A problem about blackbody radiation

This means it will emit less radiation at the same temperature as a black-body. The condition for a black-body is that it must absorb all incident electromagnetic waves and radiate at the same rate. This is why it is "black" because it absorbs all colors of light. A grey body, on the other hand, absorbs only a portion of the incident light and reflects the rest. This results in a lower emission of radiation compared to a black-body at the same temperature. This information can be found in the Stefan-Boltzmann law, which states that the emission rate (J) is proportional to the temperature (T) to the power of 4, multiplied by a constant (k). For a non-black body, the equation includes an
  • #1
VHAHAHA
58
0
The Stefan–Boltzmann law :
J = kT^4 where k is Stefan–Boltzmann constant.

but for non black body ( or grey body)
the equ becomes
J = εkT^4
where 0<ε<1

can anyone explain y a grey body emit less radiation when compare to the black body at same T? It can found the ans in wiki or the reference book "modern physics"

thank you
 
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  • #2
What is the condition it has to meet to be considered a blackbody? (Why is a blackbody "black"?)

Compare with grey.
 
  • #3
Black body absorb all incident em wave while grey body absorbs a part of it and reflects the remains
 
  • #4
Black-body also radiates at the same rate as it absorbs.
A grey body is just the same as a black-body but it absorbs less energy - therefore it has less energy to radiate.
 

FAQ: A problem about blackbody radiation

1. What is blackbody radiation?

Blackbody radiation refers to the electromagnetic radiation emitted by an object due to its temperature. It is a form of thermal radiation that is emitted by all objects, but it is most commonly associated with objects that are heated to high temperatures, such as stars or hot metals. The term "blackbody" refers to an idealized object that absorbs all radiation incident on it and emits the maximum amount of radiation for a given temperature.

2. What causes blackbody radiation?

Blackbody radiation is caused by the thermal energy of an object's atoms and molecules. As these particles vibrate and move due to their thermal energy, they emit electromagnetic radiation. The frequency and intensity of this radiation depend on the object's temperature, with hotter objects emitting higher frequency and more intense radiation.

3. Why is blackbody radiation important in science?

Blackbody radiation is important in science because it provides a way to study and understand the thermal properties of objects. By analyzing the frequency and intensity of the radiation emitted by an object, scientists can determine its temperature and other thermal properties. This is especially useful in astrophysics, where blackbody radiation is used to study the temperatures and compositions of stars and other celestial bodies.

4. How is blackbody radiation related to the laws of thermodynamics?

The laws of thermodynamics, specifically the second law, state that heat will naturally flow from a hotter object to a cooler object until they reach thermal equilibrium. Blackbody radiation is a manifestation of this law, as objects emit radiation in order to reach thermal equilibrium with their surroundings. This process is also known as thermal radiation or heat transfer through radiation.

5. How does the concept of blackbody radiation relate to quantum mechanics?

The study of blackbody radiation played a crucial role in the development of quantum mechanics. In the early 20th century, scientists observed that the classical theory of blackbody radiation, which predicted an infinite amount of radiation at high frequencies, did not match experimental data. This led to the development of quantum mechanics, which provided a more accurate explanation of blackbody radiation and paved the way for many other advancements in physics.

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