# Emission Spectrum

1. Sep 6, 2014

Are each of the color lines in the emission spectrum correspondant with a single band in the atomic orbital? For example if an electron rises to the 3p band it will emit one line of color only.

2. Sep 6, 2014

### Staff: Mentor

I really don't understand your use of the work "band" here.

If you're talking about hydrogen, an atom in the 3p state can decay to 2s or 1s, so two different wavelengths can be emitted.

3. Sep 6, 2014

### Staff: Mentor

Each wavelength corresponds to a transition from a specific higher energy orbital to a specific lower energy orbital. For example, an electron falling from the 3s to the 2s orbital in a hydrogen atom emits light with a wavelength of 656 nm. No other combination of orbitals will emit this wavelength for hydrogen, only the 3s to the 2s transition will do. Atoms of other elements may be able to emit at 656 nm (I think, not sure), but the specific orbitals involved in the transition will necessarily be different.

4. Sep 6, 2014

### Staff: Mentor

3s to 2s is a dipole forbidden transition. It will normally not be observed in a spectrum.

5. Sep 6, 2014

### Staff: Mentor

Ah, I see. The balmer series is based only on the principle quantum number, n:

The principal quantum number n represents the relative overall energy of each orbital, and the energy of each orbital increases as the distance from the nucleus increases. The sets of orbitals with the same n value are often referred to as electron shells or energy levels.

I assume that the emitted wavelength will be slightly longer or shorter than 656 depending on which specific states the electron fall from/to within these energy levels? Or is that only when you apply a magnetic field?

6. Sep 7, 2014

### Staff: Mentor

In a non-relativistic model for hydrogen, the energy is independent of the orbital quantum number $l$, and depends only on the principal quantum number $n$. In reality, other effects must be taken into account, most importantly spin-orbit coupling, which lifts the degeneracy in $l$ within one $n$ manifold, resulting in the fine structure.

The presence of a magnetic field will induce additional splitting of the lines, depending on the projection $m_j$ of the total angular momentum $j$.

7. Sep 7, 2014

### Staff: Mentor

Awesome. Thanks, DrClaude.

8. Sep 7, 2014

### Staff: Mentor

You're welcome.

I should probably add that the electric dipole selection rule is $\Delta l = \pm 1$, which is why you generally won't see a 3s → 2s transition. The 3s state will decay to 2p before it has time to decay via a "forbidden" transition (electric quadrupole, magnetic dipole, etc.).

9. Sep 7, 2014

This picture is the emission spectrum of Argon. If my understanding is correct, each line of color is synonymous with a difference in energy between the ground state and an orbital. Argon has 50+ lines of color. How is this so if there are only 20 possible orbitals?

10. Sep 7, 2014

### Staff: Mentor

A "forbidden" transition is one that usually doesn't occur because other transitions are far more likely, correct? In what cases would we see a 3s to 2s transition?

11. Sep 7, 2014

Staff Emeritus
No, it's between any two orbitals. That permits many more combinations.

12. Sep 7, 2014

### Staff: Mentor

13. Sep 7, 2014

So does this mean that more than one line can be manifested in a single transition from orbital to ground state?

14. Sep 7, 2014

### Staff: Mentor

No, each single transition is instant and emits only at one wavelength. However, an electron can transition from a higher energy orbital to the ground state by transitioning to multiple orbitals in between, emitting a specific wavelength with each transition. The electron stays in each orbital a finite amount of time before transitioning, so each of the emitted wavelengths are emitted at different times.

For example, in a helium-neon laser an electron in neon can transition from the 3s or 2s orbital to the 2p orbital and then from the 2s to the 1s orbital. http://en.wikipedia.org/wiki/Helium–neon_laser#Construction_and_operation

15. Sep 7, 2014

### Staff: Mentor

I don't think you can force it to emit only one wavelength. The best you can do is to block the wavelengths you don't want emitted from leaving the lamp. But I could be wrong.

16. Sep 8, 2014

### Staff: Mentor

Not exactly. Let me explain in more details.

Using a semi-classical theory (which works very well), we consider the electromagnetic field (external or vacuum) as classical and the atom as quantum. To first order, the interaction between the two is given by $\hat{\mu} \cdot \vec{\mathcal{E}}$, where $\hat{\mu}$ is the dipole moment operator and $\vec{\mathcal{E}}$ the EM field. The probability of a transition between two eigenstates $|1 \rangle$ and $|2 \rangle$ is proportional to $| \langle 1 | \hat{\mu} \cdot \vec{\mathcal{E}} | 2 \rangle|^2$. When $| \langle 1 | \hat{\mu} \cdot \vec{\mathcal{E}} | 2 \rangle|^2 = 0$, the transition is said to be forbidden, which is actually short for "electric dipole forbidden." This distinction is important, because there are higher order terms (electric quadrupole, magnetic dipole, etc.) that one can then consider that will make the transition possible. That said, these higher order terms are much smaller than the dipole terms, such that the transition probability is much smaller. When a state can decay through both a dipole (e.g., 3s → 2p) and a higher order transition (e.g., 3s → 2s), only the allowed, faster transition will generally be observed.

There are cases where no dipole transitions exists. A good example is the 2s state of hydrigen, which can only decay to 1s, corresponding to a forbidden transition. This decay eventually happens, but the lifetime of the 2s state is so long that it is considered a metastable state.

The selection rules are basically the cases when $| \langle 1 | \hat{\mu} \cdot \vec{\mathcal{E}} | 2 \rangle|^2 \neq 0$, and often have simple physical explanations. For instance, $\Delta l = \pm 1$ is simply conservation of momentum. The angular momentum of the photon (spin 1) has to be taken into account, so the atom has to change its angular momentum upon absorption or emission of a photon.

17. Sep 8, 2014

### Staff: Mentor

You usually don't even need to block other wavelengths: one transition will dominate (at least in the visible range). A good example is sodium: it can emit light of many colors, but in normal conditions the two D lines are much brighter, and the light looks yellow.

18. Sep 8, 2014

### Staff: Mentor

True, but the OP has asked about emitting one, and only one, wavelength at a time.