Question:easy

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?

Show Hint

Kirchhoff's law of radiation equates spectral emissivity with spectral absorptivity for a body in thermal equilibrium.
Updated On: Aug 10, 2026
  • absorptivity of the body at the same wavelength
  • reflectivity of the body at the same wavelength
  • Stefan-Boltzmann constant
  • zero
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The Correct Option is A

Solution and Explanation

Place any small object inside a cavity whose walls are perfectly black and held at a uniform temperature T. Because the walls are black, they flood the cavity with isotropic blackbody radiation of spectral intensity I_b,lambda(T), independent of what is placed inside. Let the small object have spectral absorptivity alpha_lambda and spectral emissivity epsilon_lambda. The power it absorbs per unit wavelength interval is alpha_lambda I_b,lambda(T), and the power it emits per unit wavelength interval is epsilon_lambda I_b,lambda(T). Once the object reaches thermal equilibrium with the enclosure, its temperature stops changing, which means it cannot be a net emitter or a net absorber at any wavelength band. So, wavelength by wavelength: epsilon_lambda = alpha_lambda. Check the four options against this result: reflectivity is unrelated in general; the Stefan-Boltzmann constant is a universal constant unrelated to any specific surface; zero would mean the body neither absorbs nor emits at all, contradicting it reaching thermal equilibrium by radiative exchange. \[ \boxed{\text{Answer: absorptivity of the body at the same wavelength}} \]
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