Kvant Physics Problem 137

The radiation intensity $I$ decreases along a propagation path due to absorption by methane molecules.

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Problem

Infrared radiation of a certain wavelength is absorbed by methane ($\mathrm{CH}_4$). Under normal conditions, a 1 cm layer of pure methane absorbs 98% of the radiation energy. By what factor will such 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 intensity $I$ decreases along a propagation path due to absorption by methane molecules. The dependence on path length is described by an exponential law characterized by an absorption coefficient proportional to the number of methane molecules encountered.

For pure methane at normal conditions, a layer thickness $x_0 = 1,\mathrm{cm} = 10^{-2},\mathrm{m}$ reduces the radiation energy by $98%$, so the transmitted fraction is $I/I_0 = 0.02$.

The mass fraction of methane in the atmosphere is $w = 1.4 \cdot 10^{-9}$.

The atmospheric pressure is taken as $p = 1.013 \cdot 10^5,\mathrm{Pa}$, giving an atmospheric column mass per unit area

$\frac{M_{\mathrm{air}}}{S} = \frac{p}{g} \approx \frac{1.013 \cdot 10^5}{9.8},\mathrm{kg/m^2} \approx 1.03 \cdot 10^4,\mathrm{kg/m^2}.$

The density of methane under normal conditions is taken as $\rho_{\mathrm{CH_4}} = 0.716,\mathrm{kg/m^3}$.

Radiation propagates vertically through the atmosphere, and horizontal variations are neglected. Absorption by other gases is neglected.

Physical Principles

The attenuation of radiation in a homogeneous absorber obeys the Beer–Lambert law,

$I = I_0 e^{-\mu x},$

where $\mu$ is the absorption coefficient proportional to the number density of absorbing molecules and $x$ is the path length.

For mixtures, the effective absorption is proportional to the column density of the absorbing component. If the mass fraction of an absorber is $w$, then the absorber mass per unit area is $w$ times the total atmospheric column mass per unit area.

The equivalence between different absorbers follows from proportionality of absorption to the number of absorbing molecules, which for a fixed substance corresponds to proportionality to its mass density under identical thermodynamic conditions.

Derivation

For pure methane of thickness $x_0 = 1,\mathrm{cm}$,

$\frac{I}{I_0} = 0.02 = e^{-\mu_{\mathrm{CH_4}} x_0}.$

Taking the natural logarithm gives

$\mu_{\mathrm{CH_4}} = -\frac{\ln(0.02)}{x_0} = \frac{\ln(50)}{1,\mathrm{cm}}.$

The optical depth of a medium is defined as

$\tau = \mu_{\mathrm{CH_4}} x_{\mathrm{eq}},$

where $x_{\mathrm{eq}}$ is the equivalent thickness of pure methane producing the same absorption.

The methane column mass in the atmosphere is

$\frac{M_{\mathrm{CH_4}}}{S} = w \frac{M_{\mathrm{air}}}{S} = 1.4 \cdot 10^{-9} \cdot 1.03 \cdot 10^4,\mathrm{kg/m^2} = 1.442 \cdot 10^{-5},\mathrm{kg/m^2}.$

The mass per unit area of a $1,\mathrm{cm}$ layer of pure methane is

$\frac{M_{1,\mathrm{cm}}}{S} = \rho_{\mathrm{CH_4}} x_0 = 0.716,\mathrm{kg/m^3} \cdot 10^{-2},\mathrm{m} = 7.16 \cdot 10^{-3},\mathrm{kg/m^2}.$

The equivalent methane thickness is therefore

$x_{\mathrm{eq}} = x_0 \frac{M_{\mathrm{CH_4}}/S}{M_{1,\mathrm{cm}}/S} = 1,\mathrm{cm} \cdot \frac{1.442 \cdot 10^{-5}}{7.16 \cdot 10^{-3}} = 2.01 \cdot 10^{-3},\mathrm{cm}.$

The transmission through the atmosphere becomes

$\frac{I}{I_0} = e^{-\mu_{\mathrm{CH_4}} x_{\mathrm{eq}}} = \left(e^{-\mu_{\mathrm{CH_4}} x_0}\right)^{x_{\mathrm{eq}}/x_0} = 0.02^{2.01 \cdot 10^{-3}}.$

Using logarithms,

$\ln\left(\frac{I}{I_0}\right) = (2.01 \cdot 10^{-3}) \ln(0.02) = (2.01 \cdot 10^{-3})(-3.912) = -7.87 \cdot 10^{-3},$

so

$\frac{I}{I_0} = e^{-7.87 \cdot 10^{-3}} = 0.9922.$

The attenuation factor is

$\frac{I_0}{I} = 1.0079.$

Result

The transmitted fraction is

$\frac{I}{I_0} = 0.02^{2.01 \cdot 10^{-3}} = 0.9922.$

The attenuation factor is

$\boxed{\frac{I_0}{I} \approx 1.008}.$

Sanity Checks

The exponent $2.01 \cdot 10^{-3}$ is dimensionless because it is a ratio of equivalent methane thicknesses, ensuring consistency with the Beer–Lambert law.

The result lies very close to unity because the atmospheric methane column is many orders of magnitude smaller than a $1,\mathrm{cm}$ layer of pure methane, so only a weak attenuation is expected.

A potential source of error would be inconsistent treatment of column mass units when converting atmospheric pressure to $\mathrm{kg/m^2}$ or mixing $\mathrm{cm}$ and $\mathrm{m}$ in the methane density conversion.