# Absorption and emission power

1. Mar 6, 2014

### LagrangeEuler

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?

2. Mar 6, 2014

### dauto

Yes, emission power and absorption power must be equal.

3. Mar 6, 2014

### LagrangeEuler

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

4. Mar 7, 2014

### Philip Wood

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: Mar 7, 2014
5. Mar 10, 2014

### Philip Wood

LE: Are you clear now?