Kirchhoff's Law - Practice Questions & MCQ

• Emissive power-

        If the temperature of a body is more than it's surrounding then body emits thermal radiation.

1. Spectral emissive power- For a given surface Spectral emissive power is defined as the radiant energy emitted per sec per unit area of the

   surface within a unit wavelength around $\lambda$ (i.e. lying between $\left(\lambda-\frac{1}{2}\right)$ to $\left(\lambda+\frac{1}{2}\right)$.Spectral Emissive power for particular wavelength $(\lambda)_{\text {is }}$ denoted by $e_\lambda$

   $
   e_\lambda=\frac{\text { Energy }}{\text { Area } \times \text { time } \times \text { wavelength }}
   $

   So If
   e wavelength is changed then the value of Spectral Emissive power will also change.
   2. Total Emissive Power (e) -Total Emissive power is defined as the total amount of thermal energy emitted per unit time, per unit area of the body for all possible wavelengths.

   $
   =\int_0^{\infty} e_\lambda d \lambda
   $

   - Absorptive Power
   1. Spectral Absorptive power- It is defined as the ratio of the amount of the energy absorbed in a certain time to the total heat energy incident upon it in the same time, both in the unit wavelength interval. It is denoted by $a_\lambda$
   2. Total Absorptive Power (a)- It is defined as the total amount of thermal energy absorbed per unit time, per unit area of the body for all possible wavelengths.

   $
   a=\int_0^{\infty} a_\lambda d \lambda
   $

   - Emissivity $(\epsilon)$.

The emissivity of a body at a given temperature is defined as the ratio of the total emissive power of the body (e) to the total emissive power of a perfectly black body (E).     

And it is given by

$
\varepsilon=\frac{e}{E}
$

$\varepsilon=1$ - for a perfectly black body
$\varepsilon=0$ - for polished body
$(0<\varepsilon<1)$ - for practical bodies
- Kirchhoff's law

According to Kirchhoff's law, the ratio of emissive power to absorptive power is the same for all surfaces at the same temperature and is equal to the emissive power of a perfectly black body at that temperature.
l.e $\frac{e_1}{a_1}=\frac{e_2}{a_2} \ldots \ldots=\frac{E}{A}$

And as for perfectly black body $\mathrm{A}=1$
So $\quad \frac{e_1}{a_1}=\frac{e_2}{a_2} \ldots . .=\frac{E}{A}$

$
\begin{aligned}
\frac{e_1}{a_1} & =\frac{e_2}{a_2} \ldots \ldots=E \\
\text { or } \frac{e}{a} & =E
\end{aligned}
$

If emissive and absorptive powers are considered for a particular wavelength $(\lambda)$
then $\frac{e_\lambda}{a_\lambda}=E_\lambda$
This law also implies that a good absorber is a good emitter.