Why cold gas is used for producing an absorption line spectrum
I am assuming hot/cold is defined by the population proportion in the relevant energy levels that correspond to the absorption/emission lines. Each line correspond to the difference between 2 energy levels, a higher and a lower level.
In "hot" gases, more molecules1 are at the the higher energy level. Over time they decay to the lower level and give you the emission lines. If you try to make absorption lines in "hot" gases then your absorption lines will be very weak because there isn't a lot of molecules at the lower level for you to excite into the higher level.
To get a nice absorption spectrum you want the gas to be "cold," so you get lots of molecules at the lower level. When the white light source goes through the gas, a lot of molecules absorb light at relevant frequencies and get bumped up to higher levels.
1 I say it this way to make it easier to read but it isn't exactly true. It is the electrons in the molecules that move between the energy levels, not the molecules themselves.
I suppose that if that cold gas was hot and we passed white light through it, we will get the complete spectrum
The main difference between the emission and absorption situation is that in the first case you don't send a beam of light through the gas. It emits by itself, due to the excitation produced by thermal motion. If the temperature is high enough, the average thermal energy is enough to excite some atoms on higher energy level. If it's too cold, this won't happen.
Now, if you send a beam of light through the hot gas, you still have emission of the gas itself and some of the external light is absorbed. If the gas is cold you eliminate most emission.
Thanks guys, your answers are perfect
I don't really agree with your self-correction.
In the case of light absorption, it is 'convenient' to talk in terms of the electron's energy levels but, in truth, it is the energy levels of the whole system that count. It just happens that the relative masses of an electron and a nucleus are so different that we tend to ignore the tail wagging the dog - but the nucleus is still a part of what happens. In some cases of molecular absorption, two nuclei will be of very similar (or the same) mass. But, of course, the frequencies involved will be much lower than optical.
Thanks for the correction. I wasn't aware of this :)
If solid was heated instead of a gas will the result be continuous spectrum ??
In condensed matter, the Pauli Exclusion principle makes its presence felt by turning lines (transitions between discrete energy levels) into a continuum (transitions between the continuum of states in a band). This is the logical progression from 'line splitting' in gas spectra.
It's a pretty old topic, but when revising atomic spectra some new questions came up in my mind, so I liked to just do it on this topic instead of a new one, however,
Why doesn't discrete spectra (emission - absorption ) appear in gases of relatively high pressure , liquids, solids, it seams that low pressure gases are the only suitable thing, why is that?
In short, "pressure broadening" ... the atoms are set to moving faster, but as always, in random directions. This results in Doppler shifts in the spectrum.
What you describe would actually be 'temperature broadening' and is a sort of semi-classical explanation. (Whatever the pressure, it is the temperature that governs the Kinetic Energy of the Atoms and, hence, any Doppler effect)
I learned that it is a quantum effect and due to the Pauli Exclusion Principle. The energy states in an isolated atom have 'exact' values as the state of the atom is described with a minimum of quantum numbers. Two or more atoms that occupy the same limited volume of space need more eigenvalues to describe their state and, as no two can have exactly the same state, this results in clusters of states around the basic states of a single atom. The closer together the atoms are, the more their mutual effect and so the greater the spread around the basic levels - this will give broader and broader lines. In the solid state, the lines spread into bands due to the extreme amounts of interaction between all the atoms.
You mean that atoms in a solid have speeds larger than in a gas in normal conditions?
I think that most of them don't. Typical phonon frequencies are of the order 10^12-10^13 Hz. Amplitudes are a fraction of angstrom, so maximum vibration speeds are 100-1000 m/s. About the same as in air at room temperature.
This is for a gas, nasu. And whether you call it pressure broadening, or temperature broadening - it has the same root cause.
Hardly the same "root cause". Temperature and density are two different quantities and they are independent of one another. A gas at very low temperature will still show line broadening if it is at a high enough pressure. "Temperature broadening" is essentially due to KE whilst Pressure Broadening is a Potential Energy Effect. The Pressure / density broadening carries on all the way through the states of matter.
He asked about high liquids and solids, besides the high pressure gases.
You did not mention that your answer is aimed at high pressure gases only.
The energy bands that appear due to the collective behavior of atoms closely packed in condensed matter is not due to Doppler. Possible that we look at the question from different points of view.
"Pressure broadening" seems to be a series of phenomena not quite related to Doppler shifts (from what they say on wikipedia). But I suppose you may be more familiar with this field.
Very different from.
Doppler depends upon temperature (AKA KE). A single atom in a box would exhibit doppler shift due to its movement. 'Pressure broadening' involves proximity of other atoms.
A low pressure gas is usually a low density gas. A gas capable of radiating in the visible spectrum is quite hot, so if it is low pressure, it is necessarily a low density gas. The photons emitted by a low-density gas are very unlikely to interact with the gas again. Not so in a higher density gas. In a higher density gas (more so, in liquids and solids), emitted light interacts with matter, scattering off its molecules, atoms and electrons, which changes its frequency. An extreme example is our Sun: the thermonuclear reactions in its core produce gamma-range photons, but the surface of the Sun radiates mostly in the visible part of the spectrum.
Depending upon where you place the dividing lines, solar radiation at the outside of the Earth's atmosphere is roughly 46% infrared, 9% ultraviolet, and only 45% visible.
Thanks for the correction. I really meant to say that the maximum intensity is in the visible part.
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