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Prove Kirchhoff's law of radiation theoretically. |
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Answer» Solution :KIRCHHOFF's law of RADIATION : Kirchhoff's law of radioation states that the coefficient of absorption of a body is EQUAL to its coefficient of emission at any given tempereature. It can be proved hy Thought Experiment'. Let us consider an ordinary body O and perfectly blackbody B with same surface area A, are placed in the uniform temperature enclosure.  Both the bodies will have same temperature by thermal exchange which is a zeroth law on thermodynamics. Let E be emissive power of ordinary body O, `E_(b)` be emissive power of perfectly BLACK body B, a be the coefficient of absorption, e be the coefficient of emission of O, and Q be the radiant eanrgy incident per unit time per unit area on each body. Now, total radiant energy incident per unit time on B=AQ And quantity of energy emitted per unit time by `B=AE_(b)` Energy emitted per unit time by B= Energy absorbed per unit time by B. `:.AE_(b)=AQ` `E_(b)=Q"".......(i)` The total radiant energy incident per unit time on O=AQ. Quantity of energy emitted per unit time by O=AE And quantity of energy absorbed per unit time by `O=aAQ`. Energy emitted per unit time by O = Energy absorbed per unit time by O `:.AE=aAQ` `:.E=aQ` `:.(E)/(a)=Q"".........(ii)` From EQUATIONS (i) and (ii), we get `(E)/(a)=E_(b)` `:.(E)/(E_(b))=a` But `(E)/(E_(b))=e` (coefficient of emission) `:.a=e` Coefficient of absorption =Coefficient of emissions |
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