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The adjoining diagram shows the spectral energy density distribution $E_{λ}$ of a black body at two different temperatures. If the areas under the curves are in the ratio 256 : 1, the value of temperature T is

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16,000 K

32,000 K

8,000 K

4,000 K

answer is C.

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Detailed Solution

$\frac{A_{T}}{A_{2000}} = \frac{256}{1}$ (given)

Area under $e_{λ} - λ$ curve represents the emissive power of body and $emissive power \propto T^{4}$
(Hence area under $e_{λ} - λ$ curve) $\propto T^{4}$

$\Rightarrow \frac{A_{T}}{A_{2000}} = {(\frac{T}{2000})}^{4} \Rightarrow \frac{256}{1} = {(\frac{T}{2000})}^{4} \Rightarrow T = 8000 K .$

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