Mean free path for an ideal gas is to be observed 20 μm while average speed of molecules of gas is observed

Question

Mean free path for an ideal gas is to be observed \(20 \ \mu m\) while average speed of molecules of gas is observed to be 600 m/s. Then frequency of collision is nearly

(1) \(4\times 10^7\)

(2) \(1.2\times 10^7\)

(3) \(3\times 10^7\)

(4) \(2\times 10^{-7}\)

Answer 1

Correct option is : (3) \(3\times 10^7\)

Mean free time \(\tau = \frac{\lambda}{v}\Rightarrow f = \frac{v}{\lambda} = \frac{600}{20\times 10^{-6}}\)

\( = 3\times 10^7\)