Calculate the mean free path of molecule of a gas having number density (number of molecules per `cm^3`) `2xx10^(8)` and the diameter of the molecule is `10^(-5)` cm.

Explanation

To calculate the mean free path (λ) of a molecule in a gas, we can use the formula:

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

where:
- \( \lambda \) = mean free path
- \( d \) = diameter of the molecule
- \( n \) = number density (number of molecules per cm³)

Given:
- Number density \( n = 2 \times 10^8 \, \text{cm}^{-3} \)
- Diameter \( d = 10^{-5} \, \text{cm} \)

### Step 1: Calculate \( d^2 \)
First, we need to calculate \( d^2 \):

\[ d^2 = (10^{-5})^2 = 10^{-10} \, \text{cm}^2 \]

### Step 2: Substitute values into the mean free path formula
Now we can substitute the values of \( d^2 \) and \( n \) into the mean free path formula:

\[ \lambda = \frac{1}{\sqrt{2} \cdot \pi \cdot (10^{-10}) \cdot (2 \times 10^8)} \]

### Step 3: Calculate the denominator
Now calculate the denominator step by step:

1. Calculate \( \sqrt{2} \):
\[ \sqrt{2} \approx 1.414 \]

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

3. Now calculate \( \sqrt{2} \cdot \pi \):
\[ \sqrt{2} \cdot \pi \approx 1.414 \cdot 3.14 \approx 4.442 \]

4. Now multiply this by \( d^2 \) and \( n \):
\[ 4.442 \cdot (10^{-10}) \cdot (2 \times 10^8) = 4.442 \cdot 2 \cdot 10^{-2} = 8.884 \cdot 10^{-2} \]

### Step 4: Calculate the mean free path
Now we can calculate \( \lambda \):

\[ \lambda = \frac{1}{8.884 \times 10^{-2}} \approx 11.26 \, \text{cm} \]

### Final Answer
Thus, the mean free path of the molecule is approximately:

\[ \lambda \approx 11.26 \, \text{cm} \]

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