Kvant Physics Problem 137
The radiation of a fixed infrared wavelength propagates through methane according to the exponential attenuation law.
Verified: no
Verdicts: UNKNOWN + UNKNOWN
Solve time: 7m08s
Source on kvant.digital
Problem
Infrared radiation of a certain wavelength is absorbed by methane ($\mathrm{CH}_4$). Under normal conditions, a 1 cm thick layer of pure methane absorbs 98% of the radiation energy. By what factor will this radiation be attenuated when passing vertically through the Earth’s atmosphere?
For the calculation, take the mass fraction of methane in the atmosphere to be $1{,}4\cdot 10^{-9}$.
S. M. Kozel
Setup and Assumptions
The radiation of a fixed infrared wavelength propagates through methane according to the exponential attenuation law. In a pure methane layer of thickness $d_0 = 1,\text{cm}$ at normal conditions, the transmitted energy is $2%$ of the incident energy.
The corresponding absorption coefficient of methane under these conditions is denoted by $\mu_0$ with units $\text{cm}^{-1}$.
The Earth’s atmosphere is treated as a uniform mixture of gases at normal conditions of temperature and pressure, so that the number density of molecules in air is taken equal to that in a pure gas at the same conditions. Methane is assumed to be uniformly mixed with mass fraction $w = 1.4 \cdot 10^{-9}$ throughout the atmospheric column.
Absorption is assumed to depend only on the number of methane molecules encountered along the path, so that the effective absorption coefficient in air is reduced in proportion to the methane fraction. Scattering, spectral broadening, and altitude variation of composition and density are neglected.
The vertical atmospheric column is modeled as having the same total molecular column density as a gas layer of thickness $H = 10^6,\text{cm}$ at normal conditions, corresponding to an atmospheric height of approximately $10,\text{km}$ expressed in gas-equivalent thickness.
The unknown quantity is the attenuation factor $I/I_0$ for vertical propagation through the atmosphere.
Physical Principles
The attenuation of radiation in a medium with uniform absorbing species is governed by the Beer–Lambert law,
$$I = I_0 \exp(-\mu d),$$
where $\mu$ is the absorption coefficient and $d$ is the path length.
For a mixture, the absorption coefficient is proportional to the number density of absorbing molecules, so that
$$\mu_{\text{mix}} = \mu_0 \frac{n_{\text{CH}_4}}{n_0},$$
where $n_0$ is the number density of pure methane under the reference conditions and $n_{\text{CH}_4}$ is the methane number density in air.
For gases at the same temperature and pressure, number density is proportional to volume fraction, and for dilute trace gases the volume fraction is proportional to the mass fraction.
The effective optical depth is defined as
$$\tau = \int \mu , dx,$$
and for a uniform medium this reduces to $\tau = \mu d$.
Derivation
In pure methane, the transmitted intensity after $d_0 = 1,\text{cm}$ is
$$\frac{I}{I_0} = 0.02 = \exp(-\mu_0 \cdot 1,\text{cm}).$$
Taking the natural logarithm gives
$$\mu_0 = -\ln(0.02),\text{cm}^{-1} = \ln(50),\text{cm}^{-1}.$$
Numerically,
$$\mu_0 \approx 3.912,\text{cm}^{-1}.$$
In the atmosphere, methane is present only as a fraction $w = 1.4 \cdot 10^{-9}$, so the effective absorption coefficient becomes
$$\mu_{\text{atm}} = w \mu_0.$$
The vertical atmospheric column is modeled as an equivalent gas layer of thickness $H = 10^6,\text{cm}$ at normal conditions, so the optical depth is
$$\tau_{\text{atm}} = \mu_{\text{atm}} H = w \mu_0 H.$$
Substituting the expression for $\mu_0$ gives
$$\tau_{\text{atm}} = (1.4 \cdot 10^{-9}) \ln(50) \cdot 10^6.$$
The numerical value is
$$\tau_{\text{atm}} = 1.4 \cdot 10^{-3} \cdot 3.912 \approx 5.48 \cdot 10^{-3}.$$
The transmitted fraction is
$$\frac{I}{I_0} = \exp(-\tau_{\text{atm}}) \approx \exp(-5.48 \cdot 10^{-3}).$$
Expanding the exponential to first order in a small quantity,
$$\exp(-x) \approx 1 - x,$$
gives
$$\frac{I}{I_0} \approx 1 - 5.48 \cdot 10^{-3} = 0.99452.$$
The attenuation factor, defined as $I_0/I$, is
$$\frac{I_0}{I} \approx \frac{1}{0.99452} \approx 1.0055.$$
Result
The absorption coefficient in pure methane is
$$\mu_0 = \ln(50),\text{cm}^{-1}.$$
The atmospheric attenuation factor is
$$\frac{I}{I_0} = \exp!\big(-w \ln(50),H\big).$$
Substituting $w = 1.4 \cdot 10^{-9}$ and $H = 10^6,\text{cm}$,
$$\frac{I}{I_0} = \exp!\big(-(1.4 \cdot 10^{-9})(\ln 50)(10^6)\big) \approx 0.9945.$$
The factor by which the radiation is attenuated is
$$\boxed{\frac{I_0}{I} \approx 1.0055}.$$
Sanity Checks
The absorption coefficient $\mu_0$ has units $\text{cm}^{-1}$, and multiplication by a length in centimeters produces a dimensionless optical depth, so the exponential argument is consistent.
The small value of the optical depth $\tau_{\text{atm}} \approx 5.5 \cdot 10^{-3}$ is consistent with the extremely small methane fraction $10^{-9}$, which reduces the effective absorbing column by roughly six orders of magnitude compared with pure methane.
The limiting behavior $w \to 0$ gives $\tau_{\text{atm}} \to 0$ and therefore $I/I_0 \to 1$, matching the expectation of negligible attenuation in the absence of methane. The opposite limit $w \to 1$ with $H = 1,\text{cm}$ recovers the given condition $I/I_0 = 0.02$, confirming consistency with the calibration data.