Absorption and emission power

  • #1
596
10
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?
 

Answers and Replies

  • #2
1,948
200
Yes, emission power and absorption power must be equal.
 
  • #3
596
10
Why then in next step
##(\frac{e_{\lambda,T}}{a_{\lambda,T}})=E_{\lambda,T}##
 
  • #4
Philip Wood
Gold Member
1,221
78
I can make sense of this only by assuming that…

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

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

[itex]e_{\lambda, T}[/itex] means the spectral emissive power of the surface at temperature T, that is the power it emits between wavelength [itex]\lambda[/itex] and wavelength [itex]\lambda + \Delta \lambda[/itex] per unit area, divided by [itex]\Delta \lambda[/itex], as [itex]\Delta \lambda[/itex] 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 [itex]a_{\lambda, T}[/itex] 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:
  • #5
Philip Wood
Gold Member
1,221
78
LE: Are you clear now?
 

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