Question

Between two wavelengths $500A^{\circ }$ and $1000A^{\circ },a_{\lambda }$ is the absorptive power and $e_{\lambda }$ is the emissive power of a body. If $E_{\lambda }$ is the emissive power of a perfectly black body then according to Kirchhoff's law,

• A. $e_{\lambda }=a_{\lambda }=E_{\lambda }$
• B. $e_{\lambda }=a_{\lambda }{E}_{\lambda }$
• C. $e_{\lambda }{a}_{\lambda }{E}_{\lambda }=\text{constant}$
• D. $\frac{a_{\lambda }}{e_{\lambda }}=\frac{1}{2}{E}_{\lambda }$

Answer 1

The correct option is B $e_{\lambda }=a_{\lambda }{E}_{\lambda }$

According to Kirchhoff's law, at a given temperature the ratio of emissivity ($e$) to absorptivity ($a$) of all bodies is fixed or constant and is equal to the emissivity of black body $E_b$.
$\Rightarrow \frac{e}{a}=E_b$ (for a block body)
According to question
$\Rightarrow \frac{e_{\lambda }}{a_{\lambda }}=E_{\lambda }$
$\Rightarrow e_{\lambda }=a_{\lambda }{E}_{\lambda }$