 Still, I guess around 10 students to come. Till when you had school today? Around 30. Till 2-10. 2-10, 2-10. Okay, and what is going on in school, in physics? So dual nature is just solving the exercise questions. Okay, that is in NPS. Other schools? Mahul? In the other class, we started atoms. Atoms. Atoms. Atoms we are done. So like we just finished first two chapters of electrostatins, that's it. Only that much? Yes, since our school started one and a half months back. So we've been doing this. Okay, that's a little bit of luck. Okay, you cannot be that much behind with others. Okay, so don't follow the school speed of schools. Okay, follow what is going on in symptom. Because you have to finish the curriculum early so that you could revise the 11th portions. Okay, and I, as I already told you, within a month or so, our physics at least will get over. So by mid of, by end of September, definitely it will be over. So entire October we'll be doing a little bit of revision of 11th. And when even math and chemistry gets over, math and chemistry will get over, I guess, by first week of November. So we will have around more than a month for, you know, to revise or revisit some chapters of 11th. And then we'll start the crash course. You know, right? Crash course, what will happen? First week of November. Any idea? No one has any idea? We will have daily classes. Okay, daily will have classes of three to four hours. We will solve only problems then. So Monday to Saturday, we'll be doing that. And on Sunday, you have to write the test. That is what is planned. That plan could get changed a little bit here and there, but that is what we intend to do. Okay. And where are others? Very few people have joined. Anyways. Okay, so you guys know that we are done with dual nature and atoms. And when you might be doing problems on these two chapters, did you encounter questions on x-ray? By any chance? Looks like you haven't even started practicing questions from these two chapters. Okay, so x-ray is an important portion. Okay, that is not in your NCIT book. When you solve problems, you will see that the x-ray chapter will be a lot of questions from x-ray chapter will be there. Okay, in previous year, J question or any other community exam you take it for some reason, but x-ray is not there in the NCIT. So nevertheless, we will cover it x-ray. So today's agenda is we will complete the x-ray topic. Then we will start solving questions on these two chapters. And if suppose little bit time remains at the end, then we'll be like what we do, then we'll take up some mechanics questions. Okay, so I have set off those questions also. Alright, so all of you write down x-ray. So we won't be starting nuclear today. No, you have seen the reminder in the group. I thought nuclear was written. Nuclear was written. Okay, okay, let's see. But before nuclear, many other things were written. So let's complete that first. Are you in a hurry to finish? No sir, just asking. We are anyway very far ahead in physics. So let's see. If only little bit time is remaining, then no point starting a new chapter. Let's say only half an hour remains. Why should we start a new chapter? So let's see. Okay, so write down x-ray. So, x-ray, you might be knowing little bit about the x-ray. The discovery of x-ray. Who has done the discovery of x-ray? Know it? No. William Rongton was doing some experiment. He was into the experiments related to cathode ray, William Rongton. Cathode ray is nothing but the electrons passing through an evacuated tube. So he was doing his usual experiments. And one day what he did was that he coated the evacuated tube with some metallic powder or something. And after the cathode ray got switched off, then also he observed that some sort of glow is coming out. And he concluded by some calculations that this is some sort of electromagnetic wave. And during that time, when the discovery of photons came in and electromagnetic wave was a very new phenomena, the discovery was very much interested in identifying these EM waves. So when radio waves were discovered, then the people got to know that even the visible light is an electromagnetic wave. So different types of waves were coming up. The people were aware that ultraviolet waves can be utilized to kill the germs. Then every new discovery of EM wave come, they will be looking as if it can be used for something which is of importance to them. So X-ray came up and then a little bit of analysis was done. So before the further analysis of the X-ray, William Rongton and Bragg, they already used X-ray to study the crystal lattice. So Bragg's name you might have heard in chemistry in the solid state. So there is a Bragg law also there. So these guys did some study of the crystal lattice and they got Nobel Prize for it. But they never systematically studied the X-ray. It was a Coolidge who discovered the Coolidge tube, which was used for careful study of what is X-ray and how it can be further analyzed. And when the discovery of X-ray happened and it was found out that the X-rays can be absorbed by the calcium, which is prominently present in our bones. So then this news spread like a fire. Within an year, this news got spread in the entire world. So during that point in time, there was no communication device as such, which could communicate one part of the world to the other. But then also this news spread so fast. And that point in time, there was a lot of gang war used to happen. So it has its own importance in that era. So the X-ray was utilized to look at the gunshots, bone fractures, and of course kidney stones, solid objects. So all of this were, I mean, within an year itself, within a year of discovery of X-ray, all of these were started. So X-ray has its own importance in our day to day life and it is still used. So I'm sure many of you might have already done X-ray some point of your life. So only lucky few escapes any bone fracture. Most of the time at least once in your life you'll have bone fracture. Anyways, so Coolidge was the one who has carefully designed an apparatus to study the X-ray. So let us first design or let us see what is the design of the Coolidge tube and then we will get into further aspects of the discovery and the analysis of the X-ray. So let me first draw a little bit of it. I'll let you know when you can start drawing and then you can copy because I'll be erasing some things before you start drawing. So do you had a bone fracture like some point in your life already? Yes, sir. No, sir. No, sir. No, sir. Oh, so many, so many no's. Okay, that's rare. Oh, you are pretty young also and you don't take that much risk. During our time there was no mobile phone as such. So we have to go out to play. Then you can play on the mobile phone now. And when you play outside, a lot of accidents happen. You can start drawing. All of you. Can anyone identify what it is which I have just drawn? Any guesses? To transform? No. Any other guesses? Anyone? It is an electron gun. Okay. Yes, cathode rate you can say. Fine. So it is an electron gun which I have drawn. This we have used in the, what is that? There was this German experiment. German Davidson experiment. Okay. There we have used it. So I have not drawn complicated electron gun. But it is an electron gun. Okay. There is a cylindrical coating. This entire thing is made up of glass. And then there is a coating like this. This yellow thing. This is some metal coating. You can take any metal. But they had chosen molybdenum coating. You can use anything copper also. It is fine. Okay. Then you have one more big element over here. This entire thing is made up of copper. There is blue structure. It is made up of copper. Okay. And at the tip of it over here. This is a molybdenum coating. This red thing. It is a coating. Fine. So this is LT means lower tension that the battery has lesser potential. So there has to be high tension somewhere. It is a relative term. So the high tension battery is over here. You had this molybdenum cylindrical coating that and this copper. These two things in between them. There is a potential difference. Okay. Molybdenum coating is having negative potential. And just to show that it is high potential. We are connecting three batteries like this. So this is high tension. Let's say potential is V. All right. Now electron beam will be produced by this tungsten filament. This is tungsten. It will get heated up. And it will emit electron. This is called thermionic emission. We have learned about it, right? So the electron will absorb a lot of heat. Heat energy and will come out. It will generate a lot of electrons. And what will happen is that this molybdenum coating, which has negative potential. It will ripple the electron beam from all the sides. It is a circular cross section. So there is no escaping for it. So it can't go up or down. So what happens is that the beam which gets produced converges into a narrow beam like this. Okay. And then it hits the molybdenum coating, which is in front of it like this. These are the electron beam that is created. Okay. And then from when the electron beam hits, it is observed that EM wave is generated. Okay. The EM wave that is generated like this, you can pass this EM wave through the, through a very, let's say, a very thin aluminum sheet so that lesser energy photons are not able to cross it. Only higher energy will pass through it, like a filter. So high frequency will come out of it. And we call it as X-ray. The filter is not shown here. And you need not know also just that I'm telling you. So X-ray is generated like this. Okay. Now we have to get into the details of what is happening and why X-ray is getting generated over here. So this is what we have observed. Now, first question, which you can answer right now itself is why it is made up of copper. This thing is blue portion. Why it is made up of copper? No one, no one has answered. So you don't want any resistance because you want that battery's voltage to be really high. Copper gives thing that X-ray is made up of. X-ray is created by the molybdenum coating, not by the copper. Okay. So why I do not, I should not be having entire thing made up of molybdenum or any other metal because the copper has a very good thermal conductivity. Copper can take away efficiently. Okay. What happens is that 99% of the energy of these electrons which are hitting the molybdenum coating, it goes as heat. Okay. 99% energy it converts into heat and only 1% is converted into X-ray. It's not a very efficient process generating X-ray because photons, it will be very, the molybdenum coating will be very selective about which energy of the electron to absorb which one to reject and it has to transfer it effectively. So these electrons which are coming from electron gun, it collides, it generates a lot of heat which should be taken away. Otherwise the metal will start melting and even though you made up with copper, what will happen is that slowly and slowly the glass on the glass molybdenum will get deposited and then you could start seeing the spark very early between molybdenum and the glass itself. So slowly and slowly, anyway decays, you want to make sure that it has a longer life. So you want to cool it down quickly. So you can as well throw water in it. You can throw some water on this copper rod which could take heat away from it. Cooling is another very important aspect of it. That's why it is made up of copper. So let us watch a small, this thing, a video of this experiment so that it is little bit more clear. I don't want any confusion related to it. So today also we used the same kind of setup in normal X-rays. See today's X-ray I haven't studied actually, how exactly it is created. But definitely there will be a cathode ray and there will be a metal on which the electron should hit. That will be there. But whether it is Coolidge tube or any other advanced mechanism, I haven't checked that. Also X-rays when they produce is that thing instantaneous like as soon as the electron hits it gets released. We'll discuss that. But first let us see, let's understand the observation part of it. Conclusion and why it happens we'll study. All of you listen to this video everyone. Production of X-rays. The lamp that is shown here was designed by Coolidge and hence it is called Coolidge tube. It contains a highly evacuated glass bulb containing a cathode which is a tungsten filament and anode which is a target made up of a metal of high atomic weight. The pressure inside the bulb is maintained at 10 to the minus 6 millimeters mercury. When the cathode is heated by passing a current through it from a low-tension battery it emits electrons. The filament is surrounded by a molybdenum cylinder which is kept at negative potential to the filament. This repels the electrons further and hence the electrons form a fine pencil beam at the center of the cylinder. This pencil beam is also accelerated by the cylinder towards the target. It is made up of a copper block on which a piece of molybdenum is fixed. High thermal conductivity is used to carry away the heat generated to the cooling setup. The target is also titled to an angle of 45 degrees causing the electrons to heat at an angle. This produces X-rays almost perpendicular to the electron beam. A very high potential of 20 kilovolt is applied between the source and the target. Due to the high potential difference the electrons are accelerated towards the anode and it hits the metal. When it hits the kinetic energy is transferred to produce X-rays. The intensity of X-rays depends on the number of electrons striking the metal target. So if we can increase by increasing the current through the filament then we can increase the intensity of X-rays. So one last point which... What was that other machine that was like pumping something else? It was water. It is just pumping water. Nothing else. It was used to cool. Hello? Yes sir. So one last thing that this guy mentioned was that the intensity of the X-ray depends on the number of electrons hitting this molybdenum coating. And that makes sense because one electron can generate only one photon. So if you want to increase the intensity of X-ray which is number of photons you have to increase the number of electrons. So that is what was claimed. And then he has also mentioned that this coating should be of higher atomic number. We will see why it is required when we will study it mathematically. So let us see what is causing the X-ray to get emitted when the electrons are hitting the molybdenum coating. What exactly happens? Now we are getting into the reasoning part of it. So molybdenum is of course a very high atomic number atom. So we will be drawing a representative atom for the molybdenum. So it will have multiple shells. So let me draw all that. All of you draw with me. Let's say it has a K shell and let's say this is L. Another one. Okay. I have just drawn three. K, L, M. Everyone draw this. So I am just drawing the electrons. There can be multiple electrons. So I am just drawing few of them as a representation sake. Nothing else. So this is the atom of the molybdenum. And when electron from outside the electron gun when it is targeting the electron is interacting with the atom of the molybdenum only. So let us see exactly what happens when electron comes near this atom. What happens that one electron let's say this is an electron. This electron is targeted inside the molybdenum. This electron will travel and most of the electron will miss the target. As in target as in they will not touch a single electron or the nucleus. Let us go like this. Okay. But what will happen in this process is that just because they are not touching any electron or the nucleus doesn't mean that they are not interacting. There is electrostatic force that is there. So what will happen is that initial kind energy is let's say K1. Final kind energy could decrease. Let's say K2. Okay. So if energy is decreased K2 is less than K1. What will happen to the balance of the energy that is K1 minus K2? What do you think? The electron came with K1. Its kind energy becomes K2. It has decreased. Balance of energy where it goes. It goes into some other electron. No, it is not touching. It will go into the X-ray. It will create a photon of H mu. Okay. So an EM wave will get generated because you know these electron as long as they are orbiting in a stable orbit, they will have fixed velocities. They are orbiting electrons kind energy won't change. Nothing will happen. Only the K1 minus K2 has to release and that only way to release is by releasing a photon. So K1 minus K2 will release a photon of energy H mu. Okay. Now, if K1 is fixed, K1 is fixed. How many values of K2 can exist? K2 can be how many? K2 can have how many values? Sir, anything, right? Infinite values. Infinite values. Okay, starting from zero to K1 itself. Infinite. Infinite. So this creates a continuous spectrum of frequency. All of you able to understand continuous spectrum of frequency will be emitted. So starting from the maximum possible frequency to minimum possible frequency, everything will get emitted. Now, can you tell me what is the minimum? Sorry, what is the maximum possible frequency? How will you get the maximum possible frequency? So K2 is zero. K2 is zero. So K1 will be equal to H into mu max. Right? So K1 can be written as Hc by I can say lambda threshold. Lambda will be minimum. So I'm calling it as threshold. When it is minimum, we call it like a threshold like this. Now, what is K1? Can you find K1? Look at the diagram which you have drawn here. What will be the kinetic energy of electron? Can you tell me? So E is EV. How much? EV. EV. You can see electron is accelerated from this potential to that potential. The potential difference between the molybdenum coating and the copper is V. So the energy is EV. Fine. So K1 is E into V. Equating this, you will get the maximum possible frequency or minimum possible wavelength. So does it depend on what is the material over here as in whether it is molybdenum coating here or some other steel is there, iron is there. Does it matter to the maximum possible frequency? It depends on lambda threshold. It doesn't matter because it depends on the kinetic energy. The entire kinetic energy when it gets converted into the photon, that will emit the maximum frequency. So this particular thing doesn't depend on the material. So this is independent of the material or independent of this atom itself through which the electron is passing the maximum possible frequency. So apart from this, once in a while electron, let's call it as outside electron. This outside electron once in a while can knock off an orbital electron. Suppose it knocks off this electron. This electron, it knocks off and because of which there is a vacancy here. So what will happen now? Anyone? If there is a vacancy there, what will happen? The kinetic energy you get transferred. See, I am interested only in the photon, EM wave. Let's not talk about the energy transfer here. Tell me how photons will be generated now. If electron is knocked off from K shell, after that what will happen? Suppose it get knocked off after that? It will enter some other shell. So will this electron have intense incentive to jump from L to K? Can it do that? Yes sir. Which energy is higher? Energy of K shell or L shell? L shell. L is higher. So definitely it is an incentivized thing because everything in this universe wants to decrease their energy. That's why from L it will be ready to jump to the K. So if electron gets knocked off from K, electron will jump from L to K. And when electron jumps from L to K, let me write it down. From L or higher shell. Param, why are you not writing advanced tests? You are preparing only for mains? Jump from... Sir, I didn't study enough for that one. I spent like my entire time studying for the university. No, then you are cut out for advance. If you have these kind of excuses, you have to find time. Everybody else, those who have written, they have got time. Only you did not get the time. How is that possible? Okay, electron from L or higher shell can jump to K. K shell. Now this jumping of higher shell to K shell, when it jumps, energy of the electron is decreased when it goes from L to K. And that difference energy gets emitted as a photon of high energy, which is of X-ray frequency. So let's say energy of L minus energy of K, when electron is jumping from L to K will generate a frequency, which is of X-ray. This is called K-alpha radiation. If you are jumping from M to K, these are the names, that's all. This radiation, this frequency, when it comes out, it is called K-beta radiation. Okay, now coming back to the point why the atomic number of the material should be high, because then only the difference in L and K will be very high. Difference in the two levels should be very high. And you know, energy of a particle level is proportional to atomic number square sort of thing, right? 13.6 by N square into Z square. So if it is a heavier mass, then the difference in energy levels will be high. So when electron jumps from L to K, higher frequency will get emitted. So that is the reason why, okay? Anyone has any doubt? In the beginning, why did you assume that kinetic energy increases? Where I assume kinetic energy increases? It decreases, of course. We said that K2 is less than K1. What would happen if K2 becomes more? How K2 will become more? No, there is no chance it can become more. Tell me why it will become more. But it is just interaction, right? So interaction can produce positive or negative. So if K2 increases, the energy of the electrons in the atom should decrease. Yes. And if energy of the electron decreases, its velocity should decrease. And if velocity decreases, it is basically changing the orbitals. But that is not happening. We are not assuming electrons getting knocked off. Okay, sir. Yes. So in the second case, what happens to the outside electron? Outside electron will create continuous spectrum. I mean, that is anyway there. Whatever kinetic energy it will come out, that frequency will anyway get emitted. But apart from that, this also is there. Okay. Now, this frequency, do you think it depends on the outside electron's energy final initial or it is independent of that. This frequency that get emitted. What do you think? So it is only depending on EL and EK. So it does not depend on K1. Correct. It does not depend on the outside electron's kinetic energy. It only depends on the shell's individual energies. So can I say that this frequency depends on what material I am using to generate x-ray. Right? I can say that. So this frequency is sort of a characteristic of the material which I am using. So this x-ray which gets generated when electron gets knocked off. This x-ray is called characteristic x-ray. Okay. And this x-ray is called continuous x-ray. Okay. Both are x-ray only. But this one is called continuous x-ray which doesn't depend on the material. It only depends on the kinetic energy of the electron before and after. This one is characteristic x-ray which depends on the shell's energy. So if you draw a graph of... Okay. Before drawing a graph, I will just quickly summarize this. K alpha is basically L to K from Compton number 2 to 1. Okay. Rajdeep, any doubt? Awesome. K beta is from M to K. N equal to 3 to 1. Okay. These are K alpha K beta. Similarly, you will have L alpha L beta. So what do you think L alpha will be? From where to where? From M to L. M to L. And this will be? N to L. N to L. Okay. So rather than using KLM, it will be better if we start using the principle Compton number from 3 to 2. And this one is from N equal to 4 to 2. Okay. But then in our study, most of the time we'll be talking about K alpha and K beta. We are not worried about these, but then sometimes these could also come in. At max, L alpha will come in. Okay. But we should be knowing that L beta also exists. So if you draw a graph, now we can talk about a graph. If you draw a graph where on the x-axis, this is the wavelength and on the y-axis is let's say the number of photons that are coming, number of photons of the x-ray that could come out. Okay. Or you can say intensity of the photons that are coming out for the x-ray. You're talking about that on the y-axis. So if you draw the graph, now tell me, will there be a lambda minimum? We have found out, right? There will be a lambda minimum for maximum frequency. For this one, if you create, you'll get a lambda threshold. Lesser than that wavelength is not possible. Okay. So the kind of graph you see is this. This kind of graph you see. Now, what can you make out of it? What does these two peaks represent? Any guesses? A alpha k beta. Correct. They are the aberration to the continuous spectrum which you are getting. The way they are produced is different from the way continuous spectrum is produced. So that's why in the graph also it is clearly seen. So this is the continuous x-ray which you get. There is a lambda threshold over here. So this characteristic will be only produced at these points. Even continuous is being produced at the same point. Yes. Yes. It can be. But like this characteristic thing, why does that even need these electrons? Because in that, in the characteristic, you're losing energy. So, and here the electron is losing energy. So shouldn't it become more different? It doesn't move. It is just like the electron should be knocked off and you need an external electron to knock it off. It will hit the k-shell. k-shell electron goes away and then only transition will happen. Once there is a vacancy. Otherwise, why transitioning will happen? Okay. All right. So now for the continuous spectrum, we have a readymade equation which is k1-k2 is equal to h mu. And I can say that k1 is equal to hc by lambda threshold, which is equal to e times v. So for the continuous spectrum, no problem. But for the characteristic x-ray, there has to be some formula. And that formula looks very, very similar to the Lyman-Bammer and all those series which we have learned already. We have learned for the hydrogen-like atom we have learned, but definitely it is not a hydrogen-like atom. There are multiple electrons in the orbit because of which the atomic number will get screened away. And there is a screening factor also. Do you know what is a screening effect? Have you heard of it? No sir. Others? No sir. What is a screening effect? Okay. You remind Gaurav about it. We have a screening effect. What it is? So those numbers are 0.3 something. What? 3 something? Basically. Something like shielding effect or something. Like a shielding effect. You did it last year. Poor shielding of DNF electrons and all that. So the outer shell electrons were not, the outer shell electrons, they are not very, it is not very easy for them to interact with the nucleus. Because they get shielded away by the other electrons. So the atomic number behaves as if number of protons inside the nucleus behave as if it is lesser than what it is actually. Due to the shielding of the inner shell electrons. And hence the effective atomic number is decreased. And that atomic number can be written as Z minus B. Okay. B is the screening constant. B is the screening constant. So I will just tell you the values of the screening factor. For K alpha, the screening constant B is 1. Okay. And you can use it for K beta also. It is very similar. The values of K alpha and K beta. B is 1. For L alpha, the value of B is 7.4, which is hardly used. You can keep it. Okay. This is the formula for the characteristic X-ray if you have to find the wavelength of it. Okay. Now let us work around and see what we can do further with this formula. 1 by lambda is Z minus B whole square, Rikbuck constant. 1 by N1 square minus 1 by N2 square. N1 and N2 are the common number between which the transitioning is happening. Okay. C by lambda is your frequency, which is equal to CZ minus B, the whole square, Rikbuck constant. 1 by N1 square minus 1 by N2 square. Okay. So for a fixed N1 and N2, let us say I am talking about only K alpha. Okay. K alpha, let us say I am talking about K beta. So there is a fixed transition between the two quantum states. So mu is equal to some constant times Z minus B whole square. Or I can say root of mu is equal to a constant times Z minus B. Okay. This is the final outcome and this is called the Moosley's law. This spelling, this is Moosley's law. If you draw a graph between these two things, root of frequency and atomic number Z, what you will get? A straight line. Yes. Tell me. No sir. Sorry sir. I thought I had to. Tell me. Anyone has any doubt? This is the complete theory of the X-ray. No one has any doubt? All N1 and N2, right? What? Say it again. We can't control N1 and N2, right? You cannot control N1 and N2, but first K alpha will come, then K beta will come. Okay. So if we go to the previous slide for one second, I think I will use something. Right. Let us take some numericals now. Okay. On X-ray only. You are able to read it clearly. Let me know if it is not readable. Okay. No one? So one second. One second. Vikas is not well. He is not there. Anybody else who did not come? But tell me the answer. What is the expression? So I don't know how to find that square without using a calculator. That is what they are checking, whether you can find it. Concept everybody knows. Should we do it now? Okay. Fine. So potential difference, should you want me to tell the final answer first? Let me check if I can get hold of the final answer quickly. Friends, final answer is 1 by 20. All right. So let's do this. We have a potential difference of this much applied on the X-ray tube. That coolest tube is also called X-ray tube. Okay. Take it at that way. Deep root wave length to the shortest wavelength of the X-ray produced. Shortest wavelength of X-ray will be for the maximum frequency for which the entire kinetic energy of the electron gets converted into the photon. So lambda threshold is equal to hc by e times 1000. Okay. This is the minimum wavelength. Now for the, what is that? Deep Broglie wavelength. Deep Broglie wavelength is h by momentum, m into v. This is velocity. And mass time momentum, you can write it as, oh, so you could have directly used that formula. Do you remember that 1.22 divided by root v nanometers? Do you remember this, everyone? Yes, sir. Okay. This is the Deep Broglie wavelength. So you can get it as 1.22 into 10 s of minus 9 divided by root 1000. Anybody knows root 1000? How much? 31.7. 31.7. 31.7. You know, right? 30 into 30 is 900. Okay. So like that, you can arrive at it. You can write 31, 32 like that. And then take the ratio of these two wavelengths. You get the answer. Root 10 is almost pi. All right. So this is the way you have to do this particular question. Again, J, this question came in J. All right, is how it comes. Do this. Assume that screening factor B is equal to 1 for all cases, whether it is K alpha, L alpha, K beta, screening factor is this only. You need to find the wavelength of K beta in terms of wavelength of K alpha and wavelength of L alpha. Okay. So let me write this as lambda one and this as lambda two. Don't use the values. So I've made the calculation simpler for you. All of you have written this. Yes. So how is lambda one, lambda two, lambda three? They are related. Just look at the equation. How are they related? What do you think? Can you look at some pattern here? 1 by lambda. So I go to 1 by lambda KB or 1 by lambda three is equal to 1 by lambda one plus 1 by lambda two. Look at this. If you add these two equations, you'll get a third equation. Isn't it? So 1 by lambda three is equal to 1 by lambda one plus 1 by lambda two. You can get this or you can just divide these two equation. The first and second you can divide. You can get lambda two in terms of, sorry, you can divide this equation with that. You can get lambda three in terms of lambda one or lambda three in terms of lambda two also you can get by dividing it. But this one is straightforward. The first relation which we have got. Just add two equations. You get the third one. This is the last question on x-ray. This one at least all of you should get it. It's a straightforward question. All right. The answer is 8 into 10 ratio power 18 hertz. How it comes, for example, root of frequency is some constant into Z minus one. It is K alpha. So frequency is some constant into Z minus one. Whole square. So for copper, we have a frequency of two into 10 ratio power 18. This is equal to constant into Z minus one. That is 28 square. So for the other element, we have K into 57 minus one. That is 56 square. That is four times K into 28 square. That is four times of the given frequency. Fine. So this is the introduction of x-ray. Now you are fully equipped to take any questions from anywhere on these two chapters. So let us see whether you can take it up some of the other questions anyone has any doubts till now from anywhere on these two chapters quickly ask. So could you repeat the values of B for K alpha K beta again? I took down 0.7. It is there on the PDF. I'll share it. Okay. Don't worry. I have already told you know K alpha. So K alpha K beta is one third. For L alpha, it is 7.4. Okay. Anyone has any doubt that was not a doubt. No one should we start taking numericals? We have already solved many numericals though, but then I thought today we could be doing more as these chapters are very important. Okay. First question, CBD. Three different answers. People are getting. Why is that? Okay. Should we do it now? All right. It looks like everybody has attempted the first question. We'll be doing the first one. Intensity of this much intensity, you know intensity is power per unit area. This is the intensity, which is falling perpendicularity over this surface. It reflects half the energy and absorbs half. So for electron, how much power will be there? Whatever is falling on it divided by two. This much power is available for electron to absorb because half of it getting reflected away. So electron can absorb only half the power. Okay. So power that is coming up is 750 into, this is per meter square. Okay. So you need to convert in meter 50 into 100 into 10 is power minus four. So four zeros are gone. You'll get five into 775. That is 375, right? 375 watts. This is the power which is falling. Half of it is available for the electrons to absorb. 375 by two in one minute. How much energy will be absorbed this into 60. So it will be around 11.3 kilojoules. I guess. Okay. It's D, D for Delhi, not C. This one, everyone. Second one. Do it carefully. There is no race or in a hurry. Every time you'll make a silly error. Better to get zero than minus one. Okay. Should we do it now? I can see many of you are answering B. B for Bombay. All right. So let us see that total momentum transferred to the surface in one minute. Okay. So the power that is absorbed absorbed power is 375 by two. And this is the power reflected also. Okay. So power which is coming up coming towards is equal to Dn by Dt into H into mu. So at C by lambda. Okay. So we have Dn by Dt into H by lambda. This is equal to power divided by speed of light. Now, can you tell me what is this left hand side? What is this momentum? No, it's H by lambda is momentum. What is Dn by Dt H by lambda? The rate of change of momentum. It is rate of change of momentum. Okay. City total number of photons into H by lambda is total momentum. When you differentiate it with time, you get rate of change of momentum. So Dn by Dt into H by lambda is rate of change of momentum. So it will help if you remember that rate of change of momentum is power divided by speed of light. Okay. This is for a given power. Now because of the reflection, how much momentum will be transferred? If incoming momentum is let's say X. If it reflects off, so total momentum transferred will be what? 2x. 2x. If it is absorbed only one times. Okay. So the rate at which momentum is transferred is for the reflected one. You can say power reflected divided by C into 2 plus power absorbed divided by C. This is the rate at which momentum is transferred with per second. That into 60 seconds is a total momentum that is transferred to the surface. Okay. Good thing is that power reflected and power absorbed they are equal. So you can take it out. So you'll have 3 into 3 into 375 by 2 which is power absorbed divided by speed of light 3 into 10 to the power 8 into 60. This is your power that is transferred. Can anyone quickly calculate what it is? It will come beyond this one. It's coming. I'm assuming you are correct calculation wise. Fine. So this is how you have to solve this particular question. Proceed now. Anyone has any doubt? So doesn't it come like the same answer as the previous one? Like the calculation. What for say it again? It comes as the same one as the energy. So the option is D. You're getting D. Yeah, it's D. You're getting D if you do this calculation. So C is like the energy calculation. So I can't trust you on calculation. Somebody is saying something. Somebody else is saying something else. In this slightly different method that's why I got... No, no, you get B only. Wait, wait. 375 into 60 into 3 by 3. You get D. You get 11000123410 to the power minus 4. You get D. You don't get D. Calculation of that. I mean that is the least I can expect. You can calculate it properly. What is the final answer of that calculation? I asked that. I didn't ask what do you get. This one. This is what you can do if you are in the space. So I didn't understand it. How does it go? Momentum. We'll create a force. So with flashlight. Flashlight. There's a reflector of the flashlight. The reflector has a light eye. It'll hit the reflector and bounce off like that. That hitting on this reflector. Create a force this way. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Take a picture. Okay, the sound is coming somewhere. Anyways, okay, shall we do this? Reflector of the flashlight is this much area. The power is this. So, rate of change of momentum, P is momentum actually. I don't know why in physics they keep on using the same symbol for everything as you understand, right? This is the force, dp by dt, which is power divided by speed of light. We just derived it, isn't it? So, power is 1000 divided by 3 into 10 to the power 8. This is force. So, pressure, pressure is also P. Pressure, power and momentum, everything is P. Momentum should be m, but m is mass. So, pressure is this 10 is power minus 5 divided by 3 divided by 0.5. How much this is? 1 divided by 1.5 into 10. Option A? Option A, yes, looks like. All right, do this. No, wait, wait, wait. Option A is not correct. There is a silly error. Can you identify where is that? What is that in the solution? No one. Obvious silly error. Param will tell, Param. You don't know. Ruchir will tell, Ruchir Parekh and Ruchir Singh, both Ruchir's. So, it should be twice of that. It should be twice. It is getting reflected. So, the option A is not correct. B is correct. So, but why do you need to make it? Oh, yeah. It is reflecting. Momentum transfer will be two times of the initial momentum. Okay, do this now. Is it a constant force? Yes, sir. So, expression will be constant. So, I can use the kinematics equation. Yes, sir. Karo use 150 kg. Ajagir is also there. This is the acceleration. You just need to estimate how many seconds and which formula will you use? s equal to ut plus half 80 square in each velocity is 0. Time taken will be two times distance divided by acceleration. Distance is 100 kilometers. 2 into 10 raise to power 5 divided by acceleration. It is 2. 3 goes here. 150 goes there. 2 into 10 is power minus 5. 10 is power 5 comes out. 2 is gone. So, it is root 450. Root of 450 into 10 raise to power 5. Or you can say root of 4.5 into 10 raise to power 6. What is root of 4.5? Around 2.1, 2.2. Okay. So, 2.2 10 raise to power 6. So, how many days it is? One month. How many seconds? Can you tell me, one year has how many seconds? How many seconds one year has? No, no, no. One year has how many seconds? You should be knowing all these things. Pi into 10 is power 7 seconds, roughly one year. So, one month will have divided by 12. So, pi divided by 12 is around 1 by 4. So, 2.5 into 10 by 6. 2.5 into 10 by 6. So, it will be less than a month. No. No. Yeah. So, less than a month. Option A. Okay. So, you can see here, they're checking how good you are with the numbers. How can you navigate to the answer rather than your concept. So, everything counts. You should not be just focusing on, oh, I know the concept. I don't need to calibrate anything. You will be faltering where you expect the least. Okay. So, all these things should be, you should be sharp in all these calculation related things as well. A simple question. Let's do for you. Tell me the answer. Chaiti got the answer. Got two answers. No, that was a previous one. Okay. Should I do it? Nobody else is getting answer. Here, what are you getting? Ruchar Singh got something. Krithika, your answer. Everybody's waiting for your answer. Krithika. Sir, I'm actually not writing anything because my hand is injured. So, I'm just listening to the problem solving and everything. What happened? Like, I think my position or something was wrong. So, like, I got a nerve compression and I can't write. So, I'm just listening. That's just one hand, right? Other hand, you can't use. The one which is the main hand, the nerve got compressed there. Anyways, so the answer is C. Let us see. Deep ruby wavelength of electron and orbit of hydrogen atom is this. Principal quantum number for this electron is what? Param immediately changed the answer and the moment. No, sir. Yeah, I found it. No, excuse now. So, deep ruby wavelength is h by mv. This is equal to 10s power minus 9. Okay. And we know mvr is nh by 2 pi. So, the value of m into v, we can substitute over here. You will get h divided by nh by 2 pi r. This is equal to 10s power minus 9 or 2 pi r divided by n. This is equal to 10s power minus 9. This is a hydrogen atom, right? So, r is equal to r0. r0 is 0.53 into 10s power minus 10, 0.53 Armstrong. You should know this. Is equal to 10s power minus 9. Okay. This becomes 10. So, the value of n, the value of n will be 2 pi into 0.53 divided by 10, is it? Factor of 10, am I missing? Sir, the radius should be n square, n square. Yeah, yeah, yeah, yeah. So, n square will be there. r is equal to r0 into n square. So, nn. The value of n is 10 divided by 2 pi into 0.53. This will be roughly 3. I hope everybody understood. Abiram. Okay. Everybody is getting D. Good. D Broglie wavelength in the ground state should be equal to the circumference. Okay. We know that 2 pi r should be n lambda, right? That we have learned about it when electron is a wave. It goes like this. So, 2 pi r should be equal to n lambda and if n equal to 1, that is ground state. So, lambda should be equal to 2 pi r. So, 2 pi into 0.53 Armstrong, which you will get it as B. So, in ground state, why is n1? Principal contour number, that is the k shell. It is a definition of it. Any confusion? Sir, I thought you use the principal quantum number to find which radius it is in. No, no. Principal quantum number tells you the radius, its velocity, its energy, everything it tells. Everything is a function of n, not only just the radius, everything. Energy is 13.6 by n square. When n is 1, it has the least energy. That is the ground state. As you meant to be a Young's double set sort of situation wherein the capital D is very large. No one is close to the answer. Sir, I am 1. Where is 55 degree in this diagram? So, the red line is 55. This is 55. How do you find the path difference? You drop a perpendicular here. This is your path difference. How much it is? Technometry? You can use here. What it will be? There is a path difference. How much it will be? D sin 53. This is the path difference. How much this should be for the second order maxima? How much this should be equal to? 2 lambda. Okay, get the answer quickly now. Then, Richard Parek got something. He got something. Others? lambda is d sin 53 by 2. lambda can be written as h by mv. So, velocity is 2h divided by md sin of 53. How much you get by substituting the values? Around 600. Precisely, it will be 589 meter per second. This is the velocity which is 5 times of 111. The answer for k is 5. How did you get 53 degrees of it? 55, sorry. 53, why did I write it? 55 it is. One thing someone told at the start that J advance test, physics was very easy. Is that correct? Others also felt like that. Richard Parek, Richard Singh, did you find physics much easier than chemistry and maths in advance test? Yes, it was the first test. The second test was harder. First test was simpler. And second test, which was the simplest out of three subjects? So, they are not equally so. I will take special care now. I will make harder only. Who else wrote? Richard Singh, Richard Parek. Then, who else from this? He also wrote, right? He wrote. Then, who else? I wrote. Chaitanya wrote. Akshet also wrote. Mahit also wrote. Fine. So, you know, right? The topper, what is the marks he got? Who is the topper? Trippan. So, you have checked it, right? After these guys have written the test. So, Trippan has got the highest marks. In paper one, he got, I guess, 131 he got in paper one. Paper two also he topped 104. And don't worry about your marks in J-advanced. Suppose you write J-advanced paper and you're worried that, oh, it is too less. It is useless for me to prepare for J-advanced. So, all those thought don't let that come in because when you prepare for J-advanced, every other exam automatically get prepared. So, you are, if you're only preparing for advance, suppose you don't get selected in advance, but then in the process of preparing for advance, you have learned many, many new things, which you can utilize in every other exam you write, okay? Don't get bogged down upon it. So, like, when you're, if you're writing J-advanced, write all the J-advanced tests and also prepare for it, you know, without preparation, don't write also. But for school UTs, if you miss this, this is a crime like what someone in our group has done. Do this. You know, these questions are designed in a way that if you're in a hurry, you will definitely pick a wrong answer. I'm getting all sorts of answers. B, C, D. All right, let us discuss. Let us discuss the answer. I'll tell the answer later. Definitely B is not the answer. Those who got B and in a hurry, that is wrong. Okay, fine. So, in a photo-emissive cell, what is photo-emissive cell? The photons when it is hit will create electrons with exciting wavelength lambda, the maximum kinetic energy is k. So, we can write k is equal to h C by lambda minus work function, okay? If exciting wavelength is changed to 3 lambda by 4, then the kinetic energy will be what? Let's say it is k1, this is k2. Now, it is 3 lambda by 4, so it will become 4 h C 4 by 3, this thing, minus 5, okay? I want to compare k1 and k2. How will I compare? I'll just subtract k1 and k2. So, I will get k2 minus k1 to be equal to, they want to compare with 4k by 3, right? So, I will multiply 4k by 3, I multiply 4 by 3 on the upper equation, 4 by 3 like this. Then I can compare k2 with 4k1 by 3. So, 4k1 by 3 minus k2 is equal to, this is minus, minus, this is plus, minus 5 by 3, okay? So, it looks like k2 is more than 4k1 by 3. That is the reason why when I subtract k2 from 4k1 by 3, I am getting negative. So, that's an option D. Everybody understood those who told B is the answer? Yes, sir. So, you are in a hurry. You are not even seeing that there is a minus over here, this thing. Okay. So, everyone is getting C. Let us see. So, we have k equal to h C by lambda minus 5. 5 depends on the surface which remains same for both, for all three. So, k is E times V. So, E times V1 is h C by lambda 1 minus 5 E times V2 h C by lambda 2 minus 5 E times V3 is equal to h C by lambda 3 minus 5. So, we know that over here, it is V1, V2, V3 are in ultrometric progression. So, 2 times of V2 is equal to V1 plus V2. All right? So, if I add this, sorry, V1 plus V3, if I add these two, it will be 2 times of this left-hand side. So, you will also get 2 times of lambda 2 is equal to 1 by lambda 1 plus 1 by lambda 3 which is harmonic mean. So, yes, you guys are correct, option C. So, Ruchir has answered something quickly. Okay. Which option? I think I made a mistake. Option pick. Okay. No one, no one got anything. Should I do it? Sir, can you give some more time? Anybody close to the answer? You got stuck to us and asked. I don't know what's wrong. That's like, no. That's very frustrating, right? An objective question. Spent so much time. Okay, Ruchir Parik got some option. Chaitanya got something. Anush got something. Sai, why are you shouting? Sorry, sir, my gap's locked. All right. Let me do it now. Many of you have already answered. Light beam has total intensity of this. False on an iron piece of area this. So, we'll find out power quickly. Power supplied is 10 is power minus 6 watt. Stop typing your answers. Now, I'm solving it. Work function is given. So, this question is pretty long, right? So, while you're reading, you should write down these things. Iron sample reflects 96% of light. Okay. So, only 4% of this power is absorbed for photoelectric emission. So, only 4% is absorbed. So, I'll multiply with 0.04. That says 3%. Only 3% causes photoelectric emission. 4% of intensity is, the 96 is reflected away. 4% is absorbed. So, this much power is absorbed. Out of this absorbed thing, only 3% causes photoelectric emission. It has a mix of wavelengths. It doesn't have a single wavelength. Okay. So, 3% of it causes the photoelectric emission. So, this is the effective power of the radiation, which is creating photoelectric emission. All of you understand this? Everyone? Sir, can you repeat it? Okay. 12 into 10 raised to power minus, minus 10. So, this is 1.2 into 10 raised to power minus 9. Read the question carefully and look at what I have written and then ask me doubts. Don't tell me to repeat. So, like how do you get that 4 and then 73? So, read the question first. It reflects 96%. How much it will absorb? 4%. So, this is 4%. Then, when it absorbs it, only 3% of the absorbed energy is causing photoelectric emission. So, this is the absorbed energy. 3% of it causes photoelectric. So, I have to multiply that also. Understood? Yes, sir. So, this is the effective power which creates photoelectric emission. All right. And what we need to find? Number of electrons emitted. Everyone is clear about it, right? Dn by dt is equal to power, effective power, which is this 1.2 into 10 is power minus 9 divided by, divided by the hc by lambda. Okay. Wavelength is 250 nanometer. So, I can use that electron volt thing. So, 1, 2, 4, 2 divided by 250 electron volt. So, 1.6 into 10 to the power minus 19 convert that into joules. How much this? I want to subtract the work function there. Work function, why you have to subtract? From where you get that? Why you have to subtract? So, because we need to check the total energy consumed. So, the number of electrons emitted, sir, that's why. How is the number of electrons emitted will be number of photons that are hitting 1 photon, 1 electron? What is the problem? You can check whether the wavelength of this has enough energy than that. If it is enough energy, then it will create photoelectric emission. If it is not, answer will be zero, but zero answer is not in the option. So, you don't need to check. So, dn by dt is what? Do this. Everybody understood, right? So, I'm still confused in the last. Roughly, you can tell. 1.5 into 10 to power 10. 10 for 10 or 10 is for 9? It's not just about knowing the concept, okay? So, this is the answer. So, the last step, how did you get it? I didn't understand that. This is power, right? Yes, sir. Okay, got it, sir. So, this is the time to take a break. Now, 608 will meet after 15 minutes, 623. Come on time. Bye for now. Anyone? The first answer? So, I got 6, but then I'm thinking it's n squared by z, right? That will mean one answer or just message me your answer, okay? Don't speak out. Everybody else is also doing it. Fine. So, it looks like everybody has tried. The angular momentum is 3x by 2 pi. So, content number is how much? n is equal to? 3. 3, because angular momentum should be nx by 2 pi. So, n is 3. The deep-duty variant of electron is this, where a0 is a Bohr's radius and it is a lithium plus. The radius is equal to r0 into what? n square by z. You remember this? Yes, sir. Okay. So, r0 is basically a0 here, a0 n square by z. This is r, deep-duty wavelength of the electron in this state. So, deep-duty wavelength is equal to, in this state, 2 pi r should be equal to n lambda over here, n is 3. So, this is 3. So, lambda is 2 pi r by 3. Okay. You can also get deep-duty wavelength as h by mv, like that you can proceed and multiply r numerator and denominator. So, mvr is nx by 2 pi. So, you'll get this. mvr, you multiply r numerator and denominator. mvr is nx by 2 pi. So, like this, you can arrive at wavelength to be this. You don't need to find out the velocity and all. Okay. This is your wavelength, lambda. And r is basically a0 n square, which is 3 square by z. z is what? z is 3, isn't it? So, it will get cancelled a little like this. So, lambda is equal to 2 pi a0. So, the value of p is 2. Just go through it once and let me know any doubts. Many of you, I think, got 6. Okay. Understood, right? This one? Yes, sir. Second question. What is the answer? Good one. Slow, same. Slow, same. Yes, sir. Right? Yes, sir. If there is any doubt, don't guess. Better you do it your own. Like this, k is equal to h mu minus phi, k is e times stopping potential, h mu minus phi, you'll get v is equal to h by e times mu minus phi by e. So, you can see slope is h by e only, no matter what is the value of phi. So, it is slope will be same for both. That's why one is to one. These are j advance questions from this chapter. Actual questions. Do this. You might be knowing how to solve it, but here it is about accuracy and speed. So, let's see who get the correct answer first. All right, Ruchir Singh got the correct answer. Congratulations, Ruchir. Only Ruchir Singh got it correct till now. No, no, Param, that's not correct. hc by lambda is 1242 divided by 200. 6.21 electron volt, Param. Do your calculation correct. 1242 by lambda. Let's do that. He got it correct. Ruchir Parekh now got it correct. Chaitanya got it correct. Okay, good. Photons energy is more than 6 electron volts corresponding to this. Photoelectric emission will happen to start with, but it will stop because the isolated sphere will slowly and slowly lose electron and become positive charge. So, if potential which sphere generate is v, it will act like a stopping potential. So, when e times v becomes equal to h mu minus phi, the photo electrons will stop emitting. h mu can be written as 1242 divided by 200 electron volt minus 4.7 that into e. This should be equal to e times v. So, good thing is that e get cancelled away. So, v is what 1242 divided by 200 is how much? 6.21 minus 4.7. This much should be the potential. This is equal to number of electrons n into charge of an electron 1.6 into 10 to the power minus 19 divided by 1 by 4 pipes and not which is 9 into 10 to the power 9 divided by the radius which is 10 to the power minus 2. So, when you simplify all of this, you will get n to be equal to 7 into, no, wait, you will get like this. You don't need to even get the exact expression. You should just make sure that the thing a is between 1 and 10. So, I am sure you might be getting something like this inside the bracket into 10 to the power 7. So, the value of z is 7. Okay. Everybody understood? We will proceed forward. Param, how many times you change your answer? So, I didn't see electron and photon. I thought both were the same. Oh no, I am asking how many times today have you changed the answer? At least like eight times. Almost every question. So, this time I changed it fast, like even in the paper, I would have done. Okay. But just to tell you the earlier answer was correct. What you choose later on was wrong. Okay. Others, Roger Parek answered. Anybody else? No one else? Should we solve it then? Abhiram, what is your answer? Right. So, let's proceed. A photon collides with a stationary hydrogen atom in a ground state in elastically. Inelastically means what? The photon will get absorbed inside the hydrogen atom. Okay. And the mass of hydrogen atom is so big that you can ignore the final velocity after the collision. Okay. As if nothing is there. It's just a single photon H by lambda. The H by lambda is so less. Energy of colliding photon is given 10.2 electron volt. After a time interval of order of microsecond, another photon collides. But before that, what will happen? If 10.2 electron volt is colliding, it will go from n equal to one to two or not. The electron inside the hydrogen atom will do this or not? First collision. As soon as it does this, immediately it will be back from n equal to two to n equal to one and emit a photon. It will not even wait for a microsecond. It happens in the order of 10 is power minus eight second around. Okay. It is written in your NCRT book somewhere. But anyway. Sir, in this question, why aren't we using that thing that we did last class? We only say that the energy which is absorbed is half of the incoming of the energy of the photon which is coming. No, there it was, you know, number of photons, let's say R5, 6, sorry, then out of 6, 3 will be reflected and 3 will be absorbed. Inelastically collision with only a single photon, it means that it absorbs. What is inelastic collision? It absorbs. Okay. And it's a single photon only. Earlier, there were multiple photons. Half of it will be absorbed. That was actually neutron, sir. And they said same mass. Yeah, I just, yeah, I saw that. Oh, you're talking about some other question, is it? Which one? The last class we were doing. Oh, last class. Yeah, yeah, yeah. So the neutron was hitting hydrogen atom, almost same mass. Yes, sir. So yeah, there you need to, you can't assume finally everything is at rest. Here you can assume. All right. So it'll emit a photon of 10.2 electron volts. That is for sure. So now what happens? It comes to ground state and then absorbs 15 electron volts. What will happen if it absorbs 15 electron volts? So 13.6 will get wasted to come out of the atom and then after that, 1.4 electron volts. Kinetic energy of the emitted electron is 1.4 electron volts. So a photon and one electron of this much kinetic energy. So option C. Fine. Let's proceed. This is done. This is again a J-advanced question, came in 2005. You can see they are testing how you calculate, not the concept. No one, even close to the answer. I got something that can't be correct. How much? Message? Did I got? Yeah, fine. Anybody else close to the answer? Nobody else? Should we solve? Next. Others? Okay. So nth line of Lyman series. The wavelength is what? 1 by lambda is Rigberg constant into. It is a hydrogen like element. Okay. So, So, can I change the answer? I didn't see nth line I put n value. 1 by 1 square minus 1 by n square. This is 1 by lambda. Okay. This is 1 by lambda and lambda is equal to h by mv also. It should be the deep ruby wavelength of electron, initial orbit of hydrogen. Initial orbit is n equal to 1. So, lambda is equal to I multiply r both sides mvr. And it is a first orbit. So, mvr is n h by 2 pi n is 1. So, h by 2 pi. So, lambda is 2 pi r. Okay. Now, r, you know what is it? r in the first orbit is 0.53 into 10 is power minus 10. Okay. This is your r. This is your lambda. Isn't lambda minus 1n by n? So, it should be nh by 2 pi, right? This is initial orbit, right? n is equal to 1. But the radius is going to be this divided by 11, right? Oh, n by z, sorry. Correct, correct. Radius is that n by z. Okay. So, what did I do? Is this correct? This is how it is, right? So, it's n square by z, but n is 1. n is 1. So, this divided by 11, correct? This is our lambda. So, I am going to substitute this lambda over there, put the value of Rittberg constant and then simplify the expression. And anybody good with the numbers will be able to find out that finally it will come out to be like this. n minus 1 by n is roughly equal to 25. This is what you will get. And from this equation, n is close to 25 only, right? So, but answer will be 24, right? They ask n at line. Answer is 25. n is 25. Find the value of n. So, but in lineman series, when we say like first line, that's 2 to 1. Second line, third to 1. So, nth line will be n plus 1 to 1, right? nth line of lineman series. Like here we found the number of the shell. No, no, you're correct. What you're trying to say is that first line you're saying is n equal to 2 to 1. Second line is 3 to 1. So, I'm saying n to 1. So, this should be n minus 1. n minus 1 should be 25. Wait, sir, how did it become 25? It is approximate, Mehul. This is, there is no problem. Here there is no problem. What he's saying that for nth line, the value of n should be n plus 1. That's what he's saying, right? Yes, sir. And we found the shell number. So, nth will be 24, I think. Okay, I'll just check and come back to you. What is the meaning of nth line of lineman series? Because the first line itself will have transition from 2 to 1, like what you're saying. But then nomenclature wise, I don't know how it is named because usually there is no logic as such. You know, you can name anything. For example, central minima doesn't exist. Okay, so the nomenclature, I have to see how it is written. So, this is something I'll come back. Okay, but then logic wise, this is how you proceed. Yes, I just checked the value of n. Here you should write n plus 1. So, the value of n will be equal to 24 then. So, the nth series of lineman series means you are transitioning from n plus 1 to 1. Okay, not from n to 1. Just make a note of it. Write it somewhere. nth series of lineman series means n plus 1 to 1. That's very important. Oh, sorry, I didn't see the options. Okay, anyone? That is not in the option Rahul. Yes, sir. That was for electron, I'll change again. What's the mass of a proton again? 1.67, 10 is power minus 20. All right, Ruchir got something. Nobody else? Fine. Anyways, I'll do it now. It makes a transition, a recoil speed. So, what will happen here? Conversion of momentum because photon will be emitted. Initial momentum of everything is zero. When photon is emitted with a momentum of h by lambda, the recoil of hydrogen atom will be there. Its momentum should be, let's say, mv will be opposite direction of this. So, mass time velocity should be equal to h by lambda. And I know 1 by lambda is Rigbert constant, 1 by 1 minus 1 by 5 square. Okay. So, velocity, you will get it as Rigbert constant, Rh into blank constant divided by mass of the proton. I'm ignoring mass of electron, which is nothing compared to the proton. So, this is 24 by 25. 24 by 25 also you can assume close to 1. You can do some approximation because options are very far away from each other. Okay. So, when you do all of this, what do you get? C, right? Yes. So, you also multiply with 10 because this formula that we used is for centimeter inverse. Which formula we have used? 1 by lambda is equal to Rh into whatever. Oh. So, that 109677 is for centimeter inverse. No, I'm not writing that. I haven't even put in the value of Rh. Rh you should use in the SI units. What is the value of Rh in SI unit? That into 10 is power? Seven. Seven. Yeah. Use it. Yeah. I think, yeah, in chemistry probably they are using centimeter, but it doesn't matter. Even in chemistry you can use SI units. Advocating it. Abhiram, they are not responding. Do this. Ruchir got something. Others. Atomic number you might have got. Markings, you know, right? Plus 5 minus 2. One wrong option, you're gone. Understand, Param? Yes. So, you picked one wrong option. All right. Others, here what is your answer? Akansha is also there. Akansha, what is your answer? Mehul, did you get anything? Mehul. Okay, okay, okay. Will you do it now? Particular hydrogen like atom has its ground state binding energy of this. All right. So, ground state binding energy is this, which we know should be equal to 13.6 N square by Z. No, wait. Z square by N square. This is the energy. Binding energy is when N is equal to 1. This is given as 122.4. So, look at the option. You can directly use it. Check whether Z equal to 3 satisfies this and it will. So, option A is correct. Okay. Now, look at option D. Electron of kinetic energy this much may emerge when 125 electron volt collide. Is this correct? Option D. It should be right. And that's what I'm asking. Is it correct? Yeah, it is correct. So, A and D two options are correct. Fine. So, binding energy is this much. This energy is absorbed and balance will be emitted as a kinetic energy. Okay. Now, let's see with B and C which one is correct. So, let's look at what they're asking actually. One of them should be correct because they are different types. Anyways, electron of this much electron volt kinetic energy can be brought almost to rest by this atom. And electron of this can excite the atom at the room temperature. No, they are similar. Similar kind of options are there. So, when electron is coming up, the mass of the electron is so less that you can assume that entire energy of electron will be absorbed. After collision, the velocity is almost zero. Okay. So, if electron of some energy is absorbed or it excites it, the difference in energy level should be equal to 19.8 electron volt. So, let us see energy levels. What are they? So, energy at the ground state is 122.4 electron volt. Right. Energy at the second state, how much it is? How much it is? This divided by two square. How much? Somewhere around 30. Okay, exactly. So, 30.6. 30.6 electron volt. What is the difference between E1 and E2? So, 91.8 electron volt. So, it can absorb this much kinetic energy and it will excite and it will be at rest because entire energy is transferred. So, this is where C is also correct. Okay. Now, is B correct? Electron of 90 volt. 90 volt as in 90 electron volt energy can excite. No, it is not possible because minimum energy required is 91.8 to go to the next level from ground state. Right. So, that is why options A, C, D, they are correct. Understood, Param? Yes. So, I always remember the difference in E1 and E2 as 10. So, I just multiplied with 9. Yeah, 10.2. Correct. Got it wrong, no? Yes, sir. Lack of exam practice. Yes. That is how you can blame it on something else. Lack of exam practice and maybe a little bit of overconfidence is there. Confidence is good but make sure you are grounded. Otherwise, you will make a lot of sillier errors. And these questions are like trap. The moment you are slack little bit, you are gone. So, you might have counted number of sillier errors you have made today. Have you? Many, sir. It is easy to count number of correct ones. Even though you know how to solve, see how slippery these questions are. First one? First one. By the way, this is N1 by N to the power 3 by 2. I think that this is what they meant, N to the power 3 divided by 2. Yes, sir, but I got it. Because I am getting Q by N cube. Yeah. Oh, no, it is power 3 by 2. Whenever they say proportional, they will not include any constant. Which one? A, B, C, D? Sir, according to what you are saying, then D. Yeah, then D. Okay. Why? I mean, if it is proportional to 2 by N cube, it is proportional to 1 by N cube also. Both are correct. Yes, sir. So, but like in chemistry, there is something similar to this and it serves like the closer option, more accurate option is the correct one. I don't know what it is. You can check again with Gaurav or you tell the question what it is exactly. But here, it cannot be 2 by N cube by N1 by N cube. Both options are correct. 2 is a constant. Proportional is with variable, not with the constant. Anyways, so all of you got D, no? Anyone who did not get D? Okay. So, H mu is equal to hydrogen regatomic transition with content number N to N minus 1. So, H mu is 13.6 electron volt Z square 1 by N square minus, no, 1 by N minus 1 whole square minus 1 by N square. So, you'll get frequency is equal to some constant times 1 by N minus 1 whole square this. So, frequency is that 2N minus 1 divided by N into N minus 1 whole square. 1 is very less compared to N. So, I'll ignore this and this. So, it will be K 2N N divided by N to the power 4. So, it will be 2K is another constant. That's a constant is C divided by N cube. So, C. This one, it was simple, we have done it. NC2, 6C2. Not 6, 4C2. Yeah, 4C2. Okay. No doubts, right? We'll proceed. Sir, is it N minus 1C2, sir? No. If there is any confusion, take example of N equal to 3 and draw the number of lines. You'll find it is accordingly. And if you take 3, then it should be 1 N minus 1C2. Oh, yes, yes, sorry sir. Answer for the first one, different answers people are giving. Okay, I'm getting multiple answers. A, B, C. Nobody 40? How much energy? How much energy hydrogen can give right now? Hydrogen can go from N equal to hydrogen transfer the total energy to helium plus. So, hydrogen can go from 2 to 1 and transfer its energy. How much energy it can transfer? 10.2 electron volt it can transfer. Okay, right now helium is at energy level of helium correspond to N equal to 2. Okay, so the energy of the helium in N equal to 2 is how much? 13.6 divided by N square, which is 4 into Z square. Z is 2, this is also 4. So, at the second level itself, its energy is this much. Energy at the... If you are judicious enough, you'll understand that if you put N equal to 4, you'll get energy like this, 13.6 divided by, this is 16 and this is Z square is 4. So, this is how much? 3.4. 3.4 electron volt. Okay, so you'll get a, you know, if you're good with numbers, you'll get a fair idea that E4 will be this much and difference will be 10.2 electron volt. Between E2 and E4. So, hydrogen, if it gives this much energy to helium, it can jump from 2 to 4. You can also, you know, find out E3 in between and check. The answer is this. Now do this. Okay, can you tell me, from, it is right now N equal to 4, right? So, between which level to which level, if it goes, it can give the visible spectrum. What do you think? From 4 to, what is E3 by the way? E3 is how much? Anybody calculate it? There's how many electron volts? 3.6. Oh, no. No, no, sorry. 4.5. Yeah, 4.5 electron volt. Okay, so jumping from 4.5 to 2, it's like a Lyman series. It'll emit ultraviolet spectrum. So, the gap is big. It is almost like 10, okay? But then this jump could lead to visible spectrum. 4 to 3. So, you need to, you know, analyze the situation. You can't just blindly go there. Otherwise, you know, the kind of calculation will become mad. So, E4 minus E3 can emit the visible spectrum because this gap, 4.5 and 3.4, this gap looks like a Lyman series gap, which is very less. So, here I get the wavelength, which I guess, I'm not very sure about the answer though. 100% sure. I have given the same question to other batch. There was a tie between C and D. So, just check, okay? E4 minus E3, whatever wavelength they emit, that is the answer, right? Now, do this one. Hmm. Now, kinetic energy will be how much in terms of total energy? What will be the kinetic energy? Minus total energy. Minus total energy. So, basically, magnitude of total energy is the kinetic energy only, right? So, the formula 13.6 z square by n square is the formula for the kinetic energy as well. Fine? So, for n equal to 2, kinetic energy for hydrogen atom is hydrogen atom's electron. Is this 2 square? And for helium, it is 13.6 2 square into z square, which is again 2 square only. So, the answer is 1 is to 4. Everybody understood? Now, I have exhausted all the questions, all right? On these two chapters. So, I have made sure that you are exposed to each and every type of problem. Now, you'll not see any new kind of problem unless it is like out of the world kind of question. We have done every possible kind of question from these two chapters. Anyone has any doubt from these two chapters? No, sir. No, sir? No. No, sir. No. So, now we have only 30 minutes left. So, it is not a good time to start a new chapter, all right? So, I would think that maybe, you know, we can solve some mechanics question for half an hour. And next class, we can start nuclei. What do you think? So, what I'll do is that I have few questions which I know they are slightly different types. So, you may learn something new, probably. Let us see. First, I'll give, I'll take a little bit of time to identify. Okay, tell the answer. Correct. You don't need to even solve it. Option A is correct. Yes. Okay, remember these things that in oblique collision, what is oblique collision? So, like two dimensions. Oblique collision is when the line of velocities and the line of common normal when the collision happens, these two lines are not parallel to each other. So, for example, this, this ball comes and collides with that ball. So, they will exert force on each other in this direction. But suppose line of velocity for this ball is this way. Okay. So, this line of velocity is not along the line of common normal. So, after collision, this ball will go like this and the other ball will start moving in other direction. That is what you do in carambord. So, in oblique collision, the oblique collision between equal masses if elastic, then line of final velocities will be 90 degree to each other. And if it is head-on, then what we can say? In head-on collision, what we can say? So, when the velocity is exchanged in head-on, if I keep forgetting elastic word, head-on elastic collision, velocities get exchanged. Okay. Remember these things. I thought I could just remind you. We have done, though, all of this. You can check your notes now. But we keep forgetting that this one. Okay. Ruchir answered something. Ruchir Singh. Others? So, that is for some old question, sir. What? So, that answer I had sent, that is some old question, I think. I didn't send anything. We didn't send anything. Okay, okay. Send something. No one got the answer till now. Okay. Param got something. I hope Param correct this time. This time I'm pretty sure. Oh, my God. Let us see. This is the situation. Ruchir Parekh got something. All right. All of you now focus here, everyone. So, the second is fear B. B is at rest. Okay. A is moving. This is A. This is B. Right. This guy, can I miss? Advic, correct. Advic. Akanksha. Are you guys there? Sir, I don't think Advic can speak. No, Advic can't type. But Advic was not there for at least half an hour in his place. Akanksha is not there, looks like. I don't know when this corona will get over. Anyways, let's continue. Now, B is at rest. A is moving with velocity V. Now, when A is hitting B, which direction A will apply force on B? In this direction or not? Everyone. Yes, sir. Yes, sir, we can hear you. After collision, Rajid is back now. So, after collision, B will go in this direction or not? Because it was at rest. So, whichever direction B will experience force, it will go in that direction. There's a velocity of B after collision. Which direction A will go after collision? So, that same angle, but opposite direction. Papandicular. Just now we have learned. Papandicular. It will travel like this. This is the velocity of A. Okay. Now, this angle theta. This is theta. How to get that theta? This Papandicular distance is given to us as r by 2. So, I can get sine of theta. How much it will be? r by 2 divided by? This distance center to center, which is 2r. So, sine theta is 1 by 4. Okay. Now, can I conserve momentum along the red line? Because I want to find out velocity of B. So, I'll conserve momentum along the red line. Can I do that? Yes. Right? There is no external force. So, initial momentum along the red line is mv cos theta. This should be equal to Now, component of A's momentum along the red line is how much? After collision? Zero, sir. Zero. This is equal to m into vb. So, vb is equal to v cos theta. So, sine theta is 1 by 4. So, cos theta is root 15 by 4. This is what the answer is. So, you can get it quickly also. No need to write some elaborate equations. If you have these kind of ideas. I hope everybody understood. So, this perpendicular is not because of both the masses being equal. It's for any case, right? No, when masses are equal. We have derived it. The last question, in fact, last question of your NCRT textbook on work by energy chapter is this only. Oblique collision, oblique elastic collision between two equal masses. Theta 1 and theta 2 added up to become 90 degrees. We have proved it. Masses has to be equal. Okay, now I have to find something else. Abhiram, you can hear us. For one and a half hour, you are not there. All of you solve this question. All right. What is the trick here? So, the velocity in the x direction is same. Yeah. Velocity in x direction will be constant because there is no friction. And horizontally, the ball doesn't experience any force. So, the momentum horizontally will be unchanged. So, we can't hear you. What? Say it again. Oh, I thought so I couldn't hear any. So, he's speaking to the. No. I also didn't hear anything. You're not speaking? No, sir. I'm able to hear you. Mehul is able to hear me. Okay. Can you talk louder? So, now along the horizontal direction, along the x axis, momentum is unchanged. So, Mu sin 45 is a momentum horizontally. This should be equal to MV sin theta. So, we have V sin theta equal to U by root 2. U sin 45. This is our first equation. And along the normal, I can use coefficient of restitution. Right. So, half should be equal to velocity of separation. How much it is? V by root. U by root. Velocity of separation, I'm asking. Half is equal to velocity of separation by velocity. V cos theta. V cos theta. V cos theta. After collision comes for separation velocity. Before collision is approach. Approach is U cos 45. Okay. So, you will get U cos, sorry. V cos theta is equal to U divided by 2 root 2. Now, they are asking about the kinetic energy. For that, you need to find V. So, just square and add first and two equation. You get V square equal to U square by 2. 1 plus 1 by 4. So, V square is basically 5 by 4 times, no. 5 by 8 times. Yes. V square is 5 by 8 times U square. So, kinetic energy also 5 by 8 times the initial kinetic energy. So, K2 is 5 by 8 times K1. So, I have to find out K2 minus K1 divided by K1. Right? So, it will be 3 by 8. No doubt, right? So, I have a doubt. So, let me go back to the previous question. No, no, you ask. I remember all that. You copied, no? Okay. Yes, sir. So, like if a ball strikes at a particular angle, isn't it with respect to the floor, the angle? Like in that case, it was equal, but generally. Equal ball strike at an angle of 45. No, they should mention actually. Here, since it is equal, they haven't mentioned it. Otherwise, in the question, they must mention the angle is with what? Got it? Yes, sir. There should not be any ambiguity. Okay, it was a simple, simple. Okay, let's do this. Okay. So, just write down the equations. You know, if you are not getting the calculations, tell me what all we can do here. By the time it comes down, what happens to the ring? Ring will start moving or not? Yeah, yeah. Let's go like, let's say this goes with V1, ring goes with V2. What I can do? Can it concern momentum horizontally? Yes, sir. Right? I can concern momentum or at least zero, which was initial momentum will be equal to M into V2 minus 2M into V1. Originally, there is no external force. So, V2 is equal to 2 times of V1. This is our first equation. What else I can do here? Energy. So, you can conserve energy, right? Energy conservation. Work energy theorem I can use. This I can assume gravitation potential is zero. I'll write W equal to U2 plus K2 minus U1 plus K1. Of course, dissipative forces are not there. W is zero. But it is better to write a generic formula like this, rather than U2 plus K2 is equal to U1 plus K1. So, this is equal to final potential is minus of 2MGL. Final kinetic energy half M into V2 square plus half 2M into V1 square. Minus U1 is zero, K1 is zero. This will give us the answer, these two equations. Got it? Yes, sir. Yes, sir. Find some other. Have we done this question? Have you seen this before? Yes, sir. Yes, sir. Yes, sir. This one. I don't know whether it will take time, but we can, you know, discuss it out. Let's see whether, wait, wait. Hold on. This is the complete question. Read the question and tell me your views. We don't have enough time to attempt it. Just read it. Okay. We will discuss it out as in the ideas and after the class, you can attempt and post the solution on the group. All of you read the question. Let me know once you have read the question. I'm not asking you to solve. There are two parts of the question. First part is the above paragraph. Next part is the below paragraph. Sir, I did. All of you read it, right? Okay. Tell me how to get the first part. First part looks to be simple. So use project time, right? That's a very, very broad way of telling. So the total velocity is going to be 5 root 3 plus the components, like 5 root 3 along x. No, no. So horizontal velocity will be like the x component. This velocity is absolute velocity. Oh, okay, sir. Otherwise it would have been written. If it is not written, every time I assume it to be absolute. How will you get t0 now? Tell me. So it is becoming more simple if it is absolute velocity. Yes, sir. So how will you get t0? So you take the x component and you add it to 5 root 3. That would be the horizontal velocity. Just now we discussed 10 meter per second is the absolute velocity. Right. How will you get t0? So then we find the minimum distance for which the maximum distance for which it will hit. Have they given us AB? Sir? No, no, no. That's not how you should do it. A to C is given. Can you use A to C? No idea. Okay. So we can find the time period for this project time. And then it's like a drop after that vertical drop. You're unnecessarily complicating it so much. Displacement in the y direction is given to us. Minus 120 meters. Just use s equal to ut plus half it is square along y direction. ut plus half it is square along the y axis. S y, ui, ay. S y is minus 120. U y is angle is given. So 100 sine 30 into t, ay is minus 10. So it will be 5 into t square. You'll get the value of t for this much displacement C to A. But this is the answer for the first part. Everybody understood? Sir, we're just assuming that the ball passes just over C. Someone is experiencing lag. Now is it clear, Kritika? Yes. We are assuming that, what? Say it again. It passes over C as in. So like. No, no, no, Rajdev, I think you got it wrong. See what will happen is that definitely it will go up and then come down. Vertically this is what the motion is. Displacement is how much? In this formula, s y is not the distance. It is displacement. Displacement is always this much only. No matter how much above it goes. Yes, sir, I got it. Come back. Okay, so Rajdev, the immediate feedback is revise the projectile motion. Tell me. Sir, this time from this equation, don't we add it to the total time, like the time period for that projectile? Because this is just the time for the fall from C to A. I think we take the bigger one. I don't know why you're getting so much. No, this is this is it. That's all. Because you. This is the answer. So there is two times, right? One time. Whatever it is, there might be one value which is not even possible. You may get t equal to minus something. Okay, forget about it. But why there is a confusion that you will not get the answer for t when it goes from here to here? Vertical displacement is minus 120. And you put it, you'll get the value of t. What is the problem in this equation? Yes, sir. Okay, looks like you guys have forgotten all the motion in 2D stuff. Okay. And I don't know whether I should proceed discussing this question now. Sir, I got my mistake. Okay. I'll just give a little bit of idea about the second part. The second cannon bond is fired. Assume resistive. Second cannon ball is fired after what time at t naught. So by the time cannon ball goes and hit here, you'll get t naught from here. The car will move forward. So it moves forward by a distance of x, which is equal to five root three times t naught. There's a distance that will move forward. And again, I'm telling you, I'm not solving it completely. I'm just giving you hints. Okay. You have to solve it and put it on the group. Assume the resistive force between the rails and the carriage is constant. So there is a friction force, which is constant, and it is non-impulsive. You keep on acting on it and it will decelerate this carriage velocity. So what you have to do is, first time it will go and hit here. You need to conserve the momentum horizontally. Horizontal velocity is constant for the cannon. So you'll get the carriage velocity immediately after collision. Okay. Then again, the second cannon ball will also take exactly same time because vertical displacement is again minus 120 only. So in that time, second cannon ball has to fall on the carriage. Fine. But then carriage was decelerating. But still you can conserve momentum just before and after collision horizontally because the frictional force is non-impulsive. It is big. It is a constant force of friction. Okay. So now I have given you a lot of hints in this particular question. So I want you to struggle through it, get the answer and put it on the group. Who will try it and put it today itself? Ruchar Pare, Ruchar Singh Paramp. Who else? I'll try, sir. I don't care whether you say right answer or wrong answer. Okay. I do not care about right or wrong answer. Only thing I want is you guys struggling through it, spend some time and do not look at the solution. Please don't. If you look at the solution, you are wasting a very good question. You've wasted a good question. You've wasted your time. You've wasted many things. Okay. So struggle and even if you don't get it, it is fine. Whatever you do, just, you know, suppose someone posts the answer on the group, you can discuss it out and just discuss it on the group, what, how it can be done and what is wrong, what is right. All right. Why carriage is decelerating? Because it is written. The carriage is experiencing force of friction. That is written. Registive force, assume there is a resistive force between rails and the carriage written. Got it? All right. So that's it for today. We'll meet next week with something new. We'll do nuclei basically. All right. Bye. Thank you, sir. Thank you, sir. Thank you, sir.