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Question: The spectral emissive power \({E_\lambda }\) for a body at temperature T1 is plotted against the wav...

The spectral emissive power Eλ{E_\lambda } for a body at temperature T1 is plotted against the wavelength and area under the curve is found to be A. At a different temperature T2 the area is found to be 9A. Then λ1λ2\dfrac{{{\lambda _1}}}{{{\lambda _2}}} is

(A) 3
(B) 13\dfrac{1}{3}
(C) 13\dfrac{1}{{\sqrt 3 }}
(D) 3\sqrt 3

Explanation

Solution

The area under enclosed by the graph formed between the quantities Eλ{E_\lambda }and λ\lambda denotes the total emissive power radiated by the body. And total emissive power radiated by a body can be calculated by stefan’s law, E=eσT4E = e\sigma {T^4}, where E denotes total emissive power, e denotes emissivity of the body, σ is the stefan’s constant and T denotes absolute temperature of the body. The wavelength at which spectral emissive power is maximum relates to the temperature according to wein’s displacement law i.e. λ1T1=λ2T2{\lambda _1}{T_1} = {\lambda _2}{T_2} .

Complete step by step solution:
The total area under the graph formed between the quantities Eλ{E_\lambda } and λ\lambda denotes total emissive power radiated by a body.
Given that total emissive power at temperature T1 is A,

Using Stefan's law we can say that : A=eσT14A = e\sigma {T_1}^4 .
Given that total emissive power at temperature T2 is 9A,

Using Stefan's law we can say that 9A=σT249A = \sigma {T_2}^4.
Dividing these two equations, we get :
A9A=σT14σT24\dfrac{A}{{9A}} = \dfrac{{\sigma {T_1}^4}}{{\sigma {T_2}^4}} ,
Cancelling A and σ on both sides, we get :
19=T14T24\dfrac{1}{9} = \dfrac{{{T_1}^4}}{{{T_2}^4}},
Cross multiplication gives,
T24=9T14{T_2}^4 = 9{T_1}^4,
Raising power 14\dfrac{1}{4} on both sides, we get :
T2=T13{T_2}^{} = {T_1}^{}\sqrt 3
Now, using wien's displacement law :
λ1T1=λ2T2{\lambda _1}{T_1} = {\lambda _2}{T_2},
Substituting the value of T2, we get :
λ1T1=λ2T13{\lambda _1}{T_1} = {\lambda _2}{T_1}\sqrt 3 ,
Cancelling T1 on both sides :
λ1=λ23{\lambda _1} = {\lambda _2}\sqrt 3 ,
λ1λ2=3\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \sqrt 3

Therefore the correct answer is option : (D)

Note: Before solving the problem, the student needs to be able to understand the different concepts of Radiation like Stefan’s law, Wien's Displacement law and the terms like total emissive power and spectral emissive power. A common mistake that students make while solving questions involving concepts of radiation is not being able to differentiate between spectral emissive power and total emissive power. While total emissive power is the intensity of radiation emitted by a body, spectral emissive power is the intensity of radiation emitted by a body per unit wavelength.