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Question: On observing light from three different stars \(P,\;Q,\;R\), it was found that intensity of violet c...

On observing light from three different stars P,  Q,  RP,\;Q,\;R, it was found that intensity of violet colour is maximum in the spectrum of PP, the intensity of green colour is maximum in the spectrum of RR and the intensity of red colour is maximum in the spectrum of QQ. If TP,TQT_{P},T_{Q} and TRT_{R} are the respective absolute temperature of P,  QP,\;Q and RR, then it can be concluded from the above observation that:

& A.{{T}_{P}}<{{T}_{R}}<{{T}_{Q}} \\\ & B.{{T}_{P}}<{{T}_{Q}}<{{T}_{R}} \\\ & C.{{T}_{P}}>{{T}_{Q}}>{{T}_{R}} \\\ & D.{{T}_{P}}>{{T}_{R}}>{{T}_{Q}} \\\ \end{aligned}$$
Explanation

Solution

We know that the rate at which a body absorbs heat is given by the Stefan–Boltzmann law. However, Wien’s law of displacement, that the radiation due to a body varies with temperature. Since we need the temperature due to maximum wavelength, we are using the Wien’s law here.

Formula used:
T=1λmT=\dfrac{1}{\lambda_{m}}

Complete step-by-step answer:
Wien’s law of displacement, gives the temperature of the maximum wavelength from the black-body radiation curve. Then, we have, λmT=b\lambda_{m}T=b. where, λm\lambda_{m} is the maximum wavelength, whose corresponding temperature is TT and bb is some consonant.
Clearly, we have, T=1λmT=\dfrac{1}{\lambda_{m}}
We know that the red has the highest wavelength, then star Q will have the minimum wavelength. Followed by green, then the star R will have the second largest value of temperature. And violet has the lowest wavelength or P has the highest temperature.
Hence the correct option is D.TP>TR>TQD.{{T}_{P}}>{{T}_{R}}>{{T}_{Q}}

So, the correct answer is “Option D”.

Additional Information: Stefan-Boltzmann constant or the Stefan’s constant is denoted by σ\sigma, it is a constant of proportionality, used in the Stefan–Boltzmann law of blackbody radiation; which is stated as the total intensity radiated over the wavelength, it is proportional the temperature. It is given as Eab=σAT4E_{ab}=\sigma AT^{4} where EabE_{ab} is the energy absorbed by the body here the star, whose surface area is AA and temperature is TT.

Note: A German physicist Max Planck in the 1900, gave the theory of black body radiations, he stated that the spectral lines of a hypothetical black body would emit radiations in small, discrete packets called the quanta of energy, and not as continuous radiation as expected. Here, we are assuming the stars to be a black body.