In a cubic unit cell atoms 'A' occupies the corners and centre, atoms 'B' occupies the face centres, and atoms 'C' occupies the edge centres. If all the atoms on one of the planes which is bisecting the unit cell is to two equal halves, are removed. Calculate the total number of atoms (X), present in the unit cell. What is the value of (X/4) ? (Given that, the plane is not a diagonal plane)

Explanation

To solve the problem, we need to calculate the total number of atoms present in the cubic unit cell before and after removing the atoms from the specified plane. Let's break down the solution step by step.

### Step 1: Calculate the contribution of atoms A
Atoms A occupy the corners and the center of the cubic unit cell.

- There are 8 corners in a cube, and each corner atom contributes \( \frac{1}{8} \) to the unit cell.
- The contribution from the corners is:
\[ \text{Contribution from corners} = 8 \times \frac{1}{8} = 1 \]
- There is 1 atom at the center, which contributes 1 fully:
\[ \text{Contribution from center} = 1 \]
- Therefore, the total contribution from atoms A is:
\[ \text{Total from A} = 1 + 1 = 2 \]

### Step 2: Calculate the contribution of atoms B
Atoms B occupy the face centers of the cubic unit cell.

- There are 6 faces in a cube, and each face center atom contributes \( \frac{1}{2} \) to the unit cell.
- The contribution from the face centers is:
\[ \text{Contribution from face centers} = 6 \times \frac{1}{2} = 3 \]

### Step 3: Calculate the contribution of atoms C
Atoms C occupy the edge centers of the cubic unit cell.

- There are 12 edges in a cube, and each edge center atom contributes \( \frac{1}{4} \) to the unit cell.
- The contribution from the edge centers is:
\[ \text{Contribution from edge centers} = 12 \times \frac{1}{4} = 3 \]

### Step 4: Calculate the total number of atoms before removal
Now, we can sum the contributions from all types of atoms:
\[ \text{Total atoms before removal} = \text{Total from A} + \text{Total from B} + \text{Total from C} = 2 + 3 + 3 = 8 \]

### Step 5: Determine the number of atoms removed
According to the problem, we need to remove the atoms on a plane bisecting the unit cell.

- The plane will cut through the center atom (1 atom removed).
- It will also cut through 4 face center atoms (each contributing \( \frac{1}{2} \)), so:
\[ \text{Contribution from face centers removed} = 4 \times \frac{1}{2} = 2 \]
- Additionally, it will cut through 4 edge center atoms (each contributing \( \frac{1}{4} \)), so:
\[ \text{Contribution from edge centers removed} = 4 \times \frac{1}{4} = 1 \]

### Step 6: Calculate the total number of atoms removed
Now, we can sum the contributions of the removed atoms:
\[ \text{Total atoms removed} = 1 + 2 + 1 = 4 \]

### Step 7: Calculate the total number of atoms remaining
Now, we subtract the number of atoms removed from the total number of atoms before removal:
\[ \text{Total atoms remaining} = 8 - 4 = 4 \]

### Step 8: Calculate the value of \( \frac{X}{4} \)
Let \( X \) be the total number of atoms remaining, which we found to be 4. Therefore:
\[ \frac{X}{4} = \frac{4}{4} = 1 \]

### Final Answer
The total number of atoms \( X \) present in the unit cell is 4, and the value of \( \frac{X}{4} \) is 1.

---