The adjoining diagram shows the spectral energy density distribution ${E_\lambda }$of a black body at two different temperatures. If the areas under the curves are in the ratio $16 : 1$ , the value of temperature $T$ is ........ $K$
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(d) $\frac{{{A_T}}}{{{A_{2000}}}} = \frac{{16}}{1}$ (given)
Area under ${e_\lambda } - \lambda $ curve represents the emissive power of body and emissive power $ \propto {T^4}$
(Hence area under ${e_\lambda } - \lambda $ curve) $ \propto {T^4}$
Area under ${e_\lambda } - \lambda $ curve represents the emissive power of body and emissive power $ \propto {T^4}$
(Hence area under ${e_\lambda } - \lambda $ curve) $ \propto {T^4}$
==>$\frac{{AT}}{{{A_{2000}}}} = {\left( {\frac{T}{{2000}}} \right)^4}$
==> $\frac{{16}}{1} = {\left( {\frac{T}{{2000}}} \right)^4}$==> $T = 4000K.$
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