Calculate themean 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 (L) of a molecule of a gas, we can use the formula:

\[ L = \frac{1}{\sqrt{2} \pi D^2 n} \]

where:
- \(D\) is the diameter of the molecule,
- \(n\) is the number density (number of molecules per cm³).

**Step 1: Identify the given values.**
- Number density, \(n = 2 \times 10^8 \, \text{cm}^{-3}\)
- Diameter of the molecule, \(D = 10^{-5} \, \text{cm}\)

**Step 2: Calculate \(D^2\).**
\[ D^2 = (10^{-5})^2 = 10^{-10} \, \text{cm}^2 \]

**Step 3: Substitute the values into the mean free path formula.**
\[ L = \frac{1}{\sqrt{2} \pi (10^{-10}) (2 \times 10^8)} \]

**Step 4: Calculate \(\sqrt{2}\) and \(\pi\).**
- \(\sqrt{2} \approx 1.414\)
- \(\pi \approx 3.14\)

**Step 5: Substitute these values into the equation.**
\[ L = \frac{1}{1.414 \times 3.14 \times 10^{-10} \times 2 \times 10^8} \]

**Step 6: Simplify the expression.**
\[ L = \frac{1}{1.414 \times 3.14 \times 2 \times 10^{-2}} \]

**Step 7: Calculate the denominator.**
\[ 1.414 \times 3.14 \approx 4.442 \]
\[ 4.442 \times 2 \approx 8.884 \]

**Step 8: Substitute back into the equation.**
\[ L \approx \frac{1}{8.884 \times 10^{-2}} = \frac{100}{8.884} \approx 11.26 \, \text{cm} \]

**Step 9: Round the answer.**
Thus, the mean free path \(L \approx 11.3 \, \text{cm}\).

### Final Answer:
The mean free path of the molecule is approximately \(11.3 \, \text{cm}\).

---