Black bodies A and B radiate maximum energy with wavelength difference 4. The absolute temperature of body A is 3 times that of B. The wavelength at which body B radiates maximum energy is (a) 4 (b) 6 (c) 2 (d) 8

Answer: (b). Wien's law: λm T = b, so λA = (λB)/(3) since TA = 3TB. λB - (λB)/(3) = 4 ⇒ 2/3λB = 4 λB = 6, so the answer is (b).

Black bodies A and B radiate maximum energy with wavelength… | MHT CET 2025 (22 Apr, Shift 2)

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Thermal Radiation: Black Body, Stefan & Wien

Black bodies A and B radiate maximum energy with wavelength difference $4\,\mu\text{m}$ . The absolute temperature of body A is 3 times that of B. The wavelength at which body $B$ radiates maximum energy is

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Answer: (b)

Wien's law: $\lambda_m T = b$ , so $\lambda_A = \frac{\lambda_B}{3}$ since $T_A = 3T_B$ .
$\lambda_B - \frac{\lambda_B}{3} = 4\,\mu\text{m} \Rightarrow \frac{2}{3}\lambda_B = 4\,\mu\text{m}$
$\lambda_B = 6\,\mu\text{m}$ , so the answer is (b).

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Thermal Radiation: Black Body, Stefan & Wien question from MHT CET 2025 (22 Apr, Shift 2)

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