
The energy spectrum of a black body exhibits a maximum energy around a wavelength \[{\lambda _o}\] . The temperature of the black body is now changed such that the energy is maximum around a wavelength \[\dfrac{{3{\lambda _o}}}{4}\] .The power radiated by the black body will now increase by a factor of:
(A) \[\dfrac{{256}}{{81}}\]
(B) \[\dfrac{{64}}{{27}}\]
(C) \[\dfrac{{16}}{9}\]
(D) \[\dfrac{4}{3}\]
Answer
133.5k+ views
Hint We first use wien's displacement law to calculate the ratio of temperatures of both the bodies as energy and wavelength present so this law is applicable.
Then we use Stefan’s law as when there is power radiated, we use this law. By the use of Stefan’s law, we will find energy radiated per unit time per unit area i.e. E.
Then we will find the ratio of energy radiated for both the bodies and equate with the other values obtained and find the correct option.
Complete step by step solution
We know that when there is maximum energy and wavelength present, we use the concept of Wein’s displacement law.
According to Wien's displacement law, wavelength corresponding to maximum energy decreases when the temperature of black body increases i.e. \[\lambda T = b\] , where b is constant, \[\lambda \] is the wavelength and T is the temperature.
So, we can write Wien's displacement law for both the temperature i.e. \[{\lambda _1}{T_1} = {\lambda _2}{T_2}\]
So \[\dfrac{{{T_2}}}{{{T_1}}} = \dfrac{{{\lambda _1}}}{{{\lambda _2}}}\] , now we will put the value of wavelength given in the question.
\[\dfrac{{{T_2}}}{{{T_1}}} = \dfrac{{{\lambda _o}}}{{\dfrac{{3{\lambda _o}}}{4}}}\] , so \[\dfrac{{{T_2}}}{{{T_1}}} = \dfrac{4}{3}\] .
And also, we know when there comes power radiated, we use Stefan’s law:
Energy radiated per unit time per unit area i.e. \[E = \sigma {T^4}\] , where \[\sigma \] is Stephan-Boltzmann constant. T is the temperature.
So, for first body \[{E_1} = \sigma {T_1}^4\] and for second body \[{E_2} = \sigma {T_2}^4\]
It is said that power radiated by the black body will now increase by a factor in the question. So, we get \[\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{T_1}^4}}{{{T_2}^4}}\] ;
\[\dfrac{{{E_1}}}{{{E_2}}} = {\left( {\dfrac{{{T_1}}}{{{T_2}}}} \right)^4}\] . Now we put the value \[\dfrac{{{T_2}}}{{{T_1}}} = \dfrac{4}{3}\]
So, we get \[\dfrac{{{E_2}}}{{{E_1}}} = {\left( {\dfrac{4}{3}} \right)^4} = \dfrac{{256}}{{81}}\] .
So, the correct option is A
Note Remember the formula of both Wien's displacement law and Stefan's Law and also remember when to apply these laws.
Note that Stefan's Law clarifies that total radiant heat energy emerging from a surface is proportional to the fourth power of its absolute temperature.
Also remember Stefan-Boltzmann law applies only to blackbodies, surfaces that absorb all incident heat radiation
Then we use Stefan’s law as when there is power radiated, we use this law. By the use of Stefan’s law, we will find energy radiated per unit time per unit area i.e. E.
Then we will find the ratio of energy radiated for both the bodies and equate with the other values obtained and find the correct option.
Complete step by step solution
We know that when there is maximum energy and wavelength present, we use the concept of Wein’s displacement law.
According to Wien's displacement law, wavelength corresponding to maximum energy decreases when the temperature of black body increases i.e. \[\lambda T = b\] , where b is constant, \[\lambda \] is the wavelength and T is the temperature.
So, we can write Wien's displacement law for both the temperature i.e. \[{\lambda _1}{T_1} = {\lambda _2}{T_2}\]
So \[\dfrac{{{T_2}}}{{{T_1}}} = \dfrac{{{\lambda _1}}}{{{\lambda _2}}}\] , now we will put the value of wavelength given in the question.
\[\dfrac{{{T_2}}}{{{T_1}}} = \dfrac{{{\lambda _o}}}{{\dfrac{{3{\lambda _o}}}{4}}}\] , so \[\dfrac{{{T_2}}}{{{T_1}}} = \dfrac{4}{3}\] .
And also, we know when there comes power radiated, we use Stefan’s law:
Energy radiated per unit time per unit area i.e. \[E = \sigma {T^4}\] , where \[\sigma \] is Stephan-Boltzmann constant. T is the temperature.
So, for first body \[{E_1} = \sigma {T_1}^4\] and for second body \[{E_2} = \sigma {T_2}^4\]
It is said that power radiated by the black body will now increase by a factor in the question. So, we get \[\dfrac{{{E_1}}}{{{E_2}}} = \dfrac{{{T_1}^4}}{{{T_2}^4}}\] ;
\[\dfrac{{{E_1}}}{{{E_2}}} = {\left( {\dfrac{{{T_1}}}{{{T_2}}}} \right)^4}\] . Now we put the value \[\dfrac{{{T_2}}}{{{T_1}}} = \dfrac{4}{3}\]
So, we get \[\dfrac{{{E_2}}}{{{E_1}}} = {\left( {\dfrac{4}{3}} \right)^4} = \dfrac{{256}}{{81}}\] .
So, the correct option is A
Note Remember the formula of both Wien's displacement law and Stefan's Law and also remember when to apply these laws.
Note that Stefan's Law clarifies that total radiant heat energy emerging from a surface is proportional to the fourth power of its absolute temperature.
Also remember Stefan-Boltzmann law applies only to blackbodies, surfaces that absorb all incident heat radiation
Recently Updated Pages
JEE Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

Sign up for JEE Main 2025 Live Classes - Vedantu

JEE Main Books 2023-24: Best JEE Main Books for Physics, Chemistry and Maths

JEE Main 2023 April 13 Shift 1 Question Paper with Answer Key

JEE Main 2023 April 11 Shift 2 Question Paper with Answer Key

JEE Main 2023 April 10 Shift 2 Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Wheatstone Bridge for JEE Main Physics 2025

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Diffraction of Light - Young’s Single Slit Experiment

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

If a wire of resistance R is stretched to double of class 12 physics JEE_Main

JEE Advanced 2024 Syllabus Weightage

Current Loop as Magnetic Dipole and Its Derivation for JEE
