
On observing light from three different stars $P,\;Q,\;R$, it was found that intensity of violet colour is maximum in the spectrum of $P$, the intensity of green colour is maximum in the spectrum of $R$ and the intensity of red colour is maximum in the spectrum of $Q$. If $T_{P},T_{Q}$ and $T_{R}$ are the respective absolute temperature of $P,\;Q$ and $R$, then it can be concluded from the above observation that:
\[\begin{align}
& A.{{T}_{P}}<{{T}_{R}}<{{T}_{Q}} \\
& B.{{T}_{P}}<{{T}_{Q}}<{{T}_{R}} \\
& C.{{T}_{P}}>{{T}_{Q}}>{{T}_{R}} \\
& D.{{T}_{P}}>{{T}_{R}}>{{T}_{Q}} \\
\end{align}\]
Answer
553.2k+ views
Hint: 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:
$T=\dfrac{1}{\lambda_{m}}$
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, $\lambda_{m}T=b$. where, $\lambda_{m}$ is the maximum wavelength, whose corresponding temperature is $T$ and $b$ is some consonant.
Clearly, we have, $T=\dfrac{1}{\lambda_{m}}$
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 \[D.{{T}_{P}}>{{T}_{R}}>{{T}_{Q}}\]
So, the correct answer is “Option D”.
Additional Information: Stefan-Boltzmann constant or the Stefan’s constant is denoted by $\sigma$, 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 $E_{ab}=\sigma AT^{4}$ where $E_{ab}$ is the energy absorbed by the body here the star, whose surface area is $A$ and temperature is $T$.
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.
Formula used:
$T=\dfrac{1}{\lambda_{m}}$
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, $\lambda_{m}T=b$. where, $\lambda_{m}$ is the maximum wavelength, whose corresponding temperature is $T$ and $b$ is some consonant.
Clearly, we have, $T=\dfrac{1}{\lambda_{m}}$
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 \[D.{{T}_{P}}>{{T}_{R}}>{{T}_{Q}}\]
So, the correct answer is “Option D”.
Additional Information: Stefan-Boltzmann constant or the Stefan’s constant is denoted by $\sigma$, 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 $E_{ab}=\sigma AT^{4}$ where $E_{ab}$ is the energy absorbed by the body here the star, whose surface area is $A$ and temperature is $T$.
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.
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