence, the correct answer is option: 'B'.