If the total volume of a simple cubic unit cell is `6.817×10^(-23) cm^3`, what is the volume occupied by particles in unit cell?

Answer

To find the volume occupied by particles in a simple cubic unit cell, we can follow these steps:

### Step 1: Understand the structure of a simple cubic unit cell
In a simple cubic unit cell, there is one atom located at each of the eight corners of the cube. However, each corner atom is shared among eight adjacent unit cells. Therefore, the contribution of each corner atom to the unit cell is \( \frac{1}{8} \).

### Step 2: Calculate the effective number of atoms (Z) in the unit cell
Since there are 8 corners in a simple cubic unit cell and each contributes \( \frac{1}{8} \) of an atom, the effective number of atoms (Z) in the unit cell is:
\[ Z = 8 \times \frac{1}{8} = 1 \]

### Step 3: Calculate the volume occupied by the particles in the unit cell
The volume occupied by the particles in the unit cell can be calculated using the formula for the volume of a sphere:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the atom.

### Step 4: Relate the edge length (a) to the radius (r)
In a simple cubic structure, the edge length \( a \) is related to the radius \( r \) of the atom by the equation:
\[ a = 2r \]

### Step 5: Calculate the volume of the unit cell
The total volume of the unit cell is given as \( 6.817 \times 10^{-23} \, \text{cm}^3 \).

### Step 6: Calculate the volume occupied by the particles
Since we know that there is only one effective atom in the unit cell, the volume occupied by the particles in the unit cell is equal to the volume of one atom:
\[ V_{\text{occupied}} = \frac{4}{3} \pi r^3 \]

### Step 7: Find the radius using the total volume
Since we know the total volume of the unit cell, we can express it in terms of the edge length:
\[ V_{\text{cell}} = a^3 = (2r)^3 = 8r^3 \]
Setting this equal to the total volume:
\[ 8r^3 = 6.817 \times 10^{-23} \]
Now, solve for \( r^3 \):
\[ r^3 = \frac{6.817 \times 10^{-23}}{8} = 8.52125 \times 10^{-24} \]

### Step 8: Calculate \( r \)
Now, take the cube root to find \( r \):
\[ r = (8.52125 \times 10^{-24})^{1/3} \approx 2.027 \times 10^{-8} \, \text{cm} \]

### Step 9: Calculate the volume occupied by the particles
Now substitute \( r \) back into the volume formula for one atom:
\[ V_{\text{occupied}} = \frac{4}{3} \pi (2.027 \times 10^{-8})^3 \]
Calculating this gives:
\[ V_{\text{occupied}} \approx 3.48 \times 10^{-24} \, \text{cm}^3 \]

### Final Answer
The volume occupied by particles in the unit cell is approximately \( 3.48 \times 10^{-24} \, \text{cm}^3 \).

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