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2006-07-24 02:34:47 · 3 answers · asked by Jitendra P 1 in Science & Mathematics Chemistry

3 answers

It has to do with the energy levels of the valence electrons of the elements, and the sensitivity of the human eye to light.

The process involves adding heat to the atom, which causes an electron to be "pumped" up into the next higher energy level, and then it "relaxes", releasing that energy. The released energy emitted is what you are looking for, and must be of just the right quantum to be able to excite the eye in order to be visible.

In some cases, the energy levels are too close together, so the energy emitted is in the infrared region -- which would be masked by the flame you are using, even if it was visible to the eye.

In other cases, the levels are so far apart that, when the energy is emitted as the electron drops back down is in the ultraviolet -- again, not visible to the human eye.

Only certain elements have just the right separation of energy levels to give a characteristic, visible emission of light, such as the brilliant yellow-orange of sodium, or the red of strontium.

2006-07-24 02:45:12 · answer #1 · answered by Dave_Stark 7 · 0 1

You cannot flame test an element or compound whose combustion temperature is higher than the temperature of the flame.

2006-07-24 09:39:46 · answer #2 · answered by tbolling2 4 · 0 0

well sodium is everywhere and is easier to see than the tiny transitions other elements go through

2006-07-24 14:01:16 · answer #3 · answered by shiara_blade 6 · 0 0

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