Intensity of the radiation from the surface of a black body is maximum at a wavelength of $$5760A0$$. If Wien’s constant is $$0.288$$ $$kelvin-cm$$. Then the power radiated per unit area of the surface of a black body is

At maximum intensity applying Wien's displacement law, $$5760$$ × $$T=0.288$$ × $$108$$ ⇒ $$T=5000K$$∴Power radiated per unit surface area σ$$T4=5.67$$× $$10$$−$$8$$ ×$$(5000)4W/m2=35.4$$ $$MW/m2$$

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Question

Intensity of the radiation from the surface of a black body is maximum at a wavelength of $5760 \overset{0}{A}$ . If Wien’s constant is 0.288 kelvin-cm. Then the power radiated per unit area of the surface of a black body is

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Solution

At maximum intensity applying Wien's displacement law,
$5760 \times T = 0 .288 \times 10^{8} \Rightarrow T = 5000 K$
$∴$ Power radiated per unit surface area ${σT}^{4} = 5 .67 \times 10^{- 8} \times {(5000)}^{4} W / m^{2} = 35.4 MW / m^{2}$

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