Thank you guys.
I learned another way to think about this in the following video:
where they specify that "transitions where in the excited state J value is one unit greater than that of the lower energy state gives R lines, and in if they are one unit of J less than in the lower energy state...
Sorry I think you missed the point (and pain) of my post.
I am student, trying to understand rovibrational transitions from books and university lessons, not experimentalist working in a lab.
In rovibrational transitions we have following selection rules
$$ \Delta v = \pm 1 $$
$$ \Delta J = \pm 1 $$
where ##v## is the vibrational quantum number and ##J## is rotational quantum number.
Now based on whether ##\Delta J## changes to +1 or -1 we have two branches of spectroscopic lines...
I am learning about lasers and trying to clear up some blurry concepts in my head so I am not necessary using heavy math for description here so please bear with me.
Let's take EM field in a rectangular box with walls that are perfect conductors (optical cavity). EM field distribution inside...
Thank you all for great answers, still working my way to conceptualize this process :)
This is what confuses me. Can you explain a bit more why is necessary to compress the refrigerant (which increases its temperature) just in order to cool it down to the room temperature in the next step...
I am trying to understand how refrigerator works from the thermodynamic perspective. I've read many articles and watched several youtube videos but still trying to figure out following stuff.
1. What exactly is the purpose of compressor? Ok it compresses vapor and thus increases it's pressure...
I think I am starting to get it now...
So in my example there are 9 transitions but they are grouped such that only 3 different lines are present, right?
If I group transitions that belong to the same line they would be like this:
line 1:
##D_2 \rightarrow P_1##
##D_1 \rightarrow P_0##
##D_0...
I know that normal Zeeman effect happens when total spin of electrons in an atom is S=0.
This means that energy levels splitting in external magnetic field is done only on orbital angular momentum.
So for example P orbital will split into three sub-levels with slightly different energies. Now...
So to sum it up, if there is no spin-orbit coupling (or we neglect it) then there is no fine structure in He?
By fine structure I mean spectral lines split in more than one component (with no external magnetic field present) if we look at it with high enough resolution.
For example on this...
But we have spin-orbit interaction taken into account for hydrogen atom fine structure and hydrogen is even lighter than helium.
Here http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydfin.html they say:
I was looking at grotrian diagrams for helium and I see that there is no splitting of energy levels due to spin-orbit coupling. In my book it is said that spin-orbit coupling in helium is small and can be neglected but no further explanation is given. At the same time we do spin-spin coupling...
Can you elaborate how to look at this kind of problem when solving?
How exactly do you plug in ##S=1## in this equation and get only 0,1,2 as solutions for L?