Question

What is the relation between emissive power of a real body (E) and emissive power of a black body ($E_b$) when both are at same temperature? ; α = Absorptivity
(A) \[\dfrac{{{E_b}}}{E} = 1\]
(B) $\dfrac{{{E_b}}}{E} = \alpha $
(C) $\dfrac{E}{{{E_b}}} = 1$
(D) $\dfrac{E}{{{E_b}}} = \alpha $

Answer 1

Hint: The ratio of emissive power for a body to that of a black body is to be considered as absorptivity because the relation required is at same temperature.

Complete step-by-step solution
Emissive power of a body at a particular temperature can be defined as the energy emitted per second per unit surface area of the body within a unit wavelength range.
Absorptive power of a body at a particular temperature can be defined as the ratio of the amount of energy absorbed in a given time by the surface to the amount of energy incident on the surface at the same time.
According to Kirchhoff's law, it states that the ratio of emissive power to the absorptive power for a given wavelength at a given temperature is the same for all the bodies and is equal to the emissive power of a perfectly black body at that temperature.
$\dfrac{E}{\alpha } = {E_b}$
The condition is stated as the same temperature hence the ratio of E and $E_b$ is called absorptivity.
Hence the relation between them is
$\dfrac{E}{{{E_b}}} = \alpha $

And the correct option is D.

Note: If ε is the emissivity of the body,
1. For a perfectly black body ε=1
2. For highly polished body ε=0
3. For practical bodies it lies between zero and one.