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JEE Mains · Physics · STD 11 - Kinetic theory of gases
  • JEE Mains 2023
  • Medium
  • MCQ

The number of air molecules per \(\mathrm{cm}^3\) increased from \(3 \times 10^{19}\) to \(12 \times 10^{19}\). The ratio of collision frequency of air molecules before and after the increase in number respectively is \(.........\)

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Answer & Solution

Correct Answer

(B) \(0.25\)

Step-by-step Solution

Detailed explanation

Collision frequency,

\(\mathrm{f}=\frac{\mathrm{V}}{\lambda}=\frac{\mathrm{V}}{\left(\frac{1}{\sqrt{2} \pi \mathrm{d}^2 \mathrm{n}_\psi}\right)}=\sqrt{2} \pi \mathrm{d}^2 \mathrm{vn}_v\)

\(\therefore \mathrm{f} \propto \mathrm{n}_{\mathrm{v}}, \mathrm{n}_{\mathrm{v}}\) is number density

\(\frac{\mathrm{f}_1}{\mathrm{f}_2}=\frac{\mathrm{n}_{\mathrm{r}_1}}{\mathrm{n}_{r_2}}=\frac{3 \times 10^{19}}{12 \times 10^{-19}}=0.25\)