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

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Aicharya · Accepted Answer

$\lambda_{2}-\lambda_{1}=4 \mu \mathrm{m} \quad \quad \mathrm{T}_{1}=3 \mathrm{~T}_{2}$ $\lambda_{1} \mathrm{~T}_{1}=\lambda_{2} \mathrm{~T}_{2} \quad 	herefore \quad \lambda_{1} 3 \mathrm{~T}_{2}=\lambda_{2} \mathrm{~T}_{2}$ $\lambda_{2}=3 \lambda_{1}$ $3 \lambda_{1}-\lambda_{1}=4 \mu \mathrm{m}$ $2 \lambda_{1}=4 \mu \mathrm{m} \quad 	herefore \quad \lambda_{2}=6 \mu \mathrm{m}$ $\lambda_{1}=2 \mu \mathrm{m} \quad$

Ordinary bodies 'A' and 'B' radiate maximum energy with wavelength difference $4 \mu \mathrm{m}$. The…

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