 That's all the problem on photoelectric effect. We're given light of wavelength 430 nanometer is incident on all the metals in the table. Which metal would show photoelectric emission? So light of this much wavelength is incident on all these metals and we have to predict which of these will show photoelectric effect. In which of these cases electrons are gonna come out. So where do we begin? Now there are so many different kinds of numericals that can be asked on photoelectric effect that I used to think that oh my god for every numerical now I have to remember how to solve it. But no for any numerical all you need is Einstein's photoelectric equation. And it's not enough if you just remember the equation you have to understand it. So that's where it will start. Let's start by quickly recalling what the whole Einstein's photoelectric equation was. The whole idea is if you have an electron trapped inside the metal and you want to release it it needs energy. And the energy is coming from light which is made of photons, particle nature of light. And each photon has some energy which is given by the Planck's equation H times F. F let's use green for F. And when that electron absorbs the photon it uses its energy for two things. It'll use it a part of it to escape from the metal. So part of it is used to escape from the metal and that energy is often called the work function of the metal. And that's what's given over here. How much energy is needed? How much minimum energy is needed to escape from the metal? And the rest of the energy will go out as kinetic energy. And therefore the photoelectric equation from energy conservation we can say is the energy of the photon is the work function plus the kinetic energy. And of course some electron, most electrons will not be lucky. They might also lose energy internally. And so only those few lucky electrons will come out with the maximum kinetic energy. And so that's why we say this is K max. So this is the first step for any numerical on photoelectric effect. This is the heart of solving any problem on photoelectric effect. All right, so where do we go from here? So we're asked to predict which of these will show photoelectric effect. So I'm thinking why won't some of them will not show photoelectric effect? Why? Well think about it, if the energy of the photon is less than the work function itself then there will be no photoelectric effect because work function is the minimum energy needed. So for this to show photoelectric effect the photon needs to have at least 2.14 electron volts of energy, for this at least 2.3 and so on. So imagine if my photon's energy was, I don't know maybe let's say two electron volt. Then I know none of them will show photoelectric effect. If my photon's energy was five electron volt as an example then I know all of them will show photoelectric effect. So you see what we have to do? We have to figure out that means what is the energy of the photon and then see if it is more than the work function. If it is, photoelectric effect will happen. And how do we calculate the energy of the photon? E is equal to H times F. So now would be a great idea to pause the video and see if you can continue and try and solve the answer. Try and solve the problem. All right, let's do this. So the main thing we have to find is what's the energy of the photon and we know how to calculate that. That's going to be the Planck's constant H times the frequency of light. But here's the thing, frequency is not given to us. So here's the frequency. Frequency is not given to us. We are given the wavelength of light. How do you calculate frequency from wavelength? What's the connection between them? Now again, you can remember a formula but I don't like to. I like to think of this logically. So again, if you have forgotten it, don't worry. Go back to your waves. Again, very quickly, let me show you how I do it. What I do is I take example. So let's say the frequency of a wave was five hertz as an example. And let's say it was traveling at some speed, I don't know, maybe 100 meters per second or something. 100 meters per second. Then the way I think about this is here's the source. If I wait for one second, then I know in one second, five waves will come out. So it'll be one wave, wave number two, three, four, five. And this would be the first wave. This would be second, third, fourth and fifth. So wave is going to the right. And that first wave in that one second would have traveled 100 meters, right? Because it's going 100 meters per second. And so I now know five waves occupy a space of 100 meters. So one wave will occupy a space of 100 divided by five. And that's our wavelength. So wavelength would be 100 divided by five. But what are we doing? In general, we are dividing the velocity by the frequency. And there you go. No confusions. You don't have to remember anything. That's how I like to do this. So this will be h times frequency would be the velocity. Velocity of light, and we know velocity of light. That's c. So velocity of light is c divided by the wavelength. The value of h, we know that. It's 6.63, 10 to the power minus 34. Velocity of light is three times 10 power eight. The wavelength is given, so we can substitute. Let's do that. So this will be 6.63 times 10 to the power minus 34 times three times 10 to the power eight divided by lambda, which is 430 nanometers. So that will be 430. Nano is 10 to the power minus nine. And so now all we have to do is plug all of that in our calculator. I'm gonna quickly do that. Let me just calculate these numbers. So here's my calculator. 6.63 times three divided by 430 gives me 0.0462. So let me write that. 0.0462 times 10 to the power. How much do I get? So I have a minus 34. Then this nine comes on the top and gets plus nine. Plus eight plus nine is 17. So minus 34 plus 17 gives me minus 17. And if I shift to zeros, I will get 18, 19. So I get 4.62 times 10 to the power minus 19. And what's the unit? Joules. This is the energy of the photon. And if this energy is more than the work function, photoelectric effect will happen, otherwise it won't. But now we see another problem. This energy is in joules. And the minimum energy that's given here, work function energy is given in electron volts. And so I can't directly compare them. So either I have to convert all of this in joules, which I won't do. Or we can convert this into electron volt. And that brings us now to the last question. What exactly is an electron volt and how do you convert them between joules? And again, this could be confusing. Should I multiply? Should I divide by something? Well, again, here's how I think about it. An electron volt, one electron volt, if I want to convert it into joules, just substitute for that E. There's an E over here. So it's gonna be one times, E value we know is 1.6 times 10 to the power minus 19 coulomb and volt is volt. Now this coulomb volt is joules. This itself is joules. How you ask? Well, remember, voltage or potential is work done per charge. So joules is just C times V. So this is how you convert. So one electron volt is 1.6 times 10 power minus 19 joules, but I want to convert joules to electron volt. So what is one joule equal to? It'll be one divided by. So one joule will be one divided by 1.6 times 10 to the power minus 19 electron volt. And so now I can convert this. So in energy of the photon, in electron volt will be 4.62 times 10 to the power minus 19. And to convert joule into electron volt, I'll divide. I'm gonna write that over here. 1.6 times 10 to the power minus 19 electron volt. Does that make sense? So I convert it joules to electron volt, and this cancels out. And if I again bring back my calculator, there we go. So now I'll do 4.62 divided by 1.6. And that gives me 2.887. I'll round it off to 2.89. Let's do that. So this will be 2.89 electron volt. That's the energy of our photon. Now, let's look at this. When I insert in this on cesium, will there be photoelectric effect? Yes, because the minimum energy is 2.14, less than 2.89 will happen. What about here? It will happen. It will happen. Here, it won't happen because the minimum energy is 3.2. Photon doesn't have enough energy, it won't happen. Here also, it won't happen. So these are the three metals that will show photoelectric effect.