A Why Do LS and jj Coupling Differ in Spectral Line Transitions?

[bentzy]

My question is divided into two:
a. One of Hg spectral lines is a strong one, 253.7 nm, emitted by transition from 3P1 (triplet) to 1S0 (singlet). I'm afraid I don't read the 1st state right, since it says S=1 (due to 2S+1=3) & J=1, hence (?) L=0. However, P means L=1, isn't it ? What am I missing here ?
b. In LS coupling (mainly light atoms) we add all Ls & all Ss separately 1st, and only then add the sums to get J , while for very heavy atoms (e.g., Hg), the right model is jj coupling, hence the order of addition differs: we combine 1st pairs of L & S, then add the sums together to get J. I know that the reason behind the different procedures is the change in the relative strength between LL/SS interactions & LS interaction (the latter being stronger in heavy atoms). What I don't understand is why it is so. Namely, why one type of interaction necessitates the specific procedure used, but the other is wrong ? What's the relation between a dominant interaction & its appropriate way of vector addition ?
Thanks, BC

 

[drvrm]

One of Hg spectral lines is a strong one, 253.7 nm, emitted by transition from 3P1 (triplet) to 1S0 (singlet). I'm afraid I don't read the 1st state right, since it says S=1 (due to 2S+1=3) & J=1, hence (?) L=0. However, P means L=1, isn't it ? What am I missing here ?

Hg- strong emission lines occur at 546, 436, and 545 nm. There is substantial variability in the intensity of emissions at 611, 544, 574, 546, 436, 545, 578, 437 and 530 nm.

for vapour lamp The primary emission lines are at 546 and 578 nm. Secondary emission lines are at 366, 403, 435, 1,012, 1,125, 1,362, 1,525, 1,688 and 1,692 nm.

i was wondering about you data..
the above source is

<https://www.researchgate.net/figure/Emission-spectrum-of-a-mercury-vapor-lamp_fig5>

 

[drvrm]

What I don't understand is why it is so. Namely, why one type of interaction necessitates the specific procedure used, but the other is wrong ? What's the relation between a dominant interaction & its appropriate way of vector addition ?

In light atoms the electrostatic interaction between the outer electrons are stronger than spin -orbital interaction;the spin-spin interaction is

also much larger than (ls) interaction therefore Russel Saunders coupling is used.the weak coupling leads to L +S giving J values.

But in heavy atoms the Valence electrons are placed after a great volume of electronic shells so the electrostatic interaction as well as

spin-spin interaction term is weaker than ls term ,therefore the jj coupling is used. you can get a graphical comparison in the ref. given below

<http://www.pci.tu-bs.de/aggericke/PC3e_osv/Kap_V/Russel.htm>

 

[bentzy]

Both.

 

[bentzy]

Hg- strong emission lines occur at 546, 436, and 545 nm. There is substantial variability in the intensity of emissions at 611, 544, 574, 546, 436, 545, 578, 437 and 530 nm.

for vapour lamp The primary emission lines are at 546 and 578 nm. Secondary emission lines are at 366, 403, 435, 1,012, 1,125, 1,362, 1,525, 1,688 and 1,692 nm.

i was wondering about you data..
the above source is

<https://www.researchgate.net/figure/Emission-spectrum-of-a-mercury-vapor-lamp_fig5>

Hg- strong emission lines occur at 546, 436, and 545 nm. There is substantial variability in the intensity of emissions at 611, 544, 574, 546, 436, 545, 578, 437 and 530 nm.

for vapour lamp The primary emission lines are at 546 and 578 nm. Secondary emission lines are at 366, 403, 435, 1,012, 1,125, 1,362, 1,525, 1,688 and 1,692 nm.

i was wondering about you data..
the above source is

<https://www.researchgate.net/figure/Emission-spectrum-of-a-mercury-vapor-lamp_fig5>

I'll study the reference you quote, but these don't contradict the well-known UV line I'm talking about, and I'm surprised by your wondering.
1st: there's the transition described in my original question, which is allowed by jj-coupling for very heavy atoms, hence 253/7 nm line.
2nd: historically, this line is the line observed in Franck-Hertz famous experiment.

 

[bentzy]

I'll study the reference you quote, but these don't contradict the well-known UV line I'm talking about, and I'm surprised by your wondering.
1st: there's the transition described in my original question, which is allowed by jj-coupling for very heavy atoms, hence 253/7 nm line.
2nd: historically, this line is the line observed in Franck-Hertz famous experiment.

Now, regarding your 1st reply: I know about the relative interaction strengths, which is evident from my very question, and I'm afraid you missed my main 2 emphases, and didn't refer or answer them: (a) explaining the alleged contradiction in the term symbol quoted, & (b) why in the case of light atoms the addition procedure used is physicslly correct, while for (very) heavy atoms the 2nd procedure is the right one ? I hope that my refocused questions are clearer now.

 

[PeterDonis]

I'm afraid I don't read the 1st state right, since it says S=1 (due to 2S+1=3) & J=1, hence (?) L=0.

No, that's not correct. The 3 refers to the 3rd energy level, P means L = 1, and 1 means (I think) m = 1 (for L = 1 there are three possible orbitals, denoted by m = 1, 0, and -1).

 

[bentzy]

No, the symbol didn't include n, the main quantum number, which is written in plain script. The 3 is in superscript, and denotes the multiplicity of the state which is 3 in this case (triplet), hence total S of the state is 1 (due to 2S+1=3>>S=1).

Are you familiar with the notions of multiplicity and its context ?

 

[PeterDonis]

The 3 is in superscript

Is that how it was in the source you got this from? Can you give a reference?

 

[bentzy]

I'm not just quoting, I'm working on this.

These subscripts are well-known. especially in the context of electronic spectroscopy.

You can find this in every advanced book on the subject, e.g., Quantum Physics _{of atoms, molecules etc}, by R Eisberg & R Resnick, and Concepts of Modern Physics, by A Beiser.

 

[PeterDonis]

I'm not just quoting, I'm working on this.

Understood. But your OP did not have the subscripts so it was easy to misinterpret. There are two ways to do subscripts here on PF: you can use BBCode or LaTeX. The results will look like these:

³P₁

${}^3 P_1$

I've used BBCode in the quote from your post below, and LaTeX in my responses below.

One of Hg spectral lines is a strong one, 253.7 nm, emitted by transition from ³P₁ (triplet) to ¹S₀ (singlet).

Just to expand on this, the Hg atom has two electrons in the highest $n = 6$ energy level. As I understand it (see below), the ${}^3 P_1$ state has one of these electrons in the $6\text{p}$ orbital instead of the $6\text{s}$ orbital (the ground state ${}^1 S_0$ has both electrons in the $6\text{s}$ orbital).

S=1 (due to 2S+1=3) & J=1, hence (?) L=0. However, P means L=1

Yes, the ${}^3 P_1$ state has $L = 1$. The reason is, as above, that one of the two $n = 6$ electrons is in a $\text{p}$ orbital, while the other is in an $\text{s}$ orbital. So the total orbital angular momentum is $1 + 0 = 1$.

The fact that $S = 1$ indicates that the two electrons have parallel spins.

The reason $J = 1$ is possible is that $J$ does not have to be the algebraic sum $L + S$. The allowed magnitudes of $J$ are $L + S$, $L + S - 1$, and so on down to $| L - S |$. In this case, $L = S = 1$, that means $J = 0$, $J = 1$, and $J = 2$ are all allowed.

 

[DrClaude]

(a) explaining the alleged contradiction in the term symbol quoted

@PeterDonis has answered that in #11 above.

(b) why in the case of light atoms the addition procedure used is physicslly correct, while for (very) heavy atoms the 2nd procedure is the right one ?

That was answered by @drvrm in post #3.

 

[bentzy]

@PeterDonis has answered that in #11 above.That was answered by @drvrm in post #3.

Thank you for the discussion so far. Nevertheless, my question regarding the difference in the procedure of addition remained (I've read the ref you quoted), and I'll rephrase it: in LS coupling the Ls/Ss are coupled since due to the relatively strong electric intercation between pairs of Ls or Ss. My question is, focused on this extreme, why do we have to add the Ls/Ss first ? Why preceding adding pairs of L & S first, in this case, is wrong ?
Let me also add the following: why do we consider in this case the strength of the electric interaction between the Ls etc ? where's the magnetic interaction between them ? is it just because it's much weaker ?
Same question goes for jj-coupling, but the opposite is valid here: we consider the magnetic interaction between an L & an S, and neglecting the electric one.

That's all, so far,
BC