The valence electron in sodium atom gets excited and moves to higher orbital say $3P$ and it then comes to the lower energy state $3S$ thus there should be only one line in spectrum (regarding this particular transition). But my book says since $3P$ can exist in 2 states $3P_{1/2}$ and $3P_{3/2}$ so there are doublet in the spectrum but how can only one electron make two transitions – from $3P_{1/2}$ to $3S$ and from $3P_{3/2}$ to $3S$ ?

# [Physics] How to only one valence electron in sodium cause doublet in spectrum

atomic-physicsatomsspectroscopy

#### Related Solutions

There's an infinite number of orbitals, and thus an infinite number of possible state transitions. However, the energy of the orbitals asymptotes to a finite value - for example, a hydrogen atom has energy levels given by the formula $E_n = -\frac{13.6\text{ eV}}{n^2}$ (ignoring some very tiny quantum corrections). If you let $n$ go up to infinity, the energy approaches zero. So a photon with enough energy, namely 13.6 eV, will knock the electron past all those infinite energy levels and out of the atom entirely.

Furthermore, if you look at some of the numerical values that come out of that formula: -13.6 eV, -3.4 eV, -1.51 eV, -0.85 eV, -0.54 eV, etc., you'll notice that they get closer and closer together as $n$ gets higher. At some point, the difference between one energy level and the next becomes so small that you can't measure it.

All instances of a particular atom are exactly the same, and so a given transition will always have the same energy for a particular kind of atom (including the isotope). This is how we are able to identify elements by their spectral lines. However, different atoms have different energy levels, because of the different nuclear charge and mass, and also because of multiple electron interactions, so the energy will vary from one atom to another, even for the same transition.

To address the question in your title, there is a list of the spectral lines of most elements in the CRC Handbook of Chemistry and Physics. I'm not sure if it lists the corresponding state transitions, though. Unfortunately the CRC website itself is restricted, so you'd need to pay to access it, but the table itself is probably available elsewhere online. It's quite extensive, listing hundreds of spectral lines for each of the elements.

Based on some of your comments, I think what might be tripping you up is the first statement you started with:

From the Bohr's atomic model, it is clear that electron can have only certain definite energy levels.

and

...If suppose, we assume electron losses total energy, electron can't stay in any particular shell, as it would not have that particular value of energy.

That may be true for Bohr's atomic model, but Bohr's atomic model is wrong. And electron does not have to be in a particular, definite shell or energy level. Rather, any electron state is a *superposition* of states of definite energy level (energy eigenstates).

That means the expectation value of a hydrogen electron state is going to look like
$$\langle E\rangle = \sum_n |a_n|^2 E_n\text{,}$$
where $\{a_n\}$ are arbitrary complex values with $\sum_{n>0}|a_n|^2 = 1$ and $E_n$ are energy levels in increasing order. Because of the sum-to-$1$ condition, taking any portion along the other energy eigenstates will *increase* energy compared to the ground state.

In other words, even if the electron state does not have a definite energy, you still can't go lower than the ground state.

Suppose, I have a cup of hot coffee on the table. It will be continuously losing energy in the form of heat, but it stays on the table, though there was a energy loss. Now, all of a sudden, I take off the table, the cup of coffee converts it potential energy into kinetic energy to come down.

If you don't shake the table, the coffee cup will sit there, forever. Similarly, nothing perturbs the electron in an excited energy eigenstate, then it simply will never decay. It cannot: energy eigenstates are stationary; they do not evolve into anything other than themselves.

**However**, being completely without external perturbation is actually impossible. The uncertainty principle provides the electromagnetic field with vacuum fluctuations, which will perturb the electron even if nothing else in the environment does. In your analogy, this (or something else) provides the "shaking of the table" for the electron. Once the electron state gains even a tiny component in some other energy eigenstate, the state can evolve in time.

In other words, one can think of spontaneous emission as a particular type of stimulated emission where it's the vacuum that does the perturbing.

## Best Answer

It is not one electron making the transitions. For each atom electrons can exist in two P states and if a transition to an S state is possible the one of quantum jumps occurs. So when you see the spectrum doublet it is the result of these quantum jumps by electrons in different atoms.