# How can atomic spectra be only certin frequencies?

1. Mar 16, 2005

### blah

I don't know a whole lot about quantum mechanics and there is something that dosen't make sense to me. An atom absorbes radiation in only certin frequencies. Do these frequencies have to be exact? Can they vary over a small range like say 1.000 to 1.001 or do they have to be perfectly exact? If they are exact then it would seem to me that if radiation which continuously ranged over many frequencies was exposed on atoms, the atoms would absorb none of it. But this is not the case because the sun emits radiation which continuously ranges over many frequencies (right?) and gaps are observed in the spectrum where certin atoms absorbe it.
So where am I wrong?

Last edited: Mar 16, 2005
2. Mar 16, 2005

### ZapperZ

Staff Emeritus
You are wrong because you think EM radiation can only be created via atomic transition. It doesn't. When you heat something, like tungsten, the THERMAL vibration of the atoms and molecules can generate heat and light (ordinary light bulb). This requires NO atomic transition to produce light. Secondly, when a charge particle undergoes an acceleration, it again produces EM radiation (light). Again, no atomic transition. Light generated from such sources do not have quantized energy scale. You can practically dial in whatever frequency you want within the range of operation.

Zz.

3. Mar 16, 2005

### blah

I know this

I don't understand how atoms can absorb EM radiation which ranges continuously if they can only absorb EM radiation at exact frequencies. For example, say there was some EM radiation that ranged continuously from frequency a to b. And lets say the radiation was shining on atoms that could absorb radiation ranging from frequency c to d. And c and d are between a and b. Then the amount of radiation absorbed by the atoms would be proportional to (c-d)/(a-b). If the atoms only absorbed exact frequencies then c-d would be zero and no radiation will be absorbed.

Last edited: Mar 16, 2005
4. Mar 16, 2005

### ZapperZ

Staff Emeritus
OK, so now *I* am confused to what you are asking for.

Atomic absorptionn is a well-known process. Atoms DO absorb only discrete lines if the incident light has a continuous spectrum. That's why we have the so-called absorption spectra - the "missing" discrete lines in an otherwise continuous incident light.

If you are asking about the WIDTH of these absorption lines, then that's a different matter.

Zz.

5. Mar 16, 2005

### Antiphon

The question makese sense and is related to the energy uncertainty of
a photon.

A quantum state that doesn't change in time such as the ground state of the
hydrogen atom has a definite energy. As the existence time of the state
gets shorter and shorter, the uncertainty in the energy of the state increases.

The big question is this: Can a single quantum which is created in a short
time have a definite energy, or must in have a broad energy spectrum?

This is not a simple question but I really hope someone has a good answer
because only a photon which takes an infinite amount of time to create has
an exact frequency. Yet nobody seems to talk about the energy spectrum
of single photons.

6. Mar 16, 2005

### blah

I am asking about width of these absorption lines. I would guess that the the amount of radiation an atom absorbed would be proportional to the width. I remember reading some german once measured the atomic spectra of a hydrogen atom very accurately and it was very slightly different from what the bohr model predicted. When bohr heard this he changed his model so it considerd the slight wobble of the nucleus which then explained the experimental results. This gave me the impression that the spectral lines were very exact frequencies. Is it the center of the spectral line that's mesured?

7. Mar 16, 2005

### ZapperZ

Staff Emeritus
ANY spectra lines being measure will have a finite width. This is due to a number of things, including the lifetime of that energy state (and this is where the uncertainty principle kicks in since $$\Delta (E) \Delta (t) \sim \hbar$$), the resolution of your instrumentation, and thermal smearing. You are not looking at an absorption spectra from just ONE atom, you are looking at the absorption from a gazillion atoms. So a statistical spread is expected around a central value.

Zz.

8. Mar 16, 2005

### blah

ahhhh qm is so complicated

9. Mar 16, 2005

### SpaceTiger

Staff Emeritus
A given atom absorbs approximately the same amount of energy, regardless of the width. A more interesting question is how the number of absorbing atoms depends on the width. It turns out that for weak lines, this number is a constant, but for stronger ones, it increases with width.

The centers of absorption lines are indeed very exact frequencies, but there are multiple mechanisms for broadening them, as ZapperZ already said. I think this is a complete list:

Natural Broadening: This arises from the quantum mechanical uncertainty in the energy of the state (related to the uncertainty principle, as ZZ mentioned). That means that photons will be absorbed within a small range of frequencies, even in the rest frame.

Pressure Broadening: The atoms rarely exist in isolation, and their energy states will be slightly altered by the presence of nearby charges. This effect turns out to be a function of the pressure of the gas, thus the name.

Thermal Broadening: Since the absorbing atoms will inevitably be undergoing thermal motion, they will all be in different reference frames relative to the light being absorbed. This means that they will all absorb the light at slightly different frequencies (as a result of the Doppler shift effect).

Turbulent Broadening: Gas can be undergoing motions that are non-thermal, so we have to take that into account as well. We lump this into a single category, but really the types of motion vary quite a lot.

Yup, even given the broadening mechanisms I mentioned above, there is still a very well-defined center to the line. It is this very fact that researchers in Australia have been exploiting to measure the time variation of the fine structure constant. Needless to say, that observation is extremely sensitive to small errors, so I wouldn't count on it being right.

10. Mar 22, 2005

### NEOclassic

Basic quantum concepts are not complicated!

Hi Blah,
Take a low dc voltage old time flashlight with a tungsten thread of a filament. The first thing that happens is that battery energy excites some of the electrons that "boil-off" to some distance from the tungsten thread. No appreciable amount of the so-called visible spectrum has yet occurred. The nature of the proton based quantum field is that when one of the excited electrons starts falling back toward its atom, it suddenly hits a different kind of wall - if it did hit a real material wall the impact energy would be absorbed by the wall in order that momentum/energy be conserved; however the wall that the quantum field demands in order for the electron to stop exactly at the radius of the ground state can only conserve that energy, that is given up by the stopped electron, in the form of a photon that radiates that photon energy.
Now then there are a gazillion of these excited electrons and for tungsten there are about 100 different quanta spread across the visible spectrum. It is because of the many of these that the flashlight beam appears to be "white-light"; not to worry; if that light were fed into a spectroscope there would be a picket fence of specral pales. With Sodium light, Mercury light, Neon etc there is a dominant color associated with those elements that are characterized by having many fewer spectral possibilities. Cheers, Jim

11. Mar 22, 2005

### Billy T

ZZ's answers and SpaceTiger's expansion of them are correct. ZZ even has what I am going to add a little more explicitly. The most narrow lines come from cold rarefied gases. Within this group the line width is proportional to the transition probably (this is essentially the inverse of the excited state lifetime or ZZ's delta T) When delta T is large, the uncertainty principle forces the energy spread to be small. The "energy spread" is small when the line is narrow.

My Ph.D. was in the opposite extreme (shorten title = "Stark broadening of Argon II ions") At the high density of my plasma sources, not only does the center of a line FROM AN ION shift (due mainly to the tendency for plasma electrons to statistically "shield" the upper state of the transition from the net positive attraction of the nucleus reduced by the inner core electrons), but for more complex reason, many lines from ions become asymmetric also.

From my dissertation at at LTE conditions, Ne = 8x10^16 /cc and T = 13,900 degree K, AII lines (All in Angstroms, and neglecting to report here the error bars.):
Line_____half widths______shifts
4806_______0.11________0.054
4736_______0.11________0.061
4348_______0.10________0.025
4426_______0.10________0.026
4880_______0.19________0.031
4609_______0.11________0.021

I mention the above because I will probably disappear from this forum - Chroot (Warren) is threating to throw me off becase I do not understand physics, have mentioned my book when very definately related to the thread (in about 2% of all my posts) Send Email to Local_Black_hole@Yahoo.com if you are interested to know more and I no longer exist here for PMs. According to him, I keep posting "crap / non-standard physics, which is not welcome here." He "doubts that I have a Ph.D in physics," etc. I can do little about this, but wanted you to know I have enjoyed exchanges with several of you.

Via "edit" the yahoo address has two single underscores between the three words, not everything as it appears here.

Last edited: Mar 22, 2005