# Emission lines of calcium

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1. Nov 3, 2014

### unscientific

1. The problem statement, all variables and given/known data

Strong emission lines from calcium were observed at $422.7nm$, $610.3nm$, $612.3nm$, $616.3nm$, $1034.9nm$, from transitions between $4s^2$, $4s5s$ and $4s4p$. The transition of $422.7nm$ was also observed at absorption. The singlet-triplet splitting of $4s5s$ is given by $177 600 m^{-1}$.

Using selection rules or otherwise, draw the energy diagrams and identify which transitions correspond to the observed emission spectrum lines.

2. Relevant equations

3. The attempt at a solution

First, I converted the emission spectra into units of /m:

$$422.7nm = 2365184 m^{-1}, 610.3nm = 1638270 m^{-1}, 612.3nm = 1632920 m^{-1}, 616.3nm = 1622323 m^{-1} ,1034.9nm = 966184 m^{-1}$$

Selection rule that is important here is: $\Delta S = 0$:

However, I only obtain 4 emission lines - not sure where the 5th comes from.

Last edited: Nov 3, 2014
2. Nov 4, 2014

### Staff: Mentor

You have to start by figuring all the term symbols that you can get from the 4s2, 4s5s and 4s4p configurations.

3. Nov 4, 2014

### unscientific

For $4s^2$, it's only $4s S_0$ and $4s S_1$

For $4s5s$, they are $4s5s S_0$ and $4s5s S_1$

For $4s4p$, they are $4s4p P_1$ and $4s4p P_0$

I now get 2 additional spectral lines with the addition of $4s S_1$:

Last edited: Nov 4, 2014
4. Nov 4, 2014

### Staff: Mentor

You results for 4s2 and 4s4p are not correct. You need to review the rules of addition of angular momenta and the Pauli exclusion principle.

5. Nov 4, 2014

### Staff: Mentor

I forgot to say that this is not correct. All selection rules are important.

6. Nov 4, 2014

### unscientific

for $4s^2$, outermost electron has $l=0$, and the system can either be singlet (s=0) or tripplet (s=1) which gives $j=1$ or $j=0$.
The same applies for $4s4p$, since $l=1$ either singlet (s=0) or triplet(s=1) which gives $j=2$ or $j=1$ so $P_1$ or $P_2$.

Last edited: Nov 4, 2014
7. Nov 4, 2014

### unscientific

The selection rules are:

1. $\Delta S = 0$

2. $\Delta L = 0, \pm 1$

3. $\Delta J = 0, \pm 1$

4. $\delta l = \pm 1$ (1 electron jump)

None of the transitions above violate these rules..

8. Nov 4, 2014

### Staff: Mentor

There are two electrons, not one. And what do singlet and triplet actually mean? (Hint: How can you get $s=1$?)

In your drawing, you for instance indicated a transition between 4s2 1S0 and 4s5s 1S0. Isn't this forbidden?

9. Nov 4, 2014

### unscientific

There is no change in S, since both are 0. There is also no change in L since both L=0. No change in J either.

10. Nov 4, 2014

### unscientific

I get $s=1$ when both spins point up or down or (down-up + up-down)

11. Nov 5, 2014

### Staff: Mentor

And what about $\Delta l$?

Is this possible in the 4s2 configuration?

12. Nov 5, 2014

### unscientific

$\Delta l = 1$, so is this incompatible with the fact that $\Delta L = 0$?

The $4s^2$ configuration only tell us that $L=0$, it says nothing or puts no restrictions on spin.

13. Nov 5, 2014

### Staff: Mentor

The two are compatible. One puts a restriction on the configurations involved, the other on the term symbol. In other words, you can't have a transition between 4s2 and 4s5s, whatever $L$ each corresponds to, as this would break the $\Delta l = \pm1$ rule.

Here it is not about selection rules, but the Pauli exclusion principle.

14. Nov 5, 2014

### unscientific

So to consider transition for $4s^2$ to $4s5s$, it implies that $\Delta L = 0$ and $\Delta l = 0$, which violates the $\delta l = \pm1$ rule.

I know the overall wavefunction must be anti-symmetric, but must the spatial part be symmetric? (does being in the same orbital imply that?)

15. Nov 5, 2014

### Staff: Mentor

The Pauli principle imposes rules on the symmetry of the wave function. What is important here is the consequence of the Pauli principle known as the Pauli exclusion principle, which states that no two fermions can occupy the same state.

16. Nov 6, 2014

### unscientific

Does this mean for $L=0$, the spins must be anti-aligned (s=0) so that they don't have the same state - defined by spin and orbital state.

17. Nov 7, 2014

### Staff: Mentor

Not exactly. It has nothing per say to do with $L=0$ (you can have 3S states). But in the 4s2 configuration, the two electrons have $n=4$, $l=0$, $m_l=0$. They only way for them not to be in the same state is for one to have $m_s=+1/2$, and the other $m_s=-1/2$.