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 `(N_(A) = 6.02 xx 10^(23) mol^(-1))`

Answer

To find the edge length of the unit cell of lithium with a body-centered cubic (BCC) structure, we can follow these steps:

### Step 1: Understand the BCC Structure
In a BCC unit cell, there are:
- 8 corner atoms, each contributing \( \frac{1}{8} \) of an atom to the unit cell.
- 1 atom at the body center contributing 1 full atom.

**Total number of atoms (Z) per unit cell:**
\[ Z = 8 \times \frac{1}{8} + 1 = 2 \]

### Step 2: Use the Density Formula
The density (\( \rho \)) of a substance is given by the formula:
\[ \rho = \frac{\text{mass of unit cell}}{\text{volume of unit cell}} \]

### Step 3: Calculate the Mass of the Unit Cell
The mass of the unit cell can be calculated using the atomic mass and the number of atoms per unit cell:
- Given atomic mass of lithium = \( 6.94 \, \text{g/mol} \)
- Convert this to kg: \( 6.94 \, \text{g/mol} = 6.94 \times 10^{-3} \, \text{kg/mol} \)

The mass of the unit cell can be calculated as:
\[ \text{Mass of unit cell} = Z \times \frac{\text{atomic mass}}{N_A} \]
Where \( N_A = 6.02 \times 10^{23} \, \text{mol}^{-1} \).

Substituting the values:
\[ \text{Mass of unit cell} = 2 \times \frac{6.94 \times 10^{-3}}{6.02 \times 10^{23}} \, \text{kg} \]

### Step 4: Calculate the Volume of the Unit Cell
The volume of the unit cell for a cubic structure is given by:
\[ \text{Volume} = a^3 \]
Where \( a \) is the edge length of the unit cell.

### Step 5: Substitute into the Density Formula
Now we can substitute the mass and volume into the density formula:
\[ \rho = \frac{2 \times \frac{6.94 \times 10^{-3}}{6.02 \times 10^{23}}}{a^3} \]
Rearranging gives:
\[ a^3 = \frac{2 \times \frac{6.94 \times 10^{-3}}{6.02 \times 10^{23}}}{\rho} \]

### Step 6: Substitute the Density Value
Given density \( \rho = 530 \, \text{kg/m}^3 \):
\[ a^3 = \frac{2 \times \frac{6.94 \times 10^{-3}}{6.02 \times 10^{23}}}{530} \]

### Step 7: Calculate \( a^3 \)
Calculating this step-by-step:
1. Calculate the numerator:
\[ 2 \times \frac{6.94 \times 10^{-3}}{6.02 \times 10^{23}} = \frac{13.88 \times 10^{-3}}{6.02 \times 10^{23}} = 2.303 \times 10^{-26} \, \text{kg} \]

2. Now substitute into the equation:
\[ a^3 = \frac{2.303 \times 10^{-26}}{530} = 4.345 \times 10^{-29} \, \text{m}^3 \]

### Step 8: Calculate the Edge Length \( a \)
Taking the cubic root:
\[ a = (4.345 \times 10^{-29})^{1/3} \approx 3.52 \times 10^{-10} \, \text{m} \]
Converting to picometers:
\[ a \approx 352 \, \text{pm} \]

### Final Answer
The edge length of the unit cell of lithium metal is approximately **352 picometers**.

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