 When light of frequency 2F0 falls on a metal, the stopping potential is found to be V. What would be the new stopping potential when the light of frequency 4F0 is incident, given that F0 is the threshold frequency? Since we're dealing with photoelectric effect, the first thing that I like to do is write down the photoelectric equation because I know that any photoelectric effect problem numerical can be solved from the photoelectric equation. So the photoelectric equation says the energy of the photon should equal the work function plus the maximum kinetic energy. Okay, now let's go back to the equation and see what we are asked. We are given the frequency of the light that is falling. It's given to be 2 times the threshold frequency and for that light we are given the stopping voltage. And we are asked what would happen to the stopping voltage when 4 times the threshold frequency light is shun on it. So how do we do that? Since frequency and stopping voltages are given, so I think we need to convert this equation into frequencies and stopping voltage. How do I do that? Well, energy of the photon we know is h times F. Thanks equation, thanks constant times the frequency. What is the threshold, sorry, what is the work function? Work function is the minimum energy. So I can write that as h times the minimum frequency, which is the threshold frequency. Plus, how can I write kinetic energy in terms of stopping voltage? Well, remember kinetic energy is equal to the stopping voltage, but in electron volts. So to be converted to Joules, it will be E times the stopping voltage. And so we are given F in two cases, once 2F0. So when F is 2F0, we are given the stopping voltage to be V. So the question is when F is 4F0, what is the stopping voltage? So I think we can build two equations and then we can compare them and then we can see what the new stopping voltage is going to be. So why don't you pause the video and see if you can plug in and solve yourself first. All right, let's do this. So when the frequency F is 2F0, in the first case, we'll get this to be 2HF0. That equals h times F0 plus E times the stopping voltage is just V. Let's just write that as V. So what I can do now is I can simplify and find and see what V is going to be. So V will be, if I just rearrange 2HF0 minus HF0, it will just give me HF0 divided by E. Now let's do the second case. In the second case, when F is equal to 4F0, let's see what happens to this equation. You get 4HF0 equals HF0 plus E times some new stopping voltage, which I need to find out. I don't know what the new stopping voltage is, so let's call that as V dash. And so if I now simplify, V dash becomes, let's see, this minus this gives me 3HF0 divided by E. And now I can look at these two equations. I can divide them or I can directly say, hey, HF0E is V. So V dash becomes 3 times this part is V. This is V. And that's our answer. There we go. So what we find is that when we doubled the stopping voltage, sorry, when we doubled the incident frequency, the stopping voltage did not double. It became triple. So this is one of the tricky questions that you can find. And you know, sometimes if you don't, we might try to do things very fast and you might say, ah, stopping, sorry, frequency doubles, maybe stopping voltage also doubles. Doesn't, you can see that. So it's better to always go back to the basics and solve it that way. Let's try another one. When photons of energy, 6 electron volt is incident on a metal. The speed of the most energetic electrons is found to be V. Now the question is, if the photons of energy, 11 electron volt is incident on the same metal, what would be the speed of the most energetic electrons? We are given the work function. I think it's a very similar question compared to, like, very similar to the previous one. So why don't you pause now and see if you can try this yourself first? All right. Let's do this. So I start with the photoelectric equation. It says the energy of the photon equals the work function plus the maximum kinetic energy. Now this time I look at the equation. A question I say, okay, energy of the photons is given. So I'm going to keep it that way. Work function is given. I'm going to keep it that way. And we are asked to calculate the most speed. So let's convert this into speed. How do I, how do I convert kinetic energy into speed? I can, I can just say K is equal to half MV squared. So this would be half MV max squared. And now I can plug in the values for the first case and second case and compare just like before. So let's do that. So in the first case, the first case for six electron volt, I get the speed to be V. So when this is six, six electron volt, the work function is given to be two. And that is the same for, I mean, in both cases, it should be the same because it's the same metal. Plus half MV max squared. Oh, in fact, they're calling that as V. Let me, let's also call that as V. So V squared. So if I, if I solve this, I'll get, I can directly calculate what V squared is. I can simplify. So I get four electron volt equals half MV squared. In fact, I'll just keep it this way because I know that I'll get something very similar. And if I divide that half and then we'll get canceled. So I'll just keep it like this. So that's my first equation. In the second case, I get energy of the photon is given to be 11 electron volt. So 11 electron volt equals same work function to electron volt plus half MV dash squared. This is the new speed that I need to figure out. And again, if I equate this, sorry, if I solve this, 11 minus two is nine. I get nine electron volt equals half MV, sorry, V dash squared. And now I can divide these two. If I divide these two, I get nine by four equals on the right hand side, half, half and mm cancels. I get V dash squared divided by V squared. And so I can now take squared on both sides. So I get three by two equals V dash by V. Or I get V dash equals three by two is 1.5, 1.5 times V. And there we go. That's the new speed.