The fraction of total volume occupied by the atom present in a simple cubic is

Options

A. `pi/4`
B. `pi/6`
C. `pi/(3sqrt2)`
D. `pi/(4sqrt2)`

Answer

To find the fraction of total volume occupied by the atoms present in a simple cubic structure, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Simple Cubic Structure**:
- In a simple cubic lattice, atoms are located at the corners of the cube.
- Each corner atom is shared by 8 adjacent cubes.

2. **Calculating the Total Number of Atoms**:
- Since there is one atom at each of the 8 corners, the total contribution of atoms per cube is:
\[ \text{Total atoms} = 8 \times \frac{1}{8} = 1 \text{ atom} \]

3. **Volume of One Atom**:
- The volume of a single atom (considering it as a sphere) is given by:
\[ V_{\text{atom}} = \frac{4}{3} \pi r^3 \]
- Here, \( r \) is the radius of the atom.

4. **Relating Edge Length and Radius**:
- In a simple cubic structure, the edge length \( a \) of the cube is related to the radius \( r \) of the atom by:
\[ a = 2r \]
- Therefore, we can express \( r \) in terms of \( a \):
\[ r = \frac{a}{2} \]

5. **Volume of the Cube**:
- The volume of the cube is given by:
\[ V_{\text{cube}} = a^3 \]

6. **Substituting for Volume of Atom**:
- Substitute \( r = \frac{a}{2} \) into the volume of the atom:
\[ V_{\text{atom}} = \frac{4}{3} \pi \left(\frac{a}{2}\right)^3 = \frac{4}{3} \pi \frac{a^3}{8} = \frac{4}{24} \pi a^3 = \frac{\pi a^3}{6} \]

7. **Calculating the Fraction of Volume Occupied**:
- The fraction of the total volume occupied by the atom is:
\[ \text{Fraction} = \frac{V_{\text{atom}}}{V_{\text{cube}}} = \frac{\frac{\pi a^3}{6}}{a^3} \]
- Simplifying this gives:
\[ \text{Fraction} = \frac{\pi}{6} \]

### Final Answer:
The fraction of total volume occupied by the atom present in a simple cubic structure is:
\[ \frac{\pi}{6} \]