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Q.

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 16 : 1, the value of temperature T is

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a

16,000 K

b

8,000 K

c

4,000 K

d

32,000 K

answer is D.

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Detailed Solution

$\frac{{{A_T}}}{{{A_{2000}}}} = \frac{{16}}{1}$ (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) $\propto\,T^4$
$\Rightarrow \frac{{AT}}{{{A_{2000}}}} = {\left( {\frac{T}{{2000}}} \right)^4} \Rightarrow \frac{{16}}{1} = {\left( {\frac{T}{{2000}}} \right)^4} \Rightarrow T = 4000K.$

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