In a certain gas the molecules are `5.0 xx 10^(-9)`m apart on average, have a mean free path of `5.0 xx 10^(-6)` m, and have an average speed of 500 m/s. The rate at which a molecule has collision with other molecules is about

Options

A. `10^(-11)//s`
B. `10^(-8)//S`
C. `1//s`
D. `10^8 //s`

Answer

To solve the problem of finding the collision frequency of gas molecules, we can follow these steps:

### Step 1: Understand the Concept of Collision Frequency
Collision frequency (f) is defined as the number of collisions per unit time for a single molecule. It can be calculated using the formula:

\[ f = \frac{V}{\lambda} \]

where:
- \( V \) is the average speed of the molecules,
- \( \lambda \) is the mean free path.

### Step 2: Identify the Given Values
From the problem, we have:
- Average speed \( V = 500 \, \text{m/s} \)
- Mean free path \( \lambda = 5.0 \times 10^{-6} \, \text{m} \)

### Step 3: Substitute the Values into the Formula
Now, we can substitute the given values into the collision frequency formula:

\[ f = \frac{500 \, \text{m/s}}{5.0 \times 10^{-6} \, \text{m}} \]

### Step 4: Perform the Calculation
Calculating the above expression:

\[ f = \frac{500}{5.0 \times 10^{-6}} \]

\[ f = 500 \times \frac{1}{5.0 \times 10^{-6}} \]

\[ f = 500 \times 2 \times 10^{5} \]

\[ f = 1000 \times 10^{5} \]

\[ f = 10^{8} \, \text{s}^{-1} \]

### Step 5: Conclusion
The collision frequency, or the rate at which a molecule collides with other molecules, is:

\[ f = 10^{8} \, \text{collisions per second} \]

### Summary
Thus, the rate at which a molecule collides with other molecules is approximately \( 10^{8} \, \text{s}^{-1} \).

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