The mean free path of nitrogen molecules at `27^(@)C` is `3xx10^(-7)m//s`. If the average speed of nitrogen molecules at the same temperature is 600 m/s then the collision frequency will be

Options

A. `10^(9)//sec`
B. `1.5xx10^(9)//sec`
C. `2xx10^(9)//sec`
D. `3xx10^(9)//sec`

Explanation

To find the collision frequency of nitrogen molecules at a temperature of 27°C, we can use the relationship between mean free path, average speed, and collision frequency. Here’s a step-by-step solution:

### Step 1: Understand the Mean Free Path
The mean free path (λ) is defined as the average distance traveled by a molecule between successive collisions. In this case, we have:
\[ \lambda = 3 \times 10^{-7} \text{ m} \]

### Step 2: Identify the Average Speed
The average speed (v) of nitrogen molecules at 27°C is given as:
\[ v = 600 \text{ m/s} \]

### Step 3: Calculate the Time Between Collisions
The time (t) between two successive collisions can be calculated using the formula:
\[ t = \frac{\lambda}{v} \]
Substituting the values we have:
\[ t = \frac{3 \times 10^{-7} \text{ m}}{600 \text{ m/s}} \]

### Step 4: Perform the Calculation
Calculating the above expression:
\[ t = \frac{3 \times 10^{-7}}{600} = 5 \times 10^{-10} \text{ seconds} \]

### Step 5: Calculate Collision Frequency
The collision frequency (ν) is the reciprocal of the time between collisions:
\[ \nu = \frac{1}{t} \]
Substituting the value of t:
\[ \nu = \frac{1}{5 \times 10^{-10}} \text{ s}^{-1} \]

### Step 6: Perform the Calculation for Collision Frequency
Calculating the collision frequency:
\[ \nu = 2 \times 10^{9} \text{ collisions per second} \]

### Final Answer
The collision frequency of nitrogen molecules at 27°C is:
\[ \nu = 2 \times 10^{9} \text{ s}^{-1} \]

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