The mean free path and the average speed of oxygen molecules at 300 K and 1 atm are $$3 \times 10^{-7}$$ m and 600 m/s, respectively. Find the frequency of its collisions.

The mean free time $$\tau$$ between two successive collisions is obtained from the relation

$$\tau = \frac{\lambda}{\bar{c}}$$

where $$\lambda$$ is the mean free path and $$\bar{c}$$ is the average (root-mean-square) speed of the molecules.

The collision frequency $$f$$, i.e. the number of collisions per second, is the reciprocal of the mean free time:

$$f = \frac{1}{\tau} = \frac{\bar{c}}{\lambda}$$

Substituting the given values $$\lambda = 3 \times 10^{-7}\,\text{m}$$ and $$\bar{c} = 600\,\text{m s}^{-1}$$:

$$f = \frac{600}{3 \times 10^{-7}}\;\text{s}^{-1}$$

$$f = \frac{600}{3}\times 10^{7}\;\text{s}^{-1}$$

$$f = 200 \times 10^{7}\;\text{s}^{-1}$$

$$f = 2 \times 10^{9}\;\text{s}^{-1}$$

Therefore, the collision frequency of oxygen molecules at 300 K and 1 atm is $$2 \times 10^{9}\,\text{s}^{-1}$$, which matches Option C.