Lithium forms a BCC lattice with an edge length of 350 pm. The experimental density of lithium is 0.53 g cm-3. What is the percentage of missing lithium atoms? (Atomic mass of Lithium = 7 amu)

Options

A. 97.7%
B. 95.4%
C. 4.6%
D. 2.3%

Answer

To solve the problem of determining the percentage of missing lithium atoms in a BCC lattice, we will follow these steps:

### Step 1: Convert the edge length from picometers to centimeters
Given:
- Edge length (a) = 350 pm

To convert picometers to centimeters:
\[ a = 350 \, \text{pm} = 350 \times 10^{-12} \, \text{m} = 3.5 \times 10^{-8} \, \text{cm} \]

### Step 2: Identify the parameters
- Atomic mass of lithium (m) = 7 amu
- Number of atoms per unit cell for BCC (Z) = 2
- Avogadro's number (N_A) = \(6.02 \times 10^{23} \, \text{atoms/mol}\)
- Experimental density (D_exp) = 0.53 g/cm³

### Step 3: Calculate the theoretical density (D_theoretical)
The formula for density is given by:
\[ D = \frac{Z \times m}{a^3 \times N_A} \]

Substituting the values:
- \(Z = 2\)
- \(m = 7 \, \text{g/mol}\) (since 1 amu = 1 g/mol)
- \(a = 3.5 \times 10^{-8} \, \text{cm}\)

Calculating \(a^3\):
\[ a^3 = (3.5 \times 10^{-8})^3 = 4.287 \times 10^{-24} \, \text{cm}^3 \]

Now substituting into the density formula:
\[ D_{theoretical} = \frac{2 \times 7}{4.287 \times 10^{-24} \times 6.02 \times 10^{23}} \]

Calculating the denominator:
\[ 4.287 \times 10^{-24} \times 6.02 \times 10^{23} \approx 2.58 \times 10^{-1} \, \text{g} \]

Now calculating the theoretical density:
\[ D_{theoretical} = \frac{14}{2.58 \times 10^{-1}} \approx 54.2 \, \text{g/cm}^3 \]

### Step 4: Calculate the percentage of occupied lithium atoms
Using the formula:
\[ \text{Percentage of occupied atoms} = \left( \frac{D_{exp}}{D_{theoretical}} \right) \times 100 \]

Substituting the values:
\[ \text{Percentage of occupied atoms} = \left( \frac{0.53}{0.542} \right) \times 100 \approx 97.7\% \]

### Step 5: Calculate the percentage of missing lithium atoms
\[ \text{Percentage of missing atoms} = 100\% - \text{Percentage of occupied atoms} \]
\[ \text{Percentage of missing atoms} = 100\% - 97.7\% = 2.3\% \]

### Final Answer
The percentage of missing lithium atoms is **2.3%**.

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