# Intro - The Quantum Atom With Max Planck's discoveries came the opportunity for many to redefine how we saw atoms and electrons as a whole. From the standard model particles to the simplest of hydrogen molecules, there's no pre-19th century atomic theory that's been uprooted by quantum mechanics. This article will focus on the **Bohr Model**, the model for the proton as a result of quantum mechanics. There is also a section about the **Rydberg Equation** - extremely useful in many fields! # Deriving the Model Naturally, to model the simplest of the atoms, we'd use hydrogen. It's like an [[Energy along an Orbit|orbit]], where in our model, we've just got a single electron orbiting a proton. ![[Pasted image 20240404155122.png]] *The Proton and the Electron - a tale as old as time!* The *Bohr Radius* we define as the radius of the atom. Since the mass of an electron is so much smaller than that of a proton ^[1], we have to take the radius by setting the charge as the discrete quantity of the electron. In our idealised setting, this orbit is circular. As a result, we can find the radius at which the electron is orbiting by finding the centripetal force of the electron: $\frac{e^2}{4\pi \epsilon_{0}r^2} = \frac{m_{e}v^2}{r}$ As this orbit is in equilibrium, a coulombic force must be present to create the equilibrium. Remember, $r$ is our approximated atomic radius and a scalar, so: $r = \frac{m_{e}v^2r^24\pi\epsilon_{0}}{e^2}$ Which is: $r = \frac{(m_{e}v^2r^2)^24\pi\epsilon_{0}}{m_{e}e^2}$ That numerator is equal to our angular momentum. According to [[Intro to QM (Nuclear & Quantum)|Quantum Mechanics,]] angular momentum must come in discrete packets. In other words, it has to be a multiple of the [[A 101 into Quantum Particles (Nuclear and Quantum)#A Dive into the Quantumverse|reduced Planck Constant]], by the Heisenberg Uncertainty Principle (LINK COMING LATER): $mvr = n\hbar = \frac{nh}{2\pi}$ If we plug that in, we get another expression for $r$: $r = \frac{4\pi\epsilon_{0}n^2\hbar^2}{m_{e}e^2} = \frac{n^2h^2\epsilon_{0}}{\pi m_{e}e^2}$ This also explains why there are orbitals - the electrons must be at discrete energy levels, given by the $n$. Alright, now commute this to memory, because the next equation's just as useful. For hydrogen, substituting the values for the ground state $n = 1$ gives us: $r = 5.292 \times 10^{-11}m$ # Finding Spectral Line Energies - The Balmer-Rydberg Equation In astronomy, spectral lines are practically everything. They tell us volumes about interstellar objects, from what they're made of to how much of each 'thing' they're actually made of. However, orbitals mean that atoms in an astronomical, high-energy environment sometimes don't fully ionise - they just end up in a higher-energy state. As a result, it's become important for us to find the specific energies of each of these jumps in order to classify interstellar bodies.^[2] As a result, we use the **Rydberg Equation** to find the energy of the photon emitted as the electrons move between orbitals. $E_{photon} = E_{0}\left( \frac{1}{n^2} - \frac{1}{q^2} \right)$ Where: - $q$ is the orbital the electron falls to - $n$ is the original orbital of the electron - $E_{0}$ is the ionisation energy of hydrogen - $E_{photon}$ is the energy of the photon emitted. However, this only really works for hydrogen. To branch out to other molecules, we've got to go back to our Bohr Model. The True Rydberg Equation is given as: $\frac{1}{\lambda} = R \left( \frac{1}{n^2} - \frac{1}{q^2} \right)$ Where: - $\lambda$ is the wavelength of the light emitted. - $R$ is the *Rydberg Constant*, given in units of $m^{-1}$. - $n$ and $q$ retain their values, being the orbital the electron falls to and the original orbital of the electron. >[!Abstract]- Proving the Rydberg Equation >Under [[Power, Work Done and Energy (Mechanics, %%marked for rewrite + NEEDS LOADS OF DIAGRAMS, ASK EXTERNAL HELP!%%)#Mechanical Energy - Conservation of Energy|Conservation of Energy,]] the total energy of the electron is going to stay the same even as its orbital (or quantum state) changes. In our case, the gravitational potential energy is negligible so we take the electric potential instead. > >$E = -\frac{me^4}{8\epsilon^2n^2h^2}$ >Now imagine that the electron falls to a lower orbital. The total energy at this orbital is: >$E_{q} = -\frac{me^4}{8\epsilon^2q^2h^2}$ >Where $q$ is any arbitrary number. When a photon emits light, the energy difference between the two orbitals is equal to the energy of the photon emitted. As a result: >$E_{q} - E = \frac{me^4}{8\epsilon^2h^2}\left( \frac{1}{n^2} - \frac{1}{q^2} \right) = hf$ >Remember that $f/c$ = $\frac{1}\lambda$. Therefore, our final step is to isolate $1/\lambda$, as shown: >$\frac{f}{c} = \frac{me^4}{8c\epsilon^2h^3} \left( \frac{1}{n^2} - \frac{1}{q^2} \right)$ >$1/\lambda = \frac{me^4}{8c\epsilon^2h^3} \left( \frac{1}{n^2} - \frac{1}{q^2} \right)$ > >This is the Rydberg Equation. Often times, to simplify the equation, we just write the constant in front of the fractions as: >$1/lambda = R \left( \frac{1}{n^2} - \frac{1}{q^2} \right)$ >Where $R$ is the *Rydberg Constant*, $1.09677 \times 10^{7}/m$. >[!Success]- Proving the Generalised Equation >$\frac{1}{\lambda} = R\left( \frac{1}{n^2} - \frac{1}{q^2} \right)$ > >There's another definition for the Rydberg constant, where it is given as: > >$R = \frac{E_{0}}{hc} = \frac{me^4}{8c e^2h^3}$ (i need to ask why this is true) >Multiplying $\frac{1}{\lambda}$ by $hc$ gives us the energy of the photon, hence why it works! 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]- Rydberg Equation Example Problem (Proof of Concept) > >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... >$\lambda = \frac{hc}{E}$ >$\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.641 \times 10^{-18}}$ >$\lambda = 1.211 \times 10^{-7}m$ 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...? [^1]: About $\frac{1}{1840}$ of the mass of a proton, to be exact. [^2]: See [[Nebulae (Astro)#H II Regions|H II Regions]].