The number of unit cells present in $39 \mathrm{~g}$ of potassium if it crystallizes as body centred cube is…
The number of unit cells present in $39 \mathrm{~g}$ of potassium if it crystallizes as body centred cube is ( $N=$ Avogadro number, At. wt. of potassium $=39$ )
$\frac{N}{4}$
$\frac{N}{2}$
$\frac{N}{3}$
$N$
Solution
Potassium has bcc system
Number of mole $=39 / 39=1$ mole
1 mole of atoms $=6.022 \times 10^{23}$ atoms $=N$
$\because$ In bcc system 2 atoms are present in 1 unit cell.
$\therefore N$ number of atoms are present in $\frac{N}{2}$ unit cells.