Lithium having a body centered cubic. Its density is 0.53 g `cm^(-3)` and its atomic mass is 7.00 g `"mol"^(-4)`. The edge length of unit cell of lithium metal is (Given: `root(3)(43.7)` = 3.53)

Options

A. 353 pm
B. 400 pm
C. 300 pm
D. 350 pm

Explanation

To find the edge length of the unit cell of lithium metal, we will follow these steps:

### Step 1: Understand the Body-Centered Cubic (BCC) Structure
In a body-centered cubic (BCC) structure, there are:
- 8 atoms at the corners of the cube, each contributing \( \frac{1}{8} \) of an atom to the unit cell.
- 1 atom at the center of the cube.

Thus, the total number of atoms \( Z \) in a BCC unit cell is:
\[ Z = 8 \times \frac{1}{8} + 1 = 2 \]

### Step 2: Use the Density Formula
The formula for density \( \rho \) is given by:
\[ \rho = \frac{Z \times M}{N_a \times a^3} \]
Where:
- \( \rho \) = density (0.53 g/cm³)
- \( Z \) = number of atoms per unit cell (2)
- \( M \) = molar mass (7.00 g/mol)
- \( N_a \) = Avogadro's number (\( 6.02 \times 10^{23} \) mol⁻¹)
- \( a \) = edge length of the unit cell (in cm)

### Step 3: Rearrange the Density Formula to Solve for Edge Length \( a \)
Rearranging the formula to find \( a \):
\[ a^3 = \frac{Z \times M}{\rho \times N_a} \]
Substituting the known values:
\[ a^3 = \frac{2 \times 7.00 \, \text{g/mol}}{0.53 \, \text{g/cm}^3 \times 6.02 \times 10^{23} \, \text{mol}^{-1}} \]

### Step 4: Calculate the Value
Calculating the numerator:
\[ 2 \times 7.00 = 14.00 \, \text{g} \]
Calculating the denominator:
\[ 0.53 \times 6.02 \times 10^{23} \approx 3.196 \times 10^{23} \, \text{g/cm}^3 \]
Now substituting these values back into the equation:
\[ a^3 = \frac{14.00}{3.196 \times 10^{23}} \approx 4.37 \times 10^{-23} \, \text{cm}^3 \]

### Step 5: Take the Cube Root to Find \( a \)
Taking the cube root of both sides:
\[ a \approx \sqrt[3]{4.37 \times 10^{-23}} \approx 3.53 \times 10^{-8} \, \text{cm} \]

### Step 6: Convert to Picometers
Since \( 1 \, \text{cm} = 10^{10} \, \text{pm} \):
\[ a \approx 3.53 \times 10^{-8} \, \text{cm} \times 10^{10} \, \text{pm/cm} = 3.53 \, \text{pm} \]

### Final Answer
The edge length of the unit cell of lithium metal is approximately:
\[ \boxed{3.53 \, \text{pm}} \]