Three black bodies at temperatures T1, T2, T3 (T1 < T2 < T3) emit radiation whose spectral intensity vs wavelength graphs have areas A1, A2, A3 respectively. According to Wien's displacement law and the Stefan-Boltzmann law, which of the following correctly states the ratio A1: A2: A3?

The total areas A1, A2, A3 are equal
A1: A2: A3 = T1²: T2²: T3²
A1: A2: A3 = T1⁴: T2⁴: T3⁴
T3 > T2 > T1

Correct answer: A1: A2: A3 = T1⁴: T2⁴: T3⁴

Solution

The area under the spectral distribution curve represents the total emissive power of the black body. By Stefan-Boltzmann law, E = sigma * T⁴, so A is proportional to T⁴. Therefore A1: A2: A3 = T1⁴: T2⁴: T3⁴.

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