1. U₁ = 0 2. U₃ = 0 3. U₁ > U₂ 4. U₂ > U₁

Answer: 4. U₂ > U₁ Key idea: Wien's law λ_m T=b gives the wavelength of peak emission. λ_m=dfrac2.88×10⁶5760=500 nm. Emission is maximum at 500 nm, so U₂ is greater than U₁ (250 nm) and U₃ . A black body emits at all wavelengths, so U₁ and U₃ are not zero.

Topic: Heat transfer: conduction, convection, radiation· Properties of Solids and Liquids

A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is U_1, at wavelength 500 nm is U_2 and that at 1000 nm is U_3. Wien's constant, b = 2.88×10^6 nmK. Which of the following is correct ?

1. \(U_1 = 0\)

2. \(U_3 = 0\)

3. \(U_1 > U_2\)

4. \(U_2 > U_1\)

Doubtly Asked 30 Oct 2026 NEET UG 2016 Phase 1 neet Save WhatsApp Report

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Answer: 4. \(U_2 > U_1\)

Key idea: Wien's law \(\lambda_m T=b\) gives the wavelength of peak emission.

\(\lambda_m=\dfrac{2.88\times10^{6}}{5760}=500\) nm.
Emission is maximum at 500 nm, so \(U_2\) is greater than \(U_1\) (250 nm) and \(U_3\).
A black body emits at all wavelengths, so \(U_1\) and \(U_3\) are not zero.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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A black body is at a temperature of 5760 K. The energy of radiation emitted by the body at wavelength 250 nm is U_1, at wavelength 500 nm is U_2 and that at 1000 nm is U_3. Wien's constant, b = 2.88×10^6 nmK. Which of the following is correct ?

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