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I don't know quantum mechanical model. So, I'm referring to just bohr's model of atom.

Any atom emits energy when it makes transition from higher excited state to lower excited state. Now, some times they say that this energy is light energy and some times heat energy.

I'm confused. What decides the emitted energy will be light or heat?

Do we have control over what kind of energy it emits? I mean can I make an atom emit light and not heat? or heat and not light? What are the factors influencing this?

EDIT: To clarify the issue of what exactly do I mean by heat. Well, I myself am not quite sure. I just thought heat as just heat. This question araised from my previous question where I discovered that in a filament bulb 98% of energy is converted into heat energy and rest into light energy. But where as in CFL about 40% to light and rest into heat. I know these conversion is because of electron transistions. But whats controlling light & heat components?

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Nic has a good point: what exactly do you mean when you say "heat"? –  David Z Oct 18 '11 at 17:54
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Atoms can't emit heat, because heat is a statistical idea. What the can do is emit light that is absorbed by a solid, emit light that is statistically random, because you don't know the original state, and lose energy by kicking electrons, which mechanically lose energy in a crystal.

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sometimes laypeople think that the infra-red spectrum corresponds to « heat » so you usually need to check what does the OP really mean... if they really meant infrared, they need a different answer than the one you gave, correct in its own way as it is. –  joseph f. johnson Jan 17 '12 at 19:11
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By heat I think you mean electromagnetic radiation in the infrared. Of course this is just another form of light. The wavelength/frequency of the emitted light is determined by how energetic the change in orbit of the electron was.

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"Heat" generally refers to kinetic or vibrational energy of atoms. –  Colin K Oct 18 '11 at 15:42
Thanks Colin, I am well aware of different definitions of heat, I was just trying to interpret the question –  Nic Oct 18 '11 at 17:50
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In the specific example of the light bulb, the current passing through the thin wire heats it to the point of incandescence. The molecules composing the wire, change energy states as they interact with the electrons of the current and emit a spectrum of electromagnetic radiation starting with infrared up to visible light in frequencies.

from the link:Unfortunately, the spectrum emitted by a blackbody radiator does not match the sensitivity characteristics of the human eye. Tungsten filaments radiate mostly infrared radiation at temperatures where they remain solid (below 3683 kelvins / 3410 °C / 6,170 °F). Donald L. Klipstein explains it this way: "An ideal thermal radiator produces visible light most efficiently at temperatures around 6300 °C (6600 K or 11,500 °F). Even at this high temperature, a lot of the radiation is either infrared or ultraviolet, and the theoretical luminous efficiency is 95 lumens per watt."[39] No known material can be used as a filament at this ideal temperature, which is hotter than the sun's surface. An upper limit for incandescent lamp luminous efficacy is around 52 lumens per watt, the theoretical value emitted by tungsten at its melting point.[34]

Now fluorescent lamps do not depend on incandescence to emit light: CFLs emit light from a mix of phosphors inside the bulb, each emitting one band of color. Modern phosphor designs balances the emitted light color, energy efficiency, and cost. Every extra phosphor added to the coating mix decreases efficiency and increases cost. Good quality consumer CFLs use three or four phosphors to achieve a "white" light with a color rendering index (CRI) of about 80, where the maximum 100 represents the appearance of colors under daylight or a black-body (depending on the correlated color temperature).

The mixture of phosphors in the fluorescent lamps exhibits the "control" you are asking for, the spectrum is weighted towards visible light and not infrared as the article in the link analyzes.

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