Question

The absolute temperature of a body A is four times that of another body B. For the two body the difference in wavelengths, at which energy radiated is maximum is $3.0\mu m.$ The wavelength at which the body B radiates maximum energy in micrometers is :

Answer 1

If the temperature of body A and B be $TA\&TB$, and the wavelength at which maximum emmision occurs be ${\lambda }_A\&{\lambda }_B$ respectively then

From wien's law :-

${\lambda }_A{T}_A={\lambda }_B{T}_B=b$ (constant)

Now from question,

${\lambda }_B-{\lambda }_A=3\mu m$ ....(1) [Due to lesser tamp B radiates longer wavelengths]

and , $T_A=4T_B$

$\Rightarrow \frac{1}{{\lambda }_A}=\frac{4}{{\lambda }_B}$

$\Rightarrow {\lambda }_B=4{\lambda }_A$ .... $\left(2\right)$

From $\left(1\right)$ & $\left(2\right)$ we get on solving

${\lambda }_B=12\mu m$