Question

A body is placed inside a perfectly black enclosure and is allowed to reach thermal equilibrium. According to the Kirchhoff's identity, which one of the following is equal to the emissivity of the body at a given wavelength?

• A. absorptivity of the body at the same wavelength
• B. reflectivity of the body at the same wavelength
• C. Stefan-Boltzmann constant
• D. zero

Hint:

Kirchhoff's law of radiation equates spectral emissivity with spectral absorptivity for a body in thermal equilibrium.

Answer 1

The Correct Option is A

Step 1: State Kirchhoff's law of thermal radiation.
Kirchhoff's law states that for a body in thermal equilibrium with its surroundings inside an enclosure, the spectral emissivity of the body at a given wavelength is exactly equal to its spectral absorptivity at that same wavelength. Mathematically, \( \varepsilon_\lambda = \alpha_\lambda \). This result follows directly from the requirement that a body inside a black enclosure at thermal equilibrium must absorb exactly as much radiant energy as it emits at every wavelength, otherwise its temperature would drift and equilibrium would be violated.
Step 2: Derive the identity from the equilibrium condition.
Consider the body placed inside a perfectly black (blackbody) enclosure at temperature T. The enclosure walls emit blackbody radiation of spectral intensity I_b,\lambda(T) in all directions. The body absorbs a fraction alpha_lambda of the incident radiation, so the energy it absorbs per unit area per unit wavelength is alpha_lambda I_b,lambda(T). At the same time, the body emits radiation given by epsilon_lambda I_b,lambda(T). For thermal equilibrium to be maintained, the absorbed energy must equal the emitted energy at every wavelength, giving epsilon_lambda = alpha_lambda.
Step 3: Evaluate each option against this identity.
Option (A) states that emissivity equals absorptivity at the same wavelength, precisely the relation derived above. Option (B), reflectivity, is a separate radiative property; for an opaque body alpha_lambda + rho_lambda = 1, so reflectivity is generally not equal to emissivity. Option (C), the Stefan-Boltzmann constant, is a fixed physical constant unrelated to any specific surface's properties. Option (D), zero, would only be true for a perfectly reflective, non-absorbing, non-emitting surface.
Step 4: Conclude.
The only option consistent with Kirchhoff's identity is that the emissivity of the body at a given wavelength equals its absorptivity at that same wavelength. \[ \boxed{\varepsilon_\lambda = \alpha_\lambda \ \text{(option A)}} \]