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**Statement:**

- The ratio of the emissive power to the coefficient of absorption is constant for all substances at a given temperature and is equal to the emissive power of a perfectly black body at that temperature.
**OR**At any given temperature, the emissivity (or coefficient of emission) of a body is equal to the coefficient of absorption **Explanation:**If ‘E’ is the emissive power of a substance and ‘a’ is its coefficient of absorption then by Kirchoff’s law

##### i.e a = e

**Theoretical Proof of Kirchoff’s Law of Radiation:**

- Let us consider two bodies A and B suspended in a constant temperature enclosure. B is a perfectly black body. After sometime both A and B will attain the same temperature as that of enclosure. By Prevost heat exchange theory In this state also every body will emit and absorb thermal radiations.
- Let E be the emissive power of A and ‘a’ be its coefficient of absorption. Let E
_{b}be the emissive power of B. Let Q be the radiant heat incident per unit time per unit area of each body.

Heat absorbed per unit time per unit area of A = a Q

Heat emitted per unit time per unit area of A = E.

As the temperature remains constant so the heat emitted will be equal to the heat absorbed

∴ E = a Q …………(1)

Perfectly black body B will absorb all the radiant heat incident on it.

Heat absorbed per unit time per unit area of B = Q.

Heat emitted per unit time per unit area of B = E_{b}

As the temperature of B remains constant so in case of B also heat emitted is equal to heat absorbed.

Eb = Q …………………..(2)

Dividing equation (1) by (2) we get,

Thus, the coefficient of emission is equal to coefficient of absorption. This proves Kirchoff’s law.

#### Example – 3:

- 512 J of radiant heat are incident on a body which absorbs 224 J. What is its coefficient of emission?
**Solution:****Given:**Radiant heat incident = Q = 512 J, radiant heat absorbed = Q_{a}= 224 J**To Find:**Coefficient of emission = e = ?

Coefficient of absorption = a = Q_{a}/Q = 224/512 = 0.4375

By Kirchoff’s law of radiation

Coefficient of emission (e) = Coefficient of absorption (a)

∴ e = 0.4375

**Ans:** Coefficient of emission = 0.4375

#### Example – 4:

- A body of surface area 15 × 10
^{-3}m² emits 1260 J in 40 s at a certain temperature. What is the emissive power of the surface at that temperature? **Solution:****Given:**Surface area = A = 15 × 10^{-3 }m², radiant heat emitted = Q = 1260 J, time taken = t = 40 s.**To Find:**Emissive power = E = ?

E = Q/At = 1260 /(15 × 10^{-3}× 40) = 21oo J/m²s

**Ans:** Emissive power of surface = 21oo J/m²s

#### Example – 5:

- The emissive power of a sphere of area 0.02 m² is 2100 J/m²s. What is the amount of heat radiated by the spherical surface in 20 seconds?
**Solution:****Given:**Surface area = A = 0.02 m², Emissive power = 2100 J/m²s, time taken = t = 20 s.**To Find:**Heat radiated = Q = ?

E = Q/At

∴ Q = E A t

∴ Q = 2100 × 0.02 × 20

**Ans:** Heat radiated = 840 J

#### Example – 6:

- The energy of 6000 J is radiated in 5 minutes by a body of surface area 100 cm2. Find the emissive power of the body.
**Solution:****Given:**Radiant heat emitted = Q = 6000 J, Time taken = 5 min = 5 × 60 = 300 s, Surface area = 100 cm² = 100 × 10^{-4}m²**To Find:**Emissive power = E = ?

E = Q/At = 6000 / (100 × 10^{-4}× 300)

∴ Q = 2000 J/m²s

**Ans:** Emissive Power = 2000 J/m²s

**Experimental Proof of Kirchoff’s Law or Ritchie’s Experiment:**

**Apparatus**:

- The apparatus consists of a U – tube manometer containing some coloured liquid. The two arms of the manometer are connected to two identical cylinders P and Q, having the same axis (co-axially arranged). The same face (either left or right) of each cylinder is coated with lamp black while the other face is kept polished. A third cylinder R can be placed between P and Q co-axially. One face of the cylinder R is coated with lamp black while the other face is kept polished. The cylinder R can be rotated about a vertical axis.

**Working:**

- The cylinder R is kept as shown in the figure such that all black surface point in the same direction. Hot water is poured into the cylinder R due to which its temperature will increase. No change will be observed in the liquid levels in the manometer. This shows that the quantity of heat absorbed by both P and Q from R is the same. Therefore pressure exerted by the air in P and Q on the liquid is the same on both sides.
- Let E and E
_{b}be the emissive powers of the polished and black surfaces and ‘a’ be the coefficient of absorption of the polished face. Let A be the area of the cross-section of each cylinder.

The amount of heat radiated per unit time by the black face of R = A E_{b}.

A part of this heat is incident on the polished face of P.

Heat incident per unit time on the polished face of P = k A E_{b},

The constant k depends upon the distance between P and R.

Heat absorbed per unit time by P per second = a k A E_{b}.

Heat radiated per unit time by the polished face of R = A E

Heat incident per unit time on the black face of Q = k A E

Heat absorbed per unit time by the black face of Q = k A E.

The level of coloured liquid in both the arms of the apparatus is the same.

Hence both P and Q absorbed same quantity of heat per unit time.

∴ a k A E_{b} = k A E

∴ a E_{b} = E

∴ a = E / E_{b} = e

Thus, the coefficient of absorption = coefficient of emission.

This is Kirchoff’s Law. Thus the Kirchoff’s Law is experimentally verified.

Science > Physics > Radiation > You are Here |

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