Radiation properties of an ideal black body: Which statement correctly characterizes an ideal black body in terms of basic radiative properties?

Introduction / Context:
An ideal black body is a theoretical surface that perfectly absorbs all incident radiation and also emits the maximum possible thermal radiation at a given temperature. These properties are foundational in radiative heat transfer analysis and serve as reference limits for real materials.

Given Data / Assumptions:

• Opaque surface (for a true black body, transmissivity is zero).
• Kirchhoff's law applies: at thermal equilibrium, emissivity equals absorptivity.
• Spectral/hemispherical averages considered.

Concept / Approach:
By definition, a black body has absorptivity α = 1, meaning it absorbs all incident radiation and reflects none (reflectivity ρ = 0) with no transmission (τ = 0). At equilibrium, emissivity ε = 1 as well, making it the perfect emitter following the Stefan–Boltzmann law q = σ T^4.

Step-by-Step Solution:
Recall defining property: α = 1 for a black body.Apply Kirchhoff's law: ε = α → ε = 1 at equilibrium.Opaque condition: τ = 0; thus ρ = 1 − α − τ = 0.Therefore, the correct statement among the options is 'Absorptivity = 1'.

Verification / Alternative check:
Blackbody radiation curves (Planck's law) represent the upper bound of emission; real surfaces emit less and have ε < 1 unless approximating black behavior in a restricted band.

Why Other Options Are Wrong:
Reflectivity = 1 contradicts α = 1; a black body reflects nothing.

Emissivity = 0 is the opposite of a perfect emitter.

Transmissivity = 1 applies to a perfectly transparent medium, not a black body.

Common Pitfalls:

• Confusing 'black' in visible light with total (spectral/hemispherical) behavior.
• Assuming high absorptivity automatically means high emissivity without the equilibrium condition; Kirchhoff's law provides the link.

Final Answer:
Absorptivity = 1