# Help with Atomic Calcium Electron Configurations

• physicisttobe
In summary, the excited state of atomic calcium has the electron configuration 1s 2 2s 2 2p6 3s 2 3p6 3d1 4f 1 . The term symbol corresponding to the lowest energy of this electron configuration is 3d.
physicisttobe
Homework Statement
deriving the term symbols
Relevant Equations
J= L+S
Hi everyone!
I need some help in a specific task.
An excited state of atomic calcium has the electron configuration 1s 2 2s 2 2p6 3s 2 3p6 3d1 4f 1 . (a) Derive all the term symbols (with the appropriate specifications of S, L, and J) for the electron configuration. (b) Which term symbol corresponds to the lowest energy of this electron configuration? (c) Consider a 3 F2 level of calcium derived from a different electron configuration than that shown above. Which of the term symbols determined in part (a) can participate in spectroscopic transitions to this 3 F2 level?

I´m not sure if I my solutions are correct. So could you take a look?
First, I start with a), then I'll post b) and c). I hope that you can help me, thanks in advance!

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DrClaude said:
Oh, you are right, I also need to consider S=0.
But what about the other term symbols? Are they correct?

physicisttobe said:
Oh, you are right, I also need to consider S=0.
But what about the other term symbols? Are they correct?
Yes.

Thank you. I post the rest of the task a) here :

Now to question b): How exactly should I solve this problem?
My idea would be: When you draw the electron configuration you‘ll see that 3d corresponds to the lower energy, and l=2 corresponds to d orbital. This means that all term symbols with the letter D have the lowest energy? Is this consideration correct?

I don't know how to do b. Maybe you are to use Hund's rule, but they actually don't apply here.

All right, I try to figure out how to solve b), but thank you!
And what´s about question c) ? How should I proceed here?

physicisttobe said:
And what´s about question c) ? How should I proceed here?
Selection rules.

This means Δmj = +/-1, 0 and Δl= +/- 1
This applies to these term symbols:

Is it correct?

physicisttobe said:
This means Δmj = +/-1, 0 and Δl= +/- 1
This applies to these term symbols:
View attachment 314769
Is it correct?
No. First, you are not told about the electronic configuration of the other state, but it is reasonable to assume that ##\Delta l = \pm 1## is satisfied. Second, ##\Delta m_j## is not in play, as you are looking at transitions between terms (which include all ##m_j##), not between levels.

What you need to consider are the selection rules for ##\Delta L##, ##\Delta J##, and ##\Delta S##.

Buy the way, I did not notice that in your answer above you had terms like 4G and 2P (this is the problem when you submit your answer as a scan of handwritten material: it can be hard to read). These terms are not correct. Revise what the superscript number corresponds to.

DrClaude said:
No. First, you are not told about the electronic configuration of the other state, but it is reasonable to assume that ##\Delta l = \pm 1## is satisfied. Second, ##\Delta m_j## is not in play, as you are looking at transitions between terms (which include all ##m_j##), not between levels.

What you need to consider are the selection rules for ##\Delta L##, ##\Delta J##, and ##\Delta S##.
I'm not sure if I understand you correctly, but then it must be the term 3G_3? Because ΔS=0, ΔL=1, Δl=1 and ΔJ=1
DrClaude said:
Buy the way, I did not notice that in your answer above you had terms like 4G and 2P (this is the problem when you submit your answer as a scan of handwritten material: it can be hard to read). These terms are not correct. Revise what the superscript number corresponds to.
Oh, yes youre right, for L= 4 and S=1 it must be 3G5, 3G4, 3G3. Is it correct?

physicisttobe said:
I'm not sure if I understand you correctly, but then it must be the term 3G_3? Because ΔS=0, ΔL=1, Δl=1 and ΔJ=1
There are many terms for which you will have transitions with 3F2.

physicisttobe said:
Oh, yes youre right, for L= 4 and S=1 it must be 3G5, 3G4, 3G3. Is it correct?
Correct for the G terms. Make sure all the others are correct (##S=0,1##, so you should get singlet and triplet terms only).

## 1. What is an atomic calcium electron configuration?

An atomic calcium electron configuration refers to the arrangement of electrons in the energy levels or orbitals of a calcium atom. It describes the distribution of electrons in the atom's electron shells and subshells.

## 2. How many electrons does a calcium atom have?

A calcium atom has 20 electrons. This is because calcium has an atomic number of 20, which indicates the number of protons in its nucleus. Since atoms are electrically neutral, the number of electrons is equal to the number of protons.

## 3. What is the electron configuration of a neutral calcium atom?

The electron configuration of a neutral calcium atom is 1s2 2s2 2p6 3s2 3p6 4s2. This means that the first energy level (or shell) has 2 electrons, the second energy level has 8 electrons, and the third energy level has 8 electrons. The remaining 2 electrons are in the fourth energy level.

## 4. How do you write the shorthand electron configuration for calcium?

The shorthand electron configuration for calcium is [Ar] 4s2. The [Ar] represents the electron configuration of argon, which is the noble gas that comes before calcium in the periodic table. This shorthand notation is used to save time and space when writing out the electron configuration of larger atoms.

## 5. Why does calcium have a stable electron configuration?

Calcium has a stable electron configuration because its outermost energy level (4s) is filled with 2 electrons. This is known as the octet rule, which states that atoms tend to gain, lose, or share electrons in order to achieve a full outer energy level with 8 electrons. Since calcium's outer energy level is already full, it is considered to have a stable electron configuration.

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