Thermal Doppler broadening in the Mossbauer effect?

[bcrowell]

There's something I don't understand about the Mossbauer effect. I understand that the Mossbauer effect has a mechanism for eliminating the energy shift due to the recoil of the nucleus, since the inertia can be the inertia of the lattice over-all, rather than the inertia of the individual nucleus. What I don't understand is what happens to the Doppler broadening that I would expect due to the thermal motion of the nucleus. In the Pound-Rebka experiment, for instance, the natural line width gives $\Delta E/E \sim 10^{-12}$, and the general-relativistic effect they were measuring was a shift of about 10^-14. On the other hand, the typical thermal velocity of an iron nucleus at room temperature is going to be about 100 m/s, which will give a random Doppler shift of about 10^-6. Can anyone explain what I'm missing? It seems as though the random thermal Doppler shifts would be many orders of magnitude bigger than all the other effects involved.

TIA!

-Ben

P.S. I'm new here. When posting math, is the idea to insert it between dollar signs in LaTeX format?

 

[Rajini]

Hi,
thermal motion of Mössbauer nuclei (or any atomic vibration) is fast i.e., it will be averaged to zero during 141 ns. So in general they are excluded
Thermal motions or atomic vibrations are in the order of 10^-12 per second. Compare this with Mössbauer mean life time for 57Fe it is 141.8 ns.
But it can be valid below mK temperature!
What you meant by iron velocity??
natural line width is 0.097 mm/s or in energy unit:4.64 neV (for iron).

 

[bcrowell]

Hi,
thermal motion of Mössbauer nuclei (or any atomic vibration) is fast i.e., it will be averaged to zero during 141 ns. So in general they are excluded
Thermal motions or atomic vibrations are in the order of 10^-12 per second. Compare this with Mössbauer mean life time for 57Fe it is 141.8 ns.
But it can be valid below mK temperature!
What you meant by iron velocity??
natural line width is 0.097 mm/s or in energy unit:4.64 neV (for iron).

Hi, Rajini,

Thanks very much for your reply! It hadn't occurred to me at all that the time scale for vibration was so short.

So let me see if I'm understanding you correctly.

In a gas, the molecules probably go for milliseconds between collisions. This is much longer than the lifetime of the nuclear state we're talking about, so you'd get the full Doppler broadening effect I estimated originally.

In a liquid, a molecule would undergo about 10^5 collisions over the lifetime of the nuclear state. These collisions are all uncorrelated, so the effect would be down by about 1/\sqrt 10^5 compared to the gas. This would still produce a Doppler broadening a lot bigger than the natural line width.

In a solid, we have ~10^5 collisions, but they're correlated, because the molecule has to oscillate around an equilibrium position. Therefore I'm guessing that the Doppler broadening effect is down by about 10^-5 compared to what I originally estimated. This produces a broadening on the order of 10^-11, which is on the same order of magnitude as the natural line width of 10^-12. Since my estimate is crude, it seems plausible that the Doppler broadening is on the same order of magnitude as the natural line width, or maybe even smaller.

Does this sound right?

Thanks!

-Ben

 

[Rajini]

In gases and liquids one cannot observe Mössbauer effect! Recoil is more (nothing related to collisions just recoil is more).
Therefore gases or liquids should be cooled..only in frozen state one can observe the effect. In any case in Mössbauer experiment, one cannot obtain a spectrum with resolution of the (Heisenberg) natural line width (0.097X2 mm/s or 4.6X2 neV) because of experimental difficulties..it is always broadened..for eg., for 57Fe the outerlines may have 0.24mm/s (but not equal to 0.09X2 mm/s).
PS: if you want to know why Mössbauer effect is not possible in liquid or gases: emission and absorption lines are separated by 6times the natural width..so no resonance...(Mössbauer effect is resonance of these 2 lines)..
So only in solid..the recoil is compensated.
In case in a Mössbauer spectrum, if you notice huge broadening of spectrum..it may be several reasons..slow spin relaxation, diffusion broadeing, etc..

 

[bcrowell]

Hi,

In gases and liquids one cannot observe Mössbauer effect! Recoil is more (nothing related to collisions just recoil is more).
Therefore gases or liquids should be cooled..only in frozen state one can observe the effect. In any case in Mössbauer experiment, one cannot obtain a spectrum with resolution of the (Heisenberg) natural line width (0.097X2 mm/s or 4.6X2 neV) because of experimental difficulties..it is always broadened..for eg., for 57Fe the outerlines may have 0.24mm/s (but not equal to 0.09X2 mm/s).
PS: if you want to know why Mössbauer effect is not possible in liquid or gases: emission and absorption lines are separated by 6times the natural width..so no resonance...(Mössbauer effect is resonance of these 2 lines)..
So only in solid..the recoil is compensated.
In case in a Mössbauer spectrum, if you notice huge broadening of spectrum..it may be several reasons..slow spin relaxation, diffusion broadeing, etc..

I understand that there are lots of reasons why you can't have the Mossbauer effect in liquids or gases. What I was trying to understand was why it's possible to have such narrow lines in solids. I understand your explanation that the Doppler effect tends to average out because the time scale for thermal oscillations is very short compared to the half-life of the nuclear transition. However, that averaging is not going to be perfect, and that was what I was trying to explore quantitatively with the comparison of solids, liquids, and gases. To get lines with widths on the order of 10^-12 of the energy in 57Fe, one needs the averaging to zero of the Doppler shifts to be good to about 1 part in 10^6. My order of magnitude estimate of the number of thermal oscillations over one nuclear half-life was 10^5, so it's not immediately obvious to me that one can get averaging to zero that's good to one part in 10^6. In my previous post I was trying to explore the statistical issues to see if I could understand why the averaging to zero would really be that good. I was using the gas, liquid, and solid as three examples of how the statistics could work out: no averaging, averaging over an uncorrelated random walk, and averaging over the highly time-correlated motion of a nucleus oscillating around the minimum in a potential.

 

[Rajini]

Hi,
why no sharp lines: it is because of Heisenberg principle, 0.09mm/s X 141.8 ns=hbar.So you cannot make sharper than this. The same principle for absorber (Mössbauer spectrum = convolution of single emission line (57Co/Rh) from source with absorption spectrum)..so twice of 0.09mm/s will be the minimum line width. so if a small broadening is due to thermal motions as you said..but in gases or liquids it is recoil effect. you should consider it.. for e.g., liquids and solids it is 1.958 meV (compare with 14.4 keV). i.e., emitter nucleus energy is shifted by 1.958 meV away from 14.4 keV and absorber (gas and liquid)also shifted..all shifting in the direction of ray...so the difference between the energy line is 6times1.958 meV. But Mössbauer effect is overlap of these 2 lines..
If you don't mind can you write just one calculation here...say for gases..?
Also when you write ..please use units...
thermal motions makes a very tiny shift of the spectrum due to second-order Doppler shifts-SOD (relativistic effects..) so isomer shifts are rarely separable from SOD. But broadening due SOD cannot be excluded.

 

[bcrowell]

If you don't mind can you write just one calculation here...say for gases..?
Also when you write ..please use units...

Hi -- Thanks for your reply. The unitless numbers I was giving were all unitless ratios. E.g., 10^-12 is the natural line width divided by the energy of the line. The units cancel. The calculation for gases is basically given in my original post: the typical thermal velocity is 100 m/s, which is ~10^-6 of the speed of light, so the Doppler broadening effect is ~10^-6. Anyway, thanks very much for your replies. I think I understand the answer to my original question reasonably well now. The averaging out of the thermal motion just hadn't occurred to me.-- Ben

 

[Rajini]

Hi okay..
for Iron =3.2X10^-13, it is actually a measure of the accuracy. Doppler effect is important for Mössbauer spectroscopy..one varies the energy around 14.4 keV by \pm 500 neV by using Doppler effect.
vE/c=shift (v-velocity E-14.4 keV)
have fun

 

[phillip75]

Hi There!
This is Phillip!
I am new in this site!
Just keep on posting Guys!
And help the other to solve their prob.
Thanks!
Good Luck!...


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