Three identical bodies are at temperature $T_{1}, T_{2}$ and $T_{3}$ having $e_{1}, e_{2}$ and $e_{3}$ as their respective emissivities. The thermal spectrum obtained for them is as shown in the diagram. Choose the correct order of temperatures and emissivities.

T_{1} > T_{2} > T_{3}

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T_{1} < T_{2} < T_{3}

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e_{1} < e_{2} < e_{3}

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e_{1} > e_{2} > e_{3}

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Solution

The correct options are
A $T_{1} > T_{2} > T_{3}$
C $e_{1} < e_{2} < e_{3}$
$λ \propto \frac{1}{T} \Rightarrow T_{1} > T_{2} > T_{3} as λ_{1} < λ_{2} < λ_{3} (Wien's law)$
Since area under the curve for $λ_{3}$ is more than that of $λ_{2}$ .
$∴ σ A e_{3} T_{3}^{4} > σ A e_{2} T_{2}^{4}$ (Area of body is same)
Now as $T_{3} < T_{2} \Rightarrow e_{3} > e_{2}$
Similarly, $e_{2} > e_{1}$

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Three identical bodies are at temperature T1,T2 and T3 having e1,e2 and e3 as their respective emissivities. The thermal spectrum obtained for them is as shown in the diagram. Choose the correct order of temperatures and emissivities.

The correct options are A T1>T2>T3 C e1<e2<e3λ∝1T⇒T1>T2>T3 as λ1<λ2<λ3 (Wien's law) Since area under the curve for λ3 is more than that of λ2. ∴σAe3T43 > σAe2T42 (Area of body is same) Now as T3<T2⇒e3>e2 Similarly, e2>e1

Three identical bodies are at temperature $T_{1}, T_{2}$ and $T_{3}$ having $e_{1}, e_{2}$ and $e_{3}$ as their respective emissivities. The thermal spectrum obtained for them is as shown in the diagram. Choose the correct order of temperatures and emissivities.