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

Collision frequency is given by 𝑍 = 𝑛𝜋𝑑 2𝑉𝑎𝑣𝑔, where n is number of molecules per unit volume.
𝑍1/ 𝑍2 = 𝑛1 /𝑛2 = 3/ 12 = 1/ 4 = 0.25

Answer 2

Collision Frequency Problem

Step 1: Collision Frequency Formula

The collision frequency (\( Z \)) is given by:

\( Z = n \pi d^2 v_{avg} \)

where \( n \) is the number of molecules per unit volume, \( d \) is the diameter of the molecules, and \( v_{avg} \) is the average speed.

Step 2: Ratio of Collision Frequencies

Let \( Z_1 \) and \( Z_2 \) be the collision frequencies before and after the increase in the number of molecules, respectively. Let \( n_1 \) and \( n_2 \) be the corresponding number densities. Since the diameter and average speed of the molecules remain constant, we can write:

\( \frac{Z_1}{Z_2} = \frac{n_1 \pi d^2 v_{avg}}{n_2 \pi d^2 v_{avg}} = \frac{n_1}{n_2} \)

Step 3: Substitute Given Values

Given \( n_1 = 3 \times 10^{19} \) and \( n_2 = 12 \times 10^{19} \):

\( \frac{Z_1}{Z_2} = \frac{3 \times 10^{19}}{12 \times 10^{19}} = \frac{3}{12} = \frac{1}{4} = 0.25 \)

Conclusion:

The ratio of collision frequencies before and after the increase is 0.25 (Option 1).