Problem 5.3.8

Solution

Part a)

Based on the given parameters (the presence of mass $m$ and temperature $T$), the problem requires estimating not just the concentration, but the "flux density" of atoms to the ground.

Let us consider a horizontal surface of unit area located at height $h$. It is crossed by two opposing fluxes of atoms.
The atoms crossing the surface from top to bottom experienced, on average, their last collision at a distance of the mean free path $\lambda$ above it, that is, at height $h + \lambda$. The density of this flux is directed downwards and is equal to:
$$W_1 \approx \frac{1}{2} n_{h+\lambda} \bar{v}_z$$
where $n_{h+\lambda} = \alpha(h + \lambda)$ is the concentration of atoms at height $(h + \lambda)$, and $\bar{v}_z \approx \sqrt{\frac{kT}{m}}$ is the characteristic thermal velocity of the directed motion of molecules along the $z$-axis.

Similarly, the flux of atoms going from bottom to top is formed by particles flying from height $h - \lambda$:
$$W_2 \approx \frac{1}{2} n_{h-\lambda} \bar{v}_z = \frac{1}{2} \alpha(h - \lambda) \sqrt{\frac{kT}{m}}$$

The resulting flux density of radioactive atoms to the Earth is the difference between these two opposing fluxes:
$$W = W_1 - W_2 \approx \frac{1}{2} \sqrt{\frac{kT}{m}} \cdot \alpha (h + \lambda - (h - \lambda))$$
$$W \approx \frac{1}{2} \sqrt{\frac{kT}{m}} \cdot 2\alpha \lambda = \alpha \lambda \sqrt{\frac{kT}{m}}$$

Part b)

Let us estimate the diffusion coefficient $D$ of water vapor in air. In the kinetic theory of gases, it is expressed as:
$$D = \frac{1}{3} \lambda \bar{v}$$
where $\bar{v} = \sqrt{\frac{8RT}{\pi \mu}}$ is the mean speed of a water molecule ($\mu = 18 \text{ g/mol}$).

The mean free path of a water molecule, taking into account the relative motion of air molecules, is calculated by the formula:
$$\lambda = \frac{1}{\sqrt{2} n \sigma} = \frac{1}{\sqrt{2} n \pi (r_1 + r_2)^2}$$
The concentration of air molecules $n$ at normal atmospheric pressure $P \approx 10^5 \text{ Pa}$ can be found as $n = \frac{P}{kT}$.
Then the final formula is:
$$D = \frac{1}{3} \cdot \frac{kT}{\sqrt{2} P \pi (r_1 + r_2)^2} \cdot \sqrt{\frac{8RT}{\pi \mu}}$$

Calculation:
Let us substitute the data: $T = 293 \text{ K}$, $r_1 + r_2 = (0.21 + 0.18) \text{ nm} = 0.39 \cdot 10^{-9} \text{ m}$.
The mean speed $\bar{v} \approx \sqrt{\frac{8 \cdot 8.31 \cdot 293}{3.14 \cdot 0.018}} \approx 587 \text{ m/s}$.
The concentration $n \approx \frac{10^5}{1.38 \cdot 10^{-23} \cdot 293} \approx 2.47 \cdot 10^{25} \text{ m}^{-3}$.
The mean free path $\lambda \approx \frac{1}{1.41 \cdot 2.47 \cdot 10^{25} \cdot 3.14 \cdot (0.39 \cdot 10^{-9})^2} \approx 6.0 \cdot 10^{-8} \text{ m}$.

Calculating the diffusion coefficient:
$$D \approx \frac{1}{3} \cdot 6.0 \cdot 10^{-8} \cdot 587 \approx 1.17 \cdot 10^{-5} \text{ m}^2/\text{s}$$
Converting to square millimeters, we obtain $D \approx 12 \text{ mm}^2/\text{s}$.

Theory Reference:
For a detailed derivation of the mean free path utilized in this solution, you can refer to Physics by Halliday, Resnick, and Krane (HRK), Volume 1, Chapter 22, Section 22-3 ("The mean free path"), page 502. It thoroughly describes the effective cross-section concept ($\sigma = \pi d^2$) and the concentration relation from the ideal gas law ($1/n = kT/P$) that form the basis of these molecular-kinetic equations.