If calcium crystallizes in bcc arrangement and the radius of Ca atom is 96 pm, and the volume of unit cell of Ca is `x xx 10^(-30)` m. Find the value of x.

Explanation

To solve the problem, we need to find the volume of the unit cell of calcium (Ca) that crystallizes in a body-centered cubic (BCC) arrangement, given the radius of the Ca atom is 96 pm (picometers).

### Step-by-Step Solution:

1. **Understanding the BCC Structure**:
In a BCC structure, there is one atom at each of the eight corners of the cube and one atom at the center of the cube. The body diagonal of the cube can be expressed in terms of the edge length (a) of the cube.

2. **Body Diagonal Calculation**:
The length of the body diagonal (d) in a cube is given by the formula:
\[ d = a\sqrt{3} \]
In a BCC structure, the body diagonal is also equal to the sum of the diameters of the atoms along that diagonal. Since there are two corner atoms and one body-centered atom, the relationship is:
\[ d = 4R \]
where \( R \) is the radius of the atom.

3. **Setting Up the Equation**:
Equating the two expressions for the body diagonal:
\[ a\sqrt{3} = 4R \]
Given that \( R = 96 \) pm, we can substitute this value into the equation:
\[ a\sqrt{3} = 4 \times 96 \text{ pm} \]
\[ a\sqrt{3} = 384 \text{ pm} \]

4. **Solving for Edge Length (a)**:
To find \( a \), we rearrange the equation:
\[ a = \frac{384 \text{ pm}}{\sqrt{3}} \approx \frac{384}{1.732} \approx 221.7 \text{ pm} \]

5. **Calculating the Volume of the Unit Cell**:
The volume \( V \) of the cubic unit cell is given by:
\[ V = a^3 \]
First, we convert \( a \) from picometers to meters:
\[ 221.7 \text{ pm} = 221.7 \times 10^{-12} \text{ m} \]
Now, we calculate the volume:
\[ V = (221.7 \times 10^{-12})^3 \text{ m}^3 \]
\[ V \approx 1.094 \times 10^{-30} \text{ m}^3 \]

6. **Finding the Value of x**:
The volume can be expressed in the form \( x \times 10^{-30} \text{ m}^3 \). From our calculation:
\[ x \approx 10.89 \]

### Final Answer:
The value of \( x \) is approximately **10.89**.