 So, let us move to the next question, question number 53, that energy of the colliding photon is, oh that much noise is coming, ok. So, energy of the colliding photon is 10.2 electron volt. After an interval, another photon collides with the same hydrogen atom inelastically So, there are two photons E p 1, E p 2, 15 electron volt, what will be observed by the detector? Where is the detector? It does not talk about the detector, but detector will detect the photons that are coming out of hydrogen atom, let us assume that, ok. So, how you are getting 3? So, because the second photon has 15 electron volt, and ionizing energy for ions. So, it obviously has to pick up one electron, because it makes. Correct. So, only one option talks about an electron. See, electron will need how much energy to get ionized? 13.6, 13.6 not minus, minus 13.6 is its energy, it requires plus 13.6 to make its energy is 0, total energy. But if it get more energy, then it will take 13.6 and rest of it will be used as kinetic energy, getting it? So, that electron that comes out will have energy of 15 minus 13.6 which is 1.4, ok. And of course, when the first photon strikes, the, you know the difference between the first energy level and the second energy level is what? The ground state and the first energy level, what are the difference? 10.2, 10.2. So, exactly this much energy will be absorbed, and the electron will jump from n equal to 1 to n equal to 2 from the first photon right. And then when it get de-excited, it will create a photon of same energy, 10.2, ok. That is why option number 3 is correct, right. Now, you know this is little confusing also, because at times you will tend to think, ok, this 15 electron volt can be absorbed by the electron which is already excited, ok. But there is a time interval. But the time interval is microsecond, it happens instantaneously. It will jump to n equal to 2 and once there is only one photon, jumps there and immediately it will come down, very, very less amount of time. So, there is a time. Start solving this. 50, here you may not know the formula for characteristic x-ray. This is the formula, ok. This is not hydrogen like atom, but the formula looks similar to that. But then there is a factor of B. B is a screening constant. What happens is that these electrons will get screened because of the inner orbital. You know what is screening effect, right. So, effectively it behaves like a lesser atomic number. So, that is a Z minus B. And for alpha, for alpha, let us say for K alpha, the value of B is equal to 1. Screening constant is 1 for K alpha radiation. So, when do we take this formula? When you have to find characteristic x-ray. K alpha, right. So, the first formula is 1 by lambda is Rigbert constant Z minus B is 11 minus 1, that is 10 square. N1 for alpha, N1 is 1, ok. This is N2 square. And for other atomic number, let us say it is this. So, it will be Rigbert constant Z minus B whole square. So, you can just divide these two. You can divide these two, you will get the value of Z, ok. The value of B over here is 1, ok. Z is equal to 6, all of you? 55. I will tell you. First focus on 55. You should watch that video, ok. Everything is there. Otherwise, we will just keep on talking in bits and pieces. It will never make sense complete. Watch that. B. Others, what is the fringe width? What happens to wavelength if velocity increases? It decreases because h by p is the wavelength. When you decrease the wavelength, fringe width will decrease. That is why 3, ok. No doubts, right? You go to the next one. 56. Option 3. 56, option 3. You move to the next one. Others. You did not get 56? Helium has to get converted into oxygen, ok. Helium has atomic number of oxygen. So, you need 4 helioms and there will be energy. Energy is nothing but kinetic energy of oxygen only, ok. So, delta M is what? 4 into 4.00. See, these types you should get it, ok. I mean, this shows that you came completely unprepared or you are unnecessary or confident about yourself. So, you should give the due respect to this chapter. 931, MEV is the energy released if one AMU gets converted into energy. So, you find out the mass between AMU and multiply it with 931, fine. 57, what is the answer? Anybody else? Any answer? Ok. Now, half-life of A is 4-day. Probability that the nucleus will decay in 2 half-life, ok. Now, what is the probability that it does not decay? Then it should be in, let us say, in 2 half-life N0 will become? N0 will become N0 by 4, ok. So, if it does not have to decay, it should be out of N0, it should be N0 by 4. So, probability that it does not decay is 1 by 4, there is probability that it does not decay. So, 1 minus of that is a probability that it will decay, that is 3 by 4. See, it is like if a particular nucleus does not decay, then it has to be these N0 by 4, one of these N0 by 4 nucleus, right. If it does not want to get decayed, what is the chance that it will be N0 by 4 out of N0, 1 by 4? That is the probability that it will not decay. There are only 2 mutually exhaustive event, whether it will decay or not decay. So, 1 minus of probability that it will not decay is probability that it will decay, right. And of course, the probability will be very high that after 2 half-lives, it will, the probability should be high. Yeah. Probability cannot be 1, right. And probability cannot be 1 by 4, very less probability and half is like exactly half, it should be half. You know, like that also you can arrive. If you start thinking in that way, but you cannot do that with each and every question. And this type of thing will be done only by those who have lot of practice. It cannot be like you just develop this, you know, sixth sense on your own, just sitting. Do lot of practice, it will come automatically. Okay, we will move to the next one. Which question is this? 58, this is 59. Again, that knowledge will be used. 16. Limon. Limon is for UV and passion is for infrared. Limon is the first quantum number should be 1 and passion, the first quantum number should be 3. Limon is this minus 1 by n square where n is greater than 1, right. Now, this is largest wavelength in UV region. Now, when lambda is largest, 1 by lambda should be smallest. So, what is the smallest possible? What is the value of n for that? 2, right. So, 1 by lambda 1 is Rigbert constant into 1 by 1 minus 1 by 2 square, which is 4. So, this will be equal to 3 by 4 Rigbert constant. This is the first equation. Okay, then this is given, 122. Now, smallest possible reason is IR zone. Now, let us say IR zone is when n1 is 3. So, this is 1 by 9 minus 1 by m square where m is greater than 3, right. Now, I want smallest possible wavelength. So, 1 by lambda should be largest. 1 by lambda infinity? So, 1 by lambda should be maximum and when 1 by lambda is maximum, this subtraction should be minimum. So, for that m should be infinity, it get ionized from there. So, this will be equal to Rigbert constant by 9. This is equation number 2. Now, just take a ratio and get lambda 1, lambda 2 relation and then like that you can do it. 59. That is a nice question or yeah, they can directly calculate also. But r value won't be more than 122. What? r value which you find will be the most expression. What will be the actual r value? This is just a hypothetical scenario then. 59. Look at the 59 question. You are able to see, right? Yeah. What is the answer? So, uranium 236 is unstable. It gives out products, right. So, it gives energy also, okay. So, E is what? Rest mass energy. Rest mass energy is E. So, in a way it denotes let's say it corresponds to the mass, okay. It corresponds to the mass of that substance. Now, reactant is more stable or product is more stable here? Product. Product is more stable. So, some of the rest masses of the product should be less than the sum of or the yeah, sum of the rest masses of the reactants, okay. So, that is the reason why option number 1 is correct. Some mass will get converted into energy, okay. What is the difference between 2, 3 and 4? Oh, okay. 2, 3 masses. Different products are coming out, okay. All right. Should we move to the next one? All of these are previous year J questions. Like long, long back starting when you were not even born. Even I was not born. Very long back I am very old. 40 years. 40 years probably close to 40. Okay, do this 60th, 61. Who do you think is older me or Akhil? Akhil is older. I mean not like that. As in, it's getting recorded. I show it. 60th. See, the X-ray questions are very frequent. You can't afford to ignore it. Simply cannot. Done. What is the answer for 60th? No answer. It can't be as simple as this, right? Dimensionally, all three are different. And H by momentum is dimension of wavelength. H by mc is wavelength. So, dimensionally, 2 is correct and 1 is correct. 3 is also correct. All are dimensionally correct. There is no shortcut. Okay. So, electrons with deep-ruling wavelength lambda fall on a target in X-ray tube. Fine. So, deep-ruling wavelength is lambda. So, momentum is what? H by lambda. You need to find cutoff wavelength of the emitted X-ray. Cutoff wavelength is minimum wavelength. Now, can you find out? Total energy of electron should get converted into photon. That's when the cutoff voltage, cutoff wavelength will come out. Should I do it? Energy of the electron is what? T square by 2 Me, which is H square by 2 mass of electron into wavelength. Okay. This should get converted into Hc by lambda naught. So, 3 comes out. So, lambda naught is 2 Me lambda square C H, option 1. What about 61? Which of the statements is wrong? Look at the second option. Is it true? Between 2 and 4, only one can be true. So, it is between 2 and 4. So, you should be intelligent enough to at least remove 2 options. Which one is wrong? 2 is wrong. Right? It depends on the energy of the electron. It doesn't depend on the atomic number. Characteristic X-ray depends on atomic number. Okay. We'll move to the next one. Not many are left. Ten out of fifteen. Sixty? Sixty-two, sixty-two, sixty-two, sixty-two. First one. See, for sixty-two, reductive sample S1 has twice the number of nuclei. So, let's say S1 has N naught, or let's say 2 N naught is for S1, then N naught will be for S2. So, let's say lambda 1 is rate constant for S1. So, lambda 1 into 2 N naught is dN by dt for S1, which is 5 micro curie. Whatever is the conversion factor, let's keep it like this only. 5 into some conversion factor K. Okay? This is lambda 1 and then lambda 2 into N naught should be equal to 10. 10 into conversion factor K. Okay? Now, divide them. You'll get 2 times lambda 1 by lambda 2 should be equal to 1 by 2. Okay? So, you'll get lambda 2 is equal to 4 times of lambda 1. Okay? Now, half-life is what? 4 and 6 times lambda 1. Half-life is proportional to 1 by lambda. Okay? So, basically 1 by lambda 1 is equal to 4 by lambda 2. Okay? So, half-life of the first one is 4 times the half-life of the second one. It should be the case. So, it says it can be. It may or may not be, but only option 1 satisfied that condition where half-life of first is 4 times the other. You can just multiply 0.693 here, 0.693 there. Okay? Okay, 63rd. 4th one. 4th one? How many? 4th? Saimir? You can't see. You can see? Your power has increased. You will become more powerful now. Come sit here, friend. Put it down. Okay? So, you have attended all of you, right? All of you think, many of you think 4, which is wrong? Z minus 1. Okay? Now, first of all, understand one thing that you have three different metal plates. Okay? Light beam has these three wavelengths. Equal intensities illuminates each of the plates. Now, you may tend to think that all intensity is same, so current is proportional to intensity. So, whatever current comes out, it should be same saturation current, but what if 550 nanometer wavelength is not good enough to take electron out from work function of 3 electron volt? So, that intensity is meaningless. Getting it? Okay? Now, if you want to check what you should check, you should just check for the longest possible wavelength and the largest work function. If this wavelength is able to take out electron from here, then everything will come out. All the wavelengths will create electrons from all the surfaces. Okay? But you will see that this will come around 2.2 something. So, this wavelength will not be able to take out electrons from Q and R. Only from P it will be able to take out. And similarly, when you check, this wavelength will not be able to take out electrons from Phi R. But from P and Q it will be able to take out. And this one will be able to take out from all three. Getting it? That is why intensity for P will be most than Q than R. Okay? You can check, let's say for 550 nanometer, what is the energy of a particular photon? 6.6, 10 is for minus 34, H, C. Current intensity represents similar thing, number. It is number only. One is number per unit area, another is number per second. Number and charge. Okay? So, H, C by charge of electron 1.6 into 10 is for minus 19 into lambda. Number 1 at C by lambda is 5.5 into 10 is to power 7, minus 7, sorry. Okay? So, this, this, this, this, gone. Okay? This is approximately 4 times, let's say 4.1. Okay? Then it will come out to be 12.3 divided by 5.5. Okay? This is definitely not 3 and not even 2.5. This is close to 2.3 electron volts. Close to that. Exactly how much it is. 2.23. Okay? So, now, this is the energy of photon. Will it be able to take out electrons from this metal? No. So, this intensity is useless for R and for Q also. It will only be taking out electrons from P. But if this is able to take out electrons from P, all the wavelengths will be able to take out electrons from P because all the wavelengths are lesser and their frequencies are higher. Similarly, you will see that, you know, so P should be highest. P should be highest. Okay? And once you get this information, then there, you know, this doesn't make sense. This doesn't make sense. P cannot be lowest. Even this is gone. Just take one and move ahead. You don't need to now check for. What? What? See, if 550 is good enough to take electrons from 3 electron volts, then all the wavelengths will be able to take out from electrons from 3 electron volts. And if all the wavelengths can take electrons from 3, all the wavelengths will be able to take out electrons from these 2 also. So, it will be between 2 and 4, it will be 2. There is a dip because electrons get decelerated. From P, kinetic energy of electron will be higher. So, they need more push to decelerate. I mean, they need more negative voltage because the kinetic energy will be higher. Fine. We'll move to the next one. You should bring a record edge and then play it after every notorious class 11th. Right? You guys were like this in 11th? Some of them are pretty good class 11th ones. Some of them, like Bharata and Sukheet. You know them? Bharata, you might be knowing. He just dances around. Everything is recorded. And not this one. Start doing. Don't wait for me to put another question. Oh, question from PM Wave. This is which question? 60? Done. 60 for option 2. Option 2 for 64. Option 1. Approximately 2. Answer is very close to option 1. Bomber series for hydrogen atom. 1 by 2 square. The first spectral line is this. And the wavelength of the second spectral line in the bomber series of singly ionized helium atom. Then the one for helium. Z is 2. So, this is 4 times second spectral line. So, 1 by 4. Just divide it. Getting close to this. 65. 65. No, I haven't taught this. I know the formula. He said E0 by V0 is equal to V0. You tell me. So, remember this. Although I will take a class on EM waves soon. Before your J mains, I will take all of you together. All 12thers online. This is speed of light. Magnitude of electric field in EM wave divided by magnitude or sorry amplitude of magnetic field in EM wave will be equal to speed of light. What does N stands for? 28. Nano. Nano. Okay. Fine. So, you will get the answer. What answer are you getting? 6. Okay. This is E0 by 20 into. This is 3 into 10 is for it. Speed of light. Okay. All right. Let's take couple of more. 66. Should the video gets edited before uploading or is it like this? Fine. Because when it gets edited, immediately after the question is shown, solutions will be shown. That pause will get removed. Pause is required, right? Because it will feel like a class you can move with the pace. So, my load is reduced. Okay. Anybody got the answer? Should I do it? They go. This formula you might be remembering and you might be knowing this formula also. Okay. So, M into V is Q Br. Okay. So, half MV square is half of Q square V square R square by M. Fine. Now, it talks about the largest possible radius. Right. Now, if R is largest possible, even the kinetic edge will be largest possible. They are directly proportional. Okay. So, from here, you get K max. Okay. We need to find work function and this H into Mu is the energy that corresponds to 3 to 2 transition in hydrogen term. Right. So, H into Mu should be equal to what? Energy in the third orbit minus energy in the second orbit. Right. So, minus 13.6 divided by 3 square minus of, I am coming to that. So, H into Mu is this? Yeah. Okay. But this is in electron volts. So, better is to find kinetic edge in electron volts. How do you find? Just divide with charge of electrons. So, basically, Q is also E. Q is E only. Right. So, one Q get cancelled off. Okay. So, like that, it will get simplified a bit and then just substitute value of H Mu and K max will get the value of phi in electron volts. Because you can look at the options also. Options are in terms of electron volts. So, it makes sense to find everything in electron volts. Okay. So, if you encounter this question initially, when you have a lot of time to attempt in question paper, leave it till the end. Don't start fighting with questions which are eating up your time. Otherwise, you'll be under a lot of pressure because time will just go away. Okay. So, immediately when you feel that some question will just eat up your time, but then you know how to solve it. Keep a note of it that if you get time towards the end, you will attempt this. Initially, you should only solve questions which you are getting easily. That will boost your confidence. Alright. Move to the next one. 63. Nice. 67. Why? Option 3, right? Because atomic number for hydrogen is same. So, lambda 1, no weight. The Reckberg constant will be double for deuterium, right? Mass is double. We discussed that. There will be other options in 9. 9, but if you consider helium and oxygen, but that also Reckberg constant should change also. Yeah, it changes, right? Yeah, that's it. Right, square is coming. But the deuterium mass is changing. So, we are not putting 2 square there, but we are putting just 2 corresponding to mass. Like for helium, it comes 4. Anyways, right now we will go with third because that's the only thing which aligns with everything else. We will come back to this. Okay? I will just put another question here itself. Question 1. Why? Because if it is borrowed bias, this side should be higher voltage than that. So, you may have a... Yeah, because signs are same. Signs are same, but sign itself is relative. You assume some potential is 0. If you assume this is 0 potential, this becomes minus 2 volts. Because 0 potential is an assumption. 68. 69. Yeah, 69. You want to see, right? You know, while I was taking J, light went off. What? Light went off. But it was a daytime. But the problem... But the problem was that it was light... I mean, there were so many lights and it was a room... It was an auditorium. I think around 1000 people were sitting there. It was closed. All right? And when light is too much and suddenly light goes off, you will feel darkness for at least for some 10 seconds, okay? And then on top of it, you're not getting any problem. It's a stressful situation. It is good idea to remember H into C value in terms of electron volt. So, you can remember 1, 2, 4, 0, electron volt, nanometer. 2.6, okay? So, you'll have two equations like this. All of you got these two? Yes. This is like... So, you just take the ratio. You get hC by lambda 1 minus 5 divided by hC by lambda 2 minus 5. This is km1 by km2. What is the ratio km1 by km2? 4. 4 is to 1. 4 is to 1? Yeah. The mass is same? Yeah. So, u1 square by u2 square is ratio of the kinetic energy. So, this is 4 is to 1, okay? Cool. So, h into C is 1, 2, 4, 0. 4. Stop talking. 4. 1 divided by lambda 1 is 248 minus 5 divided by 1, 2, 4, 0, then 310 minus 5 is equal to 4. Okay? So, just solve this. What answer are you getting? 3.4. See, first get 5 is equal to something and then simplify it. Don't start simplifying here only that what is 1 to 4 is divided by 248. First get what is 5 is equal to? 3.7. 3.7 you are getting? Yes. Okay. All right. So, last question for today. Question number 70. So, we have done good number of test questions. Again, x-ray. All the formulas I have already told you. B will be what sir? What? B will be? For alpha, for k-alpha, B is 1. You remember that formula? 1. 1 by lambda is equal to rickbook constant z minus 1. This is for k-alpha radiation. And for k-alpha, n1 is 1. k-shell. k-shell, n is 1. k-lmn. Fine? k-alpha is from l to k. k-beta is from m to k. What is l-alpha? From m to l. What that video? I think it's 2.5 minutes. You think? No, no. Thinks is what you got? Calculation is approximate. What is the answer? This is the last question. Get it right all of you. Don't use calculator and just do the hand calculation. Then only you tell me the answer. Homework is whether you have taken mock test or not. Circulate that sheet again. Last. Yeah, 2 mock tests have circulated, right? Yes. You write what you have done with those. Option 2? Yeah. Lambda cu is x-ray of copper, lambda mo. Lambda cu is rickbook constant. z-1 for copper is 28 square. n1 is 1. But then that I don't think will be used here. 1 by lambda molybdenum is rickbook constant into 42-1. 41 square. 1 by n1 square-1 by n2 square. Fine? So this is first and this is second. So lambda mo divided by lambda cu is 28 by 41 whole square. But reverse is asked. So 41 by 28 whole square. This is what is asked. Okay? Now, you know, definitely you'll see that it's not 3 and 4. It's between 1 and 2 and they're very close. It's almost 1.5 square. So you can take that as 42 and 28. So this is, you have to do hand calculation here. Because these two are very close. 7, 4, 7, 6, 7. 7, 4, what? Almost 7, 6, 7. Oh, like that. But then you should remember that whatever answer you get, it should be slightly less than that. There's 2. This is 1.5 square. 1.5 square is? 2.25. 2.25. So slightly less than that. 2.14. Yeah, it should be 2.14. Let me come from that. Yes. 2.14. Fine. So that's it for modern physics problem solving. I think we have taken all sorts of questions. You guys have to watch that x-ray video. Okay? And make sure you're modern physics. Modern physics. Then optics. Then thermodynamics. Then magnetism. EMI. These are very, very strong for you because mechanics, you may think that you are enjoying solving it, but it may ditch you at any moment. But if you are good at modern physics, 90% chance you'll get it right. Okay? Bye.