The adjoining diagram shows the spectral ener | vedclass

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(D) According to the Stefan-Boltzmann law,the total emissive power $E$ of a black body is proportional to the fourth power of its absolute temperature

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) According to the Stefan-Boltzmann law,the total emissive power $E$ of a black body is proportional to the fourth power of its absolute temperature $T$,i.e.,$E \propto T^4$.The area under the spectral energy density distribution curve ($E_{\lambda}$ versus $\lambda$) represents the total emissive power of the black body.Given that the ratio of the areas under the curves is $\frac{A_T}{A_{2000}} = \frac{16}{1}$.Since the area is proportional to $T^4$,we have:$\frac{A_T}{A_{2000}} = \left( \frac{T}{2000} ight)^4$Substituting the given ratio:$\frac{16}{1} = \left( \frac{T}{2000} ight)^4$Taking the fourth root on both sides:$\sqrt[4]{16} = \frac{T}{2000}$$2 = \frac{T}{2000}$$T = 2 	imes 2000 = 4000 	ext{ K}$.Therefore,the value of temperature $T$ is $4000 	ext{ K}$.

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 ........ $K$. (in $,000$)

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