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MHT CET 2026 (20 May, Shift 1) · Thermal Radiation: Black Body, Stefan & Wien

Thermal Radiation: Black Body, Stefan & Wien question from MHT CET 2026 (20 May, Shift 1)

Q8MHT CET 2026 · 20 MAY · SHIFT 1Easy

Thermal Radiation: Black Body, Stefan & Wien

Temperature of two stars are in the ratio $3 : 2$ . If wavelength of maximum intensity of first star is $4200$ Å, the corresponding wavelength for second star is

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Answer: (c)

Wien's law: $\lambda_m T =$ constant, so $\lambda_m \propto \dfrac{1}{T}$
$\lambda_2 = \lambda_1\dfrac{T_1}{T_2} = 4200\times\dfrac{3}{2} = 6300$ Å
So the answer is (c) 6300 Å.

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