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I have learnt that depending on the various gases those are involved in the reaction that produces fire, different colors (yellow, red or blue) of flames become visible.

I have a question .. what are the possible colors that fire can possess? and do they (how, if yes) relate to the intensity? like I know blue is more intense than red, etc..

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There are two sources of colour in flames. The first is from glowing particles of soot heated by the flame. This type of radiation is called black body radiation and just depends on the temperature, so unless the fire is unusually hot or cold you get the characteristic yellow/orange colour.

The second source of colour is from light emitted by excited atoms in the flame. This was briefly mentioned in your previous question but for those who haven't seen it, the energy of gas molecules in a flame is distributed according to the Maxwell-Boltzmann distribution and a small fraction of the gas molecules have energies high enough to excite electronic transitions. For example if you have copper atoms present in a flame they can be excited due to collisions with unusually high energy gas molecules and the excited atoms emit blue/green light as they decay to the ground state. That's why adding copper to a flame (or a firework!) produces a blue/green colour.

So the interesting colours in flames come from other atoms, typically transition metal atoms, in the flame. There is a long list of all the colours in the Wikipedia article on fireworks.

The intensity is down to the temperature of the flame and the concentration of metal ions. Remember that it's the very few molecules with high energies in the Maxwell-Boltzmann distribution that cause electronic transitions. This means if you increase the flame temperature you increase the number of molecules with high energies very rapidly. The concentration of metal ions has an obvious effect as the more metal ions there are the more light will be emitted. There isn't a direct connection to the colour.

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amazing info and link is useful as well.. – InfantPro'Aravind' Nov 20 '12 at 10:36

There are two different effects that dominate the coloration of a flame: blackbody radiation and electronic transitions.

In the first case, the temperature determines the wavelength that will dominate. A familiar example is the notion of "white hot" when heating the same object to two different temperatures. The higher the temperature, the shorter the dominant wavelength becomes, and so the "bluer" the flame becomes. In this process, the key is that the SAME MATERIAL is heated to DIFFERENT TEMPERATURES to change the color.

With electronic transition, the particular element or molecule in question becomes more relevant. When electrons (or more generally, whole molecular/etc systems) are excited to a state of higher energy, they may only shed this energy in fixed amounts, corresponding to fixed amounts of energy. The shorter the wavelength of the emitted light (the bluer it is) the more energy it carries. This explains the existence of spectral lines/bands, and underlies the use of the "burn test" in analytical chemistry. For example, when copper is placed in a flame, it will become colored green. I assume the flame itself becomes green because a very small amount of the copper is actually vaporized and excited, but I've never actually considered why the bulk of the flame becomes colored before.

In the case of electronic transition, the intensity of the flame does not play as strong of a role in general.

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