# Intro - A Brilliant Mind Einstein wasn't always the brilliant scientist we know and love today. Everyone has to start from somewhere, and for this young, plucky genius it was at a patent office. Today, we'll be diving into the one thing he won a Nobel Prize for - and no, it's not [[General Relativity - Foundations|general relativity]]! As the title might insinuate, it's the **Photoelectric Effect!** # The Photoelectric Effect - Basics By all accounts, the photoelectric effect refers to the displacement of the **ejection of an electron from a shiny metal surface by a photon incident on the metal surface.** ![[photoelectric effect diagram.png]] *Fig 1: A diagram of the photoelectric effect in action - a photon is incident on the metal surface - an electron is released.* But why the shiny surface? - **"Shiny surface"** MUST be true for all instances in the photoelectric effect as *we must account for oxidation in the metal*, as: - the photoelectric effect must only occur in a **metal** as a metal has delocalised electrons which can be removed. - if the metal oxidises, an ionic lattice is formed in which there are no free electrons, so electrons cannot be ejected in the oxidised metal The kinetic energy of the ejected electron $E_k$ can be written as the formula: $E_{k} = hf-\phi$ Where $hf$ is the *energy of the photon*, and: - $h$ is the *Planck Constant*, $6.633 \times 10^{-34}$ - $f$ is the *frequency of the photon* - $\phi$ is the *workfunction*, defined below. >[!Abstract]- Einstein's Photoelectric Effect Proof If the intensity of the beam increases, that means that more photons are hitting the same area of metal every ## The Workfunction Strictly speaking, the workfunction refers to **the minimum energy required to liberate SURFACE electrons from a metal!** Note how it's only the surface electrons that are removed, so if the photon's energy is only just above the workfunction's, electrons deeper within the metal surface can't be ejected. So in other words, since the electrons in the lattice structure of the metal have different energies, **the kinetic energy of the electrons emitted may vary as different energies are required to eject electrons at different positions in the metal!** # Stopping Voltage - A More Practical Application It's also possible for us to set up a [[Capacitance (ElectroMagnetism)|capacitor]]-esque system using the photoelectric effect, as detailed in the diagram below: ![[stoppingvoltage.png]] Here, light is incident on a metal 'emitter' plate, so electrons are ejected from the metal to the negative plate. This, in turn, creates a photoelectric current travelling from the negative to the positive terminals, as per [[Power, Work Done and Energy#Mechanical Energy - Conservation of Energy|conservation of energy]]. As the plates are oppositely charged, there will be an [[Electric Fields Basics (ElectroMagnetism)|electric field]] in between them. See that cell in the circuit? If we increase the p.d across that cell, then what happens? You'll notice that in this case, the emf across the cell will lead to a current that acts in the **opposite** direction to the photoelectric current. Now, something very odd will begin to happen on the emitter plate; as electrons are continually ejected through the photoelectric effect, then accelerated towards the collector plate by the electric field, the positive charge on the emitter plate will increase. Therefore, electrons will actually be attracted BACK towards the positive charge, which, when no electrons make it to the negative plate, result in there being no measured photoelectric effect. This, however, requires a large-enough potential energy So, in other words, **the electric potential where the photoelectric effect stops** is known as the "Stopping Voltage" - when the electric potential energy of the electrons to the positive charges on the plate is equal to their ejected kinetic energy: $V_{S} = \frac{E_{K}}{e} = \frac{hf - \phi}{e}$ Where $e$ is the elementary charge on the particle and $V_S$ is the stopping voltage. >[!Example]- IB Specimen Paper 1 2025 Edition >Monochromatic light of frequency $f_1$ is incident on the surface of a metal. The stopping voltage for this light is $V_1$. When the frequency of the radiation is changed to $f_2$, the stopping voltage is $V_2$. > >What is the quantity $\frac{V_{2}-V_{1}}{f_{2}-f_{1}}$ equal to? > >**Solution and Explanation:** >Since we know $V_S$ is $\frac{hf}{e}$ (where the workfunction for this case subtracts out), we can just write: >$\frac{V_{2}-V_{1}}{f_{2}-f_{1}} = \frac{hf}{fe} = \frac{h}{e}$ # What's Next? Return to the home page here: [[Home]] Return to the Quantum page here: [[Nuclear & Quantum Contents (Nuclear & Quantum)]]