Question

In thermal radiation for a black body (where epsilon is emissivity and alpha is absorptivity)

A. alpha=1 epsilon doesn't equal 1
B. alpha doesn't equal 1 epsilon=1
C. they both don't equal one
D. they both equal 1

Answer 1

For a black body, both emissivity and absorptivity are equal to 1, which means it is the perfect emitter and absorber of radiation. This defines a black body's ideal thermal radiation properties.

Step-by-step explanation:

In the context of thermal radiation for a black body, where ε (epsilon) is emissivity and α (alpha) is absorptivity, the correct statement is D. they both equal 1. For a black body, emissivity (ε) is 1, meaning it is an ideal emitter of radiation. Absorptivity (α) is also 1 for a black body, indicating it is an ideal absorber, capturing all the radiation that falls on it.

Therefore, a black body does not reflect or scatter any radiation but absorbs all the electromagnetic energy that it encounters. This is due to its ideal absorptive and emissive properties, which dictate how well it emits and absorbs radiation across different wavelengths. Objects like tungsten filaments have lower emissivities, such as 0.5, and are not considered black bodies.