The spectral emission power of a black body at 6000 K is maximum at 5500 A ˚ . If the temperature is increased by 10% then decrease in the value of λ m will be

From λ m ⊤ = b , ( 5500 ) ( 6000 ) = λ m ( 6000 + 600 ) ⇒ λ m = 5000 A ˚ Decrease in the value of λ m = 5500 5500 − 5000 × 100 = 11 100 = 9.09%

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The spectral emission power of a black body at 6000 K is maximum at $5500 \overset{˚}{A}$ . If the temperature is increased by $10%$ then decrease in the value of $λ_{m}$ will be

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Solution:

From $λ_{m} ⊤ = b$ ,

$(5500) (6000) = λ_{m} (6000 + 600) \Rightarrow λ_{m} = 5000 \overset{˚}{A}$

Decrease in the value of

$λ_{m} = \frac{5500 - 5000}{5500} \times 100 = \frac{100}{11} = 9.09%$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:Radiation

⏱ 2mℹ️ Source: QB

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