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)`)

Options

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

Explanation

To calculate the edge length of a unit cell of Lithium metal, we will use the formula that relates density, molar mass, and the volume of the unit cell. Here are the steps to solve the problem:

### Step 1: Understand the given data
- Density of Lithium, \( \rho = 530 \, \text{kg/m}^3 \)
- Atomic mass of Lithium, \( M = 6.94 \, \text{g/mol} \)
- Avogadro's number, \( N_A = 6.022 \times 10^{23} \, \text{mol}^{-1} \)

### Step 2: Convert the density to g/cm³
Since the atomic mass is given in grams, we should convert the density from kg/m³ to g/cm³:
\[ \rho = 530 \, \text{kg/m}^3 = 530 \times 10^{-3} \, \text{g/cm}^3 = 0.530 \, \text{g/cm}^3 \]

### Step 3: Use the formula for density in terms of unit cell parameters
For a body-centered cubic (BCC) structure, the formula for density is given by:
\[ \rho = \frac{Z \cdot M}{V \cdot N_A} \]
Where:
- \( Z \) is the number of atoms per unit cell (for BCC, \( Z = 2 \))
- \( V \) is the volume of the unit cell, which can be expressed as \( a^3 \) (where \( a \) is the edge length)

### Step 4: Rearrange the formula to find \( a \)
Substituting \( Z \) and rearranging the formula gives:
\[ V = \frac{Z \cdot M}{\rho \cdot N_A} \]
\[ a^3 = \frac{2 \cdot 6.94 \, \text{g/mol}}{0.530 \, \text{g/cm}^3 \cdot 6.022 \times 10^{23} \, \text{mol}^{-1}} \]

### Step 5: Calculate \( a^3 \)
Now, substituting the values:
\[ a^3 = \frac{2 \cdot 6.94}{0.530 \cdot 6.022 \times 10^{23}} \]
Calculating the numerator:
\[ 2 \cdot 6.94 = 13.88 \]
Calculating the denominator:
\[ 0.530 \cdot 6.022 \times 10^{23} \approx 3.197 \times 10^{23} \]
Now substituting these values:
\[ a^3 = \frac{13.88}{3.197 \times 10^{23}} \approx 4.344 \times 10^{-23} \, \text{cm}^3 \]

### Step 6: Calculate \( a \)
Taking the cube root to find \( a \):
\[ a = (4.344 \times 10^{-23})^{1/3} \approx 3.52 \times 10^{-8} \, \text{cm} = 352 \, \text{pm} \]

### Final Result
The edge length of the unit cell of Lithium metal is approximately \( 352 \, \text{pm} \).

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