let's go over the whole of AQA quantum physics and electromagnetic radiation the energy of an individual Photon which is a particle of light is given by e equal to HF where H is just planks constant the value of which is just 6.63 multiply this by 10 the^ of - 34 Jew seconds since the speed of light is just equal to frequency multipli by the wavelength I.E the frequency itself is just equal to the speed of light divided by Lambda we can substitute this back into this equation and what we're going to get is the second equation that the energy can also be expressed as HC over Lambda and those are the two equation that we're going to be using a lot in this portion e to HC over Lambda and E to HF now why did I start with this well the photoelectric effect is all about an experiment that actually proves the particle nature of light so what is the photoelectric effect well it occurs when a photon of sufficient energy releases an electron from a metal surface imagine that we have this metal surface let's draw one over here so here are some electrons that are negative and if we have a photon which is a particle of light with sufficient energy then this electron will be able to be released with a certain kinetic energy because the energy of the photon is proportional to the frequency we have a minimum frequency of a photon that's required to release an electron from the surface of a metal and this is known as the threshold frequency which is a really important definition the energy associated with that is known as the work function which is the minimum energy of a photon required to release an electron from the surface of the metal I think life would have be a little bit easier if they decided to call it the threshold energy but for historical reasons it's known as the work function the symbol for the work function is the Greek letter F typically and because e is equal to HF then the work function will be equal to H which is Plank's constant multiplied by the threshold frequency let's just call it FN uh the threshold frequency by the way is often given this symbol F not we also need to talk about the photon explanation of the threshold hold frequency so according to clco Theory the threshold frequency shouldn't actually exist if light was a wave low energy intense light should be able to release electrons what do I mean by that if I have a metal plate with some electrons if I have very intense low energy light meaning that it has a very large wavelength and very intense means a lot of power incident per unit area area so according to wave theory this should be able to release electrons but in practice no matter how long we wait for low energy very intense light does not release electrons from a metal plate the emission only depends on the energy of the incoming Photon for instance if we were to have just an individual Photon of let's say UV light which has a high frequency and a higher energy this will be able to release an electron from the surface of the metal this is because there is a onetoone interaction between a single incident Photon and a single electron that's about to be released and the emission depends only on the incident frequency SL incident energy of the photon one example of the photoelectric effect is the gold leaf electroscope so this is done by charging a gold leaf electroscope which is this device over here and it's often done with a plastic rod so charges accumulated on the surface and due to electrostatic repulsion the gold leaf Rises this because there's electrons here and there's electrons here and the two repel causing the leaf to rise similar to the way hair Rises if you were to touch a vandag graph generator and now imagine that we bring a UV lamp that's releasing some UV photons well if that happens there will be emission of electrons from this surface this means that the number of surplus electrons that will remain on the surface will decrease with time I the negative charge on this plate will end up decreasing the way we explain this in exam questions is that this occurs because there's a minimum amount of energy that's required to release an electron and photon must Supply this energy in a single interaction the UV photons can do that because their energy is just given by E is equal to HF and this is greater than the work function so electrons can be released and now imagine that we bring a UV lamp that's releasing some UV photons well if that happens there will be a mission of electrons from this surface this means that the number of surplus electrons that will remain on on the surface will decrease with time IE the negative charge on this plate will end up decreasing the way we explain this in exam questions is that this occurs because there's a minimum amount of energy that's required to release an electron and photons must Supply this energy in a single interactions the UV photons can do that because their energy is just given by the UV photons can do that because their energy is just given by e equal to HF and this is greater than the work function so electrons can be released and how do we deal with the photoelectric effect mathematically well we have the photoelectric effect equation this is actually just simple energy conservation HF is the energy before which is the energy of the incident Photon some of that energy has to be purely equal to the work function in order for the electron to be released and the remaining energy is the maximum kinetic energy of the electron that's emitted I.E on the left we have the energy before and then on the right we have the energy afterwards please note that HF is not the only equation that we know for the energy of the incident Photon we could write exactly the same equation with HC over Lambda is equal to 5 + e k Max and sometimes in a question we might directly be given the energy of the photon itself typically it might be given in electron volts but we'll do the conversion in a minute which is going to be equal to 5 + e k Max like that and these are all the possible ways of writing the same equation and let's do an example problem so we have a surface that has a word function of 2.3 electron volts and light of photons frequency 8.1 * 10^ 14 Hertz calculate the maximum kinetic energy and they want the answer in Jewels the first thing to note is that the work function is given in 2.3 electron volts and we want the final answer in Jewels so we're going to need to do some conversions but before we do that let's just write down our equation which is that HF is equal to the work function plus the maximum kinetic energy now we're looking for the maximum kinetic energy so we can directly rearrange for that and then we're going to get HF take away the work function okay so H planks constant is just 6.63 * 10us 34 our frequency already given in base units so that's going to be 8.1 or in standard units time 10 ^ 14 Hertz take away the work function which is 2.3 electron volts to Jews we multiply by 1.6 * 10-9 side note but if I can see the E I multiply by it and this is a very useful way of remembering how to convert from jewels to electron volts and vice versa if I can't see the I divide by it okay so this here is going to be equal to if we put that into a calculator to be around 1.69 multiplied by 10 ^ of - 19 Jews we can also take the problem further by calculating the maximum speed of the emitted electrons if we have their maximum kinetic energy we have their maximum speed why is that because remember EK Max is also just equal to a half M v² we already have that number so we can just rearrange for V which is going to be equal to the square < TK of 2 * the kinetic energy I'm just going to write e subscript k for the kinetic energy without a max divide that by m okay this here is going to give me 2 times our answer from before which was 1.69 10-19 the mass of an electron is given which is Tiny 9.11 * 10^ of - 31 kg giving us approximately 6.1 * 10 ^ 5 m per second and now let's talk about photo cells as a demonstration of the photoelectric effect so this typically done in a surface like this we have a photoemissive surface we have an electrode and then this is connected to the positive end of the terminal and we have a microammeter like so what a variable power supply here photo cells are essentially photoemissive surfaces and if we were to shine some light with some photons of energy higher than the work function then electrons are going to be released these electrons are then detected within the circuit by this micro meter and that just detected as current now if the intensity of the incident radiation is increased more photons per second are incident on the surface therefore more electrons are released per second from the surface and remember because current is equal to the rate of flow of charge if we have more electrons released then essentially we have more current as well more charge flowing through per second I.E the current will actually increase and what is the stopping potential well it is simply the potential difference that needs to be applied to the Circuit that will stop the electrons with maximum kinetic energy when that happens the current in the circuit will drop to zero here's a little example so let's say we have some photons that strike the surface and let's say that the surface has a work function of three electron volts if the photons have a frequency of 11 * 10 14 Hertz calculate the stopping potential our first job is to calculate the maximum kinetic energy of the electrons for that we're simply going to use the good old equation that HF will be equal to the work function plus the maximum kinetic energy so to get the maximum kinetic energy we need to do HF take away the work function so this here will be equal to Plank's constant 6.63 * 10 to^ - 34 multiply this by the frequency what is that 11 * 10 ^ 14 Hertz now we're going to take away the work function which is given as 3.0 electron volt * 1.6 * 10^ -19 and if we to plug this into a calculator we are going to get around 2.49 * 10 to the power of -19 Jew so how much voltage do we need to stop these electrons well if they have that much kinetic energy we can simply use the fact that EV is equal to a k Max so the voltage required will simply be equal to the kinetic energy in the maximum kinetic energy in Jewels divided by the elementary charge in other words 2.49 * 10 ^ -19 / by the elementary charge which is 1.6 * 10 ^ -19 which is around 1.56 volts or about 1.6 volts up to two significant figures and this here is our stopping potential please note that if we had a problem in which we were given the maximum kinetic energy in electron volts let's make up some numbers I don't know let's say the maximum kinetic energy was 5.0 electron volts because of the way the electron volt is defined the stopping potential for this will just be 5 Vols sometimes times we also get graphical questions based on the photoelectric effect equation were to plot a graph of the maximum kinetic energy against frequency what would this graph look like well because the photoelectric effect equation is HF is equal to the work function plus e k max if we were to rearrange for whatever is on the y- axis in this case this is the maximum kinetic energy what we're actually going to get is is HF take away the work function so writing just directly underneath the equation of a straight line we get that if this is on the y- AIS IF frequency is on the x axis what is left for our gradient is Plank's constant H and our negative intercept is going to be the work function in practice the graph will look look something like this it can be extrapolated like that with this point here being the negative work function this point here is the threshold frequency Additionally the gradient of the graph is simply Plank's constant H what would this graph look like if we had a higher work function well let's draw this in a different color I'm just going to use yellow if that's the case first of all our threshold frequency is going to be higher uh the gradient though should be exactly the same because that's still just given by planks constant and our negative work function is going to have a more negative value which is going to be somewhere over here so this graph here will represent a higher work function the gradient can never change because this is planks constant and if you change planks constant you probably get a Nobel Prize moving on to collisions of electrons with atoms we need to know about the ionization and excitation in a Mercury fluorescent tube please note that typically in exams we get Mercury but you may potentially get a different gas so initially electrons pass through the tube collide with electrons in the Mercury atoms they transfer energy and the Mercury electrons go to higher energy levels the excited electrons then deoxide to lower energy levels when they do that they emit a photon of equal energy difference to the energy difference between the energy levels for instance let's say we had an electron here then it might have been excited go on to this energy and then it relaxes it de exites and as it goes down it emits a photon which is equal to the the energy of which is equal to the energy level we could calculate its frequency using Delta e is equal to HF if that appears in a question so initially these photons are emitted in the UV range however the coating of the Mercury tube absorbs UV photons and electrons in the coating are excited to higher energy levels yet again however they come back to a previous lower energy level and then they emit a photon equal to that lower energy difference and hence they emit a lower energy visible Photon they are typical in the visible range this can appear in terms of a six marker or maybe an entire question which is split into a few different um Parts but generally speaking writing these points should virtually guarantee full marks so let's talk more about energy levels and Photon emission line Spectra are evidence for transitions between discrete energy levels in atoms what does that mean well here are the energy levels of electrons within an atom when an electron absorbs a photon so let's say we had one in the ground state over here what would would happen is it will go up in energy level and the energy of the photon has to be equal to the difference in energy levels in this case this would be 10.2 electron volts however when electrons de exite and go down an energy level they would release a photon equal to the energy level difference so for instance if an electron was to go down from this state back down to -3.6 it's going to release a photon of that energy we could even calculate the frequency of that Photon if we wanted to we'll just use Delta e is equal to HF for Delta e just worry about the actual magnitude so that will be minus 13.6 take away 3.4 and then the magnitude of that is just 10.2 electron volts which is equal to HF meaning that the frequency of that particular Photon would be 10.2 let's convert that the electron volts to juwel so multiply this by 1.6 * 10 ^ of -19 / by Plank's constant 6.63 * 10^ - 34 JW seconds giving us around 2.5 multip by what is it 10^ 15 Hertz sometimes the questions ask about the wavelength and if that's the case we're just going to use that Delta e is equal to HC over Lambda very often we would actually use the Delta e to be E1 take away E2 but the M schemes are not super consistent with this we want to avoid any negative answers so my advice is just to consider the absolute value of that and this is part of the remarkable nature of quantum physics only certain energy levels are allowed the electron can be in this location or it could be in this energy level or this energy level but it absolutely cannot be anywhere in between if it could once we have a look at the Spectrum essentially running the light from a mercury lamp for a defraction grating we should see a continuous Spectrum which essentially is like a rainbow but we don't see that we only see discrete energy levels that are going to relate to certain transitions so perhaps this one here might give away a photon of a certain frequency and a s Photon of a certain wavelength which might relate to this one over here then the other possible transition let's say going from here to here might give this one across here maybe this one here will correspond to this line and notice how these are discrete individual lines meaning that the energy levels within the atoms have to be discrete as well and here is something really important that comes up a lot in exams why are energy levels negative well energy levels are negative because in order to remove an electron energy has to be supplied F one's free an electron has zero energy and you can see all those negative numbers as they rise they get closer and closer to zero and now let's talk about wave particle duality so we have two sets of experiments the photoelectric effect suggests that electromagnetic waves I light have a particle nature however it was also discovered that electrons and particles in general can have a wave nature via electron defraction so the actual experiment is shown here you've probably seen this in the lab you don't have to remember all the different components in general we have a filament and then we applying quite a high voltage and this over here is the Electron Beam it passes through a graphite Target why do we use graphite well the spacing between the atoms is such that it would act as a defraction grating for the electron now if the electrons have pure particle Behavior they should produce just a pure spot with some particles scatter randomly so if this was true we would get sort of something basically like this this where most of the particles will maybe one will land over here here here maybe one will land over there but in general it will be just a very strongly centered pattern towards the center this is not what we see we see a pure defraction pattern we see bright concentric Rings which actually show pure wave property I.E defraction the bright Rings occur where constructive interference occurs and this quite similar to the defraction grading formula which is that D sin Theta is equal to n Lambda so particles actually have some wave Behavior if particles have wavelike Behavior then there should be an equation for the wavelength and that is the the Broly wavelength equation that it is equal to H which is Plank's constant divided by MV which is the momentum of the particle we can also use this equation to actually explain what would happen to the diameter of the Rings if we were to increase the voltage remember the equation for the wavelength was that the wavelength was equal to h / MV where this here is a lowercase V for the speed and this is an uppercase V for the voltage so if we were to increase the voltage the electrons that are traveling towards the graphite are going to have higher speeds so this V here will also increase because of that the de brly wavelength would decrease and the distance between the concentric Rings is actually proportional to the wavelength so this distance here would decrease let's write these down another way to think about it is that the wavelength will also now be different to the atomic spacing or more different therefore there's going to be less defraction because of it let's dra another example so we have a particle that has a kinetic energy E and A de brogi wavelength of Lambda what is the the broad wavelength if the kinetic energy of the particle is doubled so first of all let's write the expression for kinetic energy simply that this is equal to a half MV s so if I was to double the energy how will the speed actually change well V is equal to the square < TK of 2 e / M meaning that V is just directly proportional to the square root of the kinetic energy so if if we were to get another factor of two in here V is just going to go up by a factor of < tk2 but since the wavelength is given by H over MV and H and M are constant well if V increases by a factor of < tk2 then the wavelength should decrease by a factor of < tk2 therefore the new wavelength will actually be equal to the original wavelength let's call Lambda 2 will be equal to the original wavelength / by < tk2 so this was the whole of electromagnetic radiation and Quantum phenomena what you should definitely do next to give you the best chance of improving your grade is have a look at this next video which is the Hall of waves the next portion of the topic click right over here enjoy