Black Body Radiation

By Wien's Displacement Law, the maximum wavelength of radiation emitted and the surface temperature of the body are related as:

$\lambda_{max} T = b$

where $b$ is Wien's constant.

The power radiated by a body is directly proportional to the 4th power of the Surface Temperature, i.e,

$\dfrac{dQ}{dt} = k T^4$

$\Rightarrow \dfrac{dQ}{dt} = k \dfrac{1}{(\lambda_{max})^4}$

From the question,

$\dfrac{P}{\frac{256}{81}P} = \dfrac{(\lambda_{max}^{'})^{4}}{(\lambda_{max})^{4}}$

$\Rightarrow \dfrac{3}{4} = \dfrac{\lambda_{max}^{'}}{\lambda_{max}} \Rightarrow \lambda_{max}^{'} = \dfrac{3}{4} {\lambda_{max}}$

Hence the wavelength of the emitting body is reduced by $\dfrac{\lambda}{4}$

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Note:

• $\lambda_{max}^{'}$ is the changed wavelength.