The average speed of air molecules is `485 "ms"^(-1)` . At STP the number density is `2.7xx10^(25)m^(-3)` and diameter of the air molecule is `2xx10^(-10)` m . The value of mean free path for the air molecule is

Options

A. `2.5xx10^(-7)m`
B. `2.9xx10^(-7)m`
C. `3.5xx10^(-7)m`
D. `3.9xx10^(-7)m`

Explanation

To calculate the mean free path (λ) of air molecules, we can use the following formula:

\[ \lambda = \frac{1}{\sqrt{2} \pi n d^2} \]

Where:
- \( n \) is the number density of the molecules (in \( m^{-3} \)),
- \( d \) is the diameter of the molecules (in meters).

### Step 1: Write down the given data
- Average speed of air molecules, \( v = 485 \, \text{m/s} \) (not needed for this calculation)
- Number density, \( n = 2.7 \times 10^{25} \, \text{m}^{-3} \)
- Diameter of air molecules, \( d = 2 \times 10^{-10} \, \text{m} \)

### Step 2: Substitute the values into the formula
We need to calculate \( d^2 \):
\[ d^2 = (2 \times 10^{-10})^2 = 4 \times 10^{-20} \, \text{m}^2 \]

Now substitute \( n \) and \( d^2 \) into the mean free path formula:
\[ \lambda = \frac{1}{\sqrt{2} \pi (2.7 \times 10^{25}) (4 \times 10^{-20})} \]

### Step 3: Calculate the denominator
Calculate \( \sqrt{2} \):
\[ \sqrt{2} \approx 1.414 \]

Calculate \( \pi \):
\[ \pi \approx 3.14 \]

Now calculate the denominator:
\[ \text{Denominator} = \sqrt{2} \cdot \pi \cdot n \cdot d^2 = 1.414 \cdot 3.14 \cdot (2.7 \times 10^{25}) \cdot (4 \times 10^{-20}) \]

Calculating this step-by-step:
1. \( 1.414 \cdot 3.14 \approx 4.442 \)
2. \( 4.442 \cdot 2.7 \approx 12.0 \)
3. \( 12.0 \cdot 4 \approx 48.0 \)
4. \( 48.0 \cdot 10^{25} \cdot 10^{-20} = 48.0 \times 10^{5} = 4.8 \times 10^{6} \)

### Step 4: Calculate the mean free path
Now substitute this back into the formula for λ:
\[ \lambda = \frac{1}{4.8 \times 10^{6}} \approx 2.08 \times 10^{-7} \, \text{m} \]

### Final Answer
The mean free path of the air molecule is approximately:
\[ \lambda \approx 2.08 \times 10^{-7} \, \text{m} \]