A body centred cubic lattice is made up of hollow sphere of B. Sphere of solid A are present in hollow sphere of B. Radius of A is half of the radius of B. What is the ratio of total volume of sphere B unoccupied by A in unit cell and volume of unit cell?

Options

A. `(29 pi sqrt(3))/64`
B. `(7pi sqrt(3))/(64)`
C. `(19 pi sqrt(3))/(64)`
D. `(2pi sqrt(3))/64`

Answer

To solve the problem, we need to find the ratio of the total volume of sphere B unoccupied by A in a unit cell to the volume of the unit cell. Let's break down the solution step by step.

### Step 1: Understand the Structure
In a body-centered cubic (BCC) lattice:
- There are atoms at the corners and one atom at the center of the cube.
- Each corner atom contributes \( \frac{1}{8} \) of its volume to the unit cell, and the center atom contributes fully.

### Step 2: Determine the Number of Atoms
For sphere B:
- There are 8 corner atoms contributing \( 8 \times \frac{1}{8} = 1 \) atom.
- There is 1 atom at the body center.
- Therefore, the total number of atoms of B in the unit cell is \( 1 + 1 = 2 \).

### Step 3: Calculate the Volume of Sphere B
Let the radius of sphere B be \( R \). The volume of one sphere B is given by:
\[ V_B = \frac{4}{3} \pi R^3 \]
Thus, the total volume of B in the unit cell is:
\[ V_{B,\text{total}} = 2 \times \frac{4}{3} \pi R^3 = \frac{8}{3} \pi R^3 \]

### Step 4: Determine the Volume of Sphere A
Given that the radius of sphere A is half of the radius of sphere B, we have:
\[ r_A = \frac{R}{2} \]
The volume of one sphere A is:
\[ V_A = \frac{4}{3} \pi \left(\frac{R}{2}\right)^3 = \frac{4}{3} \pi \frac{R^3}{8} = \frac{1}{6} \pi R^3 \]
Since there are 2 spheres A (one in each hollow sphere B), the total volume of A in the unit cell is:
\[ V_{A,\text{total}} = 2 \times \frac{1}{6} \pi R^3 = \frac{1}{3} \pi R^3 \]

### Step 5: Calculate the Unoccupied Volume of Sphere B
The unoccupied volume of sphere B in the unit cell is:
\[ V_{\text{unoccupied}} = V_{B,\text{total}} - V_{A,\text{total}} = \frac{8}{3} \pi R^3 - \frac{1}{3} \pi R^3 = \frac{7}{3} \pi R^3 \]

### Step 6: Calculate the Volume of the Unit Cell
In a BCC structure, the relationship between the radius \( R \) and the unit cell edge length \( a \) is given by:
\[ 4R = \sqrt{3} a \quad \Rightarrow \quad a = \frac{4R}{\sqrt{3}} \]
The volume of the unit cell is:
\[ V_{\text{unit cell}} = a^3 = \left(\frac{4R}{\sqrt{3}}\right)^3 = \frac{64R^3}{3\sqrt{3}} \]

### Step 7: Find the Ratio
Now, we can find the ratio of the unoccupied volume of B to the volume of the unit cell:
\[ \text{Ratio} = \frac{V_{\text{unoccupied}}}{V_{\text{unit cell}}} = \frac{\frac{7}{3} \pi R^3}{\frac{64R^3}{3\sqrt{3}}} \]
Simplifying this gives:
\[ \text{Ratio} = \frac{7 \pi R^3}{64 R^3 / \sqrt{3}} = \frac{7 \pi \sqrt{3}}{64} \]

### Final Answer
The ratio of the total volume of sphere B unoccupied by A in the unit cell to the volume of the unit cell is:
\[ \frac{7 \pi \sqrt{3}}{64} \]