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A black body at temperature T emits radiation with a certain intensity spectrum. When the temperature is raised to a higher value T' (T' > T), which of the following statements are correct? (A) The intensity of radiation at every wavelength increases. (B) The maximum intensity occurs at a shorter wavelength. (C) The total area under the spectral intensity graph increases. (D) The total area under the graph is proportional to the fourth power of temperature.
- (A), (B) and (C) only
- (A) and (B) only
- (B), (C) and (D) only
- (A), (B), (C) and (D)
Correct answer: (A), (B), (C) and (D)
Solution
When temperature increases from T to T': (A) The spectral radiance at each wavelength increases because the Planck distribution curve shifts up everywhere for higher T. TRUE. (B) By Wien's displacement law, lambda_max = b/T, so as T increases, lambda_max decreases (shifts to shorter wavelength). TRUE. (C) The total energy radiated (area under spectral curve) is proportional to T⁴ by Stefan-Boltzmann law, so it increases. TRUE. (D) By Stefan-Boltzmann law, E = sigma*T⁴, so the area is proportional to T⁴. TRUE. All four statements are correct.
Related JEE Advanced Physics questions
- Match the temperature of a blackbody listed in Group-I to the corresponding statement in Group-II, and select the correct answer.
[Given: Wien’s constant = 2.9 × 10⁻³ m-K and hc/e = 1.24 × 10⁻⁶ V-m]
Group-I:
(P) 2000 K
(Q) 3000 K
(R) 5000 K
(S) 10000 K
Group-II:
(1) The peak wavelength of emitted radiation can cause photoelectron ejection from a metal with a work function of 4 eV.
(2) The peak wavelength of emitted radiation falls within the visible spectrum.
(3) The peak wavelength of emitted radiation produces the broadest central diffraction maximum in a single-slit setup.
(4) The energy radiated per unit area is one-sixteenth of that emitted by a blackbody at 6000 K.
(5) The peak wavelength of emitted radiation is suitable for imaging human bones.
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