Hyperfine structure of deuterium

In summary, the problem presented concerns the hyperfine splitting of the Deuterium Balmer alpha line. The task is to find the number of transition lines and their frequency shifts with respect to the transition without hyperfine interaction. The relevant equations are the fine structure equation and the reduced mass. The number of transition lines is determined to be six, and they are all allowed. However, there are difficulties in calculating the hyperfine correction, as the first term vanishes for l=0 and the wave function for n=2 in deuterium is unknown. The solution is derived from the general expression for hydrogen-like atoms given by Bethe and Salpeter, and it is equally valid for l=0 and l≠0.
  • #1
JulienB
408
12

Homework Statement



Hi everybody! Here is the problem:

Hyperfine splitting of Deuterium Balmer alpha line:
Consider the ##3p_{3/2}## to ##2s_{1/2}## transition in deuterium. Find the number of transition lines and their frequency shifts with respect to the transition without hyperfine interaction.

We were also given in a previous exercise ##g_d=0.8574## and the nuclear spin ##I=1##.

Homework Equations



Which equation is relevant is kind of the knot of my problem.

The Attempt at a Solution



So first I've determined the transition lines. I think that ##F=\lbrace I-j,I-j+1,...,I+j\rbrace## (if ##I>j##) from the other quantum numbers, but we were not given a definitive formula regarding ##F##. However this seems to have given coherent results with the hydrogen splitting and with what I found on the internet:

##3p_{3/2} \implies j=3/2 \implies F={1/2,3/2,5/2}##

##2s_{1/2} \implies j=1/2 \implies F={1/2,3/2}##

Therefore I have six transition lines. I think they are all allowed (correct me if I'm wrong), because ##I## is only nucleus-related, so we can rely on the allowed transitions of ##j## and ##l##. Here for every transition ##\Delta j=1## and ##\Delta l=1## so I imagine there is no problem with that.

Then I've calculated the transition energy without hyperfine structure using the fine structure equation and the reduced mass:

##E_nj=-\frac{m_r c^2 \alpha^2}{2} \frac{1}{n^2} \left[1+\frac{\alpha^2}{n^2} \left( \frac{n}{j+1/2} - \frac{3}{4} \right) \right]##
##\implies \Delta E=E_{3\frac{3}{2}} - E_{2\frac{1}{2}}=1.8893##eV

This gives me a wavelength ##\lambda=656.3##nm just like for hydrogen, so that seems correct. But now the only thing I know about hyperfine correction is the equation from Griffiths:

##E_{hf}^1=\frac{\mu_0 g e^2}{8 \pi m_p m_e} \langle \frac{3(\vec{S}_p \cdot \hat{r})(\vec{S}_e \cdot \hat{r})-\vec{S}_p\cdot\vec{S}_e}{r^3} \rangle + \frac{\mu_0 g e^2}{3 m_p m_e} \langle \vec{S}_p \cdot \vec{S}_e \rangle |\psi(0)|^2##

The first term luckily vanishes for ##l=0##, but unfortunately the ##3p_{3/2}## has ##l=1##. I am not sure how to calculate the expectation value of the first term in that case (it looks really hard).

My second problem is that even if ##l=0## for ##2s_{1/2}##, then I don't have the wave function of deuterium for ##n=2##, and google gave me only vague articles about approximations that looked suspicious in the context of my homework. Can I take the wave function ##\psi_{200}(0)## of hydrogen here? I know I can for ##n=1##.

All in all I think there must be another way that is not so hard. Maybe I just misunderstood the question? Griffiths says nothing about hyperfine structure for ##l\neq 0## and the Wikipedia article about hyperfine structure is very long when it comes to the 1st term, and what is done there seems foreign to me.

Any suggestion on how to tackle this problem?Thanks a lot in advance.Julien.
 
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  • #2
Suggestion would be to google the hyperfine interaction in hydrogen. The only difference would be the different reduced mass for deuterium as well as the different mass of the nucleus. You might find a source that does more of the calculations for you.
 
  • #3
@Charles Link Hi and thanks for your answer. So you say that the wave functions are the same?
Julien.
 
  • #4
If it interests somebody, at the end I took the general expression for hydrogen-like atoms of Bethe & Salpeter, derived in their book "Quantum Mechanics of One- and Two-Electron Atoms". The excerpt where the equation (eq. (22.12)) is can be found here: http://hacol13.physik.uni-freiburg....0-Positronium/Anhang/H.A.Bethe,E.Salpeter.pdf. I've tested the equation on hydrogen and deuterium, and it is equally valid for ##l=0## and ##l\neq 0##.
 

1. What is the hyperfine structure of deuterium?

The hyperfine structure of deuterium refers to the splitting of the energy levels in the deuterium atom due to the interaction between the nuclear magnetic moment of the deuterium nucleus and the magnetic field of the electron. This results in a more complex spectrum compared to the hydrogen atom.

2. How is the hyperfine structure of deuterium measured?

The hyperfine structure of deuterium is typically measured using spectroscopy techniques, such as laser spectroscopy or microwave spectroscopy. These methods involve shining a specific wavelength of light or microwave radiation onto the deuterium atom and observing the resulting energy levels and transitions.

3. What is the significance of studying the hyperfine structure of deuterium?

Studying the hyperfine structure of deuterium allows scientists to better understand the fundamental properties of the atom, such as its size and shape. It also has practical applications in fields such as astrophysics and nuclear physics.

4. How does the hyperfine structure of deuterium differ from that of hydrogen?

The hyperfine structure of deuterium is more complex than that of hydrogen due to the presence of a neutron in the nucleus. This results in additional energy levels and transitions in the deuterium atom, leading to a more intricate spectrum.

5. What are the potential future developments in the study of the hyperfine structure of deuterium?

Future developments in the study of the hyperfine structure of deuterium may involve using more advanced spectroscopy techniques to further refine our understanding of the atom's properties. It may also have implications in the search for new physics beyond the standard model.

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