Lithium has a bcc structure .its density is `530Kgm^(-1)` and its atomic mass is `6.94 g mol^(-1)`. Calculate the edge length of a unit cell of Lithium metal. `(N_(A)=6.02xx10^(23)mol^(-1))`

Options

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

Explanation

To calculate the edge length of a unit cell of lithium metal, we can follow these steps:

### Step 1: Understand the given data
- Density of lithium (d) = 530 kg/m³ = 0.530 g/cm³
- Atomic mass of lithium (M) = 6.94 g/mol
- Avogadro's number (N_A) = 6.02 × 10²³ mol⁻¹
- For a body-centered cubic (bcc) structure, the number of atoms per unit cell (Z) = 2.

### Step 2: Write the formula for density
The density (d) of a crystalline solid can be expressed using the formula:
\[ d = \frac{Z \cdot M}{A^3 \cdot N_A} \]
Where:
- \(d\) = density
- \(Z\) = number of atoms per unit cell
- \(M\) = molar mass
- \(A\) = edge length of the unit cell
- \(N_A\) = Avogadro's number

### Step 3: Rearrange the formula to find the edge length (A)
Rearranging the formula to solve for \(A\):
\[ A^3 = \frac{Z \cdot M}{d \cdot N_A} \]
\[ A = \left(\frac{Z \cdot M}{d \cdot N_A}\right)^{1/3} \]

### Step 4: Substitute the known values into the equation
Now, substituting the known values into the equation:
- \(Z = 2\)
- \(M = 6.94 \, \text{g/mol}\)
- \(d = 0.530 \, \text{g/cm}^3\)
- \(N_A = 6.02 \times 10^{23} \, \text{mol}^{-1}\)

\[ A = \left(\frac{2 \cdot 6.94}{0.530 \cdot 6.02 \times 10^{23}}\right)^{1/3} \]

### Step 5: Calculate the value inside the parentheses
Calculating the value:
\[ A = \left(\frac{13.88}{3.1966 \times 10^{23}}\right)^{1/3} \]
\[ A = \left(4.342 \times 10^{-23}\right)^{1/3} \]

### Step 6: Calculate the cube root
Calculating the cube root:
\[ A \approx 3.52 \times 10^{-8} \, \text{cm} \]

### Step 7: Convert to picometers
To convert centimeters to picometers:
\[ 1 \, \text{cm} = 10^{10} \, \text{pm} \]
\[ A \approx 3.52 \times 10^{-8} \, \text{cm} \times 10^{10} \, \text{pm/cm} = 352 \, \text{pm} \]

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

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