# Ground State - Orbitals School's lied to you for a good chunk of your life - those concentric rings you see around the nucleus don't actually exist, instead being replaced with probabilistic blobs that denote where an electron could be at any given time. These are **orbitals**, which we can express as functions of $n$. >[!Note] Orbital Definition - Ground State >The Ground State is the orbital which electrons occupy when they are not excited or perturbed. We denote this "ground state" with the number $n=1$. Subsequent orbitals would increase in ones, going from $n=1$ to $n=2$ and so on. # The Anatomy of Excitation When an electron absorbs a photon, it becomes *excited* and jumps up an orbital - from the ground state to orbital $N = 2$, from the $N = 2$ orbital to the $N = 3$ orbital, and so on! Sometimes, an electron will reemit a photon taking it back to the ground state. This photon will be the same wavelength as the one they absorbed - creating an **absorption spectrum**. This is as atoms can only absorb certain wavelengths of light to bump up orbitals. [^1] Remember, what comes up must go down - and electrons will end up emitting photons to bump themselves back down to the stable ground state. The further away from the nucleus the harder it is to get to the next orbital in line, though the ionisation energy between the rings does decrease! [^1]: Actually, there are metastable configurations where an atom can actually be bumped up into a state between orbitals - this results in a sort of delayed "photon release" mechanism, as photons are released much slower than in the unstable second or above orbitals. One such example of this is with the element [Nebulium](https://www.britannica.com/science/nebulium), which was actually the result of this metastable configuration allowing oxygen to emit a completely new spectral line! They're called forbidden lines - more on those in the extended section of this doc! # A Hydrogen Deep Dive Hydrogen. The simplest atom possible, and therefore the example we will be using. In fact, we've even generalised its absorption lines into "series" - named after the people who had the bright idea to stamp their foot down first We'll go over the first four emission lines of hydrogen. Don't try to memorise them - it's a cool party trick, but that is if your party is 'all geek no life'. Here are most of the hydrogen-alpha series - or at least, the ones that are standardised. - **Lyman-$\alpha$ series**: Emission from a higher orbital to the ground state - **Balmer-$\alpha$ series**: Emission from a higher orbital to the second orbital N = 2 - **Paschen-$\alpha$ series**: Emission from a higher orbital to the third orbital N = 3 - **Brackett-$\alpha$ series**: Emission from a higher orbital to the fourth orbital N = 4 To find the energy of the photon emitted by the hydrogen atom (or any atom, to be honest) we can use the **Rydberg Equation** as shown: $E_{photon} = E_{0}\left( \frac{1}{n^2_{1}} - \frac{1}{n^2_{2}} \right)$ Where: - $n_1$ is the lower orbital being emitted to - $n_2$ is the higher orbital the electron falls from - $E_0$ is the ionisation constant (binding energy, for hydrogen it's 13.6eV) - $E_{photon}$ is the energy of the photon emitted >[!Success]- IB Syllabus Woes >This isn't in the IB syllabus - though the IB will still tell you to find the energy! You're going to have to use your basic knowledge of [[Waves Foundations|waves]] to do this, as well as a form of the de Broglie equation $E = hf$. This value is given in [[The ElectronVolt (eV) (Nuclear & Quantum)|electron volts.]] When I did this I found it tremendously difficult to just grasp, so here's an example question I made up just for you guys! >[!Example]- Example Problem - Rydberg Equation >Given a photon has been emitted from a hydrogen molecule with the Lyman-$\alpha$ designation from orbital $N=2$ to $N=1$ find the a) Energy and b) Wavelength of the photon emitted. > >**Solution a)** >We can use the Rydberg Equation to find the energy of the photon in eV: >$E_{photon} = E_{0}\left( \frac{1}{1} - \frac{1}{4} \right)$ >Giving us: >$E_{photon} = 10.2eV$ >To carry this value onto the next question (where we use the [[Waves Foundations#Energy of a Wave|de Broglie wavelength]]) we convert this value into joules: >$E_{photon} = 10.2 \times 1.609 \times 10^-19 = 1.641 \times 10^{-18}$ >**Solution b)** >Let's plug this value into the de Broglie energy equation, as shown: >$1.641 \times 10^{-18}J = \frac{hc}{\lambda}$ >Where $h$ is the Planck Constant and $c$ is the speed of light. A quick google search gives a value of $6.626 \times 10^{-34}$ for $h$ and $3 \times 10^8$ for $c$. Now all we need to do is substitute these values... >$\lambda = \frac{hc}{E}$ >$\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.641 \times 10^{-18}}$ >$\lambda = 1.211 \times 10^{-7}m$ >Well done! You're past this now. Fire is actually a direct cause of excitations! Let's take a look at the diagram first... ![[Pasted image 20230908112604.png]] Fire lets out light. When an electron (like the one pictured above) collides with one that it can absorb (belongs to its absorption spectrum) it releases a photon - giving different fires their distinctive hues. Why do you think Uranus or Neptune are both blue...? >[!Example]- Example Question >A hydrogen electron emits a photon of energy from the Balmer Series (n=3) to the ground state, emitting in the visible red. This photon has a wavelength of 486.2 nanometres. > >*a) What is the energy of the wave emitted?* >*b)* *Where in the spectrum is this wave?* > >**Solution:** # orbitals - geometrically write orbitals as $1S^1$ --> meaning there is one electron in the S orbital electrons in an orbital can have spin up/spin down: therefore there are two+ electrons that can fit in one orbital rules of orbitals: 1) low energy orbitals filled first then high-energy orbitals (so S orbitals are filled first then the p orbitals) e.g nitrogen orbital configuration: $1S^2 + 2S^2 + 2P^3$ Here we can see that each of the S orbitals have been filled. This means the higher-energy level P orbital must be partially filled with the electrons. Chromium - funny example --> electronic configuration $[Ar]\ 4S^1\ 3d^5$ - as then all the 'blobs' in the d orbital are filled by one electron (plus it's lower level than the s orbital!) ## The S Orbital this is a big sphere ## The P Orbital 3 blobs in a triangular-ish shape ## The D Orbital clover-esque shaped orbital. 5 shells! ## The F Orbital (Extended) pain. i'll do this later. i swear. # Extended - Forbidden Lines We're going deeper into uncharted territory - how exciting! # Nucleon Shells By this point, you should be familiar with how electrons work, with the orbitals and all that. However, would you be surprised to learn that even *nucleons*, protons and neutrons, obey the same shell laws as they would have before? You know, when I learnt this I was also stupefied - we've been taught all our lives that a nucleus is a blob, but nope - [[A Beginner's Dive into Quantum Principles (Nuclear and Quantum)#Pauli Exclusion Principle|Pauli's Exclusion Principle]] applies to even this. Delighting! Instead of the standard octet shell patterns, these nucleons follow a different set of rules, where the most stable members of their *Check out my notes on subatomic particles:* [[A 101 into Quantum Particles (Nuclear and Quantum)]]