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2007-02-03 04:32:26 · 2 answers · asked by manikant k 1 in Science & Mathematics Physics

2 answers

black body is a very good absorber and since it absorbs so well it radiates so well also . it just a word used to tell us that the reflection factor is equals to 1.

2007-02-03 04:43:04 · answer #1 · answered by kater al nada 2 · 0 0

An ideal blackbody is one which absorbs 100% light falling on it.
Plank said (in 1909, I believe) that a black body absorbs or releases light in DISCRETE small packets. Each such packet is called a QUANTUM (plural: quanta). This phenomenon of radiation (emission) or absorption of light in the form of discrete packets was termed as Plank's Hypothesis of Blackbody Radiation. I don't think you need to know the mathematical calculations involved in the theory, because I don't think you've studied the mathematics required to do so. But, I'll give you the basic equation involved, which is:
The energy(E) released or absorbed by an ideal body is directly proportional to the frequency( v, the Greek letter nu) of the light emitted or absorbed.
That is, E = hv, where h is a constant of proportionality, known as Plank's constant. It has a value of 6.676 * 10^-34 Js.
This equation can also be written as: E=hc/l [since v=c/l , where c is velocity of light, & l (the Greek letter lambda is used, mostly)is wavelength of light emitted or absorbed].

Hope this helps you.

2007-02-03 17:16:10 · answer #2 · answered by Kristada 2 · 0 0

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