Orsdxy

We were taught about the emission spectra in class last year, but my teachers couldn't give me an answer to 'what determines the colour of light emitted?'. (they were giving me the answers to the worksheet, but not my question if this makes sense).
This annoyed me. So now I am here.

From what I have read so far, (and this is only my guess) when the electron jumps to the next energy level in a Hydrogen atom, it emits red light. (I am just gonna explain why I think this happens. Feel free to criticise it however you like.)
Because Hydrogen is a small atom, the electron doesn't need a lot of energy to jump to the next level. So because red light is the least energetic visible light, it is emitted. So then would (let's just say,) fluorine emit a higher energy wavelength because it is more 'effort' for the electron to jump?

I just looked at the wavelength for certain elements. And apparently, potassium emits a pink/purple colour. But copper emits a light green. So obviously, my guess is incorrect, as copper is more massive than Potassium.

I would really appreciate it if someone could answer this. And if you could, may you try and explain it a simple as possible? I sorta only got into Quantum Mechanics early last year, so I'm not familiar with all of the terms. But I am happy to put in extra research if you need to use other theories to explain this thoroughly.
Thanks.

You seem to be under the impression that each atom emits light of a single colour, but this is not the case. Generally speaking, every atom will be able to emit light over a broad array of discrete wavelengths, which include a bunch of limit points where you have an infinity of different spectral lines congregating at a limit until they cannot be resolved.

As an example, this is what the spectrum for hydrogen looks like, which is broken up into a number of different spectral series:

^{Image source}

Generally speaking, the emission spectra of most atoms cover roughly similar wavelength ranges, basically spreading over the visible range with substantial bleed over into the UV and infrared regions. As such, it is not the case that there is some simple order between the lines of the different atoms, and you cannot rank atoms in terms of "effort" for jumps or in terms or size or weight, since they all produce interlaced forests of lines, with characteristic signature but with no real relationship to each other:

Image source

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This means that your core question,

what determines the colour of light emitted?

doesn't make sense as posed. But you can still ask a very similar question, by inserting an appropriate plural:

what determines the colours of light emitted?

And here, I'm afraid, the only answer is "it's complicated". The line emission and absorption spectra of atoms can indeed be calculated from first principles, but (with the sole exception of hydrogen) this cannot be done analytically, and it requires some substantial number-crunching to figure out.

Atomic line emission and absorption spectra come from the differences between the discrete energy levels in the atom, and those are the solutions of a complex dynamical problem in quantum mechanics, involving all of the interactions between the electrons and the nucleus as well as each other. Typically, calculating the emission spectrum of hydrogen is within the reach of a 'mature' course in quantum theory in an undergraduate degree (often the second QM course within the degree), but the tools for even an approximate calculation of atomic spectra for multi-electronic atoms require a further, dedicated course on atomic physics.

It is not true that hydrogen would only emit red light. It can emit anything - even ultraviolet which is more energetic radiations than blue light. The electrons need not be at $n=1$ for hydrogen all the time. It could be in another energy level. You must have studied about the Lyman, Balmer, Paschen series. The radiation comes when those transitions occur between a higher energy level to a lower energy level. So, the electrons does not have to do any "effort". If you were doing this as an experiment, you would have heated the hydrogen gas first and recorded its line spectra. By heating, you provide the electrons with energy that gets them to elevate from $n=1$ to a higher state and then force them to release that energy because $n=1$ is more stable.

And to emit visible light, as you might remember, all electrons have to make a transition to $n=2$ before going for $n=1$ and so exactly where the electron in discussing gets excited will give the answer to what color it emits. That is, the electron can get energy for it to get elevated to $n=3,4,5,6$ or wherever. Of course, that energy it obtains should depend on the heating given to the gas in the first place.

You're guess of "electrons jumping between energy levels" is definitely on the right track (in fact, it's basically the answer to your question!).

Electrons fly around the nucleus in something called orbitals (these are not exactly the same thing as orbits, but for the sake of this explanation the difference doesn't really matter). Any given orbital has a specific energy associated with it (see note below). The crucial point is: while every atom/molecule has an infinite number of orbitals, their energies are not continuous. For example, in the case of hydrogen, the ground state orbital has an energy of -13.6 eV, while the first excited state has an energy of -3.4 eV (quick notes: (i) the energies are negative because they're binding energies; and (ii) I'm disregarding relativistic and other 'fine-structure' effects here).

Now, under normal conditions, an electron spontaneously tries to minimize its energy, and it does so by jumping to an orbital with a lower energy. In the case of hydrogen, if an electron jumps from the first excited state to the ground state, its energy decreases by $Delta E$ = 10.2 eV. However, that energy cannot just vanish (conservation of energy, right?), so instead it is used to create a photon of frequency $nu = Delta E/h$, as dictated by Planck's formula. If you make the calculation, you'll see that such a photon has a wavelength of roughly 122nm - deep into the UV region of the spectrum. Of course, this is just an example of a possible transition: we could equally have considered a jump from the second excited state to the ground state, or from the third excited state to the first excited state. In many cases, these other transitions occur in the visible part of the spectrum.

For other atoms, the principle is the same: electrons jump between orbitals and, in doing so, change their energy while emitting a photon with a frequency/wavelength corresponding to the difference.

Here's where it gets beautiful: the energies of the orbitals vary from atom to atom, and so the photons coming from different atoms have different frequencies/wavelengths. In fact, each atom has a series of spectral lines associated with it (each spectral line coming from a different transition between a couple of orbitals), and these are so unique that they are often thought of as a "fingerprint" of the atom.

To summarize:

• Each atom has many orbitals that its electrons can occupy, and the energies of these orbitals vary from atom to atom.


• Electrons "like" to jump between orbitals (namely, from higher to lower energy orbitals), and, when they do so, the atom releases a photon with a frequency/wavelength corresponding to the difference between the energy of the orbitals.


• Typically, atoms don't emit a single "colour", but rather a mixture of photons with many different wavelengths, which our eyes then interpret as a single colour (like pink/purple for Potassium, or green for Copper).

Note on orbital energies: the precise value for the energy of an orbital depends on a number of different factors, and to calculate them you need the full "machinery" of quantum mechanics.