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The black body spectrum of an object Q1 is such that its radiant intensity (i.e., intensity per unit wavelength interval) is maximum at a wavelength of 200 nm. Another object O2 has the maximum radiant intensity at 600 nm. The ratio of power emitted per unit area by source O1 to that of source O2 is
- a)1 : 81
- b)1 : 9
- c)9 : 1
- d)81 : 1
Correct answer is option 'D'. Can you explain this answer?
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The black body spectrum of an object Q1is such that its radiant intens...
From Wein's displacement law,
From Boltzmann's law
From Boltzmann's law
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The black body spectrum of an object Q1is such that its radiant intens...
Black Body Spectrum Comparison
- Object Q1 peaks at 100 nm
- Object O2 peaks at 600 nm
Calculating Power Emitted
- According to Wien's displacement law, the peak wavelength and temperature of a black body are inversely proportional. Since O1 peaks at 100 nm and O2 at 600 nm, the temperature of O1 is higher than that of O2.
- The power emitted per unit area by a black body is given by Stefan-Boltzmann law: P = σT^4, where σ is the Stefan-Boltzmann constant and T is the temperature.
- As the temperature of O1 is higher than O2, the power emitted per unit area by O1 is greater than that of O2.
Calculating the Ratio
- Since the power emitted per unit area is proportional to T^4, if the temperature of O1 is x times higher than O2, then the ratio of power emitted is (x^4) : 1.
- As the temperatures are in the inverse ratio of the peak wavelengths, O1 temperature is 6 times higher than O2 (600/100 = 6).
- Therefore, the ratio of power emitted per unit area by source O1 to that of source O2 is 6^4 : 1 = 1296 : 1, which simplifies to 81 : 1.
Therefore, the correct answer is option D) 81 : 1.
- Object Q1 peaks at 100 nm
- Object O2 peaks at 600 nm
Calculating Power Emitted
- According to Wien's displacement law, the peak wavelength and temperature of a black body are inversely proportional. Since O1 peaks at 100 nm and O2 at 600 nm, the temperature of O1 is higher than that of O2.
- The power emitted per unit area by a black body is given by Stefan-Boltzmann law: P = σT^4, where σ is the Stefan-Boltzmann constant and T is the temperature.
- As the temperature of O1 is higher than O2, the power emitted per unit area by O1 is greater than that of O2.
Calculating the Ratio
- Since the power emitted per unit area is proportional to T^4, if the temperature of O1 is x times higher than O2, then the ratio of power emitted is (x^4) : 1.
- As the temperatures are in the inverse ratio of the peak wavelengths, O1 temperature is 6 times higher than O2 (600/100 = 6).
- Therefore, the ratio of power emitted per unit area by source O1 to that of source O2 is 6^4 : 1 = 1296 : 1, which simplifies to 81 : 1.
Therefore, the correct answer is option D) 81 : 1.
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The black body spectrum of an object Q1is such that its radiant intensity (i.e., intensity per unit wavelength interval) is maximum at a wavelength of 200 nm. Another object O2has the maximum radiant intensity at 600 nm. The ratio of power emitted per unit area by source O1to that of source O2isa)1 : 81b)1 : 9c)9 : 1d)81 : 1Correct answer is option 'D'. Can you explain this answer?
Question Description
The black body spectrum of an object Q1is such that its radiant intensity (i.e., intensity per unit wavelength interval) is maximum at a wavelength of 200 nm. Another object O2has the maximum radiant intensity at 600 nm. The ratio of power emitted per unit area by source O1to that of source O2isa)1 : 81b)1 : 9c)9 : 1d)81 : 1Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The black body spectrum of an object Q1is such that its radiant intensity (i.e., intensity per unit wavelength interval) is maximum at a wavelength of 200 nm. Another object O2has the maximum radiant intensity at 600 nm. The ratio of power emitted per unit area by source O1to that of source O2isa)1 : 81b)1 : 9c)9 : 1d)81 : 1Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The black body spectrum of an object Q1is such that its radiant intensity (i.e., intensity per unit wavelength interval) is maximum at a wavelength of 200 nm. Another object O2has the maximum radiant intensity at 600 nm. The ratio of power emitted per unit area by source O1to that of source O2isa)1 : 81b)1 : 9c)9 : 1d)81 : 1Correct answer is option 'D'. Can you explain this answer?.
The black body spectrum of an object Q1is such that its radiant intensity (i.e., intensity per unit wavelength interval) is maximum at a wavelength of 200 nm. Another object O2has the maximum radiant intensity at 600 nm. The ratio of power emitted per unit area by source O1to that of source O2isa)1 : 81b)1 : 9c)9 : 1d)81 : 1Correct answer is option 'D'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The black body spectrum of an object Q1is such that its radiant intensity (i.e., intensity per unit wavelength interval) is maximum at a wavelength of 200 nm. Another object O2has the maximum radiant intensity at 600 nm. The ratio of power emitted per unit area by source O1to that of source O2isa)1 : 81b)1 : 9c)9 : 1d)81 : 1Correct answer is option 'D'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The black body spectrum of an object Q1is such that its radiant intensity (i.e., intensity per unit wavelength interval) is maximum at a wavelength of 200 nm. Another object O2has the maximum radiant intensity at 600 nm. The ratio of power emitted per unit area by source O1to that of source O2isa)1 : 81b)1 : 9c)9 : 1d)81 : 1Correct answer is option 'D'. Can you explain this answer?.
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