Question

The number of air molecules per cm³ increased from 3 × 10¹⁹ to 12 × 10¹⁹. The ratio of collision frequency of air molecules before and after the increase in the number respectively is :

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

Answer 1

The Correct Option is A

To solve this problem, we need to understand the relationship between the number of air molecules and their collision frequency. Collision frequency is directly proportional to the number of molecules in a given volume.

The collision frequency, \( Z \), can be expressed by the formula:

Z \propto n

where \( n \) is the number of molecules per unit volume.

Let's denote:

• \( n_1 = 3 \times 10^{19} \) molecules/cm³ (initial number of molecules)
• \( n_2 = 12 \times 10^{19} \) molecules/cm³ (final number of molecules)

The collision frequency ratio \( \frac{Z_1}{Z_2} \) can be found using the ratio of the number of molecules:

\frac{Z_1}{Z_2} = \frac{n_1}{n_2}

Substituting the given values:

\frac{Z_1}{Z_2} = \frac{3 \times 10^{19}}{12 \times 10^{19}}

On simplification:

\frac{Z_1}{Z_2} = \frac{3}{12} = \frac{1}{4} = 0.25

Thus, the ratio of the collision frequency of air molecules before and after the increase is 0.25.

Conclusion: The correct answer is 0.25.