1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Absorption and emission power

  1. Mar 6, 2014 #1
    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. jcsd
  3. Mar 6, 2014 #2
    Yes, emission power and absorption power must be equal.
     
  4. Mar 6, 2014 #3
    Why then in next step
    ##(\frac{e_{\lambda,T}}{a_{\lambda,T}})=E_{\lambda,T}##
     
  5. Mar 7, 2014 #4

    Philip Wood

    User Avatar
    Gold Member

    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: Mar 7, 2014
  6. Mar 10, 2014 #5

    Philip Wood

    User Avatar
    Gold Member

    LE: Are you clear now?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook