Question

What is the number of unit cells present in 3.9 g of potassium if it crystallizes in BCC structure?

• A. $\frac{N_A}{10}$
• B. $N_A \times 10$
• C. $2 N_A$
• D. $\frac{N_A}{20}$

Hint:

For any unit cell problem: $\text{Number of unit cells} = \frac{\text{Moles} \times N_A}{n}$, where $n=1$ for simple cubic, $n=2$ for BCC, and $n=4$ for FCC. Remembering this formula saves valuable calculation time!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem asks for the total number of unit cells contained within a given mass (3.9 g) of potassium metal, given that it forms a body-centered cubic (BCC) crystal lattice.

Step 2: Key Formula or Approach:
First, we find the total number of potassium atoms present in the sample using the mole concept: $$ \text{Number of atoms} = \frac{\text{Mass}}{\text{Molar mass}} \times N_A $$ A body-centered cubic (BCC) unit cell contains a net effective number of 2 atoms per unit cell ($n = 2$). Therefore, the total number of unit cells is calculated by dividing the total number of atoms by the atoms per unit cell: $$ \text{Number of unit cells} = \frac{\text{Total number of atoms}}{2} $$

Step 3: Detailed Explanation:
The atomic mass of potassium (K) is $39\ \text{g/mol}$. Let's compute the total number of atoms in $3.9\ \text{g}$ of potassium: $$ \text{Number of atoms} = \frac{3.9}{39} \times N_A = 0.1 N_A $$ Since a BCC crystal structure accommodates exactly $2$ constituent atoms inside each unit cell, we distribute these atoms into unit cells: $$ \text{Number of unit cells} = \frac{0.1 N_A}{2} = \frac{1}{20} N_A = \frac{N_A}{20} $$

Step 4: Final Answer:
The total number of unit cells present is $\frac{N_A}{20}$, corresponding to option (D).