Thermal radiation fundamentals What does Kirchhoff's law of thermal radiation state (at a given temperature and wavelength)?

Introduction / Context:
Kirchhoff’s law links emission and absorption for bodies in thermal equilibrium and is foundational for understanding gray-body behavior and black-body references used in heat-transfer calculations.

Given Data / Assumptions:

• Thermal equilibrium at a given temperature and wavelength (or narrow band).
• Definitions: emissive power E_λ and absorptivity a_λ at the same wavelength.
• Black body has a_λ = 1 and serves as the reference emitter.

Concept / Approach:
Kirchhoff’s law states E_λ / a_λ is identical for all bodies at the same T and λ and equals E_λ,black. This law underpins the concept of emissivity and guides how real surfaces are compared to an ideal black body.

Step-by-Step Solution:
Restate the law: E_λ / a_λ = E_λ,black.Interpretation: A good absorber at a given wavelength is a good emitter at that wavelength.Thus, option C is the correct description.

Verification / Alternative check:
Check limiting case: for a black body, a_λ = 1 ⇒ E_λ = E_λ,black, consistent with definition.

Why Other Options Are Wrong:
(A) is Stefan–Boltzmann law, not Kirchhoff’s. (B) expresses Wien’s displacement law. (E) is incorrect: a gray body emits a fraction ε of black-body emission.

Common Pitfalls:
Confusing the different classical radiation laws; remember each addresses a distinct property (E~T^4, λ_max T = constant, or E/α equivalence).

Final Answer:
For all bodies, the ratio of emissive power to absorptivity is the same and equals the emissive power of a perfect black body.