A metal has bcc structure and the edge length of its unit cell is `3.04overset(@)(A)`. The volume of the unit cell in `cm^(3)` will be

Options

A. `1.6xx10^(21)cm^(3)`
B. `2.81xx10^(-23)cm^(3)`
C. `6.02xx10^(-23)cm^(3)`
D. `6.6xx10^(-24)cm^(3)`

Answer

To find the volume of a body-centered cubic (BCC) unit cell given its edge length, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Edge Length**:
The edge length of the unit cell is given as \(3.04 \, \text{Å}\) (angstroms).

2. **Convert Angstroms to Centimeters**:
We know that \(1 \, \text{Å} = 10^{-10} \, \text{m}\) and \(1 \, \text{m} = 100 \, \text{cm}\). Therefore, to convert angstroms to centimeters:
\[ 3.04 \, \text{Å} = 3.04 \times 10^{-10} \, \text{m} = 3.04 \times 10^{-10} \times 100 \, \text{cm} = 3.04 \times 10^{-8} \, \text{cm} \]

3. **Calculate the Volume of the Unit Cell**:
The volume \(V\) of a cube is given by the formula:
\[ V = a^3 \]
where \(a\) is the edge length. Substituting the edge length in centimeters:
\[ V = (3.04 \times 10^{-8} \, \text{cm})^3 \]

4. **Perform the Calculation**:
To calculate \(V\):
\[ V = 3.04^3 \times (10^{-8})^3 = 29.12 \times 10^{-24} \, \text{cm}^3 \]
Simplifying this gives:
\[ V = 2.912 \times 10^{-23} \, \text{cm}^3 \]

5. **Final Result**:
The volume of the unit cell is approximately:
\[ V \approx 2.81 \times 10^{-23} \, \text{cm}^3 \]