A bcc lattice is made up of hollow spheres of `B`. Spheres of solids `A` are present in hollow spheres of `B`. The radius of `A` is half of the radius of `B`. The ratio of total volume of spheres of `B` unoccupied by `A` in a unit cell and volume of unit cell is `A xx (pisqrt(3))/(64)`. Find the value of `A`.

Options

A. `8`
B. `7`
C. `24`
D. None of these

Answer

To solve the problem step by step, let's break down the information given and derive the required value of \( A \).

### Step 1: Define the Radii
Let the radius of sphere \( B \) be \( r \). According to the problem, the radius of sphere \( A \) is half of the radius of \( B \):
\[ r_A = \frac{r}{2} \]

### Step 2: Determine the Edge Length of the BCC Unit Cell
In a Body-Centered Cubic (BCC) lattice, the relationship between the edge length \( a \) and the radius \( r \) of the spheres is given by:
\[ \sqrt{3} a = 4r \]
From this, we can express the edge length \( a \):
\[ a = \frac{4r}{\sqrt{3}} \]

### Step 3: Calculate the Volume of the Unit Cell
The volume \( V_{cell} \) of the unit cell is given by:
\[ V_{cell} = a^3 = \left(\frac{4r}{\sqrt{3}}\right)^3 = \frac{64r^3}{3\sqrt{3}} \]

### Step 4: Calculate the Volume of Spheres of \( B \)
In a BCC unit cell, there are 2 spheres of \( B \) (1 at the body center and 8 at the corners, each contributing \( \frac{1}{8} \)):
\[ V_B = 2 \times \frac{4}{3} \pi r^3 = \frac{8}{3} \pi r^3 \]

### Step 5: Calculate the Volume of Spheres of \( A \)
The volume of the spheres of \( A \) is:
\[ V_A = 2 \times \frac{4}{3} \pi \left(\frac{r}{2}\right)^3 = 2 \times \frac{4}{3} \pi \frac{r^3}{8} = \frac{1}{3} \pi r^3 \]

### Step 6: Calculate the Volume of Spheres of \( B \) Unoccupied by \( A \)
The volume of spheres of \( B \) unoccupied by \( A \) is:
\[ V_{B \text{ unoccupied}} = V_B - V_A = \frac{8}{3} \pi r^3 - \frac{1}{3} \pi r^3 = \frac{7}{3} \pi r^3 \]

### Step 7: Calculate the Ratio of Volumes
Now, we need to find the ratio of the volume of spheres of \( B \) unoccupied by \( A \) to the volume of the unit cell:
\[ \text{Ratio} = \frac{V_{B \text{ unoccupied}}}{V_{cell}} = \frac{\frac{7}{3} \pi r^3}{\frac{64r^3}{3\sqrt{3}}} \]
This simplifies to:
\[ \text{Ratio} = \frac{7 \pi r^3}{64 r^3 / \sqrt{3}} = \frac{7 \pi \sqrt{3}}{64} \]

### Step 8: Set the Ratio Equal to the Given Expression
According to the problem, this ratio is equal to:
\[ A \cdot \frac{\pi \sqrt{3}}{64} \]
Thus, we have:
\[ \frac{7 \pi \sqrt{3}}{64} = A \cdot \frac{\pi \sqrt{3}}{64} \]

### Step 9: Solve for \( A \)
Cancelling \( \frac{\pi \sqrt{3}}{64} \) from both sides gives:
\[ A = 7 \]

### Final Answer
The value of \( A \) is:
\[ \boxed{7} \]