Allowed and forbidden transitions

[PainterGuy]

Hi,

Please have a look on the attachment, or for higher resolution please check this link: https://imageshack.com/a/img923/9834/vXbpvR.jpg

Could you please help me with the queries below?

Question 1: I don't understand that why transitions like "1" to "b" are not allowed or not probable.

Don't "1", "2", "3", "4", and "5" represent the same orbital angular momentum number, i.e L=2? And don't "a", "b", and "c" represent L=1? If I'm right then for transition from "1" to "b", orbital angular momentum changes from "2" to "1", and magnetic quantum number changes from "2" to "0".

Question 2: Which are those three unique transitions which produce three spectral lines? 1 to a, 2 to b, ...?

Thank you!

 

[DrClaude]

Question 1: I don't understand that why transitions like "1" to "b" are not allowed or not probable.

Don't "1", "2", "3", "4", and "5" represent the same orbital angular momentum number, i.e L=2? And don't "a", "b", and "c" represent L=1? If I'm right then for transition from "1" to "b", orbital angular momentum changes from "2" to "1", and magnetic quantum number changes from "2" to "0".

There is a selection rule on $l$ and one on $m$, such that $\Delta l = \pm 1$ and $\Delta m = 0, \pm 1$. The "1" to "b" changes $m$ by 2, and hence is not electric-dipole allowed.

Why these transitions are not allowed is related to conservation of angular momentum. The rule for $m$ is based on the triangle rule: if you imagine the angular momentum as a classical vector, you are limited in how you can draw a vector as the sum of two angular momenta and form a closed triangle.

Question 2: Which are those three unique transitions which produce three spectral lines? 1 to a, 2 to b, ...?

The magnetic field splits the two sets of levels ($n=3$ and $n=2$) by equal amounts, such that all transitions with the same $\Delta m$ have the same energy. You get only three lines because of the selection rule on $\Delta m$.

Note that this is often referred to as the normal Zeeman effect, and that for other atoms than hydrogen, you can get what is called the anomalous Zeeman effect. In the latter case, the two sets of levels do not have the same splitting, such that the $m = 0 \rightarrow 1$ transition is not at the same energy as $m = 1 \rightarrow 2$, and so on, and you can get more than three lines.

 

[PainterGuy]

Thank you for the explanation, @DrClaude!

 

[PainterGuy]

Hi,

I have another related question so I thought I better ask it here. I'd really appreciate if you could help me with it.

Please have a look at the attachment, or for higher-resolution picture please have a look here: https://imageshack.com/a/img922/4874/V8N4en.jpg

Question 1:
Before I ask my main question, let me start with a simple scenario to make sure if I have it right.

In the hydrogen atom an electron makes a transition from 4s to 2s emitting a photon of frequency A, and another electron make a transition from 4s to 3s emitting a photon of frequency B.

In my opinion frequency A will be greater than frequency B. I'm saying this because in this case energy is proportional to frequency, and the transition 4s to 2s involves more energy.

Or, I can put it differently like this. The 2s is at lower energy level than 3s therefore the transition from 4s to 2s will result in more loss of energy in form of high frequency photon.

Do I have it correct?

Question 2:
This is my main question and I'm going to assume that what I have said in Question 1 is correct.

The text says, "The slightly higher energy state results from the electron spinning in the same direction as the orbital angular momentum, the lower state where spin is opposite, as shown in Figure 3.11.2."

I agree with the above statement. But looking at Figure 3.11.2 tells me something different.

I think that photon with wavelength 656.272 nm corresponds to the transition from 3s to 2P_3/2 and the photon of wavelength 656.285 nm results from 3s to 2P_1/2 transition.

Larger frequency results into smaller wavelength; c=f*lamda. But the wavelength 656.285 is greater than 656.272 which means that the frequency of photon resulting from 3s to 2P_1/2 is lower than the other one. In my opinion, it should have been the opposite; the frequency for photon resulting from 3s to 2P_1/2 (i.e. photon #1) should have been greater. Where am I going wrong?

Thank you for your help!

 

[DrClaude]

Question 1:
Before I ask my main question, let me start with a simple scenario to make sure if I have it right.

In the hydrogen atom an electron makes a transition from 4s to 2s emitting a photon of frequency A, and another electron make a transition from 4s to 3s emitting a photon of frequency B.

In my opinion frequency A will be greater than frequency B. I'm saying this because in this case energy is proportional to frequency, and the transition 4s to 2s involves more energy.

Or, I can put it differently like this. The 2s is at lower energy level than 3s therefore the transition from 4s to 2s will result in more loss of energy in form of high frequency photon.

Do I have it correct?

Well, s ↔ s transitions are dipole forbidden, so let's consider 4s and 3s to 2p. Then you are right that photon A has a higher frequency.

Question 2:
This is my main question and I'm going to assume that what I have said in Question 1 is correct.

The text says, "The slightly higher energy state results from the electron spinning in the same direction as the orbital angular momentum, the lower state where spin is opposite, as shown in Figure 3.11.2."

I agree with the above statement. But looking at Figure 3.11.2 tells me something different.

I think that photon with wavelength 656.272 nm corresponds to the transition from 3s to 2P_3/2 and the photon of wavelength 656.285 nm results from 3s to 2P_1/2 transition.

Larger frequency results into smaller wavelength; c=f*lamda. But the wavelength 656.285 is greater than 656.272 which means that the frequency of photon resulting from 3s to 2P_1/2 is lower than the other one. In my opinion, it should have been the opposite; the frequency for photon resulting from 3s to 2P_1/2 (i.e. photon #1) should have been greater. Where am I going wrong?

Yes, they inverted the numbers.

 

[PainterGuy]

Thank you, @DrClaude!

Question 1:
Just wanted to clarify a point from my first post, the text (yellow highlight) said, "The simple rule is that a transition is allowed (i.e., is quite probable) if the change in orbital angular momentum (l) is -1 or +1 (but not zero). A transition is also allowed if the change in the magnetic quantum number(m_l) is -1, 0, or +1."

I believe that both selected rules must be satisfied for any transition therefore writing something like "if the change in orbital angular momentum (l) is -1 or +1 (but not zero) AND also the change in the magnetic quantum number (m_l) is -1, 0, or +1" would have been more accurate, in my opinion.

Do I have it right? Or, perhaps I'm reading too much into it!

Question 2:
Earlier you said,

There is a selection rule on lll and one on mmm, such that Δl=±1Δl=±1\Delta l = \pm 1 and Δm=0,±1Δm=0,±1\Delta m = 0, \pm 1. The "1" to "b" changes mmm by 2, and hence is not electric-dipole allowed.

I was reading this discussion about helium's triplet state and it says,

In the 1s state (both electrons in the ground state), one electron must be spin up and the other spin down by the Pauli exclusion princple. In other words the electron spins are anti-parallel and the state is a singlet.

However, if only one electron is in the 1s state, and the other electron is in a higher energy level, the two electrons can have parallel or antiparallel spins. If the spins are parallel, it is a triplet state.

The lifetime of the 1s 2s triplet state is 7859 seconds. It relaxes to the ground state by a magnetic dipole (M1) transition.

I understand that for helium the transition from 2s to 1s is not allowed because the selection for orbital angular momentum is not satisfied although magnetic quantum number rule is satisfied because delta_m_l=0.

My question is about the terminology. What's the difference between electric-dipole transition and magnetic dipole (M1) transition. As the lifetime of 2s triplet state for helium is 7859 seconds (almost 131 minutes!), it means that the electron finds it extremely hard to make 'regular' photon-emission transition.

Could you please elaborate on it?

Thanks a lot for your help and time!

 

[DrClaude]

Question 1:
Just wanted to clarify a point from my first post, the text (yellow highlight) said, "The simple rule is that a transition is allowed (i.e., is quite probable) if the change in orbital angular momentum (l) is -1 or +1 (but not zero). A transition is also allowed if the change in the magnetic quantum number(m_l) is -1, 0, or +1."

I believe that both selected rules must be satisfied for any transition therefore writing something like "if the change in orbital angular momentum (l) is -1 or +1 (but not zero) AND also the change in the magnetic quantum number (m_l) is -1, 0, or +1" would have been more accurate, in my opinion.

Do I have it right? Or, perhaps I'm reading too much into it!

The text doesn't use the best phrasing. If any rule says that a transition is forbidden, then this takes precedence.

A simple way to understand it is that this all comes from integrations between the wave function of the initial state, that of the final state, and something called the transition dipole moment. That integral is almost always zero, except for special cases, and these special cases are the selection rules for $\Delta l$ and $\Delta m_l$.

Actually, in spherical coordinates, the rule for $\Delta l$ comes from the integration over $\theta$, and the rule for $\Delta m_l$ from the integration over $\phi$. If either is zero, then the entire integral is zero. All transitions are thus forbidden, unless allowed by all selection rules. The integral over $r$ is never zero (unless there would be some weird special case, but I don't know of any), which is why there is no selection rule on $n$: all $\Delta n$ are allowed.

I was reading this discussion about helium's triplet state and it says,I understand that for helium the transition from 2s to 1s is not allowed because the selection for orbital angular momentum is not satisfied although magnetic quantum number rule is satisfied because delta_m_l=0.

My question is about the terminology. What's the difference between electric-dipole transition and magnetic dipole (M1) transition. As the lifetime of 2s triplet state for helium is 7859 seconds (almost 131 minutes!), it means that the electron finds it extremely hard to make 'regular' photon-emission transition.

Could you please elaborate on it?

My guess is that you don't yet know perturbation theory, so I can't go into details, but I'll try to summarise what is going on.

As I said above, the transitions rates are calculated using an integral, something like
$$
\int \psi_f^* \hat{H}' \psi_i d\tau
$$
where $\psi_i$ is the wave function of the initial state, $\psi_f$ the wave function of the final state, $\hat{H}'$ a Hamiltonian representing the coupling between the atom and the electromagnetic field, and the integration is carried out over all degrees of freedom of the atom.

Most often, this integral is too complicated to be carried out exactly, so some approximation must be made. The most common is to consider that the wavelengths of light in play are much greater than the size of the atom, such that the EM field is constant over the atom. Then, one can do an expansion of $\hat{H}'$ into a series of terms, which turn out to correspond to the electric dipole, the magnetic dipole, the electric quadrupole, and so on. Each term is smaller than the previous one, such that it is usually sufficient to consider only the dominant term, the electric dipole. This is why these selection rules are often called the (electric) dipole selection rules. So even if a transition is dipole forbidden, it can still happen due to other terms in the series. But since these terms are weaker, the rate of transition is much smaller, or equivalently the lifetime of the excited state is much longer.

 

[PainterGuy]

Thank you, @DrClaude, once again.

I really appreciate your attempt to help me with this difficult stuff which is quite hard to explain in simple terms to someone with limited knowledge of the subject.

Anyway, I wanted to let you know that I only understand some of the concepts qualitatively including Schrodinger equation so I won't be able to delve into mathematical description of all this. I'm trying to learn these ideas at basic level as a self-study.

I was further reading that text and came across few confusing points. Could you please guide me?

Please have a look on the attachment, or for better resolution check this link: https://imageshack.com/a/img923/9712/B2HMqq.jpg

Question 1:
Is it correct where it says that "6P state, with l=0 or 1"? I understand that the capitalization of "P" denotes that it's a spectroscopic state is being discussed and not regular 6p orbital.

I understand that there is already a typo on the left where it should have been "6S" instead.

Question 2:
The electronic configuration of mercury is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f14 5d10 6s2. Which spectroscopic notation is being used here? Is it Paschen notation? In this notation the principal quantum number of being used is of valence electrons?

In this book another notation is also mentioned where the principal quantum number starts from the last filled shell in the atom.

Question 3:
If both electrons are spinning in the same direction then total spin could be -1 or +1. This is what happens in triplet state.

If both electrons are spinning in the opposite direction then total spin would be 0. That's singlet state.

But there is a contradiction where it says that in the case of triplet state total spin could also be "0". Could you please comment on it?

Question 4:
There exist three different orientations for p orbitals, p_x, p_y, p_z.

The diagram shows that two valence electrons of mercury make a transition from 7s orbital to its 6p orbital. Or, I might be totally wrong, it's just one valence electon in 7s orbital while the other is in 6p orbital. For mercury, base state for both valence electrons is 6s.

According to Hund's rule, the electrons fill the orbital one by one, so the transition 6P_2 doesn't really mean that the electrons are in the same orbital?

Thanks a lot for your time.

 

[DrClaude]

Question 1:
Is it correct where it says that "6P state, with l=0 or 1"? I understand that the capitalization of "P" denotes that it's a spectroscopic state is being discussed and not regular 6p orbital.

Yes, that's spectroscopic notation. However, I do not like the notation they use (and they are not the only one using such a notation), as specifying only "6" says nothing about the electronic configuration. Something like 6s² S or 6s6p ³P₂ would have been clearer.

Note that I don't understand what they mean by the "6P state, with $l=0$ or $1$." I take it that they use $l$ for the total orbital angular momentum (since they equate $j$ with $l$ when $s=0$), but if it is a P state, then $l=1$, since this is what the P stands for.

Question 2:
The electronic configuration of mercury is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 4f14 5d10 6s2. Which spectroscopic notation is being used here? Is it Paschen notation? In this notation the principal quantum number of being used is of valence electrons?

In this book another notation is also mentioned where the principal quantum number starts from the last filled shell in the atom.

The electronic configuration is independent of any spectroscopic notation. It comes from the Aufbau principle, where one simply fills hydrogen-like orbitals in order of energy, following the Pauli exclusion principle. A given electronic configuration will then lead to (most often) many spectroscopic terms, each represented by a term symbol. (In other words, the electronic configuration is not sufficient to define an energy level, since the relative orientation of spin and angular momentum leads to levels of different energy.)

In this book another notation is also mentioned where the principal quantum number starts from the last filled shell in the atom.

I have never heard of that.

Question 3:
If both electrons are spinning in the same direction then total spin could be -1 or +1. This is what happens in triplet state.

Total spin cannot be negative. They are confounding it with the projection of the spin on the z axis.

If both electrons are spinning in the opposite direction then total spin would be 0. That's singlet state.

But there is a contradiction where it says that in the case of triplet state total spin could also be "0". Could you please comment on it?

Simple picture: consider spin as a classical vector. You have two spin-1/2 to add together. If they both point along z, you will have a resulting spin +1 along z, if they both point along -z, then the result is a spin of -1 along z. You can also have them with one pointing up and the other pointing down,, which would sum to 0 along z, but with them pointing both along the same direction in the x-y plane, such that they still add up to a total spin of 1.

If you have a total spin of $S=1$, then there are three possible values of $M_S$, +1, 0, -1. This is what the book is referring to.

Had the book not mixed up total spin and its projection along the z axis, this would probably not have confused you.

Question 4:
There exist three different orientations for p orbitals, p_x, p_y, p_z.

The diagram shows that two valence electrons of mercury make a transition from 7s orbital to its 6p orbital. Or, I might be totally wrong, it's just one valence electon in 7s orbital while the other is in 6p orbital. For mercury, base state for both valence electrons is 6s.

The 7S₁ term comes from the 6s7s electronic configuration, while the 6P triplet originates from 6s6p. To be more precise, the transitions observed here correspond to
6s7s ³S₁ → 6s6p ³P_0,1,2

According to Hund's rule, the electrons fill the orbital one by one, so the transition 6P_2 doesn't really mean that the electrons are in the same orbital?

Not sure why you are citing Hund's rules here. Where you thinking instead of the Pauli exclusion principle? In any case, you see from what I wrote above that the two valence electrons are always occupying different orbitals in this particular transition.

 

[PainterGuy]

Thanks very much, @DrClaude!

Question 3 (cont'd):
The text says,
"Where two electrons are considered, there are two basic types of spin states, triplets and singlets. A triplet results from a group of three symmetric states for spin in which the resulting total spin is -1, 0, or +1, yielding j values of 0, 1, and 2, as stated above. A singlet results when the spin of an excited electron is antiparallel to that of the remainder of the atom or molecule: an asymmetric state where the total combined spin is 0. Consider the simplest two-electron example in helium with two electrons: If the excited electron has a spin in the direction opposite to that of the remaining ground-state electron (which stays in a 1s state), a singlet results (with the excited species called parahelium); if the spin of the excited electron is in the same direction parallel to that of the remaining ground-state electron, a triplet results (with the excited species called orthohelium)."

Total spin cannot be negative. They are confounding it with the projection of the spin on the z axis.

Simple picture: consider spin as a classical vector. You have two spin-1/2 to add together. If they both point along z, you will have a resulting spin +1 along z, if they both point along -z, then the result is a spin of -1 along z. You can also have them with one pointing up and the other pointing down,, which would sum to 0 along z, but with them pointing both along the same direction in the x-y plane, such that they still add up to a total spin of 1.

If you have a total spin of $S=1$, then there are three possible values of $M_S$, +1, 0, -1. This is what the book is referring to.

I'm sorry but I'm still struggling with it. I believe that $M_S$ stands for the sum of individual magnetic spins and $S=1$ represents sum of absolute values of individual magnetic spins.

In helium example, the text kept it simple. I still don't know how to interpret this fragment "A triplet results from a group of three symmetric states for spin in which the resulting total spin is -1, 0, or +1". Why not simply say "if the spin of the excited electron is in the same direction parallel to that of the remaining ground-state electron, a triplet results"? If you could comment, it'd be really helpful.

Question 4 (cont'd):

The 7S₁ term comes from the 6s7s electronic configuration, while the 6P triplet originates from 6s6p. To be more precise, the transitions observed here correspond to 6s7s ³S₁ → 6s6p ³P_0,1,2

I thought I better confirm it. Shouldn't it be "2" instead? Thanks.

 

[DrClaude]

I'm sorry but I'm still struggling with it. I believe that $M_S$ stands for the sum of individual magnetic spins and $S=1$ represents sum of absolute values of individual magnetic spins.

$M_S$ is the projection of the spin on the z axis. For any angular momentum, there are two quantum numbers that can be used to characterize it, its value and its projection on an arbitrary axis (the uncertainty principle forbids one from knowing the projection on more than one axis at the same time). By convention, the quantization axis is usually the z axis.

For any angular momentum (meaning $j$ can be here replaced by $l$ or $s$), a system in an eigenstate of both the angular momentum operator $\hat{J}$ and the operator $\hat{J}_z$ is represented by two quantum numbers $j$ and $m_j$, corresponding to a total angular momentum of $\sqrt{j (j+1)} \hbar$ and a component of that total angular momentum along z of $m_j \hbar$.

In helium example, the text kept it simple. I still don't know how to interpret this fragment "A triplet results from a group of three symmetric states for spin in which the resulting total spin is -1, 0, or +1". Why not simply say "if the spin of the excited electron is in the same direction parallel to that of the remaining ground-state electron, a triplet results"? If you could comment, it'd be really helpful.

It depends on what "parallel" means here. If you take it to mean "pointing in the same direction along z," then the problem is that the triplet state doesn't simply corresponds to parallel spins. When you add together two spin-1/2 particles, you can get a total spin of 0 or 1. In terms of spinors, for $\alpha$ meaning spin-up and $\beta$ meaning spin-down, then two spin-1/2 particles in a state with spin 1 can be in one of three states,
$$
\alpha(1) \alpha(2) \\
\frac{1}{\sqrt{2}} \left [ \alpha(1) \beta(2) + \beta(1) \alpha(2) \right] \\
\beta(1) \beta(2)
$$
As you see, one of these states is made up of a combination of the two particles not being parallel, but still adding up to a total spin of 1. That state is the one for which $M_S=0$.

The singlet state, corresponding to $S=0$ is
$$
\frac{1}{\sqrt{2}} \left [ \alpha(1) \beta(2) - \beta(1) \alpha(2) \right]
$$

I thought I better confirm it. Shouldn't it be "2" instead? Thanks.

No, that's a 3. These are two triplet states, meaning $S=1$ (the spectroscopic notation is ^2S+1L_J). They have to be the same, since the selection rule is $\Delta S = 0$ (in the dipole approximation, there is no coupling between the EM field and spin, so the total spin cannot change).

 

[PainterGuy]

Thank you, @DrClaude!

Question 1:
I have a point to clarify from Question 4 in the last post. Please have a look on the attachment titled "11-21-2018 4-31-09", or for better resolution check this link: https://imageshack.com/a/img923/5935/Ld25Ou.jpg

The figure shows an electron making a transition from 7S₁ to either of three energy levels of p orbital, 6P₀, 6P₁, 6P₂. The books uses the subscript "1" in 7S₁ to denote triplet state.

There are three different orientations for p orbital with the same energy, i.e. p_x, p_y, and p_z. I believe each orientation is divided into three different energy levels depending upon different interactions like spin-spin interactions, L-S coupling, etc.

Assuming my understanding is fine, the other electron of mercury could still be in ground state, i.e. 6s¹, or in one of the other orientations of 6p orbital, or it might be somewhere above 7s orbital. Do I have it correct?

Question 2:
Please have a look on the attachment titled "11-21-2018 4-37-19", or for better resolution check this link: https://imageshack.com/a/img923/9229/wqWflz.jpg

Neon's 5s¹ level is divided into four different energy levels. How do these energy levels occur? Are the following correct reasons?

1: spin and angular momentum in same direction in 5s¹
2: spin and angular momentum in opposite direction is 5s¹
3: both unpaired electrons in 2p⁵ and 5s¹ spinning in the same direction
4: both unpaired electrons in 2p⁵ and 5s¹ spinning in the same direction

Thanks a lot for you help and time!

 

[DrClaude]

The figure shows an electron making a transition from 7S₁ to either of three energy levels of p orbital, 6P₀, 6P₁, 6P₂. The books uses the subscript "1" in 7S₁ to denote triplet state.

That's not correct. The subscript indicates the value of $J$. As I wrote above, this is a triplet state, ³S₁. Since $J = L + S$ and $L=0$ (since it is a S term), then $J = S = 1$.

There are three different orientations for p orbital with the same energy, i.e. p_x, p_y, and p_z. I believe each orientation is divided into three different energy levels depending upon different interactions like spin-spin interactions, L-S coupling, etc.

The $m_l$ degeneracy (same energy for each $m_l$ value) is lifted by spin-orbit interaction. There is no spin-spin interaction being taken into account at this level of the theory.

Assuming my understanding is fine, the other electron of mercury could still be in ground state, i.e. 6s¹, or in one of the other orientations of 6p orbital, or it might be somewhere above 7s orbital. Do I have it correct?

As I wrote above, these levels correspond to the 6s6p and 6s7s electronic configurations of Hg, so yes, one electron is still in the 6s orbital.

Neon's 5s¹ level is divided into four different energy levels. How do these energy levels occur? Are the following correct reasons?

1: spin and angular momentum in same direction in 5s¹
2: spin and angular momentum in opposite direction is 5s¹
3: both unpaired electrons in 2p⁵ and 5s¹ spinning in the same direction
4: both unpaired electrons in 2p⁵ and 5s¹ spinning in the same direction

Things are way more complicated here, and I won't be able to give you a full explanation. To understand the level structure, you have to know the rules of addition of angular momentum in quantum mechanics. In addition, there is the Pauli principle that constrains the states that are available, especially in this case where you have equivalent electrons (same n and l). I discussed above the triplet vs singlet state for two electrons, where I explained that in the triplet state the spin of the electrons is not necessarily "parallel" in the simple sense of the word. Here, you can't think of it only as spin and orbital angular momenta being aligned or anti-aligned.

 

[PainterGuy]

Thanks a lot, @DrClaude!

Question 1:
You are correct that the subscript "1" in ³S₁ is J. I also do agree that the superscript "3" represents triplet state although the book isn't clear about this. However, I wanted to clarify a point about the superscript. In the book, as you can see below, a superscript is used to represent individual energy levels of an energy state such as 6P state below. So, isn't there a little confusion between ³S₁ and notation like 6³P₂? I'm sorry if it's just splitting hairs.

"Unlike sodium, mercury has two valence electrons. These two electrons may have their spins in opposite directions, in which case they effectively cancel each other out, resulting in a spin of s=0. The subscript j, then, is simply the orbital angular momentum l. If the spins are in the same direction, the contribution from spin is s=1. Consider the 6S state. l=0, but spin can contribute a value of 0 or 1, so the 3S state becomes two states: 6S₀ and 6S₁. The 6P state, with l=0 or 1, can now yield values for j of 0, 1, and 2. These levels are designated as 6³P₀ , 6³P₁ , and 6³P₂ , where the superscript 3 denotes that the multiplicity of the level is three." Source: http://imagizer.imageshack.us/a/img923/5935/Ld25Ou.jpg (first paragraph)

Question 2:
This one might be a silly question but anyway does triplet state always occur for s orbitals? For example, as you can see the highlighted statement, http://imagizer.imageshack.us/a/img923/5935/Ld25Ou.jpg, in energy level 6P₂ both electrons are spinning in the same direction but it's not considered a triplet state.

Question 3:
It's related to Question 2 but I separated it for clarity.

Also, it looks like triplet state is metastable. Are forbidden transitions, as dictated by selection rules, related to the triplet states? For example, both singlet and triplet state of helium cannot decay radioactively because these are forbidden transitions; these are metastable states with extended lifetime. The value of the singlet lifetime for both He-3 and He-4 is 19.7 msec and the lifetime of 2s triplet state for helium is 7859 seconds.

Also triplet state is less likely to occur compared to singlet state, https://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map:_Physical_Chemistry_for_the_Biosciences_(Chang)/14:_Spectroscopy/14.7:_Fluorescence_and_Phosphorescence (third para).

Question 4:
You are right about neon transitions problem but I just wanted to clarify a general point. Please have a look at the attachment. You can see that the orbital 5s is divided into 4 energy states and 4p is divided into 10 energy states. Ideally, there could be 40 different transitions between these two orbitals. Do I make sense?

The figure shows an electron making a transition from 7S₁ to either of three energy levels of p orbital, 6P₀, 6P₁, 6P₂.

There are three different orientations for p orbital with the same energy, i.e. p_x, p_y, and p_z. I believe each orientation is divided into three different energy levels depending upon different interactions like spin-spin interactions, L-S coupling, etc.

The m_l degeneracy (same energy for each m_l value) is lifted by spin-orbit interaction. There is no spin-spin interaction being taken into account at this level of the theory.

So, the equal energy level of each m_l degeneracy, p_x, p_y, and p_z, gets splitted into three unequal energy states.

I really appreciate your help and time. I wish you a Happy Thanksgiving!

 

[DrClaude]

Question 1:
You are correct that the subscript "1" in ³S₁ is J. I also do agree that the superscript "3" represents triplet state although the book isn't clear about this. However, I wanted to clarify a point about the superscript. In the book, as you can see below, a superscript is used to represent individual energy levels of an energy state such as 6P state below. So, isn't there a little confusion between ³S₁ and notation like 6³P₂? I'm sorry if it's just splitting hairs.

It's the same notation: ^2S+1_J. The author is sloppy and sometimes forgets the multiplicity (2S+1).

Question 2:
This one might be a silly question but anyway does triplet state always occur for s orbitals? For example, as you can see the highlighted statement, http://imagizer.imageshack.us/a/img923/5935/Ld25Ou.jpg, in energy level 6P₂ both electrons are spinning in the same direction but it's not considered a triplet state.

That passage is very badly written. For instance, the author writes that the energy is lowest for 6P₂, while obviously in the diagram (which is correct) the lowest energy level is 6P₀.

I don't understand your question: "does triplet state always occur for s orbitals?" To get a triplet state, meaning $S=1$, you need at least 2 electrons with parallel spin. If you can get that for s orbitals (meaning that n must be different, e.g., 3s4s), then yes there will be a triplet state. If the two electrons are in the same orbital (e.g., 3s²), then because of the Pauli exclusion principle, you can't a triplet.

Question 3:
It's related to Question 2 but I separated it for clarity.

Also, it looks like triplet state is metastable. Are forbidden transitions, as dictated by selection rules, related to the triplet states? For example, both singlet and triplet state of helium cannot decay radioactively ? because these are forbidden transitions; these are metastable states with extended lifetime. The value of the singlet lifetime for both He-3 and He-4 is 19.7 msec and the lifetime of 2s triplet state for helium is 7859 seconds.

The 1s2s ³S₁ in helium is metastable because the only level below that one is 1s^{2 1}S₀, and the transition between the two are forbidden because $\Delta S \neq 0$. That said, the 1s2p ³P_0,1,2 is not metastable because it can decay to 1s2s ³S₁. So yes, the first triplet state is metastable because it is a triplet state, but most triplet states are not metastable.

Also triplet state is less likely to occur compared to singlet state, https://chem.libretexts.org/Textbook_Maps/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map:_Physical_Chemistry_for_the_Biosciences_(Chang)/14:_Spectroscopy/14.7:_Fluorescence_and_Phosphorescence (third para).

Transitions are less likely to occur. But the case presented is when the ground state is a singlet state. There are many cases where the ground state is a triplet state.

Question 4:
You are right about neon transitions problem but I just wanted to clarify a general point. Please have a look at the attachment. You can see that the orbital 5s is divided into 4 energy states and 4p is divided into 10 energy states. Ideally, there could be 40 different transitions between these two orbitals. Do I make sense?

You have to be careful with the language. "The orbital 5s is divided into 4 energy states" makes no sense. It is the electronic configuration 2p⁵5s¹ that splits into 4 levels (those 2p⁵ electrons are important). I would be surprised if transitions between the 40 pairs of levels would all be allowed, but there are going to be quite a few of them.

So, the equal energy level of each m_l degeneracy, p_x, p_y, and p_z, gets splitted into three unequal energy states.

As I wrote previously, things are a bit more complicated than this. When you have more than one electron, you have to consider relative orientations of orbital angular momentum and spin, so you cannot think in terms of the m_l values of a single orbital. It depends on how you can arrange all those electrons on the different orbitals.

 

[PainterGuy]

Thanks a lot, @DrClaude!

I'm sorry that my terminology and language below is not really accurate.

My question was:

Question 1:
You are correct that the subscript "1" in ³S₁ is J. I also do agree that the superscript "3" represents triplet state although the book isn't clear about this. However, I wanted to clarify a point about the superscript. In the book, as you can see below, a superscript is used to represent individual energy levels of an energy state such as 6P state below. So, isn't there a little confusion between ³S₁ and notation like 6³P₂?

http://imagizer.imageshack.us/a/img923/5935/Ld25Ou.jpg (first paragraph)

Your reply:

It's the same notation: ^2S+1_J. The author is sloppy and sometimes forgets the multiplicity (2S+1).

So, the electron in energy level 7³S₁ is in a triplet state with reference to 6s¹ electron but the transition is not allowed from this level to 6s¹ according to selection rules.

Along the same lines, the electron in energy level 6³P₂ is in a triplet state with reference to 6s¹ electron and the transition is allowed from this level to 6s¹ according to selection rules. It means that triplet state could also occur between different orbitals like between 's' and 'p' orbitals.

^2S+1L_J where "S" is total spin of two electrons, L is angular momentum, and J=L+S.

My question was:

Question 3:
It's related to Question 2 but I separated it for clarity.

Also, it looks like triplet state is metastable. Are forbidden transitions, as dictated by selection rules, related to the triplet states? For example, both singlet and triplet state of helium cannot decay radioactively because these are forbidden transitions; these are metastable states with extended lifetime. The value of the singlet lifetime for both He-3 and He-4 is 19.7 msec and the lifetime of 2s triplet state for helium is 7859 seconds.

Your reply:

The 1s2s ³S₁ in helium is metastable because the only level below that one is 1s^{2 1}S₀, and the transition between the two are forbidden because $\Delta S \neq 0$. That said, the 1s2p ³P_0,1,2 is not metastable because it can decay to 1s2s ³S₁. So yes, the first triplet state is metastable because it is a triplet state, but most triplet states are not metastable.

I thought that I better clarify your choice of words so there is no misunderstanding.

You said, "$\Delta S \neq 0$". But the transition is also forbidden in view of selection rules on angular momentum and magnetic quantum number.

You said, "That said, the 1s2p ³P_0,1,2 is not metastable because it can decay to 1s2s ³S₁." Though it is a triplet state but still not a metastable one. Although the $\Delta S \neq 0$, the selection rules allow the transition.

I'm sure I have it quite wrong!

Thanks a lot.

 

[DrClaude]

I'm sorry that my terminology and language below is not really accurate.

I'm correcting the terminology because I think it is essential here to distinguish different things, otherwise it is going to be impossible to understand what this is all about.

Also, I must say that the textbook you are using is pretty awful, so I understand your confusion.

So, the electron in energy level 7³S₁ is in a triplet state with reference to 6s¹ electron but the transition is not allowed from this level to 6s¹ according to selection rules.

Along the same lines, the electron in energy level 6³P₂ is in a triplet state with reference to 6s¹ electron and the transition is allowed from this level to 6s¹ according to selection rules. It means that triplet state could also occur between different orbitals like between 's' and 'p' orbitals.

I get the feeling that you are mixing up some things. A given electron cannot be in a triplet state; it takes at least two electrons to make up a triplet state, since it is state for which S=1. In reality, you have to consider the entire electronic configuration of Hg, which in the ground state is

1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 4f¹⁴ 5s² 5p⁶ 5d¹⁰ 6s²

It is all these electrons together that are in a given level. That said, we actually need to concentrate on the valence electrons, in this case 6s², because the other electrons will not contribute anything to L or S since they are in filled sub-shells (ns², np⁶, nd¹⁰, or nf¹⁴).

So considering only the valence electrons, we consider here two excited electronic configurations, 6s6p and 6s7s. In both cases, we get a singlet and a triplet state, depending on the spin orientations (see post #11 above). The 6s6p configuration has L=1, so it gives two terms, ¹P and ³P, and adding J we get the levels
¹P₁ ³P₀ ³P₁ ³P₂

From the diagram, it is then clear that what the author designates 6P is the triplet ³P₀, ³P₁, and ³P₂

The 6s7s configuration has L=0, so it gives two terms, ¹S and ³S, and the levels are
¹S₀ ³S₁

Checking the diagram, what is label 7S₁ is thus this ³S₁ level.

All transitions in shown in the diagram are allowed.

You said, "$\Delta S \neq 0$". But the transition is also forbidden in view of selection rules on angular momentum and magnetic quantum number.

I should've been clearer here. The transition has
$\Delta l = 0$ → forbidden
$\Delta L = 0$ → generally allowed, but L=0 ↔ L= forbidden
$\Delta J = 1$ → allowed
$\Delta S = 1$ → forbidden

But here the change of spin is the "most forbidden," leading to a much longer lifetime (about 7870 s). For comparison, the 1s2s ¹S₀ level is also metastable since it breaks two selection rules (but not the one on spin), but its lifetime is only 20 ms.

You said, "That said, the 1s2p ³P_0,1,2 is not metastable because it can decay to 1s2s ³S₁." Though it is a triplet state but still not a metastable one. Although the $\Delta S \neq 0$, the selection rules allow the transition.

Here, $\Delta S=0$, which is why it is not metastable. It is a triplet state, but there is a triplet state lower in energy to which it can decay.

 

[PainterGuy]

Hi @DrClaude

I'm sorry to ask you about this after so many days. I wanted to clarify a point from my previous post.

I should've been clearer here. The transition has
$\Delta l= 0$ → forbidden
$\Delta L= 0$ → generally allowed, but L=0 ↔ L= forbidden

You first time used "L" in post #11 when you introduced this formula ^2S+1L_J. I had thought that you were simply using "L" in place of "l" where "l" is azimuthal quantum number. Also I believe that somewhere in mind I was thinking that "L" is total angular momentum of levels involved. It looks like I was wrong.

In posts #13 and #16 you are only using "l" of the upper level involved in the transition. Perhaps "L" is used instead of "l" when more than one valence electrons are involved.

This is what Wikipedia article, https://en.wikipedia.org/wiki/Azimuthal_quantum_number#Derivation (second para) says:

Could you please clarify your use of "L" in your last post and previous ones? Thanks a lot.

 

[DrClaude]

The convention I followed is that $l$ is used for single electrons and $L$ is used for many electrons. In a sense $L= \sum l$ (although one has to be careful, as there are special rules when adding up angular momentum in quantum mechanics).

This is what Wikipedia article, https://en.wikipedia.org/wiki/Azimuthal_quantum_number#Derivation (second para) says:
View attachment 235389

I completely disagree. I think this is a personal opinion that goes against Wikipedia's NPOV rule.