The density of $\mathrm{Na}$ is $0.613 \mathrm{~g} \mathrm{~cm}^{-3}$. If the edge length of unit cell of…
The density of $\mathrm{Na}$ is $0.613 \mathrm{~g} \mathrm{~cm}^{-3}$. If the edge length of unit cell of $\mathrm{Na}$ is $5 Å$, the effective number of atoms of Na per unit cell is (Atomic weight of $\mathrm{Na}=23 \mathrm{u}$ )
$8$
$1$
$2$
$4$
Solution
We know that, $d=\frac{Z \times m}{a^3 \times N_A}$
where, $d$ is density, $Z$ is the number of atoms, $m$ is the atomic mass, $a$ is edge length and $N_A$ is Avagadro's number.
$
\begin{aligned}
Z & =\frac{d \times a^3 \times N_A}{m} \\
& =\frac{0.613 \mathrm{~g} \mathrm{~cm}^{-3} \times\left(5 \times 10^{-8} \mathrm{~cm}\right)^3 \times 6.023 \times 10^{23}}{23 \mathrm{~g}}=2
\end{aligned}
$