Physics, asked by nirmalrabha100, 3 months ago

defuce Kirchhoff's law of radiation

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Answered by rangini2020
0

Answer:

Gustav Robert Kirchhoff (1824–1887) stated in 1860 that “at thermal equilibrium, the power radiated by an object must be equal to the power absorbed.” This leads to the observation that if an object absorbs 100 percent of the radiation incident upon it, it must reradiate 100 percent.

Answered by nandini9124
0

Answer:

Kirchhoff's law of radiation : At a given temperature the coefficient of absorption of a body is equal to its coefficient of emission.

Theoretical proof : Consider the following thought experiment. An ordinary body A and a perfectly black body B are enclosed in an athermanous enclosure as shown in figure.

According to the theory of heat exchanges there will be a continuous exchange of heat energy between each body and its surroundings. Hence, the two bodies, after some time, will attain the same temperature as that of enclosure. Let a and e be the coefficients of absorption and emission respectively of body A and body B.

Let E and E

b

be the emissive powers of bodies A and B respectively.

Suppose that Q is the quantity of radiant energy incident on each body per unit time per unit surface area of the body.

Body A will absorb the quantity a Q per unit time per unit surface area and radiate the quantity E per unit time per unit surface area.

Since there is no change in its temperature, we must have

a Q=E...(1)

As body B is a perfect blackbody it will absorb the quantity Q per unit time per unit surface area and radiate the quantity Eb

per unit surface area.

Since there is no change in its temperature. We must have

Q=Eb ....(2)

From Equation (1) and (2) we get

a= E/Q =E/Eb...(3)

By definition of coefficient of emission

e= E/Eb....(4)

From Equation (3) and (4) we get a=e.

Hence, Kirchhoff's law of radiation proved.

Explanation:

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