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

To calculate the edge length of a unit cell of Lithium, which has a body-centered cubic (bcc) structure, we can use the given data and apply the formulae related to solid state chemistry.

Firstly, in a bcc structure, there are 2 atoms per unit cell. The formula relating the density, atomic mass, and edge length \( a \) of the unit cell is:

Density = \frac{Z \times \text{Molar Mass}}{N_A \times a^3}

Where:

• Z = 2, the number of atoms per unit cell for bcc.
• The density of Lithium is given as 530 \, kg \, m^{-3}.
• Molar mass of Lithium = 6.94 \, g \, mol^{-1} = 6.94 \times 10^{-3} \, kg \, mol^{-1} (conversion from grams to kilograms).
• The Avogadro's number N_A = 6.02 \times 10^{23} \, mol^{-1}.
• a is the edge length in meters.

Rearranging the formula to solve for the edge length a:

a^3 = \frac{Z \times \text{Molar Mass}}{N_A \times \text{Density}}

Substituting in the known values:

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

Calculating further:

a^3 = \frac{13.88 \times 10^{-3}}{3.19106 \times 10^{26}}

a^3 = 4.349 \times 10^{-30} \, m^3

Taking the cube root to find a:

a = (4.349 \times 10^{-30})^{1/3} \, m

Solving, we find:

a \approx 3.52 \times 10^{-10} \, m

Converting meters to picometers (1 m = 10¹² pm):

a \approx 352 \, pm

Thus, the edge length of the unit cell of Lithium is 352 pm, which aligns with the given correct answer.