Kirchhoff’s law of thermal radiation relates emissivity and absorptivity at thermal equilibrium.\r To which type(s) of radiation does this law apply?

Introduction:
Kirchhoff’s law links how a surface emits and absorbs radiation. At a given temperature and wavelength, a body’s emissivity equals its absorptivity provided the body is in thermodynamic equilibrium with its surroundings. This principle underpins the use of gray and selective surface models in engineering calculations.

Given Data / Assumptions:

• Thermal equilibrium between surface and radiation field.
• Definitions of emissivity and absorptivity for both monochromatic (spectral) and total quantities.
• Diffuse behavior assumed unless directionality is specified.

Concept / Approach:
Kirchhoff’s law is most precisely stated spectrally: at each wavelength and direction, emissivity equals absorptivity. Integrating over all wavelengths yields the total (hemispherical) version. Therefore, it holds for monochromatic and for total radiation, provided the equilibrium condition is met and the same temperature is considered.

Step-by-Step Solution:
State spectral form: ε_λ = α_λ at the same λ and T.Integrate over λ to obtain total form: ε = α under equilibrium.Hence, the correct choice encompasses both monochromatic and total radiation.

Verification / Alternative check:
Blackbody reference cavities demonstrate the equality, while selective coatings show wavelength-dependent behavior consistent with the spectral statement.

Why Other Options Are Wrong:

• Only total or only monochromatic: Each is incomplete; the law is valid in both domains when stated appropriately.
• Neither: Contradicts fundamental radiative transfer theory.

Common Pitfalls:
Forgetting the equilibrium requirement; surfaces exchanging with non-equilibrium fields may not satisfy the equality without correction.

Final Answer:
both (a) & (b)