Question

How many unit cells are present in $39g$ of potassium that crystallises in body-centred cubic structure?

• A. $N_A$
• B. $\frac{N_A}{4}$
• C. $0.5N_A$
• D. $0.75N_A$

Answer 1

Answer: (C)

$0.5N_A$

Number of atoms $=\frac{Mass}{Atomicmass}\times N_A=\frac{39}{39}\times N_A=N_A$
In bcc unit cell, $Z=2$
$\therefore$ Number of unit cells $=\frac{N_A}{2}=0.5N_A$