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Three black-body discs A, B and C have radii 2 m, 4 m and 6 m respectively. The wavelengths at maximum intensity are lambda_A = 300 nm, lambda_B = 400 nm and lambda_C = 500 nm respectively. Compare the radiated powers Q_A, Q_B and Q_C.
- Q_A is maximum
- Q_B is maximum
- Q_C is maximum
- Q_A = Q_B = Q_C
Correct answer: Q_B is maximum
Solution
By Stefan-Boltzmann law Q = sigma*A*T⁴ = sigma*(pi*r²)*T⁴. By Wien's displacement law: T = b/lambda_max. So T⁴ = b⁴/lambda_max⁴. Q proportional to r²/lambda_max⁴. For A: 4/300⁴. For B: 16/400⁴. For C: 36/500⁴. Comparing Q_A: Q_B: Q_C = 4/300⁴: 16/400⁴: 36/500⁴. Multiply all by 10¹²: = 4/(81): 16/(256): 36/(625) = 0.0494: 0.0625: 0.0576. So Q_B is maximum.
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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