Density of lithium atom is 0.53 `g//cm^(3)`. The edge length of Li is `3.5 A^.The number of lithium atoms in a unit cell will be.. .
(Atomic mass of lithium is 6.94)

Options

A. 2
B. 1
C. 4
D. 6

Explanation

To find the number of lithium atoms in a unit cell (denoted as Z), we can use the relationship between density, atomic mass, and the volume of the unit cell. Here’s a step-by-step solution:

### Step 1: Convert edge length from Angstroms to centimeters
The edge length of lithium is given as \(3.5 \, \text{Å}\). To convert this to centimeters:
\[ 1 \, \text{Å} = 10^{-8} \, \text{cm} \]
Thus,
\[ \text{Edge length in cm} = 3.5 \, \text{Å} \times 10^{-8} \, \text{cm/Å} = 3.5 \times 10^{-8} \, \text{cm} \]

### Step 2: Calculate the volume of the unit cell
The volume \(V\) of the cubic unit cell can be calculated using the formula:
\[ V = a^3 \]
where \(a\) is the edge length. Substituting the value we found:
\[ V = (3.5 \times 10^{-8} \, \text{cm})^3 = 4.2875 \times 10^{-24} \, \text{cm}^3 \]

### Step 3: Use the density formula to find Z
The formula for density (\(d\)) is given by:
\[ d = \frac{Z \cdot m}{N_a \cdot V} \]
where:
- \(Z\) = number of atoms in the unit cell
- \(m\) = atomic mass of lithium (6.94 g/mol)
- \(N_a\) = Avogadro's number (\(6.022 \times 10^{23} \, \text{mol}^{-1}\))
- \(V\) = volume of the unit cell

Rearranging the formula to solve for \(Z\):
\[ Z = \frac{d \cdot N_a \cdot V}{m} \]

### Step 4: Substitute the values into the equation
Substituting the known values:
- Density \(d = 0.53 \, \text{g/cm}^3\)
- \(N_a = 6.022 \times 10^{23} \, \text{mol}^{-1}\)
- Volume \(V = 4.2875 \times 10^{-24} \, \text{cm}^3\)
- Atomic mass \(m = 6.94 \, \text{g/mol}\)

So,
\[ Z = \frac{0.53 \, \text{g/cm}^3 \times 6.022 \times 10^{23} \, \text{mol}^{-1} \times 4.2875 \times 10^{-24} \, \text{cm}^3}{6.94 \, \text{g/mol}} \]

### Step 5: Calculate Z
Calculating the numerator:
\[ 0.53 \times 6.022 \times 10^{23} \times 4.2875 \times 10^{-24} \approx 1.36 \]
Now, dividing by the atomic mass:
\[ Z \approx \frac{1.36}{6.94} \approx 0.196 \]

### Step 6: Round Z to the nearest whole number
Since the number of atoms must be a whole number, we round \(Z\) to the nearest whole number:
\[ Z \approx 2 \]

### Final Answer
The number of lithium atoms in a unit cell is approximately \(Z = 2\).

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