Kirchoff's Law
Category : JEE Main & Advanced
According to this law the ratio of emissive power to absorptive power is same for all surfaces at the same temperature and is equal to the emissive power of a perfectly black body at that temperature. Hence \[\frac{{{e}_{1}}}{{{a}_{1}}}=\frac{{{e}_{2}}}{{{a}_{2}}}=...\,\,{{\left( \frac{E}{A} \right)}_{\text{Perfectly black body}}}\]
But for perfectly black body \[A=1\] i.e. \[\frac{e}{a}=E\]
If emissive and absorptive powers are considered for a particular wavelength l, \[\left( \frac{{{e}_{\lambda }}}{{{a}_{\lambda }}} \right)={{({{E}_{\lambda }})}_{\text{black}}}\]
Now since \[{{({{E}_{\lambda }})}_{black}}\] is constant at a given temperature, according to this law if a surface is a good absorber of a particular wavelength it is also a good emitter of that wavelength.
This in turn implies that a good absorber is a good emitter (or radiator)
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