If e_lambda and a_lambda are the emissive pow | vedclass

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(C) Kirchhoff's law of thermal radiation states that for an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium,the e

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$\frac{e_\lambda}{a_\lambda} = E_\lambda$

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(C) Kirchhoff's law of thermal radiation states that for an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium,the emissivity is equal to the absorptivity.Mathematically,the ratio of the emissive power $(e_\lambda)$ of a body to its absorptive power $(a_\lambda)$ at a given wavelength and temperature is equal to the emissive power $(E_\lambda)$ of a perfectly black body at the same wavelength and temperature.Therefore,the relation is given by: $\frac{e_\lambda}{a_\lambda} = E_\lambda$.

If $e_\lambda$ and $a_\lambda$ are the emissive power and absorptive power of a body respectively,and $E_\lambda$ is the emissive power of a perfectly black body,what is the relation according to Kirchhoff's law?

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