Kirchhoff's Law of Radiation

All bodies radiate energy in the form of photons. When these photons reach another surface, they may either be absorbed, reflected or transmitted. The behavior of a surface with incident radiation is described by the following quantities:

• absorptivity ($a$) is the fraction of incident radiation absorbed
• reflectivity ($r$) is the fraction of incident radiation reflected
• transmissivity ($t$) is the fraction of incident radiation transmitted.

The energy conservation gives $a+r+t=1$. For opaque objects, transmissivity $t=0$ and hence $a+r=1$. A blackbody has absorptivity $a=1$ and absorbs all radiation incident on it. Its reflectivity is $r=0$.

Real bodies radiate less effectively than black bodies. The measurement of this is the emissitivity ($e$) defined by \begin{align} e=E/E_b,\nonumber \end{align} where $E$ is radiated power per unit area from the real body at temperature $T$, and $E_b$ is radiated power per unit area from a black-body at the same temperature $T$. The emissivity of the blackbody is 1 and that of a real body is between 0 and 1.

Consider a small real body in thermal equilibrium with its surrounding blackbody cavity i.e., $T_\mathrm{body}=T_\mathrm{cavity}$. The power emitted per unit area of the blackbody (cavity) is $E_b$. Thus, incident power per unit area of the real body is $E_b$. The power absorbed per unit area of the real body is $aE_b$. The power emitted per unit area of the real body is equal to $E=e E_b$. Since real body is in thermal equalibrium with its surrounding, energy balance gives $eE_b=a E_b$ i.e, $a=e$ . The relation $a=e$ is known as Kirchhoff's Law of radiation. It implies that good radiators are good absorbers. Note that Kirchhoff's law is valid only when body is in thermal equilibrium with its surrounding.

Solved Problems from IIT JEE

Problem from IIT JEE 2006

In a dark room with ambient temperature $T_0$, a black body is kept at temperature $T$. Keeping the temperature of the black body constant (at $T$), sun rays are allowed to fall on the black body through a hole in the roof of the dark room. Assuming that there is no change in the ambient temperature of the room, which of the following statement(s) is (are) correct?

1. The quantity of radiation absorbed by the black body in unit time will increase.
2. Since emissivity $=$ absorptivity, hence the quantity of radiation emitted by black body in unit time will increase.
3. Black body radiates more energy in unit time in the visible spectrum.
4. The reflected energy in unit time by the black body remains same.

Solution: This problem is likely to cause confusion and hence we present detailed analysis. By definition, a blackbody absorbs all radiation falling onto it. It has emissivity $e=1$, absoptivity $a=1$, and reflectivity $r=0$.

Given blackbody is not in thermal equilibrium with its surrounding as $T\neq T_0$. Initially (before sun rays fall), if $T>T_0$ then emitted power per unit area from the blackbody is more than incident power per unit area falling upon it. There is net energy flow out of the blackbody. Since blackbody's temperature is kept constant at $T$, there must be a mechanism (other then radiation) to supply energy to the blackbody to compensate for energy loss due to radiation. This is not mentioned in the problem statement.

When sun rays fall on the blackbody, it absorbs all the incident photons (by definition of blackbody). So, the net quantity of radiations absorbed by the black body in unit time increases. Option (A) is correct.

The power emitted by the blackbody depends on its temperature. Since temperature remains constant (at $T$), the emitted power remains the same. Option (B) is wrong. The energy gained by the blackbody (due to increase in absoption) is lost through a mechanism of temperature controller.

Although blackbody absorbs visible light (sun-rays), its temperature and emission spectrum remains the same. Thus, option (C) is wrong.

The reflected energy by the blackbody in unit time is zero (because $r=0$). It remains zero even if sun-rays falls on it. Hence, option (D) is correct.

Problem from IIT JEE 2003

The temperature ($T$) versus time ($t$) graphs of two bodies X and Y with equal surface areas are shown in the figure. If the emissivity and the absorptivity of X and Y are $E_x$, $E_y$ and $a_x$, $a_y$, respectively, then,

1. $E_x > E_y$ and $a_x < a_y$
2. $E_x < E_y$ and $a_x > a_y$
3. $E_x > E_y$ and $a_x > a_y$
4. $E_x < E_y$ and $a_x < a_y$

Solution: The rate of cooling for $x$ is greater than that of $y$. The cooling occurs due to heat loss by radiation. By Stefan's law, the rate of heat loss is, \begin{align} \mathrm{d}Q/\mathrm{d}t=\sigma EA(T^4-T_0^4),\nonumber \end{align} where $E$ is emissivity. Hence, $E_x > E_y$. Since good emitters are good absorbers, the absorptivity $a_x > a_y$.

Problem from IIT JEE 2002

An ideal black-body at room temperature is thrown into a furnace. It is observed that,

1. initially it is the darkest body and at later times the brightest.
2. it is the darkest body at all times.
3. it cannot be distinguished at all times.
4. initially it is the darkest body and at later times it cannot be distinguished.

Solution: Let $T$ be the temperature of the furnace and $T_0$ be the room temperature. When a black body at temperature $T_0$ is thrown into the furnace it emits radiation energy at a rate $e_\mathrm{body}\sigma AT_0^4=\sigma AT_0^4$ (since $e_\mathrm{body}=1$). On the other hand, a portion of the furnace having same area $A$ emits the radiation energy at a rate $e_\mathrm{furnace} \sigma AT^4 $. Generally, $ \sigma A T_0^4 < e_\mathrm{furnace} \sigma AT^4 $, (because $T_0\ll T$), so the black body appears as the darkest body. The temperature of the black body rises till it reaches $T$. At this temperature, the black body emits the radiation energy at a rate $\sigma AT^4$ whereas the furnace emits the radiation energy at a rate $e_\mathrm{furnace} \sigma AT^4$. Since $e_\mathrm{furnace}<1$, the rate of radiation energy emitted by the black body is more than that by the furnace so the black body appears to be the brightest. Option (A) is correct.

Questions on Kirchhoff's Law

Question 1: If between wavelength $\lambda$ and $\lambda+\mathrm{d}\lambda$, $e_\lambda$ and $a_\lambda$ be emissive and absorptive powers of a body and $E_\lambda$ be the emissive power of the perfect blackbody, then according to Kirchhoff's law

A. $e_\lambda=a_\lambda=E_\lambda$
B. $e_\lambda E_\lambda=a_\lambda$
C. $e_\lambda=a_\lambda E_\lambda$
D. $e_\lambda a_\lambda E_\lambda=$const

Question 2: Which of the following statements is wrong

A. Rough surfaces are better radiators than smooth surface
B. Highly polished mirror like surfaces are very good radiators
C. Black surfaces are better absorbers than white ones
D. Black surfaces are better radiators than white

Related

1. Stefan's-Boltzmann Law
2. Wien's Displacement Law of Radiation
3. Newton's Law of Cooling

References

1. Concepts of Physics Part 2 by HC Verma (Link to Amazon)
2. Kirchhoff's Law (MIT webiste)
3. Article on Failure of Kirchhoff's Law (pdf)
4. Detailed article on blackbody radiation (pdf)