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  • The number of air molecules per  increased from <img alt="3 \times 10^{19} \text{ to }12 \times 10^{19}" src

The number of air molecules per cm^3 increased from 3 \times 10^{19} \text{ to }12 \times 10^{19}. The ratio of collision frequency of

air molecules before and after the increase in the number respectively is :

Option: 1

0.25


Option: 2

0.75


Option: 3

1.25


Option: 4

0.50


Answers (1)

best_answer

Collision frequency is given by Z=n \pi d^2 V_{a v g}, where \mathrm{n} is number of molecules per unit volume. \frac{z_1}{z_2}=\frac{n_1}{n_2}=\frac{3}{12}=\frac{1}{4}=0.25

Posted by

Anam Khan

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