Black Body and Krichoff’s Law - JEE Important Topic

When radiation hits a body, three phenomena are possible. The radiation can be reflected back from the body, can be transmitted, or can be absorbed by the body. Radiation consists of many different wavelengths, some of which can be identified by their colour, such as visible rays. 

A black body can be defined as a physical body or a surface that absorbs all incident electromagnetic radiation irrespective of the frequency and the angle of the incident. The absorption of radiation results in an increase in the temperature of the black body. Due to an increase in temperature, the body starts emitting radiation. This radiation is known as Blackbody Radiation.

Black Body

A perfectly black body is that which absorbs completely the radiation of all wavelength’s incident on it regardless of the frequency or angle of incidence. A perfectly black body neither reflects or transmits any radiation. Therefore, absorption of a black body is unity. A black body emits maximum radiation at a given temperature as compared to any other body. When a perfectly black body is heated to a suitable high temperature, it emits radiation of all wavelengths. 

Why Is It Called a Black Body?

We know that any object of black colour is a good absorber. And since the black body has excellent absorptive properties, it is named black body. It does not mean that a black body is necessarily black in colour. Since the black body does not reflect back any radiation, it appears black in colour. But when the black body is heated, and starts emitting the radiation, then it can emit radiation of any colour. And it may appear red, blue or yellow, etc. For example, the Sun is a black body but it’s not black in colour.

Black Body Examples 

Some examples of the black body are listed below.

• Temperature of the sun is very high (6000K approx.) It emits radiation of all wavelengths, so it is an example of a black body. 

• Lamp Black is also a black body. It reflects only 1% of the incident radiation.

• A cavity with a small hole in it. The light that enters the hole undergoes several reflections inside the cavity walls such that no light is reflected outside the cavity. If the walls are painted black, making them absorptive, then the cavity acts as a perfect black body.

Kirchhoff’s Law of Heat Radiation

Kirchhoff's law of thermal radiation was formulated in 1859, but no experimental proof was given at that time. The arguments of the law were completely theoretical.

‘A good emitter is also a good absorber’.

The statement of Kirchhoff’s Law is given as ‘The ratio of emissive power to absorptive power is same for all surfaces at the same temperature and is equal to the emissive power of a perfectly black body at that temperature.’, i.e.

\[\dfrac{E}{a}=K={{E}_{b}}\]

where ‘E’ and ‘a’ are the emissive power and absorption coefficient of a body at a particular temperature, ‘K’ is a constant and E_b is the emissive power of a black body.

Proof of Kirchhoff’s Law

Consider an enclosure and assume three bodies are kept inside the enclosure, all having the same dimensions with surface area equal to ‘A’ and one of them is a block body. Let the temperature inside the enclosure be ‘T’. After some time, at thermal equilibrium, the temperatures of all bodies will be equal. 

Proof of Kirchoff’s law

Since the geometrical dimensions of all the bodies are the same, the incident heat on each of the bodies will be equal. Let ‘Q’ be the incident heat on each of the bodies. Let a1, a2 and a=1 are the absorptive powers of the body 1, 2 and black body respectively and their emissive powers be E1, E2 and Eb.

For the first body, 

Heat absorbed $=\text{ }{{a}_{1}}Q\text{                    }\left[ \because \text{Heat absorbed}=a\times \text{ Heat Incident} \right]$                                       

Heat emitted $={{E}_{1}}At\text{                     }\left[ \because \text{Heat emitted}=E\times \text{Area}\times \text{time} \right]$

We know that 

Heat absorbed = Heat emitted

Therefore, 

${{a}_{1}}Q={{E}_{1}}At$                   ….. (1)

Similarly, for the second body, 

${{a}_{2}}Q={{E}_{2}}At$                 ….. (2)

And for the Black body,

$aQ={{E}_{b}}At$, and $\left[ \because a=1 \right]$

$\Rightarrow \text{ }Q={{E}_{b}}At$             ….. (3)

From equations (1), (2) and (3), we get 

$\dfrac{{{E}_{1}}}{{{a}_{1}}}=\dfrac{Q}{At}$                      ….. (4)

$\dfrac{{{E}_{2}}}{{{a}_{2}}}=\dfrac{Q}{At}$                     ….. (5)

${{E}_{b}}=\dfrac{Q}{At}$                     ….. (6)

From equations (4), (5) and (6), we get 

$\dfrac{{{E}_{1}}}{{{a}_{1}}}=\dfrac{{{E}_{2}}}{{{a}_{2}}}={{E}_{b}}$

The above equation shows that at a given temperature and a given wavelength, the ratio of emissive power to absorption coefficient is constant, and is equal to the emissive power of the black body. Thus, Kirchhoff’s Law is proved.

Conclusion

A black body can be defined as a physical body or a surface that absorbs all incident electromagnetic radiation irrespective of the frequency and the angle of incident. A black body emits maximum radiation at a given temperature as compared to any other body. The phenomenon of emitting the radiation by a black body is termed as Black Body Radiation. Some examples of black bodies are Sun/ Stars, Lamp Black, Graphite, Platinum Black, Gold Black, and an isothermal enclosure etc. Kirchhoff’s law of thermal radiation establishes a relation between the emissive power of a body and its absorption coefficient. The statement of Kirchhoff’s law is ’The ratio of emissive power to absorptive power is same for all surfaces at the same temperature and is equal to the emissive power of a perfectly black body at that temperature.’, i.e., \[\dfrac{E}{a}=K={{E}_{b}}\]