The energy radiated by a black body at 2300 K is found to have the maximum emissive power at a wavelength 1260 nm , its emissive power being 8000 Wm − 2 . When the body is cooled to a temperature T K,

Question

The energy radiated by a black body at 2300 K is found to have the maximum emissive power at a wavelength 1260 nm , its emissive power being $8000 Wm^{- 2}$ . When the body is cooled to a temperature T K, the emissive power is found to decrease to $500 Wm^{- 2}$ . Find the wavelength (in nm) at which intensity of emission is maximum at the temperature T. If your answer is x then find the value of x/360.

Answer:

7

Solution:

$E \propto T^{4} : \frac{8000}{500} = 16 \Rightarrow T = \frac{2300}{2} = 1150 K$ .

Wien's law: $λ^{'} = 1260 \times 2 = 2520 nm$ , so $\frac{x}{360} = \frac{2520}{360} = 7$ .

TestHub · Accepted Answer

E ∝ T 4 : 500 8000 ​ = 16 ⇒ T = 2 2300 ​ = 1150 K . Wien's law: λ ′ = 1260 × 2 = 2520 nm , so 360 x ​ = 360 2520 ​ = 7 .

Physics - Thermodynamics Question with Solution | TestHub

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