Question

Lithium has a $bcc$ structure. Its density is $530 \, kg \, m^{-3}$ and its atomic mass is $6.94 \, g \, mol^{-1}$. Calculate the edge length of a unit cell of Lithium metal. ($NA = 6.02 \times 10^{23} \, mol^{-1}$)

• A. 352 pm
• B. 527 pm
• C. 264 pm
• D. 154 pm

Answer 1

The Correct Option is A

Body-Centered Cubic (BCC) Calculation 

For BCC, the formula to calculate the edge length \(a\) is:

\(d = \frac{Z \times M}{N_A \times a^{3}}\)

Where:

• Z = Number of atoms per unit cell (for BCC, Z = 2)
• M = Molar mass (for lithium, M = 6.94 g/mol)
• N_A = Avogadro's number (6.02 × 10²³ atoms/mol)
• a = Edge length of the unit cell

Now, substituting the values into the formula:

\(530 = \frac{2 \times 6.94 \times 10^{-3}}{6.02 \times 10^{23} \times a^{3}}\)

Rearranging the equation to solve for \(a^{3}\):

\(a^{3} = \frac{2 \times 694}{530 \times 6.02 \times 10^{28}}\)

After solving:

\(a^{3} = 0.435 \times 10^{-28} \, m\)

\(= 43.5 \times 10^{-30} \, m\)

\(= 43500000 \, pm\)

\(\therefore a = 352 \, pm\)

Conclusion:

The edge length \(a\) of the BCC unit cell is 352 pm.