All right. So the last unit we're covering reviewing is modern physics, right, for AP physics 2. And so there's a lot of various topics. It's not very very quantitative or analytical. So it's primarily very conceptual because a lot of these topics are pretty advanced. So let's talk about atomic structure and quantum state. So atoms have a nucleus made of protons and neutrons. They have orbiting electrons. Number of protons indicates the property of the material. That's where you use a periodic table of elements. Atoms can have different number of neutrons. call them isotopes. Mass number is the number of protons plus the number of neutrons. And the strong nuclear force is the fundamental force that allows neutrons and protons to be bound together stronger than the electrostatic force. So the bore model of the atom, electrons orbit the nucleus like planets orbit a star. So it's like thinking about like the nucleus, the protons and the neutrons and the electrons orbiting it. That's a classical model. It's not how the electrons really work, but it is an interesting model. You're basically using just the electrostatic force of attraction and centrial acceleration as a model. But the more wave the standing wave model of electrons is the more modern version. Right? Electrons orbit in different energy states. They are discrete states. They are quant right? They can only jump between specific states. So it's like if you were to think of it as like orbiting, they can only orbit at certain locations whether rather than being like a continuous spectrum of various orbits, right? So this is where we call it quantum uh states. To move into an higher energy state, in order for an electron to gain energy, it has to absorb a photon. And to go into a lower energy state, the electron must release energy by releasing a photon. So photons contain energy. Those are light particles. Uh the binding energy for an electron to be bound to an atom is the number amount of energy for the electron to escape. So if you give the electron, you hit it with a photon with the greater than the binding amount of energy, then it will escape, right? um energy classically right we normally measure energy in jewels but we some we will sometimes use electron volts to measure energy for very small energy levels and there's a conversion I'm pretty sure the equation sheets have conversions for there if you want uh mass energy equivalence is that mass itself contains energy so E= MC² where M is the mass of object E is the energy and C is the uh the uh speed of light so C^2 is a very very big number uh in a reaction you can compute how much energy is released are stored by looking at what we call the mass defect. That's the mass that's lost in it. So for example, if you have a nuclear fusion, you take two hydrogen atoms, they form a helium atom, but you lose a little bit of mass. That mass that's lost called a mass defect. Um multiply by C^ 2 is how much energy is released in the reaction. Uh so light is made up of particles. Photons, we still think of them having wave properties. Photons have an energy equal to HF which H is so they don't have any mass but they still have energy HF where F is the frequency of the photon right um they have a wavelength the bro's wavelength for that light can be found by taking plank's constant H and divided by the momentum and the momentum of the photon you can rearrange this is given by this now careful momentum of photon not MV because if you do MV the momentum would be zero so we use this to talk about the energy the momentum of the photon which is non zero. So it has a non-zero amount of photon despite the mass being zero. So we use this for the equation for the photon. Um when you're talking about the momentum of a photon photoelectric effect um this is the thing that basically says that um if you shine this is what showed that light was made up of photons, right? Because what what ended up happening is the photons um themselves needed to have a certain amount of energy. So you shine light on a metal, it at a certain energy, if you get the frequency high enough of the photons, it will eject electrons. That that's because the photons hit the binding energy amount um and hit the electrons and then they can escape. So it's not about how much light you shine, it's specifically about the frequency of the light or the amount of energy stored in the photon, right? The kinetic energy of the escaping um the escaping electron is given by the energy in the photon minus the work function. The work function is almost think of it like the binding energy. It's like how much energy it's required to cause the electron to escape and any excess energy then goes into kinetic energy of the electron. Right? Um so because this is the key thing this is why it was such an important this is what why uh it was considered groundbreaking is because it's only the photon energy is affected by frequency. shining brighter light which is more photons at the same frequency is even though you're sending more energy overall it's not interacting with the electrons to cause it to escape right it will not change the amount of energy given to the electrons it has to be the energy of the photon of the individual photons themselves that affect it compton scattering is when photons interact with a free electron so energy and momentum is transferred to the electron so the photon loses some energy losses some momentum and frequency and wavelength they all change. So what uh what what ends up happening is because you give some kinetic energy to the electron. So it hits it the electron scatters off the electron gains kinetic energy. So the photon must lose some energy and that means lower frequency or higher wavelength. Um the momentum is going to change because of scattering because it's still overall the momentum is conserved because remember photons have some momentum. Um and then the we talked about the frequency and the wavelength. The formula you're given, this is deriving it just based on using conservation of energy and conservation of momentum. But the change in the wavelength is given by h / mass of the electron times the speed of light* 1 - cosine theta where theta is the scattering angle here. Okay. So what you can do is figure out what the change in the photon is and that's going to give the wavelength. That's going to give you the change in the energy. So when it scatters it the wavelength is going to get longer because it loses some energy in there, right? because the energy in the photon is proportional to its frequency. Black body radiation is that um matter when it's when it's warm will release some thermal energy into electromagnetic energy, right? So it like lights up. So the warmer the temperature, the higher the energy the photons are emitted, right? Because warmer temperature means more energy. So the warmer the temperature, the higher the temperature, the higher the energy of the photons. What that means that's higher frequency or smaller wavelengths, right? Right? And so that's why the wavelength is inversely proportional to the temperature. The greater the temperature means the greater the energy of the photon, which means the smaller the wavelength. Smaller wavelength is higher frequency, right? Because wavelength times frequency is speed of light, right? So as the frequency goes up and the energy of the photon goes up, then the wavelength goes down. The rate at which energy emitted is given by this. Um so if you have a if you're talking about like it's the surface area times a constant times temperature. The key thing is just to understand the fourth power temperature on that dependency there. Uh last part is nuclear decay, right? And so there's different nuclear decays and they're governed by the strong nuclear force again or that's the interaction between the protons and the neutrons. You have three primary decay mechanisms. One's an alpha decay where an atom meets a helium nucleus which is two protons and two neutrons. So that is going to change significantly the mass of the atom. a beta decay or you have a beta positive where a proton's converted to a neutron and a posetron where a positively charged electron is emitted or you have a beta uh negative decay which is a neutron is converted to a proton but in order to keep charge conserved an electron is committed right so this is a positive and a negative so the net charge net change is zero and same thing here because you go from positive to neutron you're going to lose a charge so you have to have a positive charge to offset it in order to say that the because you can't create charges Right? Charges cannot be created or destroyed by conservation of charge. Um sometimes a nutri in in a beta decay a nutrino has no charge and very negligible mass is emitted to conserve momentum for the only purposes of doing momentum conservation there. You don't need understand much more than that. Gamma decay is that's where electrons are just ch are changing between states and that releases a photon. So these two change the nucleus of an atom, right? By either losing protons and neutrons or converting protons into neutrons. This one just releases energy and that's it. Um during all of these processes, momentum, energy, mass, and total charge has to be conserved during the uh these decay processes. All four of those things have to be conserved during that process. Decay is a random process. The number of nuclei random remaining in the sample is given by this equation where this uh lambda is the decay constant. N0 is the initial amount of material. So we've done half lives is where we talk about half lives where it's like the amount of time until you have half of the material remaining. Um uh is you know t1 half that's our halflife uh factor there. And so let's go look at some multiple choice and free response questions here. So at time t equals 0 sample of radioactive material with decay constant lambda has n0 nuclei. At time t= t1 there are n1 nuclear remaining. At a later time t2 there are n2 nuclei remaining. Which of the following croc rates n2 to n1? So n is going to be n0 e to the lambda t. Then we know n1 is just going to be n0 e to the negative lambda t1. And we know n2 is just going to be n0 e to the negative lambda t2. So we want to know like how to solve for n2 instead of terms of n1. What you can do is just solve for n0 here. So you get n1 over e to the lambda t1 is equal to n0 and plug that into here. So you get n2 is going to be n1 over e to the negative lambda t1 e to the negative lambda t2 because these are both the same base e. You're just going to subtract their exponents. So it's going to be negative lambda t2 minus t1 if you do that. And so that's why you're going to get something like that. And that is going to be and yeah uh because if you pull it out you subtract t2 minus t1. So, it's going to be uh let's see, it's not C. Oh, yeah. It's A here. There. There's no N0 here, right? Because we replace the N0 with this quantity here. Okay. A black body temperature T0 emits a spectrum of light with peak wavelength lambda 0. The rate at which energy emitted is P 0. The temperature of the black body is then changed to the new peak wavelength is lambda 0 over2, which if correctly indicates the rate at which energy is emitted. So remember the power is going to be um area time the surface area times that T 4th that we got that T 4th dependency and you also have to remember that the maximum wavelength is like some constant over temperature. So what happens is if you are going to change the maximum wavelength by 1/2 then that means the temperature has to be multiplied by two on the denominator right in order for the two equations to be balanced. If we multiply this by two, then this side's going to be multiplied by 2 4 * 16. So we're going to get D there. Okay. So we have an electron in an atom is initially bound state with negative E1. It absorbs a photon transition to excited state negative E2. So let's think about it like we have E1 and we're g it's going to it's going to absorb a photon. So it's going to jump up to E2, right? So it absorbs a photon and after a brief time the electronically drops to a lower energy state with negative negative E3 and so so we'll put a negative sign here if you want and then it's going to drop down to E3. Could be above or below but we're going to drop E3. It observed that the photon emitted during the second transition has a higher frequency than the photon absorbed during the first transition. Is the magnitude of the energy change during the second transition greater than less than or equal to the magnitude of the energy change in the first transition? Well, because it has a higher frequency, right? It's going to have a it's going to have more energy, right? So, this this emitted photon is more energy. So, it's going to drop down to here like this and become negative E3 like this. So, this is kind of the picture because you have a bigger energy difference, right? Which means more energy of the photon emitted here. And so, then um um because you emitted a larger photon or more energy photon, it's going to have higher frequency. So that's consistent. So it's going to be greater than and we want to do that qualitatively. And we just want to say it just like we did. We said um a higher frequency photon means that it was a larger amount of energy, more energy, larger energy photon. And that requires a greater uh energy change state change, right? Because the energy state change is going to is what impacts the um the energy of the photon. Derive an expression for the frequency of the absorbed emitted photons. Express your answer in terms of E1, E2, and E3 and fundamental constants. Okay. So um we know the energy of the photon is going to be lambda times the frequency. Sorry, it's going to be uh uh planks content times the frequency. Now for the first photon the um the emitt you always do if you want the energy of the photons it's going to be larger minus uh smaller. So it's going to be uh um let's see E2 minus E1. So E1's like a bigger number in magnitude but um like we're think because these are negative is going to that's the energy of the photon. This is the emitted photon. This is the H of the frequency uh emitted. No, sorry. This is the absorbed one. This is the first state because we're going this state. We're absorbing a photon. So, we're going up absorbed. So, F absorbed is going to be uh E2 plus E1 over H. Okay. And then we do the similar thing for the one that is emitted. Right. This is going to be um E2 minus E3 is going to be H * the F emitted. So the F emitted is going to be negative E2 plus E3 over H like that. Okay. Uh rank the values of E1, E2 and E3. And we look at this diagram here. So um it might help just because the negative sign is a little bit confusing in this question. Think of this as like -10,8 and -6. So then you think of it as like E2 is 6, E1 is 8 and E3 is 10. So then if you want to just rank those numbers um you would just say um E3 then E1 then E2, right? So it' be E3 is greater than E1 greater than E2 like that. Okay? So sometimes if you're symbolically you can't rank them, just putting numbers in here can be helpful like this, right? Okay. Um let's look at this one. Experiment. A monochromatic light of frequency f is instant on a metal surface of work function f. Electrons are emitted from the surface and enter a region of a uniform electric field directly opposite of their initial velocity. The electrons travel a distance d through the field before coming to rest. Is the stopping distance d of the electrons greater than greater when the frequency of the incident of light is increased assuming all conditions stay the same? Okay. So what's happening is you are sending photons, right? They're hitting it a metal and then we're ejecting an electron this way and we're putting an electric field. The electric fields this way because these are negative charges and it's going to it's going to apply a force on the electron opposite direction of the electric field and then eventually they're going to come to a stop here. Right? So he starts off with some kinetic some some kinetic energy and will eventually come to a stop or starts with some velocity. Now, the higher the frequency, when you have a greater frequency, what happens? You're going to have more energy in the photon. And so, then you're going to have more kinetic energy left over. More kinetic energy means goes faster. If it goes faster and you're pushing it with the same force, then it's going to come to stop at a greater distance. So, it's greater. And that's the reason you want to say, so you say greater frequency. Greater frequency of the photons means more energy. in the photon. So the resulting resulting uh kinetic energy of the electron increases which means it travels further farther further I don't remember farther in the under the in the same E field okay derive an expression for the stopping distance. So we want to think about how much kinetic energy we have is we're going to have whatever the energy of the photon is minus the work function. So that's going to be hf minus the work function. Right? So we're going to start with kinetic energy. We're going to end with zero kinetic energy. So we have to just think about like in terms of the distance I would think of as work in change in energy. Right? So we have the kinetic energy. We're going to the change in kinetic energy is going to be the work done by the electric field, right? So, we're going to end with zero kinetic energy. We're going to start with initial kinetic energy of hf minus v. And we're doing negative work on it because it's going to travel a distance d. And we're applying a force uh in the opposite direction. And that force is the electric force, which is q * the electric field uh and then times the distance to get the work. So, the negative signs can cancel. This Q we usually use the letter E to represent the charge of an electron because that's what it is. So then um the uh the distance is going to be you just divide by E over here. It's going to be HF minus V / E * capital E. You could probably use Q for the charge if you wanted to, but E is most commonly for charge of an electron. And that would be the stopping distance, right? That's our work energy process. So does that agree? Right? So if we increase f here does that increase the stopping distance and say yes because increasing f will increase the numerator hf minus v and that implies d is greater. So that's what we said. We said d would be greater when the f increases. Right? That's what we said in part a. Okay, let's look at another one here. So, we have a sample of unstable nucleus X underos a nuclear decay producing a new nucleus Y. The nuclear reaction is observed to be this where beta minus represents an emitted electron and V minus is an emitted anti-utrino. Okay. Uh I think that's supposed to be like a bar over it. At time t equals z there are it's not a negative sign. Yeah, because anti-utrinos are neutral charge, right? Time t equals 0, there are n0 nuclear present. Half life is measured t1 half. Based on the observed reaction, identify the type of radioactive decay occurring. Okay, so uh this is just basically definition. We're emitting a beta particle, a negatively charged beta car. So this is a beta minus decay because it's based on the charge of the um beta minus decay. And we are emitting an electron, right? And um an anti-utrino there. Okay. Uh suppose nx is the uh because um you're emitting oops emitting a negatively charged uh beta particle beta a negatively charged electron. Okay. Suppose nx is the number of neutrons in the nucleus. Px is the number of protons in the nucleus. How many protons and neutrons does it contain? So we got to think about um we got to think about the change. Now the total the number of neutrons ny plus the number of protons um of of py has to still equal like like we're still the total number of protons and neutrons are going to remain the same. Okay like this part we know we have to we have to know it has to be the same. However because we have a negative charge here we're going to have to have one more proton over here in order to keep the charge like balanced over here. So let's say this net charge was zero. We're going to need a positive charge to cancel out that negative charge. So what that means is a neutron had to get converted into a proton. Okay? So that means a neutron has to get converted into a proton in this reaction. So that means um uh the number of protons um in y has to be one more than the protons in x. And the number of neutrons in Y is you take the number of neutrons in X and subtract one. Right? This this shouldn't have said that. That was a it's just sort of a template kind of thing because uh hopefully you guys see how that's happening because we have to have a extra positive charge over here in order to balance out that there was no no charges that were just created. Okay. Uh derive an expression for the number of undeayed nuclear remaining as a function of t0. So we have the halflife is here. So remember our equation is n is equal to n0 e to the negative lambda t right and so what we know is that with t2 we know that we have 12 n0 is equal n0 e to the negative lambda * t2 you can derive an expression for what that lambda is going to be because that's what I want to I want to I I just I can't use it as lambda here so the n0 cancels uh you have that you take the ln of both sides ln l will get rid of e so you get ln of 1/2 the ln and e cancel so you get negative lambda t2 so your lambda is going to be negative ln of 12 over t2 right so then you can write this as n= n0 e to the uh negative and negative would technically make it positive ln of 1/2 ln of 1/2 though is a negative number though so t2 so this would be how you could do it or you could have run it as n0 0 e to the ln of 2 over t to the 1/2. If you guys know that one, that's because you can make this log rules is 1/2 is 2 to the negative 1. You can bring the negative in front and make it just um negative that. Or if you don't even want to do the exponential form, this is from pre-calculus, you could just say like, well, it's n0, it's times t, it's times 1/2. And you do that every uh every every half life. So when t is t to the 1/2, you're exactly multiplying by this. But all of these would be the same. Doesn't really matter. Any of those would be acceptable in the ways um you could do it. I think it's more inclined to like look like one of these two just based on the formulas I gave you. But honestly, I would prefer to do this. I think it's more natural for half-life to go with that equation. Okay. Suppose after T1 only 33% of the original sample remains. Derive an expression for T1 in terms of T1 half. So, for example, here we're just going to go here and say like, well, we're going to have 1/3. Actually, I should I should have just made this 1/3. So, 1/3 n0 is how much we have left is equal to n0 e to the ln of 1/2. Just use whichever equations you want. T the 12 uh oh, sorry, there should have been a t here. I forgot the t. Uh t2 uh and then this is t1 is the time, right? All right. So I plug in t1 t to the 1/2 and then we just want to the the n zeros are going to cancel. We just want to solve for t1. So we're going to take the ln of both sides. So we get ln of 1/3 is equal to ln will cancel with that e. You get ln of 12 over t to the 12 uh t1. And then you're just going to say well multiply by t to the 1/2 and divide that. So you're going to have ln of 1/3 / ln of 1/2* t to the 1/2 would give you that. Um you could put in your calculator if you want for that thing. You could make this ln of 3 over ln of two because technically it would cancel. Either of those are fine but that is acceptable. Whichever equation equation you use there that's what I would use. Okay let's do one more um F FRQ where we do some Compton scattering just to give you one example of what that looks like. So in Compton scattering experiment, a beam of photons with initial wavelength lambda strikes a stationary electron. Some photons are scattered at angle theta 1. Others are scattered angle theta 2 relative to the original photon and theta 1 is small. So theta 2 is greater than theta 1 and they're both less than 90°. Is the final energy of the photon scattered by theta 2 um greater than less than or equal to the final energy of the photon scattered by theta 1? Justify your answer qualitatively without referencing equations here. So now when we want to do this qualitatively without just looking at the Compton equation, we have to think about the principles that are applied. So we have a photon right with a wavelength and it's hitting an electron and it scatters at an angle, right? So there's an electron at rest here and then the the scattering angle is going to be like that, right? So it's going to scatter and then the electron is going to move off. Now the electron scatters at an angle theta and this this angle is some other angle some other angle. We'll just use like fe as it. So we're scattering by this angle. So in our equation for compton scattering when we're doing this which is the um gosh let me look up the equation that we had up here. [Music] Uh I know we did a lot of man so much stuff. H over right this equation is what I'm talking about. mass of the electron, speed of the light over there like that. Okay. Um that's what theta refers to this angle. Now let's talk about what happens if the angle is greater. So we have a greater angle. Now it's less than 90°. So it's going to be greater in this case, right? So two things are conserved. The momentum has to be conserved. So he's going to have um he's going to have um if you think about like this angle here theta, he's going to have less x momentum and more y momentum, right? So from a conservation of momentum point of view, this thing this thing whatever the momentum of this guy is, it has to compensate for um it has to compensate in the x and the y direction. Um there I think at this one it would be sufficient to just argue and just say like okay so the greater the angle change right the cosine of theta that x component is going to be smaller cosine of theta is going to um get smaller that's going to go down right so this overall quantity will increase so that will increase which means um um our change is going to be we're going to have a greater change in the wavelength and remember we're going to lose some energy so if we change the uh wavelength by more we're going to have and we're losing right because overall we are see let me think about this let me double check we are yeah giving some kinetic energy so we're going to lose so I'm here so this change in this wavelength means the the bigger the change in the wavelength right means we're going to lose more energy so we're going to end up with uh the final energy of photon is going to be less than and so your reference there I know we're not going to reference equations here but what we're going to talk about is a a larger uh scattering angle means a larger change lo and that means a reduction in the sorry larger not reduction larger Increase increase increase in the wavelength of the scattered photon. Okay. So we have a larger change and larger which means meaning the photon lost more energy. Okay, so derive an expression for the final energy of the scattered photon in terms of the initial wavelength scattering angle fundamental constants. So here we're going to say oh the change in the wavelength, right? This is our first equation because you're given this equation. So you're allowed to do this me * c 1 minus cosine theta. Okay. Um and then that's the change in the wavelength. So the the the initial energy is going to be you know HC over the initial wavelength and then the final energy which is going to be E it's going to be HC over the final wavelength and this delta lata is is how much longer it's getting right so um if we want to we want to solve for theta f here so the initial wavelength was just theta so we're going to move it over so theta F the final wavelength is going to be the initial wavelength plus H over ME C * 1 - cosine theta and then the energy of the photon is just going to be HC divided by the final wavelength which is going to be lambda plus the change in the wavelength right me 1 minus cosine theta okay and so we're going to have that and so does your derived expression agree with your qualitative prediction so let's talk about what's going to happen here right so as we increase theta. So as if when theta 2 is greater than theta 1 that implies cosine of theta 2. So when you increase cosine co when you increase the angle co in the first quadrant uh the cosine decreases and that implies uh 1 - cosine 1 - cosine of theta 2 is going to then be bigger because this number is going to be smaller. It's going to be greater than 1 - cosine theta 1. So this quantity is smaller. So that means the denominator is smaller. And the denominator is smaller then the energy and so the denominator is smaller which results in uh actually let me double check if that was consistent. Let me work out the reasoning again. So I said denominator is going to be larger right because you're going to um you're going to increase this quantity. So the denominator is larger which implies that E will be smaller. So that's the kind of the reason you're going to use or that's how you would connect that with your prediction up here. Okay.