If the number density of molecule of an ideal gas becomes 4 time then its mean free path becomes

Options

A. Half of its initial value
B. Double of its initial value
C. Four time of its initial value
D. One fourth of its initial value

Answer

To solve the problem, we need to understand the relationship between the mean free path (λ) of gas molecules and their number density (n). The mean free path is given by the formula:

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

where:
- \( \lambda \) is the mean free path,
- \( d \) is the diameter of the gas molecules,
- \( n \) is the number density of the gas molecules.

### Step-by-step Solution:

1. **Understand the Initial Conditions**:
- Let the initial number density be \( n \).
- The initial mean free path is \( \lambda \).

2. **Change in Number Density**:
- According to the problem, the number density becomes 4 times the initial value. Therefore, the new number density \( n' \) is:
\[ n' = 4n \]

3. **Substituting in the Mean Free Path Formula**:
- Now, we need to find the new mean free path \( \lambda' \) using the new number density \( n' \):
\[ \lambda' = \frac{1}{\sqrt{2} \pi d^2 n'} \]
- Substituting \( n' = 4n \) into the equation:
\[ \lambda' = \frac{1}{\sqrt{2} \pi d^2 (4n)} = \frac{1}{4} \cdot \frac{1}{\sqrt{2} \pi d^2 n} \]

4. **Relating New Mean Free Path to Initial Mean Free Path**:
- We know that the initial mean free path \( \lambda \) is:
\[ \lambda = \frac{1}{\sqrt{2} \pi d^2 n} \]
- Therefore, we can express \( \lambda' \) in terms of \( \lambda \):
\[ \lambda' = \frac{1}{4} \lambda \]

5. **Conclusion**:
- Thus, the new mean free path \( \lambda' \) becomes:
\[ \lambda' = \frac{\lambda}{4} \]

### Final Answer:
The mean free path becomes one-fourth of its initial value.