JEE Physics 2023 Question with Solution

Given: Number density of air molecules changes from $n_1=3\times10^{19}$ to $n_2=12\times10^{19}$ per cm³.

Find: The ratio of collision frequency before and after the increase, i.e. $\frac{Z_1}{Z_2}$.

Collision frequency is given by

$$Z=n\pi d^2 v_{avg}$$

where $n$ is the number of molecules per unit volume.

Since $d$ and $v_{avg}$ remain constant,

$$\frac{Z_1}{Z_2}=\frac{n_1\pi d^2 v_{avg}}{n_2\pi d^2 v_{avg}}=\frac{n_1}{n_2}$$

Substituting the given values,

$$\frac{Z_1}{Z_2}=\frac{3\times10^{19}}{12\times10^{19}}=\frac{3}{12}=\frac{1}{4}=0.25$$

Therefore, the ratio of collision frequencies before and after the increase is $0.25$. The correct option is A.

Given: Initial number density $n_1=3\times10^{19}$ and final number density $n_2=12\times10^{19}$.

Find: $\frac{Z_1}{Z_2}$.

From the collision frequency formula,

$$Z=n\pi d^2 v_{avg}$$

This shows that collision frequency is directly proportional to number density:

$$Z\propto n$$

Hence,

$$\frac{Z_1}{Z_2}=\frac{n_1}{n_2}$$

Now substitute:

$$\frac{Z_1}{Z_2}=\frac{3\times10^{19}}{12\times10^{19}}$$

Cancel $10^{19}$ from numerator and denominator:

$$\frac{Z_1}{Z_2}=\frac{3}{12}=\frac{1}{4}=0.25$$

Therefore, the required ratio is $0.25$, so the correct option is A.