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Here is an atomic physics problem I am trying to solve:

The carbon ground state is $[He](2s)^2(2p)^2$. Suppose to excite an electron to the configuration $[He](2s)^2(2p)(3s)$. The spectroscopic terms associated to this configurations have energies $-3.7eV$ and $-3.9eV$. The ground state has energy $-11.3eV$. Which emission lines will be measured in a decay to the ground state?

What I have done so far:

Considering the $P$ orbital associated to the ground state there are two particle with spin $s = \frac{1}{2}$, therefore $S = 1$ and $L = 1$, therefore we have a triplet $J = 0, 1, 2$. Considering the excited state, we have for the $2P$ orbital that $S = \frac{1}{2}$ and again $L = 1$, therefore we have two levels associated to $J = \frac{1}{2}, \frac{3}{2}$. For the $3S$ level we have instead $L=0$ and $S=\frac{1}{2}$, therefore $J = \frac{1}{2}$.

Now I consider the selection rule according to which $\Delta J = 0, \pm 1$. If this is true, then none of the excited levels (which have non-integer $J$) can decay to the ground state (which has integer $J$), so the answer would be: we observe no lines at all.

My question:

Is my reasoning correct? Am I doing something wrong?

Andrea
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  • By chance have you over-interpreted the question? Did they just want to calculate the energy differences and convert that into wavelegnths? – AChem Dec 08 '21 at 17:28
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    Seems you're overestimating how "forbidden" things are. https://en.wikipedia.org/wiki/Atomic_carbon#Electromagnetic_properties – Mithoron Dec 08 '21 at 22:25
  • So, yes, you're overthinking. – Mithoron Dec 08 '21 at 22:28
  • Thank you all for your replies, however I am still pretty confused. As I understand selection rules are approximately right, namely they tell us if some decay is less likely than others, but it can still happen. Then the solution would be "I observe 3 lines, 2 from the doublet $L^2_{1/2, 3/2}$ to the ground state ($S^3_0$) and one line from the singlet $S^2_{1/2}$ to the ground state"? Plus, I could tell the wavelength of the decay from the doublet from the question, but not from the singlet. Is this right? Thanks! – Andrea Dec 09 '21 at 11:06
  • Afaik, those selection rules are for a transition to be radiative and observable spectroscopically. Transitions 'forbidden' by selection rules can still happen, although non-radiatively, just that they won't be observable spectroscopically. – TheLearner Dec 09 '21 at 14:51
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    Have a look at the NIST Atomic Spectra Database and search for "C I". This gives you the configurations, term symbols and level energies and might help you to get on track. – Paul Dec 09 '21 at 17:38
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    @TheLearner Radiates just fine, only slowly - phosphorescence instead of fluorescence. – Mithoron Dec 10 '21 at 00:00

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