A black body radiates maximum energy at wavelength ' λ ' at temperature T₁ and its emissive power is E. When the temperature of the body is changed to T₂, it radiates maximum energy at wavelength (2λ)/(3), then the emissive power will become (a) 99/16E (b) 81/16E (c) 63/16E (d) 45/16E

Answer: (b). Wien: T ∝ (1)/(λm), so (T₂)/(T₁) = (λ)/(2λ/3) = 3/2. Stefan: E ∝ T⁴ ⇒ E' = (3/2)⁴ E = 81/16E So the answer is (b).

A black body radiates maximum energy at wavelength ' λ ' at… | MHT CET 2026 (18 Apr, Shift 2)

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Thermal Radiation: Black Body, Stefan & Wien

A black body radiates maximum energy at wavelength ' $\lambda$ ' at temperature $T_1$ and its emissive power is E. When the temperature of the body is changed to $T_2$ , it radiates maximum energy at wavelength $\frac{2\lambda}{3}$ , then the emissive power will become

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Answer: (b)

Wien: $T \propto \frac{1}{\lambda_m}$ , so $\frac{T_2}{T_1} = \frac{\lambda}{2\lambda/3} = \frac{3}{2}$ .
Stefan: $E \propto T^4 \Rightarrow E' = \left(\frac{3}{2}\right)^4 E = \frac{81}{16}E$
So the answer is (b).

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Thermal Radiation: Black Body, Stefan & Wien question from MHT CET 2026 (18 Apr, Shift 2)

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