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
600.9k+ 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.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Which among the following are examples of coming together class 11 social science CBSE

