The formation of blackbody radiation

In summary, the conversation discusses the formation of blackbody emission and how it relates to various sources such as incandescent lamps and electric arcs. It is explained that a nearly equilibrium state between radiation and matter is set up in these cases. The conversation also delves into the concept of transparency and how it affects the emission and absorption of radiation. This has many applications, including in combustion modeling.
  • #1
2
0
I am puzzled about the formation of blackbody emission (Planck's law). Specifically, we know that such things like incandescent lamps, an electric arc in a gas at high pressure etc. produce a nearly blackbody spectrum of corresponding temperature. Does this mean that in these cases a nearly equilibrium state between radiation and matter is set up? Or how do you explain why the spectra are close to that of a cavity?:confused:
 
Science news on Phys.org
  • #2
Consider first a cavity with a small hole: the emission is low enough as to not perturb the BB radiation.

Consider now a very large volume of gas. Since the volume is very large, most of the radiations emitted inside the volume will be reabsorbed, only a tiny fraction has chance to escape.

Consider now a small volume of a solid.
This is exactly similar to the large volume of gas, as long as this volume of solid does not become transparent (like a gold foil).
That's why hot iron from a blast furnace also shows a nice BB spectrum.

The key point is "transparency".
See the full theory of radiation heat exchanges for more details.
These notions of emissivity, absorptivity have many applications, for example in combustion modeling.

Michel
 
Last edited:

1. What is blackbody radiation?

Blackbody radiation is the electromagnetic radiation emitted by a perfect blackbody, which is a theoretical object that absorbs all radiation that falls on it and emits radiation at all wavelengths. It is a type of thermal radiation that is emitted by all objects with a temperature above absolute zero.

2. How is blackbody radiation formed?

Blackbody radiation is formed when an object's atoms and molecules are excited and emit photons. As the temperature of an object increases, the atoms and molecules move faster and emit a wider range of photons, resulting in a continuous spectrum of radiation. This process is also known as thermal radiation.

3. What is the significance of blackbody radiation?

Blackbody radiation is significant because it was one of the first pieces of evidence for the existence of quantum mechanics. It also played a crucial role in the development of the theory of electromagnetism and the understanding of the origins of the universe.

4. What factors affect the intensity and wavelength of blackbody radiation?

The intensity and wavelength of blackbody radiation are affected by the temperature and composition of the object emitting the radiation. The higher the temperature, the greater the intensity and shorter the wavelength of radiation. Additionally, the composition of the object determines the specific wavelengths of radiation that are emitted.

5. How is blackbody radiation related to Planck's law?

Blackbody radiation is related to Planck's law, which describes the amount and distribution of radiation emitted by a blackbody at a given temperature. Planck's law states that the energy of each photon emitted is directly proportional to its frequency, and the total energy emitted is determined by the temperature of the object. This law helped to explain the observed spectrum of blackbody radiation and is a fundamental aspect of quantum mechanics.

Suggested for: The formation of blackbody radiation

Replies
21
Views
2K
Replies
19
Views
1K
Replies
2
Views
1K
Replies
31
Views
2K
Replies
45
Views
2K
Replies
18
Views
1K
Replies
49
Views
5K
Replies
58
Views
2K
Back
Top