The mean free path and rms velocity of a nitrogen molecule at a temperature 17° C are 1.2× $10^{- 7}$ m and 5× $10^{2}$ m/s respectively. The time between two successive collisions

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$3.4 \times 10^{- 13} s$

$3.4 \times 10^{- 10} s$

$2.4 \times 10^{- 10} s$

$1.2 \times 10^{- 10} s$

answer is A.

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Detailed Solution

$Time between two successive collisions t^{} = \frac{meanfreepath}{V_{r m s}} = \frac{1.2 \times 10^{- 7}}{5 \times 10^{2}} = 2.4 \times 10^{- 10} seconds.$

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