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MHT CET 2024 (10 May, Shift 1) · Thermal Radiation: Black Body, Stefan & Wien

Thermal Radiation: Black Body, Stefan & Wien question from MHT CET 2024 (10 May, Shift 1)

Q66MHT CET 2024 · 10 MAY · SHIFT 1Medium

Thermal Radiation: Black Body, Stefan & Wien

Ordinary bodies P and Q radiate maximum energy with wavelength difference $3\,\mu\text{m}$ . The absolute temperature of body P is four times that of Q. The wavelength at which body Q radiates maximum energy is

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Answer: (b)

Wien's law: $\lambda_m \propto \frac{1}{T}$ , so $\lambda_P = \frac{\lambda_Q}{4}$ .
$\lambda_Q - \frac{\lambda_Q}{4} = 3\,\mu\text{m} \Rightarrow \frac{3}{4}\lambda_Q = 3\,\mu\text{m}$
$\lambda_Q = 4\,\mu\text{m}$ , so the answer is (b).

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