The number of air molecules per `cm^3` increased from `3 times 10^(19)` to `12 times10^(19)`. The ratio of collision frequency of air molecules before and after the increase in number respectively is :

Options

A. `0.75`
B. `1.25`
C. `0.50`
D. `0.25`

Answer

To solve the problem, we need to find the ratio of the collision frequency of air molecules before and after the increase in the number of molecules.

### Step-by-Step Solution:

1. **Identify the Initial and Final Number of Molecules:**
- Initial number of air molecules per cm³, \( n_1 = 3 \times 10^{19} \)
- Final number of air molecules per cm³, \( n_2 = 12 \times 10^{19} \)

2. **Understand the Relationship Between Collision Frequency and Number of Molecules:**
- The collision frequency \( f \) of air molecules is directly proportional to the number of molecules per unit volume. This can be expressed as:
\[ f \propto n \]
- Therefore, the ratio of collision frequencies before and after the increase can be expressed as:
\[ \frac{f_1}{f_2} = \frac{n_1}{n_2} \]

3. **Substitute the Values:**
- Substitute the values of \( n_1 \) and \( n_2 \) into the equation:
\[ \frac{f_1}{f_2} = \frac{3 \times 10^{19}}{12 \times 10^{19}} \]

4. **Simplify the Ratio:**
- The \( 10^{19} \) in the numerator and denominator cancels out:
\[ \frac{f_1}{f_2} = \frac{3}{12} = \frac{1}{4} \]

5. **Convert the Ratio to Decimal Form:**
- The ratio \( \frac{1}{4} \) can also be expressed as:
\[ \frac{f_1}{f_2} = 0.25 \]

6. **Conclusion:**
- The ratio of the collision frequency of air molecules before and after the increase is \( 0.25 \).

### Final Answer:
The ratio of collision frequency of air molecules before and after the increase is \( \frac{1}{4} \) or \( 0.25 \).

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