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The mean free path of nitrogen molecules at 27°C is 3 × 10^{-7} m/s. if the average speed of nitrogen molecules at the same temperature is 600 m/s then the collision frequency will be

  • Option 1)

    10^{9}/ sec

  • Option 2)

    1.5 *10^{9}/sec

  • Option 3)

    2 *10^{9}/sec

  • Option 4)

    3 *10^{9}/sec

 

Answers (1)

best_answer

As we learnt in 

Mean free path -

The average distance travelled  by a molecule between two successive collision.

-

 

 Mean free path =3\times 10^{-7}m

Speed = 600\ m/s

Collision Frequency =\frac{1}{T}=\frac{1}{(L/V)}=\frac{V}{L}=\frac{600 }{3 \times 10^{-7}}=2 \times 10^9 /sec


Option 1)

10^{9}/ sec

Incorrect

Option 2)

1.5 *10^{9}/sec

Incorrect

Option 3)

2 *10^{9}/sec

Correct

Option 4)

3 *10^{9}/sec

Incorrect

Posted by

prateek

View full answer

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