# Stokes Shift in Hydrogen Spectrum ?

1. Aug 13, 2014

### neilparker62

Would one expect the emission and absorption spectral lines for Hydrogen to be at slightly different frequencies ? So if electrons are excited to higher energy levels on absorption of photons and then fall back down again emitting photons, would there be any difference in frequency. Assuming transition is the same (eg 1s to 2s transition).

2. Aug 17, 2014

### Staff: Admin

I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?

3. Aug 18, 2014

### neilparker62

With a slightly modified Bohr formula, I am obtaining a transition energy (1s to 2s) which translates to a frequency slightly in excess of the accurately measured value. When I do a theoretical correction for electron recoil energy I improve the accuracy (as compared to measure value) from about 18 ppm to 6 ppm. Furthermore my frequency is slightly below the measured value which is what I would expect since statistically only a proportion of emissions generate recoil - ie not those emissions which happen along a radial line in which case the momentum is transferred to the electron and nucleus mass and recoil energy will be negligible.

I understand the conventional wisdom is that it is not possible for the electron to experience recoil independently of the nucleus. But if there was a hypothetical recoil energy, it would show up in a small frequency difference between the absorption line and the emission line. So I'm wondering if anyone has carefully measured Hydrogen 1s 2s emission and absorption frequencies separately or has the assumption always been that they will be the same?

4. Aug 20, 2014

### neilparker62

I should add the expected frequency difference is about 16.4GHz.

5. Aug 27, 2014

### dextercioby

I don't understand what model you're using. Try to put some formulas to your words (scan some written notes or type some LaTex code here). I never heard of the Stokes shift.

6. Aug 28, 2014

### neilparker62

Re Stokes shift: http://en.wikipedia.org/wiki/Stokes_shift

There is not supposed to be any Stokes shift for the Hydrogen 1s 2s transition. If there were it would be new theory which - by PF rules - I can't advance here. But I'm asking if anyone has actually measured to confirm whether indeed the Hydrogen 1s 2s emission and absorption frequencies/wavelengths are identical.

7. Aug 29, 2014

### M Quack

For the recoil mass, you have to use the mass of the Hydrogen atom including the nucleus - not just the electron. With that the recoil energy becomes very very small, as a proton is about 2000 times heavier than a naked electron.

A similar effect occurs in the Moessbauer effect. When a Moessbauer nucleus in a crystal lattice emits a gamma ray, the recoil energy is taken up by the entire crystal lattice, not just by the emitting nucleus. This allows relative bandwidths of Moessbauer lines in the delta E/E = 10^-15 range.

Stokes shifts are observed when excitations exist that have zero or negligible net momentum transfer, e.g. vibrational and rotational modes of molecules or crystals. Apparently the rotational modes of hydrogen molecules H2 are quite easy to observe.

If you had a hydrogen crystal then it might be possible to observe Stokes and Anti-stokes shifts corresponding to optical phonons at momentum transfer=zero.

8. Aug 31, 2014

### neilparker62

Thanks for your response above - I do appreciate that no Stokes shift is expected for the reasons that you have given. All I'm asking is has anyone actually carefully measured the Hydrogen 1s 2s absorption and emission frequencies separately (just in case!) or do they just assume that the two will be the same ? I would say if a difference was found (even at a few parts per million) it would be at odds with theory as you have outlined above.

9. Mar 24, 2015

### neilparker62

Ok - here is the formula I am using. α is the fine structure constant and for better accuracy we need to use the electron's reduced mass.

$$hf_2-hf_1 = m_e c^2 \left[\sqrt{1 - \frac{\alpha^2}{4}}-\sqrt{1-\alpha^2}\right]$$

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