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A black body at a temperature of 1640 K has wavelength corresponding to maximum emission equal to 1.75 μm. Assuming the moon to be a perfectly black body, the temperature of moon if the wavelength corresponding to maximum emission is 14.35 μm, is
- a)100 K
- b)150 K
- c)200 K
- d)250 K
Correct answer is option 'C'. Can you explain this answer?
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A black body at a temperature of 1640 K has wavelength corresponding t...
We can use Wien's displacement law to solve this problem.
Wien's displacement law states that the wavelength corresponding to the maximum emission of a black body is inversely proportional to its temperature. Mathematically, we can write:
λmax T = constant
where λmax is the wavelength corresponding to maximum emission, T is the temperature in Kelvin, and the constant is known as Wien's displacement constant and has a value of approximately 2.898 × 10^-3 m·K.
In this problem, we are given the temperature T = 1640 K and the wavelength corresponding to maximum emission λmax = 1.75 μm (convert to meters). We can plug these values into Wien's displacement law and solve for the constant:
λmax T = constant
(1.75 × 10^-6 m) (1640 K) = constant
constant ≈ 2.862 × 10^-3 m·K
Now that we have the constant, we can use Wien's displacement law to find the wavelength corresponding to maximum emission at a different temperature. For example, if we want to find the maximum emission wavelength at a temperature of 3000 K, we can write:
λmax T = constant
λmax (3000 K) = 2.862 × 10^-3 m·K
λmax ≈ 962 nm
Therefore, a black body at a temperature of 3000 K would have a maximum emission wavelength of approximately 962 nm.
Wien's displacement law states that the wavelength corresponding to the maximum emission of a black body is inversely proportional to its temperature. Mathematically, we can write:
λmax T = constant
where λmax is the wavelength corresponding to maximum emission, T is the temperature in Kelvin, and the constant is known as Wien's displacement constant and has a value of approximately 2.898 × 10^-3 m·K.
In this problem, we are given the temperature T = 1640 K and the wavelength corresponding to maximum emission λmax = 1.75 μm (convert to meters). We can plug these values into Wien's displacement law and solve for the constant:
λmax T = constant
(1.75 × 10^-6 m) (1640 K) = constant
constant ≈ 2.862 × 10^-3 m·K
Now that we have the constant, we can use Wien's displacement law to find the wavelength corresponding to maximum emission at a different temperature. For example, if we want to find the maximum emission wavelength at a temperature of 3000 K, we can write:
λmax T = constant
λmax (3000 K) = 2.862 × 10^-3 m·K
λmax ≈ 962 nm
Therefore, a black body at a temperature of 3000 K would have a maximum emission wavelength of approximately 962 nm.
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A black body at a temperature of 1640 K has wavelength corresponding to maximum emission equal to 1.75 μm. Assuming the moon to be a perfectly black body, the temperature of moon if the wavelength corresponding to maximum emission is 14.35 μm, isa)100 Kb)150 Kc)200 Kd)250 KCorrect answer is option 'C'. Can you explain this answer?
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A black body at a temperature of 1640 K has wavelength corresponding to maximum emission equal to 1.75 μm. Assuming the moon to be a perfectly black body, the temperature of moon if the wavelength corresponding to maximum emission is 14.35 μm, isa)100 Kb)150 Kc)200 Kd)250 KCorrect answer is option 'C'. 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 A black body at a temperature of 1640 K has wavelength corresponding to maximum emission equal to 1.75 μm. Assuming the moon to be a perfectly black body, the temperature of moon if the wavelength corresponding to maximum emission is 14.35 μm, isa)100 Kb)150 Kc)200 Kd)250 KCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A black body at a temperature of 1640 K has wavelength corresponding to maximum emission equal to 1.75 μm. Assuming the moon to be a perfectly black body, the temperature of moon if the wavelength corresponding to maximum emission is 14.35 μm, isa)100 Kb)150 Kc)200 Kd)250 KCorrect answer is option 'C'. Can you explain this answer?.
A black body at a temperature of 1640 K has wavelength corresponding to maximum emission equal to 1.75 μm. Assuming the moon to be a perfectly black body, the temperature of moon if the wavelength corresponding to maximum emission is 14.35 μm, isa)100 Kb)150 Kc)200 Kd)250 KCorrect answer is option 'C'. 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 A black body at a temperature of 1640 K has wavelength corresponding to maximum emission equal to 1.75 μm. Assuming the moon to be a perfectly black body, the temperature of moon if the wavelength corresponding to maximum emission is 14.35 μm, isa)100 Kb)150 Kc)200 Kd)250 KCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A black body at a temperature of 1640 K has wavelength corresponding to maximum emission equal to 1.75 μm. Assuming the moon to be a perfectly black body, the temperature of moon if the wavelength corresponding to maximum emission is 14.35 μm, isa)100 Kb)150 Kc)200 Kd)250 KCorrect answer is option 'C'. Can you explain this answer?.
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