
The temperature of the sun can be found out by using-
(A) Wien's displacement law
(B) Kepler's law of motion
(C) Stefan - Boltzmann law
(D) Planck's law
Answer
572.4k+ views
Hint
Stefan-Boltzmann law, statement that the total radiant heat power emitted from a surface is proportional to the fourth power of its absolute temperature. The law applies only to blackbodies, theoretical surfaces that absorb all incident heat radiation.
Complete step by step answer
We know solar constant (heat flux falling on the earth), $S = 1350 \, W/m^2$.
Let, the energy emitted by the sun per second per unit area is $= E$.
Then, the total energy emitted by the sun is one second $4πR^2E$.
This energy is falling on a sphere of radius equal to the radius of the Earth's orbit around the sun i.e., on a sphere of surface area $4πr^2$.
Now, energy falling per unit area = $ \dfrac{{4\pi {R^2}E}}{{4\pi {r^2}}} = \dfrac{{E{R^2}}}{{{r^2}}} $.
Now,let, $ \dfrac{{E{R^2}}}{{{r^2}}} =S$ … (1)
So, using stefan boltzman’s law we get,
$ E = \sigma {T^4} $
Putting the value of E from (1)
$ \Rightarrow T = {(\dfrac{{S \times {R^2}}}{{\sigma \times {r^2}}})^{\dfrac{1}{4}}} = 5760K $
Option (C) is correct.
Note
Planck's law: Planck's law describes that the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature $T$, when there is no net flow of matter or energy between the body and its environment.
Stefan-Boltzmann law, statement that the total radiant heat power emitted from a surface is proportional to the fourth power of its absolute temperature. The law applies only to blackbodies, theoretical surfaces that absorb all incident heat radiation.
Complete step by step answer
We know solar constant (heat flux falling on the earth), $S = 1350 \, W/m^2$.
Let, the energy emitted by the sun per second per unit area is $= E$.
Then, the total energy emitted by the sun is one second $4πR^2E$.
This energy is falling on a sphere of radius equal to the radius of the Earth's orbit around the sun i.e., on a sphere of surface area $4πr^2$.
Now, energy falling per unit area = $ \dfrac{{4\pi {R^2}E}}{{4\pi {r^2}}} = \dfrac{{E{R^2}}}{{{r^2}}} $.
Now,let, $ \dfrac{{E{R^2}}}{{{r^2}}} =S$ … (1)
So, using stefan boltzman’s law we get,
$ E = \sigma {T^4} $
Putting the value of E from (1)
$ \Rightarrow T = {(\dfrac{{S \times {R^2}}}{{\sigma \times {r^2}}})^{\dfrac{1}{4}}} = 5760K $
Option (C) is correct.
Note
Planck's law: Planck's law describes that the spectral density of electromagnetic radiation emitted by a black body in thermal equilibrium at a given temperature $T$, when there is no net flow of matter or energy between the body and its environment.
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