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en/5.3.8.md
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| + | ### Statement | ||
| + | |||
| + | $5.3.8.$ [Insert the problem statement] | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | ### Statement | ||
| + | |||
| + | $5.3.8.$ | ||
| + | a. The relative content of radioactive atoms in a gas is small. Their number per unit volume increases linearly with height: $n = \alpha h$. The mass of an atom is $m$, its mean free path is $\lambda$, and the temperature is $T$. Estimate the density of these atoms on the ground. | ||
| + | b. Estimate the diffusion coefficient of water vapor in air at 20°C. The radius of water molecules is 0.21 nm. The radius of nitrogen and oxygen molecules is 0.18 nm. | ||
| + | |||
| + | ### Solution | ||
| + | |||
| + | <b>Part a)</b> | ||
| + | |||
| + | 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: | ||
| + | $$W = W_1 - W_2 \approx \frac{1}{2} \sqrt{\frac{kT}{m}} \cdot \alpha (h + \lambda - (h - \lambda))$$ | ||
| + | $$W \approx \frac{1}{2} \sqrt{\frac{kT}{m}} \cdot 2\alpha \lambda = \alpha \lambda \sqrt{\frac{kT}{m}}$$ | ||
| + | |||
| + | <b>Part b)</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: | ||
| + | $$D = \frac{1}{3} \cdot \frac{kT}{\sqrt{2} P \pi (r_1 + r_2)^2} \cdot \sqrt{\frac{8RT}{\pi \mu}}$$ | ||
| + | |||
| + | <b>Calculation:</b> | ||
| + | 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 $D \approx 12 \text{ mm}^2/\text{s}$. | ||
| + | |||
| + | #### Answer | ||
| + | a. $W = W_1 - W_2 \approx \alpha \lambda \sqrt{kT/m}$ | ||
| + | |||
| + | b. $D \approx 12 \text{ mm}^2/\text{s}$ | ||
| + | |||
| + | #### Answer | ||
| + | |||
| + | [Insert a concise answer or boxed result] | ||
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| ### Statement | |||
| $5.3.8.$ [Insert the problem statement] | |||
| ### Solution | |||
| ### Statement | |||
| $5.3.8.$ | |||
| a. The relative content of radioactive atoms in a gas is small. Their number per unit volume increases linearly with height: $n = \alpha h$. The mass of an atom is $m$, its mean free path is $\lambda$, and the temperature is $T$. Estimate the density of these atoms on the ground. | |||
| b. Estimate the diffusion coefficient of water vapor in air at 20°C. The radius of water molecules is 0.21 nm. The radius of nitrogen and oxygen molecules is 0.18 nm. | |||
| ### Solution | |||
| <b>Part a)</b> | |||
| 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: | |||
| $$W = W_1 - W_2 \approx \frac{1}{2} \sqrt{\frac{kT}{m}} \cdot \alpha (h + \lambda - (h - \lambda))$$ | |||
| $$W \approx \frac{1}{2} \sqrt{\frac{kT}{m}} \cdot 2\alpha \lambda = \alpha \lambda \sqrt{\frac{kT}{m}}$$ | |||
| <b>Part b)</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: | |||
| $$D = \frac{1}{3} \cdot \frac{kT}{\sqrt{2} P \pi (r_1 + r_2)^2} \cdot \sqrt{\frac{8RT}{\pi \mu}}$$ | |||
| <b>Calculation:</b> | |||
| 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 $D \approx 12 \text{ mm}^2/\text{s}$. | |||
| #### Answer | |||
| a. $W = W_1 - W_2 \approx \alpha \lambda \sqrt{kT/m}$ | |||
| b. $D \approx 12 \text{ mm}^2/\text{s}$ | |||
| #### Answer | |||
| [Insert a concise answer or boxed result] | |||