# Absorption and emission power

Three bodies 1,2,3 are in closed region. Region is at temperature ##T##.
##e_{\lambda,T}## - spectral emission power
##a_{\lambda,T}## - spectral absorption power

In experiment
##(\frac{e_{\lambda,T}}{a_{\lambda,T}})_1=(\frac{e_{\lambda,T}}{a_{\lambda,T}})_2=(\frac{e_{\lambda,T}}{a_{\lambda,T}})_3##

I am confused. Does it perhaps ##(\frac{e_{\lambda,T}}{a_{\lambda,T}})_1=1##? How much body absorbs and emits so much. Right?

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Yes, emission power and absorption power must be equal.

Why then in next step
##(\frac{e_{\lambda,T}}{a_{\lambda,T}})=E_{\lambda,T}##

Philip Wood
Gold Member
I can make sense of this only by assuming that…

$a_{\lambda, T}$ means the spectral absorptivity of a surface at temperature T, that is the fraction of radiation of wavelength $\lambda$ which it absorbs. A number (≤ 1) without units.

$E_{\lambda, T}$ means the spectral emissive power of a black body at temperature T, that is the power it emits between wavelength $\lambda$ and wavelength $\lambda + \Delta \lambda$ per unit area, divided by $\Delta \lambda$, as $\Delta \lambda$ approaches zero.

$e_{\lambda, T}$ means the spectral emissive power of the surface at temperature T, that is the power it emits between wavelength $\lambda$ and wavelength $\lambda + \Delta \lambda$ per unit area, divided by $\Delta \lambda$, as $\Delta \lambda$ approaches zero.

The relationship you've just quoted would then express Kirchhoff's Law, crudely summarised as 'good absorbers are good emitters', since it states that $a_{\lambda, T}$ is equal to the ratio of the spectral emissive power of the surface in question to that of a black body (which has the greatest spectral emissive power, at all wavelengths, of ANY surface at the same temperature).

Last edited:
Philip Wood
Gold Member
LE: Are you clear now?