If between wavelength lambda and lambda + dla | vedclass

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Question

(C) According to Kirchhoff's law of thermal radiation,for any body in thermal equilibrium,the ratio of its emissive power $(e_\lambda)$ to its absorpt

Vedclass · Accepted Answer

$e_\lambda = a_\lambda E_\lambda$

Vedclass · Answer

(C) According to Kirchhoff's law of thermal radiation,for any body in thermal equilibrium,the ratio of its emissive power $(e_\lambda)$ to its absorptive power $(a_\lambda)$ at a given wavelength $\lambda$ and temperature $T$ is equal to the emissive power $(E_\lambda)$ of a perfectly black body at the same wavelength and temperature.Mathematically,this is expressed as:$\frac{e_\lambda}{a_\lambda} = E_\lambda$Rearranging this equation,we get:$e_\lambda = a_\lambda E_\lambda$Therefore,option $C$ is the correct statement.

If between wavelength $\lambda$ and $\lambda + d\lambda$,$e_\lambda$ and $a_\lambda$ are the emissive and absorptive powers of a body,respectively,and $E_\lambda$ is the emissive power of a perfectly black body,then according to Kirchhoff's law,which of the following is true?

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