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I have noticed when I turn on the light switch in my house light comes from the bulb. How is this light created?(process occurring in the bulb) and why is this small amount of electricity enough to create light?

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The answer to this question will depend on the type of light you are asking about. Incandescent: Blackbody radiation. Fluorescent: gas discharge and fluorescence. LED: electroluminescence (electron-hole recombination). –  tmac May 15 '12 at 6:50

4 Answers 4

up vote 3 down vote accepted

An electron or high-energy photon "striking" an atom imparts energy to it, which causes one or more of the electrons in orbit around the nucleus to jump from a lower shell to a higher one. As the electron returns to the lower state it releases a photon whose wavelength corresponds to the energy differential between the orbits. The energy needed to cause the electrons to jump is dependent on the starting and ending shells, and different elements in different configurations have different potential shell energy differentials.

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Is there a more rigarous answer as that implies that photons can have different energies and therefore different relative mass –  Argus May 15 '12 at 2:17
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The energy a photon has determines its frequency (and hence color if within the visible band), not its speed or mass. –  Ignacio Vazquez-Abrams May 15 '12 at 2:20
    
Perhaps @Argus would like to say "relativistic mass", which would still incite protests here. Nonetheless, there is a quantity in mass units you may associate with photons, what you may call it, I do not know. –  Alan Rominger May 15 '12 at 2:31
    
@IgnacioVazquez-Abrams: thanks for the clarification –  Argus May 15 '12 at 4:06

The filament in a light bulb is approximately a black body radiator and the amount and wavelength of the light it emits depends on it's temperature.

The filament is heated by resistive losses of the current flowing through it. The key to bulb efficiency is to get the maximum heating from the minimum amount of current, and generally this means making the filament very thin. However if you make the filament too thin it's fragile and burns out quickly, so todays light bulbs are a compomise between efficiency and bulb life.

Ignacio is quite correct in describing one mechanism for light emission, and indeed this is the main mechanism in fluorescent lights. However for conventional light bulbs the light is emitted from dipole oscillations and not from electronic transitions. This is why we get a black body spectrum instead of the sharp lines you'd expect from electronic transitions. See http://en.wikipedia.org/wiki/Thermal_radiation for more info on this mechanism.

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I would like to note that the reason fluorescent lights use less wattage is because black body radiation has the bulk of photons in infrared, not in visible light, while fluorescence is the result of electon transitions and has less waste. –  anna v May 15 '12 at 6:48

To complete Ignacio's answer one may add that the energy released in the return of the electron to the lower state.

"Red photons carry about 1.8 electron volts (eV) of energy, while each blue photon transmits about 3.1 eV."

To be compared to 10^12 electron volts which is about the energy of a flying mosquito.

It is enough to create light because each transition has such a small energy so that the collective light seen can be built up easily.

1 ev is 1 eV = 1.602×10^−19 joules ( 1 watt is 1 joule/second)

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To answer the question in the title: It's because photons have no mass.

The work required to accelerate any object to a certain speed is:

$ W = \Delta E_K = {1\over2}mv^2 $

In the case of a photon, the velocity is the speed of light ($c $), but the mass is $ 0 $. So we have:

$ W = {1\over2}\times0c^2 = 0 $

Which shows that accelerating a photon to the speed of light takes no energy.

This is a bit of an oversimplification though; when you start talking about photons and stuff, all sorts of funny quantum wibbly-wobblyness becomes involved.

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Welcome to physics SE! Your Equations may be right for an object. A photon has no mass without motion, however it gains mass according to $E=mc^2$. This concludes that it "gets harder to accelerate" the photon. Sorry, but -1. Read further onRelativity theory: $E=mc^2/\sqrt{1 - v^2/c^2}$ –  Stefan Bischof Jun 3 '13 at 20:12
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Although the formulae and some points in this answer are wrong because they should take into account relativity, the spirit of the answer is on the right lines and makes a relevant point that the other answers missed. I.e. the energy required to accelerate a particle to a high speed is proportional to its rest mass. The photon has no rest mass so it always travels at the speed of light, the fastest speed possible. This is the answer to the question in the title and that may be what the questioner was most interested in. –  Philip Gibbs Jun 3 '13 at 20:49

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