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Question: Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperature \(T_1\) and \(T_2\), re...

Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperature T1T_1 and T2T_2, respectively. The maximum intensity in the emission spectrum of A is at 500 nm and in that of B is at 1500 nm. Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of B?
A. 9
B. 3
C. 2
D. 4

Explanation

Solution

Stefan-Boltzmann law of black body radiation has to be used to determine ratio of the energies. We can replace the temperatures by inverse of respective wavelengths as per Wien's displacement law in finding the ratios.

Formula used:
Wien's law: In the spectra of a black body, the wavelength having maximum intensity is related to absolute temperature of the black body as:
λm=bT\lambda_m = \dfrac{b}{T}
where b is Wien's constant.
Stefan-Boltzmann law:
The rate of heat energy emitted by a perfect black body can be written as:
H=σT4AH = \sigma T^4 A
where A is the surface area of the body and σ\sigma is Stefan-Boltzmann constant.

Complete answer:
Using the Stefan-Boltzmann law, we can write the ratio of rate of total energy radiated by the two spherical bodies (black bodies) as:
EAEB=σTA4AAσTB4AB\dfrac{E_A}{E_B} = \dfrac{\sigma T^4_A A_A}{\sigma T^4_B A_B}.

Now, the ratios of the areas can be written as:
AAAB=4πRA24πRB2=(RARB)2\dfrac{A_A}{A_B} = \dfrac{4 \pi R^2_A}{4 \pi R^2_B} = \left( \dfrac{R_A}{R_B} \right)^2.
The radius of sphere A is given to be 6 cm the radius of sphere B is 18 cm
    AAAB=(RARB)2=(13)2\implies \dfrac{A_A}{A_B} = \left( \dfrac{R_A}{R_B} \right)^2 = \left( \dfrac{1}{3} \right)^2.

Now, we are also given the maximum intensity wavelength emitted in the spectrum of the two bodies. By Wien's displacement law, the maximum intensity wavelength is inversely proportional to the temperature. So we can write:
TATB=λBλA=1500500=3\dfrac{T_A}{T_B} = \dfrac{\lambda_B}{\lambda_A} = \dfrac{1500}{500} = 3

Now, the ratio of the energies for the two spheres now becomes:
EAEB=(λBλA)4(RARB)2=(3)4(13)2=9\dfrac{E_A}{E_B} = \left( \dfrac{\lambda_B}{\lambda_A} \right)^4 \left( \dfrac{R_A}{R_B} \right)^2 = (3)^4 \left( \dfrac{1}{3} \right)^2 = 9

So, the correct answer is “Option A”.

Note:
The question does not clearly mention that the value of wavelength has been given to us. As we know nm stands for nanometer which is a unit of wavelength and as wavelength is a characteristic for a spectrum we must be able to guess what to do with 1500 nm and 500 nm based on this.