If calcium crystallizes in `bc c` arrangement and the radius of `Ca` atom is `96 pm`, then the volume of unit cell of `Ca` is

Options

A. `10.9 xx 10^(-36) m^(3)`
B. `10.9 xx 10^(-30) m^(3)`
C. `21.8 xx 10^(-30) m^(3)`
D. `21.8 xx 10^(-36) m^(3)`

Explanation

To find the volume of the unit cell of calcium (Ca) crystallizing in a body-centered cubic (BCC) arrangement, we will follow these steps:

### Step 1: Understand the relationship between the edge length (A) and the radius (R) in a BCC structure.
In a BCC structure, the body diagonal of the cube is equal to \(4R\) (where R is the radius of the atom). The body diagonal can also be expressed in terms of the edge length (A) as \(\sqrt{3}A\).

### Step 2: Set up the equation.
From the relationship established, we can write:
\[ \sqrt{3}A = 4R \]

### Step 3: Solve for the edge length (A).
Rearranging the equation gives:
\[ A = \frac{4R}{\sqrt{3}} \]
Substituting \(R = 96 \, \text{pm} = 96 \times 10^{-12} \, \text{m}\):
\[ A = \frac{4 \times 96 \times 10^{-12}}{\sqrt{3}} \approx \frac{384 \times 10^{-12}}{1.732} \approx 221.7 \times 10^{-12} \, \text{m} \]

### Step 4: Calculate the volume of the unit cell.
The volume (V) of the cubic unit cell is given by:
\[ V = A^3 \]
Substituting the value of A:
\[ V = (221.7 \times 10^{-12})^3 \]
Calculating this gives:
\[ V \approx 10.9 \times 10^{-30} \, \text{m}^3 \]

### Final Answer:
The volume of the unit cell of calcium is approximately \(10.9 \times 10^{-30} \, \text{m}^3\).

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