Key Concepts in Modern Atomic Physics

Hi students, we are going to start modern physics from this checklist, last topic of physics. And if we talk about JEE main need, then almost every school in coachings this topic is taught at the end. And many times students are not able to remain that much oriented and modern physics is somewhere weak. 4 or 5 chapters are in this modern physics checklist. And today we are starting with the very first topic of atomic physics. Atomic physics may The section of atomic structure is extremely important for all of us. And this is the main basis of atomic structure that you can understand an atom well. Variety of questions are made. Even in chemistry, some sections of atomic structure are common. But in physics, the atomic structure we study goes a little in-depth. And conceptual understanding is a little more than chemistry. Rather, in chemistry, the syllabus is a lot. Even we reach the Schrodinger equation. In physics, there is no Schrodinger equation in the course. but still atomic structure. On this, variety and different type of questions which are asked in JEE Advanced, these are beyond the level of chemistry. Although we will cover all the points in JEE Main Neat and even up to JEE Advanced level in Quick Revision Checklist. So, let's do it one by one. And I would say that those students who have not yet made a properly planned checklist, they should thoroughly prepare their concept checklist by previous year questions as well as with the concept checklist I am giving you. It is a small checklist but quickly. In coming time, we will cover all the concepts in this checklist which can be covered in questions JEE, MANIA, NEET. We are going to cover. Stay tuned with the checklist points. So, students, we will start the first point in atomic structure, structure of atom. There are three popular models of atom structure, you must have read in your basic theory. If you have any doubt, you can go to the physics galaxy concept videos and read in detail. The three models were Thomson's model, Rutherford's model and Bohr's model. Quickly, let's take a review. The Thomson model was the plumb-poding model in which Thomson suggested that an atom is like a spherical ball in which the whole positive charge is uniformly scattered and the negative charges are randomly distributed like seeds in watermelon. And these negative charges are termed as electrons, negative charged particles and positive charge is fully distributed. Until Thomson model, there was no concept of nucleus. This has become a popular model because it was explaining the neutral behavior of an atom. As well as few more factors it has explained. We don't need to get into much depth. Second model, the Rutherford model that was based on alpha scattering experiment. You all know about alpha experiment. In which alpha particle scattering was found on gold foil. majority of space in an atom is empty and electrons are available around the central point where highly positive charge is concentrated and they keep rotating around it. Which was mentioned in Rutherford model. And that was the discovery of nucleus that at the center point maximum mass and charge is concentrated in an atom. So Rutherford model was saying that electrons revolve around positive nucleus but according to Maxwell's theory When charge is accelerated, it radiates energy. So, it was assumed that if electron is in circular motion, which is an accelerated motion, and this energy will radiate, then it will slowly come and collapse in the nucleus. Its path will be in spiral form. It will be continuously radiating and losing energy. The answer to this was not available in other Ford models. That's why other Ford models got failed. But in Bohr model, all these things were explained. And that was the first most successful model of its time. which is still used to explain many classical theories. So Bohr model was explaining everything, all the concepts or drawbacks of Rutherford model. Bohr model was given in the form of three postulates. In the next point, I will quickly cover Bohr model in detail so that you get the basic idea of Bohr model. And we will also talk in upcoming points about the questions which are framed on Bohr model. So let's move ahead. Poji, please clear it. Thank you, Bhoji. So, students, now in this point we are going to talk about Bohr's atomic model. If you have read the basic theory and concept videos, then I have already explained that Bohr's atomic model was explained in three postulates. And if we combine all three postulates, then we get a clear picture of the atom. We will review all three postulates one by one so that it will give you a better idea about Bohr's atomic model. So, first we talk about... first postulate of Bohr model which was related to the circulation of electron in the surrounding of nucleus. If we take nuclear charge plus ZD then in its surrounding an electron is continuously revolving in a radius r. Now Bohr said that electrons continuously rotate around the nuclear charge without radiating energy. Why doesn't energy radiate? This was clear in the second postulate and third postulate. Here it was said that electrons are rotating in different orbits and if an electron is in nth orbit whose radius is rn and its velocity is vn So, here the force of nuclear charge that is the Coulombic electric force Fe, this is given as KQ1Q2 by R square. So, here it becomes KZE square by RN square. ZE is the charge of nucleus, E is the charge of electron. This force will be providing the necessary centripetal force for a stable circular motion. Or in the frame of electron, we can assume that centrifugal force is balanced with this force. Either way we can say. So, we can write kz square by Rn is equals to mvn square by Rn where Rn gets this Rn square. So, Rn gets cancelled out. So, this gives mvn square is equals to kz square over Rn. This is the first postulate equation which was given as Bohr atomic models basis. Electron continuously stable circular motion may whomta rata in the. positive electric field of nucleus and this is the first equation for Bohr's first postulate. Let's see about the second postulate. Poji please clear the board. Thank you Poji. So student in second postulate Bohr said that when electron rotates around the nucleus then it cannot revolve in any orbit. It revolves in some specific orbits and those orbits are called quantized orbits or stable energy levels. So the basic sense of second postulate was It was only that electrons can revolve around nucleus only in some stable orbits. These are called energy levels and these are stable orbits. The condition of stable orbits was given as m vn rn is equals to n h over 2 pi where n belongs to set of integers and m vn rn is the angular momentum of electron in a specific orbit. Yani second postulate ye kehati hai ki electron jis orbit mein stable revolve kar raha hai wahan pe uska angular momentum h over 2 pi ka integral multiple hona chahiye that's it. This is the second condition although iska derivation b hai Which was done by De Broglie's hypothesis that when electron rotates around nucleus at high speed, so in that case electron can be considered as an energy wave. And if electron is revolving and behaving like a stable energy wave, so there should be a formation of stationary wave. I will quickly take a label here just to explain you. If there is an orbit of r radius and electron is moving at v velocity, So here wavelength... of electron wave if we take out. We will cover De Broglie hypothesis in next checklist in matter waves. So, here wavelength of electron wave lambda is given as h over mv. This most of you might be aware of that wavelength is given as h over momentum. So, its orbit length is 2 pi r for stationary wave formation. This was analyzed 2 pi r the length must be equals to n lambda when you substitute the value of lambda this 2 pi r is equals to n h over m v and here you can see this relation is leading to the derivation of Bohr's second postulate which is m v r is n h over 2 pi. So, if electron is rotating in some circle, rotating in some orbit and due to high speed if it is considered to be behaving like an energy wave. So, in that case electron that if the energy wave is made a stationary wave in orbit, then no energy will radiate. That was what Bohr has explained. But Bohr was not able to directly give the reply. This was analyzed later after de Broglie's hypothesis correlated with Bohr's atomic model. And then this relation was deduced. Whereas postulate means Bohr deduced these experimental results which were theoretically verified later by de Broglie's hypothesis. So you just need to keep this result in mind as of now as second postulate. Let's move on to the next point. please clear karo Thank you Pooji. So students, in third postulate, it was told that the energy emitted from any atom, how it is emitted. So third postulate was related to electron transition between orbits or energy levels in the surrounding of nucleus. So third postulate says that if there are two energy levels, n1 and n2, and if electron is in n2 energy level, and if it makes a transition to the lower energy level. Transition will happen only when lower level is vacant. If lower level is vacant which is a more stable state. Because the lower energy level is near the nucleus. So electron will be more stable. So where there is more stability there energy is less. So when electron will transit from higher energy level to lower energy level. So it will radiate energy. And this consider energy is radiated in the form of a single electron. electromagnetic radiation photon. This is the most useful thing to keep in mind. If electron transits from a higher orbit to a lower orbit, then it always throws the difference of the energy as much as it is missing in the form of a single packet, that is a single photon. And here the photon energy, this photon energy can be directly called h nu or hc by lambda And this equals to E n2 minus. E n1 where E n2 is the energy of higher energy level, E n1 is the energy of lower energy level. This is what is generally considered and this is the equation 3 of Bohr's third postulate. I will tell you again to understand that only one thing is that electron energy emits when it comes from higher to lower level. That is third postulate. But energy will come out in the form of a single photon, you must not consider that stream of photons will come out. If one electron jumps from higher to lower orbit, then it will produce one photon whose energy will be En2 minus En1, whatever is the difference of energies. So these three postulates, if you combine and think all together, it gives you a clear picture about an atom, what is an atom, and what is the behavior of an electron in it. Let's move on to the next points. Poji, please clear it. Thank you, Poji. So students, in the next point we will talk about the properties of electrons. If we consider the Bohr model in an atom, then what basic properties of electrons can be there? Because in J-main, NEET, Olympiad, even J-advanced, all places the questions of electron properties are always asked. I am taking basic properties first and keep in mind that all the properties in the three postulates of Bohr, P1, P2 and P3, whose three equations I have written here, These three postulates are deduced from the equation. I am not taking the derivations here. If you have confusion, if you want to see it from the point of view of GE Advanced, then you can see all the derivations in the concept videos. So, if the first point is to be taken out, then you can eliminate Vn from P1 and P2 and take out the radius of nth orbit. The expression comes out to be n square h square by 4 pi square kz square m. Yeah, but and... orbit number and z atomic number because we generally assume that Bohr model is valid only for one electron system. We will talk about Bohr model failure later. But everyone knows that Bohr model is most commonly used for hydrogenic atoms or one electron systems. So, if we remove n and z, then the value of all the constants is 0.529 into n square by Z Engstrom. Keep in mind that radius is proportional to orbit square and inversely proportional to atomic number. These results are more useful. In this way, if you put this Rn value in P2, then you will get the second property, velocity of electron in nth orbit. That comes out to be 2 pi k z square over nh. Important point to be understood here, that you see the radius of nth orbit. It depends on the mass of electron but the velocity of electron in nth orbit does not depend on mass of electron. Many times questions come in this way that instead of electron, some other particle is given having charge equal to that of electron but mass double that of electron. So in that case, what will be the difference between its radius and velocity? So you can say that if mass is doubled then radius will be half. But if we talk about velocity, then velocity will not change on any orbit or atomic number. This is important to be noted that velocity of nth orbit, when we talk about velocity of particle which is revolving around the nucleus in nth orbit, this does not depend on mass of electron. Rest, we will add value of constraints other than z and n. This comes out to be 2.18 into 10 to power 6 multiplied by z by n meters per second that you need to keep in mind. We will keep proportionality in mind. That's why I have written the result. You can see in derivation concept videos. Or if you want, then in physics galaxy book volume 4, all these things are explained in great detail. There also you can go through it with variety and lot of illustrations and examples. So, if we talk about angular velocity in nth orbit. Now, if you remember Rn Vn expression, then you can write omega n directly. This is Vn by Rn. You don't need to remember the result. You just see the proportionality result. How is the dependency on n and z? This is Vn by Rn. So, this is proportional to z square by n cube. Radian per second. You have to keep in mind that when we compare the angular speed of two atoms, then angular speed of electron in nth orbit is proportional to z square and inversely proportional to n cube. Similarly, if we talk about frequency of revolution of electron fn, you can write as omega n upon 2 pi. So, proportional relation remains same. This is z square over n cube. This is per second only. We are talking about revolution per second. One point if we talk about time period of electron revolution in earth orbit. So time period of electron revolution is reciprocal of frequency. So the constant multiplied by n cube by z square seconds. So these proportional relations work. Most of the questions come like this. That if in hydrogen and helium plus. Helium plus is a one electron system. If we apply Bohr model. and compare it with the time period of revolution in hydrogen's second orbit and helium's fourth orbit, then you can directly find out the ratio of n cubed by z square. That you need to keep in mind. Few more properties we need to take up. Let's proceed further. Poji, please clear it. Thank you Poji. So few more properties here we are going to take up. We had studied basic properties like radius, velocity, angular velocity, frequency, and time period of electron in nth orbit. Now when electron is revolving in an anath orbit, then revolving electron is like a revolving charge. And if charge is rotating at a faster speed in a circle, then it behaves like a circular current or current carrying circular coil. So if electron is negative charge, then we always take current opposite to negative charge flow. So we can also define current in anath orbit. If we see current in anath orbit, then it is a direct indication that current is given as. Always keep in mind that whenever a charge is rotating in a circular orbit, this thing... Magnetic effects checklist also mentioned that if the charge is rotating in circle then current is given by simply charge multiplied by revolution frequency. This is the way how we define current. So, here charge is E and revolution frequency is Fn which we have covered in previous slide. So, if we just substitute the value of Fn. So, this gives the proportional relation. If you will pay attention to Fn then we had done it is Z square by n cube's proportional. So, the same relation will be here it is z square by n cube this relationship will hold true. Similarly, if there is a current flow in a circle, then it will always produce a magnetic induction at its center. So, here we talk about magnetic induction at nucleus due to nth orbit. And B produced at nucleus you know it is given by mu naught I by 2R. So, I put IN and RN. IN will be the circular current of nth orbit and RN will be the radius of nth orbit. So if the current is I-flow in the circle, then the magnetic field at the center is taken out by Tesla. It is mu naught I by 2R. You can substitute the values by taking them out in the derivation book or concept video. Here the proportional relation is important. What will be the ratio of magnetic induction at the center, i.e. at the nucleus, due to the third and fifth orbits of Li plus 2 and hydrogen? We only mean by Z and N, we don't mean by other things here because most of the questions are asked in this way. In NEET, you need to keep in mind that proportional relationship is the most important. So, here you see mu naught over 2 is constant. I n's proportionality is z square by n cube. And R n's proportionality is n square by z. So, the final proportional relation will be z cube by n to power 5. That you need to keep in mind. So, if we move ahead, if we have magnetic moment of nth orbit, then magnetic moment of circular coil can be defined. multiplied by the area enclosed by the current is pi rn square. Magnetic field you have read that any current carrying, close coil magnetic moment is current multiplied by the area enclosed by the coil. So here in circular orbit if there is i current, so the area enclosed by it is pi r square. If you substitute the values, final result will be Enh over 4 pi m. I am taking direct result. Derivation you can do in your copy after this class. or you can see in the book or concert video. So you can see it does not depend on z. So here when we take out for n equals to 1, this is very important thing. For n equals to 1, this gives magnetic moment for the first orbit is E h over 4 pi m, this is defined as Bohr magneton, which is used as a fundamental unit to measure magnetic moment. The least magnetic moment is That's what it is termed as Fundamental Unit to Measure Bohr Magnetron. That means if the magnetic moment of any electron is taken out in any orbit, then it will be taken out in terms of this unit. This is the minimum magnetic moment which can be possible. This is E h over 4 pi m. So, you will just keep attention for the terminology. Last important property of electron is that the remaining checklist or chapter, So this entire chapter depends on electron energy and on different questions based on it. So we will discuss all that later. So when we talk about energy of electron, in nth orbit energy is generally defined as kinetic plus potential. But when we look at kinetic energy, then from first postulate, half m Vn square will be taken out to be kZd square by to Rn. So, energy we can write as kinetic energy plus potential energy. So, this comes out to be, directly we write this as half mvn square which we will write as kz square over Rn. And potential energy because in electron and nucleus attractive forces are it will be minus, k z e square. over Rn which gives us minus half kZE square by Rn. This is the final result of total energy which we define. This is minus half kZE square over Rn. Or this is also following the common relation where ever a particle in inverse square force rotates in circular orbit. So, We have read that magnitude of total energy is equal to magnitude of kinetic energy and that is half of the magnitude of potential energy. So, that is what is being followed over here. Or as result me ap R and K value dal denge to this gives you the expression for an earth orbit energy that is also important you need to keep in mind, this minus 2 pi k z square. 2 pi square k square z square. Because if you add r value, then this is n square h square by 4 pi square k square. So, all those things will be pi square k square. z will also be multiplied. Multiply it by e to power 4 m by n square h square. I have just substituted the expression of rn here. And this is the result I am getting. So, if we separate z and n. if we separate them, then the final result is minus 13.6 into Z square by N square in electron volt. You have to keep this result in mind. Because the things I am going to take from the next point are associated with all the energies. This result is going to be used couple of times in analyzing different type of questions which are frequently asked in J main, J advanced, NEET, all the exams. So, we will keep these expressions in mind. Let's move on to the next point. Poji, please clear it. Thank you, Poji. So students, here we will talk about energy of different orbits because many times the numerical questions are asked, if you remember the direct energy in those questions, then that is going to help you in quick solving such questions. We are talking about energies of different energy levels in hydrogenic atom and these energies also give a clear picture for our thinking that how energy levels are distributed in an atom. If we calculate the energies, the expression is 13.6 z square by n square electron volt. So, n co 1. 2, 3, 4, 5 so the energies for first orbit will be 13.6, for second orbit 3.4, 1.5, 1.85, 0.54. I prefer that you remember these 5 energies on tips, this is going to help you out in many cases. So if you are able to recall these 5 energies, so you can see that the difference between 13.6 and 3.4, the gap between these two is 10.2 eV. If we look at the gap between If we look at the gap of 2 and 3 carefully, So gap of 3.4 and 1.5 is 1.89 eV. The gap of 3 and 4 is 0.66 eV. If you see the gap of 4 and 5, this is 0.31 eV. As we are moving away from the nucleus, the gap between the energy levels is decreasing. That means the energies are coming relatively close. So when we will make energy level diagram in terms of energy, the diagram which is in front of you is called energy level diagram. That means we talk about energies on y axis and x axis is energy level. So first orbit energy is minus 13.6 z square. This is the least possible energy, the most stable state closest to nucleus. After that n equal to 2, 3, 4, 5. As we are going to higher orbit, I am reducing the gap between the energy levels of these two. So, energy levels, we can say, as they move away from the nucleus, energy levels keep coming closer. This gives a clear idea about how energy electrons are distributed in different energy levels. Now what is the benefit of this? So when we talk about the benefit, we read the third postulate of Bohr's that when an electron transits from higher level to lower level, then it releases a photon equal to the difference of the energies. So if the energies of the energy levels are fixed, these are the only 5 energies, because if there is any question beyond the 5th orbit, we will calculate from this expression, but generally it does not come. The questions that come for the first 5 orbits are asked in GE mania NEET. So if you have these 5 energies on your mind, and I have written the difference as I remember because students keep asking questions and we keep solving them. So whenever you solve 30-40 questions and I am very sure those who have done modern physics practice will remember these values. The biggest benefit of remembering is that as soon as questions come, the spark in the brain is that which transition is related to it and how much energy is emitting. So you just need to keep in mind that from 5th orbit to 4th orbit, 5th to 3rd, 5th to 2nd, 5th to 1st, if an electron transits in any orbit, So, the electron transition will emit photons. What will be the energy of the energy emitted? That is what you need to keep in mind. And if you remember these energies, then it will be removed. As we directly talk, energy is transmitted from the 4th orbit to the 1st orbit. So, the energy of the 4th orbit is 0.85 first year minus 13.6. All the values are negative. So, the difference will be, the photon emission here will be delta E. That will be 12.75 electron volt. 13.6 minus 0.85 directly comes out. Similarly, if transition is happening in 3 to 1, then the energy of 3 to 1, this is 4 to 1, this comes out to be, you can see, 1 is 13.63 is 1.51. So, if you take the difference, this gives you 12.09 electron volt. So, direct calculation will be done. 2 to 1 already it is quite obvious. Delta E if we take out 2 to 1 then it is 10.2 electron volt. That is the direct energy which we can write 2 or 1's difference. Apart from this also you can see all the possible energies. If you take out 4 to 2 then it is 0.85 or 13.6's difference. I am writing for the sample. Rest I am leaving it to you. You can directly calculate. Delta E 4 to 2 will be 2.55 electron volt. So, these direct values. It gives you quick spark while solving the question. Calculation time is saved a lot. There is no need of reverse calculation. So I still strongly suggest that keep in mind basic difference and energy values and keep energy level diagram in mind. Let's move ahead towards some other points. Poji please clear it. Thank you Poji. So students at this point we will talk about Excitation and ionization of an atom. By name all of you are well aware of When we give energy from outside to an atom, and that electron absorbs energy from lower orbit and excites in upper orbit, or transits in upper orbit, this phenomenon we call excitation of an atom. And if we give a lot of energy, if we give more energy than its ionization potential, then that electron comes out of the atom. So such energy which is required to eject an electron from an atom is called ionization energy, and the phenomenon is called ionization of an atom. I'm so... As far as name is concerned that's quite obvious. Excitation is basically called as transition from lower to higher orbit. Or ionization is from any orbit to ejection. Ejection means it is transition to N tend to infinity. When we talk about infinity, then energy is considered approximately zero because energy is minus 13.6 z square by n square. If you see in the previous slide, I did not state it separately, but it was written in the diagram, that if N tends to infinity, then the final energy level of the outermost energy level is considered zero. So up to E equal to zero energy level, i.e. outermost energy level, the excitation of that phenomenon is called ionization. But here one or two things are very important to understand. Say if this is n1 energy level and this is n2 energy level. For example, let's assume that n is equal to 1 and n is equal to 3. n1 is lower and n2 is higher so in between n equal to 2 will also exist. And let's assume that electron exists in n equal to 1. We know well that delta E from 1 to 3 is 12. point zero nine electron volt previous slide may I'm a bad career you one or three kajo energy level car difference of twelve point zero nine electron volt or delta even two is ten point two electron volt so I got bars M energy supply cut a electron code there are two ways in which energy can be supplied that also you need to keep in mind so yeah Pam look let and there are Two ways in which energy can be supplied for excitation or ionization. For excitation or ionization, two types of energy can be supplied. First method is by photon i.e. electromagnetic radiation. And the second method is by collision. I will cover the collision method in the last checklist. If you have read the concept videos, then in the end of the concept videos of atomic structure, we discussed how a particle is supplied with energy by collision. Here, for revision in the checklist, I am going to state the basic idea. Later, we will discuss it in detail. But the collision phenomenon... In general, GE is asked in advance. GE mains have many elementary cases. NEET generally is not asked. But still, because GE mains are involved in cases, we will definitely cover. So, when we supply photon, say if there is a photon, and we supply this electron, then electron will be excited by absorbing photon. But here, there is a condition of excitation, that photon energy has to be equal to exact difference of the two energy levels, otherwise it will not be absorbed. nahin hoga. Let's take two cases. Here we have case 1 which is photon energy. Energy of photon here is 12.09 electron volt. And we take case 2 here that is photon energy dash is equal to 11 electron volt. Two types of photon are incident here. So if we supply photons of 12.09 energy, then 12.09 is the difference from 1 to 3. So this electron will excite and jump here. And it will transit to n equals to 3 state. But which is this? Case 1. If we see case 2, then what will happen when we give 11 electron volt energy? If it was of 10.2, then it would jump from 1 to 2. If it is 11, then it should reach somewhere between 2 and 3, where there is a stable energy level. It means if the energy level is not stable then electron transit cannot happen. It means the important point to be understood is this photon will not be absorbed. You have to keep in mind that this photon will not be absorbed. That means if this photon is of case 2 then this photon will come out as it is. This is the case 2 photon. which is of energy 11 electron volt. It will not be absorbed. Absorbed will be the same photon which is equal to the exact difference. And you have to always keep in mind, when we talk about a photon, then photon is such an energy packet which is either absorbed or not absorbed. That's important for all of you to understand. So let us proceed further. You have to keep this in mind. When we talk about ionization, then ionization is for n equals to infinity. More than that, the energy you give will be absorbed. So, for ionization you have to keep in mind if, here we will write a small note, if supplied energy is more than ionization potential, which will be 13.6 electron volt for n equals to 1. So, if supplied energy is more than ionization potential then excess energy will become, Kinetic energy of ejected electron. We have to keep this in mind. So, if photon is in n equals to 1, then its energy is minus 13.6 electron volt. If this photon's energy is 13.6 electron volt, then its total energy will be zero when added, and it will go out. If you give it 15 electron volt, then its excess energy is 1.4 electron volt, so electron will go out with kinetic energy 1.4 electron volt. That's what you need to keep in mind. So, in ionization, it is not necessary to have exact energy. More than exact energy will serve the purpose. N equals to 3 energy is minus 1.51 electron volt. That means, if any electron is in third energy level, then theoretically, practically it is not possible. Why? Watch this thing in the concept video. That will explain you. But, keep in mind that if any electron is in 1.51 electron volt negative energy level, if we put it in 1.51 electron volt energy level, energy supply, then its total energy will be zero. So, this can be ionized. And the more energy it has, that will transform into its kinetic energy. That is a theoretical point. Practically, there are many aspects. We will see now, what questions can be made and how cases can come in front of us. So, this excitation-ionization part should be very clear to all of you. I have not covered one part. That is excitation and ionization by collision. I will cover this as a point at the end of this checklist. Let's move on to the next point. Please clear it. Thank you Poji. So, student next point we will talk about frequency and wavelength of emitted radiation. So, first thing when radiation emits? We have talked about this in third postulate that whenever electron transitions from higher energy level to lower energy level, then radiation emits in form of a single photon. I told you that only one photon will emit whose energy is equal to the difference of higher and lower energy level which is involved in transition. That means if electron is transitioning from N2 energy level to N1 energy level. So from here a single photon will emit. And this photon energy if we talk about. So here the photon energy will be. This delta E energy is equal to EN2 minus EN1. And this is equal to taken as HC by lambda. Or H nu where nu is the frequency of radiation emitted. Lambda is the wavelength. Here, if we put values of n2 and n1, then it is minus 13.6 z square by n2 square minus of minus 13.6 z square by n1 square is equal to hc by lambda. Just if you have a look here, it gives 1 by lambda is equal to 13.6 by hc multiplied by z square times 1 by n1 square minus 1 by n2 square. This is a very familiar and popular relation thirteen point six by h c is a constant. This constant is termed as Rydberg's constant. So, the expression becomes one by lambda is r z square one by n one square minus one by n two square. This is called Rydberg's formula for calculation of emitted wavelength. So, you have to keep in mind that whenever an electron is transitioned from any n2 orbit to n1 orbit, then the wavelength of the radiation emitted by electromagnetic radiation is given by this relation. n1 and n2 are integers. We can write the value of r as 10967800 per meter. Or it is approximately written as 10 to power 7 per meter. This is also sometimes taken as. So, you have to take this into consideration. 10 to the power 7 per meter. You take it. And numerical value I have written. It comes from different experiments. It comes from different. But in exam always the value of Rydberg constant will be given. That you can consider. This is an approximate value that we consider. So, we can get wavelength from Rydberg's formula. But there is one more way to get wavelength emitted. Here the 1 by lambda written here is also called new bar. That is wave number. Just for the sake of term I have written it. This number of waves per unit length. This wavelength has a direct relation to it. We can do this. We have already covered the values of energies in the previous slide. Of energies from 1 to 5. So if you have clear energies, you can directly substitute. So I am telling you the alternative method here. This is also very useful. Alternative way to calculate lambda. And here this is very important. Because in many cases this will directly help you and quickly you will be able to evaluate. Because we have delta E we can write as hc by lambda. So delta E is measured in electron volt. If we have values in electron volt, then directly you can write lambda's value. This is hc by delta E. And delta E we take here in electron volt values. So if you divide E's value by hc. So, the value will come out to be 1, 2, 4, 3, 1 divided by delta E and directly it gives you the value in angstrom but this energy is to be taken in electron volt. In the concept video, I have derived the entire value by calculating it. Here, just for the sake of clarity, I am explaining that if you have any question given to you, let's take an example here that there is a hydrogen atom. In hydrogen atom, n equals to 4 say n equals to 2. electron transition on 2. So, the photon that is produced, we have to find the wavelength of this photon. So, in this case, this wavelength which you want to calculate, you can directly write it as 12431 divided by 2.55 angstrom. This is the direct answer to this question. Where does 2.55 come from? You can go back 2-3 slides and see. I have told you about the energies of different energy levels. was mentioned there. N equals to 4 energy is 0.85. negative me n equals to 2 ka minus 3.4 difference is 2.55 toh agar difference yad hai toh ab directly ek step me division karke aap emitted radiation ki wavelength nikal sakte hai. Vice versa is also true kai baar wavelength of incident radiation given hai energy nikal na hai toh iska ulta bhi aap kar sakte hai. Delta E in electron volt can be calculated as 1, 2, 4, 3, 1 by lambda which you can write in angstrom. So, this is also an equal relation. Sometimes if wavelength is given, you can use relation directly to get energy. There should never be any confusion. So, to get this reciprocal, to square, to calculate, the whole process is bypassed. And this is the easiest and fastest way to evaluate energy from wavelength or wavelength from energy that you need to keep in mind. Let's move on to the next point. Poji, please clear it. Thank you, Poji. So, student, another important point from the point of view of questions, that is de-excitation of an atom. We have talked about excitation. We have talked about excitation. If any atom is supplied with energy from outside, then if that energy is in electromagnetic form and that energy is equal to the difference of a lower and higher energy level, then electron transits at a higher energy level. This is what we call excitation. Now we talk about de-excitation, this is also a very important point. If we assume that this is N1 energy level and this is higher N2 energy level, and there can be many energy levels in between, we are not right now bothered about it. And we assume that from a lower energy level, electron excites and reaches higher energy level, and it has absorbed some energy which is equal to the difference of N1 and N2. Now, if we talk about this higher energy level, then it came because it was given energy. But lower levels are also vacant. All of you are well aware about the capacity of every energy level or a shell. 2, 8, 18, you have already read that there is a maximum capacity of an energy level where it can retain electrons. After that, when electrons come from outside, they will go to higher shells. So if electrons go up after getting excited, but the lower levels are vacant, it means lower level, lower energy, this is more stable state, so it will not stay in higher energy level for long. When it gets excited, it is always for a short time. If you fill all the lower levels in the same time, it will remain above. Otherwise, after a very short time, it will transit back to the lower level. Now, that short time is called lifetime of excitation. So, we can write here, after lifetime of excitation, an electron transit to any of lower energy levels by releasing photons. Photons of which energy? Jis level mein jaega, uska jo difference hoga utti energy ka photon nikal dega. A manlo yeh upar toh paunch gaya after a very short time of lifetime of excitation of the order of. 10 to the power minus 8 second. Many of you are aware of it. If you want to study it in depth, then we will cover some advanced classes from Olympiad point of view. The series of Olympiad will also come in the coming time. Then it will explain you. But right now it is important 10 to the power minus 8 second is the maximum lifetime of excitation. This is the only time it transits back. Now where does it do random phenomena? Below 3, 4, 5 orbits are empty. So it can come from here to here. It can come from here to here. It can go anywhere because all the states are more stable compared to higher level. So it can go anywhere. Wherever the road is visible, it will be seen. We are not looking inside. It is a random phenomenon. So all possibilities exist. That's what we keep in mind. And then it came here. And then it came here. So it can come here. It can come here. It can come here. It can come here. There are many possibilities. Till the time it does transit, the lowest energy level, i.e. n equals to 1, does not come. Till then this transition will continue. I am saying n equals to 1 because in Bohr model we had discussed that this model is valid only for one electron system. And if there is only one electron, then for that most stable state would be n equals to 1 state. So after final transitions it will come to 1 and start will also be from 1. But when there are transitions in higher state, we will theoretically consider it and can ask questions on it. So in this case, after reaching n2 of all possible orbits, So, we can say that electron can transit back to n equals to 1 by following all possible transition between any two states. So, we can say that number of possible photons or photon energies emitted due to de-excitation from N equals to N2 state are Here we call it number of lines or spectral lines. Because if photons are emitted passing through spectrometer, then spectral lines are plotted in spectrum. So sometimes it is called number of spectral lines. Every individual photon has a spectral line. Every photon energy is corresponding to a separate spectral line. This value will be N2C2. You need to just count the number of combinations. Ki n2 orbit se finally 1 tak aana hai, to jitni bhi possible is straight on mein 2 2 ke combinations laya lo, nn minus 1 by 2. So this will be nn minus 1 by 2. These are the number of possible transitions. Yaani n equals to 4 mein hai to 4 into 3 by 2, total 6 lines aasakti hai. 5 mein hai to 5 into 4 by 2, 10 lines. Yaani 10 type of photons an atom can emit in different types of possible transitions that you need to keep in mind. One more thing is used here. Many times it is asked that how much maximum line can a single electron submit. So in the next point I will explain it separately. We will not mix it. Otherwise conceptual confusion can happen. So in the next point I will review this thing too. And there is a separate concept that also I am taking up. Stay tuned with that. So Pooji please clear it. Thank you Pooji. So students at this point we will talk about maximum spectral line submitted by an atom. The last point we discussed was that if from n equals to 1, someone transits to some n state, then all the states in between will be involved in de-excitation. Or if there is an electron in n state, then from various paths, it can finally reach n equals to 1. So here I repeat, from state n, maximum possible Spectral lines which can be emitted. It doesn't need maximum. Possible spectral lines which can be emitted that you need to keep in mind. And these are NC2 which we just talked about. But here one thing to understand is that when this NC, suppose if it bypassed one in between and came forward, and then came out, then number of lines will be reduced because in between one skippy happened. So, that possible transition cannot come whereas in NC2 we have counted all possible transition cases. So, when will this happen? When there will be a lot of atoms in a gas and all of them will be excited on n. So, in different different paths different atoms will produce photons. So, all of them will cumulatively produce these lines. So, from a gas in which there are millions trillions of atoms are there and all are excited to n equals to nth state or n2th state. but in In emitted radiation which comes due to de-excitation of all the atoms and gas, NC2 type of spectral lines will be emitted. i.e. if we make its spectrum, we will get NC2 number of lines. But now let's talk about just one atom. If a single atom is in N state, then you tell me how many maximum possible or minimum possible radiations can come out? Minimum you can directly say that if it comes directly from N to 1, This means that only one photon will come out from here, whose energy will be delta E from n to 1. That's it. And it will sit in 1 because there is one atom, one electron, and no other photon will come out. So, there is no minimum possible photon, only one photon will come out. Now, if we talk about the maximum number of photons, then you do the stepping. It came from n to n-1, then from n-1 to n-2, and reached 1 while doing it. So, what will be the corresponding of every transition? One photon will emit here. So, in this case here we can state in this situation for every step one photon is emitted. If it bypasses one step then one photon will be reduced. So, if there are n layers then maximum number of possible photons will be how many? Maximum possible spectral lines. Now, we do not use the term spectral line because whenever a spectrum is made it is always made by gas and not by any atom. We will write maximum possible photons. So I have removed maximum possible photons emitted by one atom. So here what will be the answer for one atom? It will be n minus 1. If it is in 5th state, then 5 to 4, 4 to 3, 3 to 2, 2 to 1, 4 will be emitted. If it is in n equals to 2, then 1 n equals to 3, 2 is there. So if it is in nth state, then n minus 1 maximum possible number of photons which can be emitted if it comes down to n equals to 1 at every step. If we include every step, then it will be maximum. Apart from this, the number of photons will always be less than n-1 and more than 1 because at least one photon is not released. Let's move on to further points. So, Pooji, please clear it. Thank you, Pooji. So, as to our next point, we will talk about Hydrogen Spectrum and Spectral Series. We talk about spectrum, so, all types of radiations that come out of hydrogen gas, that means, if hydrogen gas is put in a discharge tube and all the atoms are excited at high energy levels, then when they de-excite, all the radiation that comes out, The combinations that we discussed, as much radiation as possible can be released from the electronic transition of hydrogen gas, they should be passed through a spectrometer and a spectrum plot is made, that is, spectral lines should be arranged in a low wavelength to high wavelength or in the order of high energy to low energy. So, the spectral lines that are found in a photographic film are called hydrogen spectrum. And these spectral lines, i.e. the energy of photons emitted by hydrogen gas, they have a specific order. From that order, the energy of photons or spectral lines are classified into different groups, which we call spectral series. So, these are the two things we have to discuss quickly, hydrogen spectrum and spectral series. Let's first see the picture of hydrogen spectrum. Every line that you see here is depicting a specific wavelength or a specific energy radiation. If you look carefully, the line with the highest energy or lowest wavelength is on the left-most side because where the wavelength is lowest, you can say energy will be very high. And as we increase the wavelength, the energies will be relatively low. Group of lines are visible, these group of lines are called spectral series. If we want to see the corresponding transition of these spectral series, we can see in energy level diagram. You can see the second diagram below. In this diagram, the transition from any energy level from infinite to n equals to 1, which are considered as high energy transitions. So, in high energy transition, correspondingly, Lyman series is called. That means, if an electron jumps from anywhere from any energy level to n equals to 1, then the radiation emitted is and all the radiation is in a group as Lyman series. In this way, from anywhere from any infinity to n equal to 3, when electron jumps to n equal to 2, so this radiation we call as Balmer series. Similarly for n equal to 3, 4, 5, Lyman-Balmer, Parson, Brackett, Fund series, all these spectral series are obtained which are simply arranged in a group of energy or spectral lines. So these spectral series can be obtained and for every spectral series, John Spector series ki baat. So, in every spectral series there is a first line and a last line. First line is of spectral series corresponding to lowest energy level. When electron transit from immediate energy level above that, then the radiation that comes out is called first line. If we talk about Lyman series, then in case of Lyman series, the first line is corresponding to transition 2 to 1. And last line is corresponding to transition infinity to 1. Similarly, if we talk about Balmer series, in case of Balmer series, first line will be corresponding to 3 to 2 and last will be corresponding to infinity to 2. And so on, we can write all series lines. So, if we want to get wavelength of spectral series, if we want to get energy, then already in previous slide, I have told you, either you will get it from Redberg formula, or the direct relation we discussed, corresponding to energy. We can calculate wavelength 12431 by energy in electron volt. Either method you can use. So, this is also an important relation because many questions are directly asked in the name of spectral series. For first and last line, corresponding wavelength or energy comes out. So, directly you can handle these. So, let's move on to the next point. Poji, please clear it. Thank you, Poji. So one more point which is very crucial for GE advance aspirants, GE main may be up to now one or two times questions have been asked on this. Neat I don't think I can't recall any question asked on this concept, but it is important for GE advance, it is also important for GE main. And as GE main questions have been asked in NEET many times, so from that point of view it is useful for all students, that is effect of nuclear mass on Bohr model. The Bohr model that we discussed was on the basis of a nucleus and electron revolving around it. This was the concept we considered in an orbit radius Rn and electron velocity Vn. And we discussed a lot of electron properties, that is, its velocity, radius, angular speed, time, current, we took out all those parameters. But when we talk about practical consideration, then the case we are talking about is not practically like that. This is considered because nuclear mass is considered to be very heavy compared to electron. So we consider nucleus as rest and electron is considered to be rotating around it. But what happens practically is that since there is no external force in this system, then this system will always revolve about its center of mass. about revolve. Practically it is not possible that in absence of external force and trauma mass acceleration. And if the nucleus is fixed and the electron is rotating, even though the electron is very lightweight and the center of mass is outside the nucleus, it will be moving in a circular path which is not practically possible. So the actual situation is that if this is a nucleus, this is an electron, center of mass is located somewhere here, so the nucleus revolves about the center of mass and the electron revolves about this. So, the situation is always like this. The position at which the nucleus will be, the electron will always be on its opposite side such that the center of mass will remain at rest. And this is a kind of binary star system which we have talked about in gravitation. In case of a binary star, we always use reduced mass. So practically, if we want to take the mass of the nucleus into consideration or account it, then all the properties that we had taken out for the electron in Bohr model, All those properties will change. In which properties electron's mass is expressed. Like in radius of nth orbit we talked about mass is expressed in denominator. Radius of nth orbit was given as This was n square h square by 4 pi square k z e square m. This was the mass of electron. Whereas in case of expression of velocity it was independent of mass. This means that if the effect of mass is not on velocity, then the velocity will remain same. But when we talk about radius, this radius is taken out by considering nucleus at rest and electron moving around it. And if nucleus is also moving, electron is also moving, such that center of mass is at rest, then in this case, what we have to do first is to transform it in this inertial frame. For that, we use the reduced mass concept. If we consider nucleus at rest, So mass of electron will be replaced with the reduced mass of electron given as M E M N by M E plus M N. This reduced mass derivation we already discussed in concept videos. Now you want to see in concept videos or in the book of Physics Galaxy Volume 4, its full derivation is given. But rather than that, I want you to keep in mind as a checklist revision point. If ever the mass of nucleus is given in front of us in questions, which have come in J-Advance. So, in that case, we will replace mass of electron with reduced mass of electron. To replace this means that we have analyzed nucleus by considering it on rest. So, the difference will be that here as the radius is there, radius of 1th orbit will be given as n h square h square by 4 pi square k z e square. Here we will use mu e that is m e m n and in numerator it is m e plus m n. So, in this formula if this was Rn0 then Rn will be given as Rn0 multiplied by, here mE was already there so this will be taken as mE plus mN divided by mN. This is the only correction you need to introduce if you need to account nuclear mass in calculation of any of the electron property. In calculation of any of the electron property, if you want to account the mass of nucleus then you have to account this correction. Sometimes mass will come in the numerator like in the expression of energy. So what you have to do is multiply mn by me plus mn in the numerator. So you will get the difference. Now where do we need such cases? Once a question was asked in Olympiad or something was asked in G-Advance. So you need to be careful. Like I take a basic question. We have to calculate the difference of hydrogen and deuterium ionization energy. So, there will be no difference between Z and N in hydrogen and deuterium. The difference is in the mass of the nucleus. So, there we will have to make a correction in the expression of energy. Once we take the mass of hydrogen, once we take the mass of deuterium. And the difference in the values will be the difference of hydrogen and deuterium ionization energy. It will not come out directly. So, such cases are often required. You need to be careful. Let's move ahead. Poji, please clear it. Thank you, Pooji. So, the next point is the use of Bohr model in different potential energy systems. The Bohr model we have studied is in the electric field of a nucleus, in electrical potential energy, an electron is revolving around it. And we have discussed the entire analysis related to its properties and transitions in this chapter of atomic structure. Many times, such questions arise that instead of a nucleus, is a system given in which electricity electron or a charged particle is revolving and its potential energy expression is given. And we are told to use Bohr model to analyze that situation. I would like to handle the situation in the form of an example. I just explain it with the help of an illustration. given in a system a charge has potential energy given as Here, we have given potential energy as u is equal to k e logarithm of r. There is already a question which has earlier been asked in J-Advanced. So, for explanation, this is one of the best examples. But, anything can happen. We can take a sample here that potential energy as a function of distance f r is given. Anything can happen. Potential energy is given for a charged particle. E is the charge on the charged particle. and we are told to use Bohr model find radius of nth orbit for this charge. That means if we use Bohr model, then Bohr's second postulate will be applicable which states cannot revolve in every orbit of any radius. Here quantization of angular momentum i.e. mvr equals to nh over 2 pi will be applicable. So, here it is given that in this system a charge has potential energy given as this. We can also say more precisely a revolving charge. i.e. a picture is clear that we are talking about an orbital motion. Potential energy has a dependency on radius. So, from here we can get this. If we go for the solution. We can force on charge here will be given as minus of du by dr. Potential energy already negative hogi because it is in revolving stage i.e. attractive forces influence me. So, force ka jo magnitude aayega if you differentiate it this gives ke by r. So, this must be balancing mv square by r for the circular motion. Boor ki first postulate modify hogi. Differentiate karenge natural log of r ka derivative is 1 by r. So, this gives us m v n square is equal to equals to k e and by p two any second postulate we use we have m v n r n is equals to n h over two pi. So, in this case here the radius of n th orbit we are getting is n h over two pi m v n you can write as root of k e by m. So, this gives us this m also gets cancelled load So this gives nH over 2 pi root of mke. This is the radius of nth orbit. Same question can be asked, what will be the energy of charge in nth orbit? So we will get kinetic energy plus potential energy. So in any given case, the consideration will be only this much. That you will differentiate the energy given and get the force. Equated to mv square by r. This will be the modified equation. This is the modified equation of P1. You can say that the modified equation of the first postulate will come in front of us. So you need to be careful for the second postulate remains same. And you can analyze and calculate the things that have been asked in the question. So just be careful for such type of questions. It has been asked many times. Let's move ahead. Pooji, please clear it. Thank you, Pooji. So, we have reached the last point of this checklist, that is atomic collisions. This point I have already spoken earlier when I told you about excitation, that when any atom is supplied with energy, there are two ways for the excitation of an electron. One is electromagnetic radiation, i.e. giving energy through photons, about which we had talked that if supplied energy is equal to the exact difference of two energy levels, then the electron gets excite and goes to higher energy level. The second method was by collision. That's what I am going to discuss in this topic, atomic collision. When energy is supplied by collision, then what happens? I will take some elementary cases which are important for your JEE main and NEET. These things are also useful in JEE Advanced. But if you want to go on detailed analysis, so I would prefer that you go to Physics Galaxy Volume 4, in the first chapter, the last topic is given. With many examples, including the cases which have already been asked in JEE Advanced, you can review from there also. So if we talk practically, there is an N1 orbit and an N2 orbit and an electron exists in N1. If we give it energy from the outside with photons, we had said that it will be excited if the supplied energy is equal to the exact difference of N1 and N2. Otherwise, suppose an electron is coming from outside and finally it is coming and colliding with this electron. So, now we will discuss this. If the electron energy is supplied by the collision, then depending upon the impact parameter, the amount of supplied energy can be different. If this collision is considered perfectly inelastic, then in that case, inelastic means that all the energy is lost here. Because electron is such a small elementary particle, we don't worry much about its momentum. That means this electron can supply its entire energy to this electron. Or If it comes from some other direction, i.e. the impact parameter varies a little, so what happens in that case? In that case, the supplied energy will be less than the maximum energy it contains. Means, there will be a little less energy supply, any fraction of that energy can also be supplied. That is what you need to account for atomic collision. When this electron is coming and its kinetic energy is given to us is E0. So, we will assume that, this is a note you need to keep in mind, I will take an example also just to explain this specific note. In above collision, the supplied energy to atomic electron, this is the atomic electron, this is the external electron which is projected onto this particle. So, supplied energy to atomic electron can be Any fraction of E0, depending on impact parameter, that's not necessarily written over here. It is not necessary to write. But we will assume that any fraction of supplied energy can be absorbed. If it comes from above, from below, how much energy will it give in collision? Because collision is a kind of elastic phenomenon in which supplied energy can be any fraction of the maximum possible energy loss by the incoming particle. Let's talk about an example. We can see that in hydrogen, say if it is n equals to 2 and this is n equals to 4 and 2 has an electron. Or let's take 1. 1 has an electron. If we talk about 1 to 4. So 1 to 4, the energy difference is delta E14 is 12.75 electron volt. That means if we supply energy from photon, then that energy should be exactly 12.75. then only it will get excited or else it will not get excited. But, if we assume that an electron is coming from here and electron is coming and colliding to this electron and this electron's energy is 12.9 electron volts, say for example. So, if this was a photon, then it would not absorb electron energy because it is not equal to exact. And the difference of 1 to 5 is more than 12.9 and electron cannot stay in the middle. That means to absorb photon has to be of energy exactly 12.75 electron volt for this transition to occur. But if electron is coming, then we just talked that it can supply any fraction of its energy in collision. And 12.75 is less than this. This means it can give any energy between 0 to 12.9. It is not necessary that it will give. If there are many collisions, then 12.75 will be transferred in some collision. And what will happen? This electron will transit. That means at least one possible case is there. So we can say that if energy is being supplied through collision, in case of the colliding particle is also an electron, then the incoming kinetic energy of that electron can absorb less energy than that. Now you will say that 12.9 is more than 10.2, then it is absolutely correct. This means when an electron hits a wall, it may cause the electron to collapse. in n equals to 2, it can be in n equals to 3 or 4, there are possibilities in all three. So, there is a possibility in 4, but there is also a possibility in 2 or 3 because they are also less than this. So, depending upon the impact parameter of collision, i.e. the electron collides with it head-on or at which point, depending on the internal mechanism of collision, it can be excited in 2, 3 or 4. So, that's what you need to keep in mind. This is the basis factor in atomic collision. How do we get questions in GE Advanced? Instead of the coming electron, there can be an atom. then we will not be able to ignore its momentum and we need to account that when there is collision in both atoms, then maximum possible energy loss i.e. the coefficient of restitution should be 0. So, any fraction of maximum possible energy loss can be absorbed by electron for the purpose of transition. Here I am not analyzing it numerically, I have explained it. You can see it numerically in PG Concept videos, in the book, in Atomic Collision. There is a complete analysis and many illustrations. You can follow the same. So here I have taken up almost all the concepts related to atomic structure which are in J main, NEET and even in G advance. These are the only related questions. So if you are clear about all the things, then it is very very sure that this topic's question will not be wrong in J main, NEET. So I prefer you all to do your concept checklist properly. Review the previous year questions thoroughly. Make a checklist of all the things. Review the exams with the students who have them. Review them 2-3 times. Review them slowly. But make the checklist at your concept level. Make a copy of all the points. And map the checklist that I am covering here. Wish you all the best.

4 or 5 chapters are in this modern physics checklist. And today we are starting with the very first topic of atomic physics. Atomic physics may The section of atomic structure is extremely important for all of us. And this is the main basis of atomic structure that you can understand an atom well.

Variety of questions are made. Even in chemistry, some sections of atomic structure are common. But in physics, the atomic structure we study goes a little in-depth. And conceptual understanding is a little more than chemistry.

Rather, in chemistry, the syllabus is a lot. Even we reach the Schrodinger equation. In physics, there is no Schrodinger equation in the course. but still atomic structure. On this, variety and different type of questions which are asked in JEE Advanced, these are beyond the level of chemistry.

Although we will cover all the points in JEE Main Neat and even up to JEE Advanced level in Quick Revision Checklist. So, let's do it one by one. And I would say that those students who have not yet made a properly planned checklist, they should thoroughly prepare their concept checklist by previous year questions as well as with the concept checklist I am giving you. It is a small checklist but quickly. In coming time, we will cover all the concepts in this checklist which can be covered in questions JEE, MANIA, NEET.

We are going to cover. Stay tuned with the checklist points. So, students, we will start the first point in atomic structure, structure of atom.

There are three popular models of atom structure, you must have read in your basic theory. If you have any doubt, you can go to the physics galaxy concept videos and read in detail. The three models were Thomson's model, Rutherford's model and Bohr's model.

Quickly, let's take a review. The Thomson model was the plumb-poding model in which Thomson suggested that an atom is like a spherical ball in which the whole positive charge is uniformly scattered and the negative charges are randomly distributed like seeds in watermelon. And these negative charges are termed as electrons, negative charged particles and positive charge is fully distributed.

Until Thomson model, there was no concept of nucleus. This has become a popular model because it was explaining the neutral behavior of an atom. As well as few more factors it has explained. We don't need to get into much depth.

Second model, the Rutherford model that was based on alpha scattering experiment. You all know about alpha experiment. In which alpha particle scattering was found on gold foil. majority of space in an atom is empty and electrons are available around the central point where highly positive charge is concentrated and they keep rotating around it. Which was mentioned in Rutherford model.

And that was the discovery of nucleus that at the center point maximum mass and charge is concentrated in an atom. So Rutherford model was saying that electrons revolve around positive nucleus but according to Maxwell's theory When charge is accelerated, it radiates energy. So, it was assumed that if electron is in circular motion, which is an accelerated motion, and this energy will radiate, then it will slowly come and collapse in the nucleus.

Its path will be in spiral form. It will be continuously radiating and losing energy. The answer to this was not available in other Ford models. That's why other Ford models got failed.

But in Bohr model, all these things were explained. And that was the first most successful model of its time. which is still used to explain many classical theories. So Bohr model was explaining everything, all the concepts or drawbacks of Rutherford model.

Bohr model was given in the form of three postulates. In the next point, I will quickly cover Bohr model in detail so that you get the basic idea of Bohr model. And we will also talk in upcoming points about the questions which are framed on Bohr model.

So let's move ahead. Poji, please clear it. Thank you, Bhoji. So, students, now in this point we are going to talk about Bohr's atomic model.

If you have read the basic theory and concept videos, then I have already explained that Bohr's atomic model was explained in three postulates. And if we combine all three postulates, then we get a clear picture of the atom. We will review all three postulates one by one so that it will give you a better idea about Bohr's atomic model. So, first we talk about...

first postulate of Bohr model which was related to the circulation of electron in the surrounding of nucleus. If we take nuclear charge plus ZD then in its surrounding an electron is continuously revolving in a radius r. Now Bohr said that electrons continuously rotate around the nuclear charge without radiating energy.

Why doesn't energy radiate? This was clear in the second postulate and third postulate. Here it was said that electrons are rotating in different orbits and if an electron is in nth orbit whose radius is rn and its velocity is vn So, here the force of nuclear charge that is the Coulombic electric force Fe, this is given as KQ1Q2 by R square. So, here it becomes KZE square by RN square.

ZE is the charge of nucleus, E is the charge of electron. This force will be providing the necessary centripetal force for a stable circular motion. Or in the frame of electron, we can assume that centrifugal force is balanced with this force. Either way we can say. So, we can write kz square by Rn is equals to mvn square by Rn where Rn gets this Rn square.

So, Rn gets cancelled out. So, this gives mvn square is equals to kz square over Rn. This is the first postulate equation which was given as Bohr atomic models basis.

Electron continuously stable circular motion may whomta rata in the. positive electric field of nucleus and this is the first equation for Bohr's first postulate. Let's see about the second postulate. Poji please clear the board.

Thank you Poji. So student in second postulate Bohr said that when electron rotates around the nucleus then it cannot revolve in any orbit. It revolves in some specific orbits and those orbits are called quantized orbits or stable energy levels. So the basic sense of second postulate was It was only that electrons can revolve around nucleus only in some stable orbits. These are called energy levels and these are stable orbits.

The condition of stable orbits was given as m vn rn is equals to n h over 2 pi where n belongs to set of integers and m vn rn is the angular momentum of electron in a specific orbit. Yani second postulate ye kehati hai ki electron jis orbit mein stable revolve kar raha hai wahan pe uska angular momentum h over 2 pi ka integral multiple hona chahiye that's it. This is the second condition although iska derivation b hai Which was done by De Broglie's hypothesis that when electron rotates around nucleus at high speed, so in that case electron can be considered as an energy wave.

And if electron is revolving and behaving like a stable energy wave, so there should be a formation of stationary wave. I will quickly take a label here just to explain you. If there is an orbit of r radius and electron is moving at v velocity, So here wavelength...

of electron wave if we take out. We will cover De Broglie hypothesis in next checklist in matter waves. So, here wavelength of electron wave lambda is given as h over mv.

This most of you might be aware of that wavelength is given as h over momentum. So, its orbit length is 2 pi r for stationary wave formation. This was analyzed 2 pi r the length must be equals to n lambda when you substitute the value of lambda this 2 pi r is equals to n h over m v and here you can see this relation is leading to the derivation of Bohr's second postulate which is m v r is n h over 2 pi.

So, if electron is rotating in some circle, rotating in some orbit and due to high speed if it is considered to be behaving like an energy wave. So, in that case electron that if the energy wave is made a stationary wave in orbit, then no energy will radiate. That was what Bohr has explained. But Bohr was not able to directly give the reply.

This was analyzed later after de Broglie's hypothesis correlated with Bohr's atomic model. And then this relation was deduced. Whereas postulate means Bohr deduced these experimental results which were theoretically verified later by de Broglie's hypothesis.

So you just need to keep this result in mind as of now as second postulate. Let's move on to the next point. please clear karo Thank you Pooji.

So students, in third postulate, it was told that the energy emitted from any atom, how it is emitted. So third postulate was related to electron transition between orbits or energy levels in the surrounding of nucleus. So third postulate says that if there are two energy levels, n1 and n2, and if electron is in n2 energy level, and if it makes a transition to the lower energy level. Transition will happen only when lower level is vacant.

If lower level is vacant which is a more stable state. Because the lower energy level is near the nucleus. So electron will be more stable. So where there is more stability there energy is less. So when electron will transit from higher energy level to lower energy level.

So it will radiate energy. And this consider energy is radiated in the form of a single electron. electromagnetic radiation photon.

This is the most useful thing to keep in mind. If electron transits from a higher orbit to a lower orbit, then it always throws the difference of the energy as much as it is missing in the form of a single packet, that is a single photon. And here the photon energy, this photon energy can be directly called h nu or hc by lambda And this equals to E n2 minus. E n1 where E n2 is the energy of higher energy level, E n1 is the energy of lower energy level.

This is what is generally considered and this is the equation 3 of Bohr's third postulate. I will tell you again to understand that only one thing is that electron energy emits when it comes from higher to lower level. That is third postulate. But energy will come out in the form of a single photon, you must not consider that stream of photons will come out. If one electron jumps from higher to lower orbit, then it will produce one photon whose energy will be En2 minus En1, whatever is the difference of energies.

So these three postulates, if you combine and think all together, it gives you a clear picture about an atom, what is an atom, and what is the behavior of an electron in it. Let's move on to the next points. Poji, please clear it. Thank you, Poji.

So students, in the next point we will talk about the properties of electrons. If we consider the Bohr model in an atom, then what basic properties of electrons can be there? Because in J-main, NEET, Olympiad, even J-advanced, all places the questions of electron properties are always asked. I am taking basic properties first and keep in mind that all the properties in the three postulates of Bohr, P1, P2 and P3, whose three equations I have written here, These three postulates are deduced from the equation.

I am not taking the derivations here. If you have confusion, if you want to see it from the point of view of GE Advanced, then you can see all the derivations in the concept videos. So, if the first point is to be taken out, then you can eliminate Vn from P1 and P2 and take out the radius of nth orbit. The expression comes out to be n square h square by 4 pi square kz square m.

Yeah, but and... orbit number and z atomic number because we generally assume that Bohr model is valid only for one electron system. We will talk about Bohr model failure later. But everyone knows that Bohr model is most commonly used for hydrogenic atoms or one electron systems. So, if we remove n and z, then the value of all the constants is 0.529 into n square by Z Engstrom.

Keep in mind that radius is proportional to orbit square and inversely proportional to atomic number. These results are more useful. In this way, if you put this Rn value in P2, then you will get the second property, velocity of electron in nth orbit.

That comes out to be 2 pi k z square over nh. Important point to be understood here, that you see the radius of nth orbit. It depends on the mass of electron but the velocity of electron in nth orbit does not depend on mass of electron.

Many times questions come in this way that instead of electron, some other particle is given having charge equal to that of electron but mass double that of electron. So in that case, what will be the difference between its radius and velocity? So you can say that if mass is doubled then radius will be half. But if we talk about velocity, then velocity will not change on any orbit or atomic number. This is important to be noted that velocity of nth orbit, when we talk about velocity of particle which is revolving around the nucleus in nth orbit, this does not depend on mass of electron.

Rest, we will add value of constraints other than z and n. This comes out to be 2.18 into 10 to power 6 multiplied by z by n meters per second that you need to keep in mind. We will keep proportionality in mind. That's why I have written the result. You can see in derivation concept videos.

Or if you want, then in physics galaxy book volume 4, all these things are explained in great detail. There also you can go through it with variety and lot of illustrations and examples. So, if we talk about angular velocity in nth orbit. Now, if you remember Rn Vn expression, then you can write omega n directly. This is Vn by Rn.

You don't need to remember the result. You just see the proportionality result. How is the dependency on n and z? This is Vn by Rn.

So, this is proportional to z square by n cube. Radian per second. You have to keep in mind that when we compare the angular speed of two atoms, then angular speed of electron in nth orbit is proportional to z square and inversely proportional to n cube.

Similarly, if we talk about frequency of revolution of electron fn, you can write as omega n upon 2 pi. So, proportional relation remains same. This is z square over n cube.

This is per second only. We are talking about revolution per second. One point if we talk about time period of electron revolution in earth orbit. So time period of electron revolution is reciprocal of frequency. So the constant multiplied by n cube by z square seconds.

So these proportional relations work. Most of the questions come like this. That if in hydrogen and helium plus.

Helium plus is a one electron system. If we apply Bohr model. and compare it with the time period of revolution in hydrogen's second orbit and helium's fourth orbit, then you can directly find out the ratio of n cubed by z square. That you need to keep in mind. Few more properties we need to take up.

Let's proceed further. Poji, please clear it. Thank you Poji. So few more properties here we are going to take up. We had studied basic properties like radius, velocity, angular velocity, frequency, and time period of electron in nth orbit.

Now when electron is revolving in an anath orbit, then revolving electron is like a revolving charge. And if charge is rotating at a faster speed in a circle, then it behaves like a circular current or current carrying circular coil. So if electron is negative charge, then we always take current opposite to negative charge flow.

So we can also define current in anath orbit. If we see current in anath orbit, then it is a direct indication that current is given as. Always keep in mind that whenever a charge is rotating in a circular orbit, this thing...

Magnetic effects checklist also mentioned that if the charge is rotating in circle then current is given by simply charge multiplied by revolution frequency. This is the way how we define current. So, here charge is E and revolution frequency is Fn which we have covered in previous slide. So, if we just substitute the value of Fn.

So, this gives the proportional relation. If you will pay attention to Fn then we had done it is Z square by n cube's proportional. So, the same relation will be here it is z square by n cube this relationship will hold true. Similarly, if there is a current flow in a circle, then it will always produce a magnetic induction at its center.

So, here we talk about magnetic induction at nucleus due to nth orbit. And B produced at nucleus you know it is given by mu naught I by 2R. So, I put IN and RN.

IN will be the circular current of nth orbit and RN will be the radius of nth orbit. So if the current is I-flow in the circle, then the magnetic field at the center is taken out by Tesla. It is mu naught I by 2R. You can substitute the values by taking them out in the derivation book or concept video. Here the proportional relation is important.

What will be the ratio of magnetic induction at the center, i.e. at the nucleus, due to the third and fifth orbits of Li plus 2 and hydrogen? We only mean by Z and N, we don't mean by other things here because most of the questions are asked in this way. In NEET, you need to keep in mind that proportional relationship is the most important.

So, here you see mu naught over 2 is constant. I n's proportionality is z square by n cube. And R n's proportionality is n square by z.

So, the final proportional relation will be z cube by n to power 5. That you need to keep in mind. So, if we move ahead, if we have magnetic moment of nth orbit, then magnetic moment of circular coil can be defined. multiplied by the area enclosed by the current is pi rn square.

Magnetic field you have read that any current carrying, close coil magnetic moment is current multiplied by the area enclosed by the coil. So here in circular orbit if there is i current, so the area enclosed by it is pi r square. If you substitute the values, final result will be Enh over 4 pi m.

I am taking direct result. Derivation you can do in your copy after this class. or you can see in the book or concert video. So you can see it does not depend on z.

So here when we take out for n equals to 1, this is very important thing. For n equals to 1, this gives magnetic moment for the first orbit is E h over 4 pi m, this is defined as Bohr magneton, which is used as a fundamental unit to measure magnetic moment. The least magnetic moment is That's what it is termed as Fundamental Unit to Measure Bohr Magnetron. That means if the magnetic moment of any electron is taken out in any orbit, then it will be taken out in terms of this unit.

This is the minimum magnetic moment which can be possible. This is E h over 4 pi m. So, you will just keep attention for the terminology. Last important property of electron is that the remaining checklist or chapter, So this entire chapter depends on electron energy and on different questions based on it.

So we will discuss all that later. So when we talk about energy of electron, in nth orbit energy is generally defined as kinetic plus potential. But when we look at kinetic energy, then from first postulate, half m Vn square will be taken out to be kZd square by to Rn.

So, energy we can write as kinetic energy plus potential energy. So, this comes out to be, directly we write this as half mvn square which we will write as kz square over Rn. And potential energy because in electron and nucleus attractive forces are it will be minus, k z e square.

over Rn which gives us minus half kZE square by Rn. This is the final result of total energy which we define. This is minus half kZE square over Rn.

Or this is also following the common relation where ever a particle in inverse square force rotates in circular orbit. So, We have read that magnitude of total energy is equal to magnitude of kinetic energy and that is half of the magnitude of potential energy. So, that is what is being followed over here.

Or as result me ap R and K value dal denge to this gives you the expression for an earth orbit energy that is also important you need to keep in mind, this minus 2 pi k z square. 2 pi square k square z square. Because if you add r value, then this is n square h square by 4 pi square k square. So, all those things will be pi square k square.

z will also be multiplied. Multiply it by e to power 4 m by n square h square. I have just substituted the expression of rn here.

And this is the result I am getting. So, if we separate z and n. if we separate them, then the final result is minus 13.6 into Z square by N square in electron volt.

You have to keep this result in mind. Because the things I am going to take from the next point are associated with all the energies. This result is going to be used couple of times in analyzing different type of questions which are frequently asked in J main, J advanced, NEET, all the exams.

So, we will keep these expressions in mind. Let's move on to the next point. Poji, please clear it.

Thank you, Poji. So students, here we will talk about energy of different orbits because many times the numerical questions are asked, if you remember the direct energy in those questions, then that is going to help you in quick solving such questions. We are talking about energies of different energy levels in hydrogenic atom and these energies also give a clear picture for our thinking that how energy levels are distributed in an atom.

If we calculate the energies, the expression is 13.6 z square by n square electron volt. So, n co 1. 2, 3, 4, 5 so the energies for first orbit will be 13.6, for second orbit 3.4, 1.5, 1.85, 0.54. I prefer that you remember these 5 energies on tips, this is going to help you out in many cases. So if you are able to recall these 5 energies, so you can see that the difference between 13.6 and 3.4, the gap between these two is 10.2 eV. If we look at the gap between If we look at the gap of 2 and 3 carefully, So gap of 3.4 and 1.5 is 1.89 eV.

The gap of 3 and 4 is 0.66 eV. If you see the gap of 4 and 5, this is 0.31 eV. As we are moving away from the nucleus, the gap between the energy levels is decreasing.

That means the energies are coming relatively close. So when we will make energy level diagram in terms of energy, the diagram which is in front of you is called energy level diagram. That means we talk about energies on y axis and x axis is energy level.

So first orbit energy is minus 13.6 z square. This is the least possible energy, the most stable state closest to nucleus. After that n equal to 2, 3, 4, 5. As we are going to higher orbit, I am reducing the gap between the energy levels of these two.

So, energy levels, we can say, as they move away from the nucleus, energy levels keep coming closer. This gives a clear idea about how energy electrons are distributed in different energy levels. Now what is the benefit of this? So when we talk about the benefit, we read the third postulate of Bohr's that when an electron transits from higher level to lower level, then it releases a photon equal to the difference of the energies. So if the energies of the energy levels are fixed, these are the only 5 energies, because if there is any question beyond the 5th orbit, we will calculate from this expression, but generally it does not come.

The questions that come for the first 5 orbits are asked in GE mania NEET. So if you have these 5 energies on your mind, and I have written the difference as I remember because students keep asking questions and we keep solving them. So whenever you solve 30-40 questions and I am very sure those who have done modern physics practice will remember these values. The biggest benefit of remembering is that as soon as questions come, the spark in the brain is that which transition is related to it and how much energy is emitting. So you just need to keep in mind that from 5th orbit to 4th orbit, 5th to 3rd, 5th to 2nd, 5th to 1st, if an electron transits in any orbit, So, the electron transition will emit photons.

What will be the energy of the energy emitted? That is what you need to keep in mind. And if you remember these energies, then it will be removed. As we directly talk, energy is transmitted from the 4th orbit to the 1st orbit.

So, the energy of the 4th orbit is 0.85 first year minus 13.6. All the values are negative. So, the difference will be, the photon emission here will be delta E. That will be 12.75 electron volt. 13.6 minus 0.85 directly comes out.

Similarly, if transition is happening in 3 to 1, then the energy of 3 to 1, this is 4 to 1, this comes out to be, you can see, 1 is 13.63 is 1.51. So, if you take the difference, this gives you 12.09 electron volt. So, direct calculation will be done. 2 to 1 already it is quite obvious. Delta E if we take out 2 to 1 then it is 10.2 electron volt.

That is the direct energy which we can write 2 or 1's difference. Apart from this also you can see all the possible energies. If you take out 4 to 2 then it is 0.85 or 13.6's difference.

I am writing for the sample. Rest I am leaving it to you. You can directly calculate.

Delta E 4 to 2 will be 2.55 electron volt. So, these direct values. It gives you quick spark while solving the question.

Calculation time is saved a lot. There is no need of reverse calculation. So I still strongly suggest that keep in mind basic difference and energy values and keep energy level diagram in mind. Let's move ahead towards some other points. Poji please clear it.

Thank you Poji. So students at this point we will talk about Excitation and ionization of an atom. By name all of you are well aware of When we give energy from outside to an atom, and that electron absorbs energy from lower orbit and excites in upper orbit, or transits in upper orbit, this phenomenon we call excitation of an atom. And if we give a lot of energy, if we give more energy than its ionization potential, then that electron comes out of the atom.

So such energy which is required to eject an electron from an atom is called ionization energy, and the phenomenon is called ionization of an atom. I'm so... As far as name is concerned that's quite obvious.

Excitation is basically called as transition from lower to higher orbit. Or ionization is from any orbit to ejection. Ejection means it is transition to N tend to infinity.

When we talk about infinity, then energy is considered approximately zero because energy is minus 13.6 z square by n square. If you see in the previous slide, I did not state it separately, but it was written in the diagram, that if N tends to infinity, then the final energy level of the outermost energy level is considered zero. So up to E equal to zero energy level, i.e. outermost energy level, the excitation of that phenomenon is called ionization.

But here one or two things are very important to understand. Say if this is n1 energy level and this is n2 energy level. For example, let's assume that n is equal to 1 and n is equal to 3. n1 is lower and n2 is higher so in between n equal to 2 will also exist.

And let's assume that electron exists in n equal to 1. We know well that delta E from 1 to 3 is 12. point zero nine electron volt previous slide may I'm a bad career you one or three kajo energy level car difference of twelve point zero nine electron volt or delta even two is ten point two electron volt so I got bars M energy supply cut a electron code there are two ways in which energy can be supplied that also you need to keep in mind so yeah Pam look let and there are Two ways in which energy can be supplied for excitation or ionization. For excitation or ionization, two types of energy can be supplied. First method is by photon i.e. electromagnetic radiation. And the second method is by collision.

I will cover the collision method in the last checklist. If you have read the concept videos, then in the end of the concept videos of atomic structure, we discussed how a particle is supplied with energy by collision. Here, for revision in the checklist, I am going to state the basic idea. Later, we will discuss it in detail.

But the collision phenomenon... In general, GE is asked in advance. GE mains have many elementary cases. NEET generally is not asked. But still, because GE mains are involved in cases, we will definitely cover.

So, when we supply photon, say if there is a photon, and we supply this electron, then electron will be excited by absorbing photon. But here, there is a condition of excitation, that photon energy has to be equal to exact difference of the two energy levels, otherwise it will not be absorbed. nahin hoga.

Let's take two cases. Here we have case 1 which is photon energy. Energy of photon here is 12.09 electron volt.

And we take case 2 here that is photon energy dash is equal to 11 electron volt. Two types of photon are incident here. So if we supply photons of 12.09 energy, then 12.09 is the difference from 1 to 3. So this electron will excite and jump here.

And it will transit to n equals to 3 state. But which is this? Case 1. If we see case 2, then what will happen when we give 11 electron volt energy?

If it was of 10.2, then it would jump from 1 to 2. If it is 11, then it should reach somewhere between 2 and 3, where there is a stable energy level. It means if the energy level is not stable then electron transit cannot happen. It means the important point to be understood is this photon will not be absorbed. You have to keep in mind that this photon will not be absorbed.

That means if this photon is of case 2 then this photon will come out as it is. This is the case 2 photon. which is of energy 11 electron volt. It will not be absorbed.

Absorbed will be the same photon which is equal to the exact difference. And you have to always keep in mind, when we talk about a photon, then photon is such an energy packet which is either absorbed or not absorbed. That's important for all of you to understand. So let us proceed further. You have to keep this in mind.

When we talk about ionization, then ionization is for n equals to infinity. More than that, the energy you give will be absorbed. So, for ionization you have to keep in mind if, here we will write a small note, if supplied energy is more than ionization potential, which will be 13.6 electron volt for n equals to 1. So, if supplied energy is more than ionization potential then excess energy will become, Kinetic energy of ejected electron. We have to keep this in mind. So, if photon is in n equals to 1, then its energy is minus 13.6 electron volt.

If this photon's energy is 13.6 electron volt, then its total energy will be zero when added, and it will go out. If you give it 15 electron volt, then its excess energy is 1.4 electron volt, so electron will go out with kinetic energy 1.4 electron volt. That's what you need to keep in mind.

So, in ionization, it is not necessary to have exact energy. More than exact energy will serve the purpose. N equals to 3 energy is minus 1.51 electron volt. That means, if any electron is in third energy level, then theoretically, practically it is not possible.

Why? Watch this thing in the concept video. That will explain you.

But, keep in mind that if any electron is in 1.51 electron volt negative energy level, if we put it in 1.51 electron volt energy level, energy supply, then its total energy will be zero. So, this can be ionized. And the more energy it has, that will transform into its kinetic energy.

That is a theoretical point. Practically, there are many aspects. We will see now, what questions can be made and how cases can come in front of us. So, this excitation-ionization part should be very clear to all of you.

I have not covered one part. That is excitation and ionization by collision. I will cover this as a point at the end of this checklist.

Let's move on to the next point. Please clear it. Thank you Poji.

So, student next point we will talk about frequency and wavelength of emitted radiation. So, first thing when radiation emits? We have talked about this in third postulate that whenever electron transitions from higher energy level to lower energy level, then radiation emits in form of a single photon. I told you that only one photon will emit whose energy is equal to the difference of higher and lower energy level which is involved in transition. That means if electron is transitioning from N2 energy level to N1 energy level.

So from here a single photon will emit. And this photon energy if we talk about. So here the photon energy will be.

This delta E energy is equal to EN2 minus EN1. And this is equal to taken as HC by lambda. Or H nu where nu is the frequency of radiation emitted.

Lambda is the wavelength. Here, if we put values of n2 and n1, then it is minus 13.6 z square by n2 square minus of minus 13.6 z square by n1 square is equal to hc by lambda. Just if you have a look here, it gives 1 by lambda is equal to 13.6 by hc multiplied by z square times 1 by n1 square minus 1 by n2 square. This is a very familiar and popular relation thirteen point six by h c is a constant.

This constant is termed as Rydberg's constant. So, the expression becomes one by lambda is r z square one by n one square minus one by n two square. This is called Rydberg's formula for calculation of emitted wavelength. So, you have to keep in mind that whenever an electron is transitioned from any n2 orbit to n1 orbit, then the wavelength of the radiation emitted by electromagnetic radiation is given by this relation.

n1 and n2 are integers. We can write the value of r as 10967800 per meter. Or it is approximately written as 10 to power 7 per meter.

This is also sometimes taken as. So, you have to take this into consideration. 10 to the power 7 per meter.

You take it. And numerical value I have written. It comes from different experiments.

It comes from different. But in exam always the value of Rydberg constant will be given. That you can consider. This is an approximate value that we consider.

So, we can get wavelength from Rydberg's formula. But there is one more way to get wavelength emitted. Here the 1 by lambda written here is also called new bar.

That is wave number. Just for the sake of term I have written it. This number of waves per unit length.

This wavelength has a direct relation to it. We can do this. We have already covered the values of energies in the previous slide.

Of energies from 1 to 5. So if you have clear energies, you can directly substitute. So I am telling you the alternative method here. This is also very useful.

Alternative way to calculate lambda. And here this is very important. Because in many cases this will directly help you and quickly you will be able to evaluate.

Because we have delta E we can write as hc by lambda. So delta E is measured in electron volt. If we have values in electron volt, then directly you can write lambda's value.

This is hc by delta E. And delta E we take here in electron volt values. So if you divide E's value by hc. So, the value will come out to be 1, 2, 4, 3, 1 divided by delta E and directly it gives you the value in angstrom but this energy is to be taken in electron volt.

In the concept video, I have derived the entire value by calculating it. Here, just for the sake of clarity, I am explaining that if you have any question given to you, let's take an example here that there is a hydrogen atom. In hydrogen atom, n equals to 4 say n equals to 2. electron transition on 2. So, the photon that is produced, we have to find the wavelength of this photon.

So, in this case, this wavelength which you want to calculate, you can directly write it as 12431 divided by 2.55 angstrom. This is the direct answer to this question. Where does 2.55 come from? You can go back 2-3 slides and see.

I have told you about the energies of different energy levels. was mentioned there. N equals to 4 energy is 0.85. negative me n equals to 2 ka minus 3.4 difference is 2.55 toh agar difference yad hai toh ab directly ek step me division karke aap emitted radiation ki wavelength nikal sakte hai. Vice versa is also true kai baar wavelength of incident radiation given hai energy nikal na hai toh iska ulta bhi aap kar sakte hai.

Delta E in electron volt can be calculated as 1, 2, 4, 3, 1 by lambda which you can write in angstrom. So, this is also an equal relation. Sometimes if wavelength is given, you can use relation directly to get energy.

There should never be any confusion. So, to get this reciprocal, to square, to calculate, the whole process is bypassed. And this is the easiest and fastest way to evaluate energy from wavelength or wavelength from energy that you need to keep in mind.

Let's move on to the next point. Poji, please clear it. Thank you, Poji.

So, student, another important point from the point of view of questions, that is de-excitation of an atom. We have talked about excitation. We have talked about excitation. If any atom is supplied with energy from outside, then if that energy is in electromagnetic form and that energy is equal to the difference of a lower and higher energy level, then electron transits at a higher energy level. This is what we call excitation.

Now we talk about de-excitation, this is also a very important point. If we assume that this is N1 energy level and this is higher N2 energy level, and there can be many energy levels in between, we are not right now bothered about it. And we assume that from a lower energy level, electron excites and reaches higher energy level, and it has absorbed some energy which is equal to the difference of N1 and N2. Now, if we talk about this higher energy level, then it came because it was given energy. But lower levels are also vacant.

All of you are well aware about the capacity of every energy level or a shell. 2, 8, 18, you have already read that there is a maximum capacity of an energy level where it can retain electrons. After that, when electrons come from outside, they will go to higher shells. So if electrons go up after getting excited, but the lower levels are vacant, it means lower level, lower energy, this is more stable state, so it will not stay in higher energy level for long. When it gets excited, it is always for a short time.

If you fill all the lower levels in the same time, it will remain above. Otherwise, after a very short time, it will transit back to the lower level. Now, that short time is called lifetime of excitation. So, we can write here, after lifetime of excitation, an electron transit to any of lower energy levels by releasing photons.

Photons of which energy? Jis level mein jaega, uska jo difference hoga utti energy ka photon nikal dega. A manlo yeh upar toh paunch gaya after a very short time of lifetime of excitation of the order of.

10 to the power minus 8 second. Many of you are aware of it. If you want to study it in depth, then we will cover some advanced classes from Olympiad point of view.

The series of Olympiad will also come in the coming time. Then it will explain you. But right now it is important 10 to the power minus 8 second is the maximum lifetime of excitation.

This is the only time it transits back. Now where does it do random phenomena? Below 3, 4, 5 orbits are empty. So it can come from here to here.

It can come from here to here. It can go anywhere because all the states are more stable compared to higher level. So it can go anywhere.

Wherever the road is visible, it will be seen. We are not looking inside. It is a random phenomenon.

So all possibilities exist. That's what we keep in mind. And then it came here.

And then it came here. So it can come here. It can come here.

It can come here. It can come here. There are many possibilities.

Till the time it does transit, the lowest energy level, i.e. n equals to 1, does not come. Till then this transition will continue. I am saying n equals to 1 because in Bohr model we had discussed that this model is valid only for one electron system.

And if there is only one electron, then for that most stable state would be n equals to 1 state. So after final transitions it will come to 1 and start will also be from 1. But when there are transitions in higher state, we will theoretically consider it and can ask questions on it. So in this case, after reaching n2 of all possible orbits, So, we can say that electron can transit back to n equals to 1 by following all possible transition between any two states. So, we can say that number of possible photons or photon energies emitted due to de-excitation from N equals to N2 state are Here we call it number of lines or spectral lines.

Because if photons are emitted passing through spectrometer, then spectral lines are plotted in spectrum. So sometimes it is called number of spectral lines. Every individual photon has a spectral line.

Every photon energy is corresponding to a separate spectral line. This value will be N2C2. You need to just count the number of combinations.

Ki n2 orbit se finally 1 tak aana hai, to jitni bhi possible is straight on mein 2 2 ke combinations laya lo, nn minus 1 by 2. So this will be nn minus 1 by 2. These are the number of possible transitions. Yaani n equals to 4 mein hai to 4 into 3 by 2, total 6 lines aasakti hai. 5 mein hai to 5 into 4 by 2, 10 lines. Yaani 10 type of photons an atom can emit in different types of possible transitions that you need to keep in mind. One more thing is used here.

Many times it is asked that how much maximum line can a single electron submit. So in the next point I will explain it separately. We will not mix it.

Otherwise conceptual confusion can happen. So in the next point I will review this thing too. And there is a separate concept that also I am taking up.

Stay tuned with that. So Pooji please clear it. Thank you Pooji. So students at this point we will talk about maximum spectral line submitted by an atom. The last point we discussed was that if from n equals to 1, someone transits to some n state, then all the states in between will be involved in de-excitation.

Or if there is an electron in n state, then from various paths, it can finally reach n equals to 1. So here I repeat, from state n, maximum possible Spectral lines which can be emitted. It doesn't need maximum. Possible spectral lines which can be emitted that you need to keep in mind. And these are NC2 which we just talked about.

But here one thing to understand is that when this NC, suppose if it bypassed one in between and came forward, and then came out, then number of lines will be reduced because in between one skippy happened. So, that possible transition cannot come whereas in NC2 we have counted all possible transition cases. So, when will this happen?

When there will be a lot of atoms in a gas and all of them will be excited on n. So, in different different paths different atoms will produce photons. So, all of them will cumulatively produce these lines. So, from a gas in which there are millions trillions of atoms are there and all are excited to n equals to nth state or n2th state. but in In emitted radiation which comes due to de-excitation of all the atoms and gas, NC2 type of spectral lines will be emitted.

i.e. if we make its spectrum, we will get NC2 number of lines. But now let's talk about just one atom. If a single atom is in N state, then you tell me how many maximum possible or minimum possible radiations can come out?

Minimum you can directly say that if it comes directly from N to 1, This means that only one photon will come out from here, whose energy will be delta E from n to 1. That's it. And it will sit in 1 because there is one atom, one electron, and no other photon will come out. So, there is no minimum possible photon, only one photon will come out. Now, if we talk about the maximum number of photons, then you do the stepping.

It came from n to n-1, then from n-1 to n-2, and reached 1 while doing it. So, what will be the corresponding of every transition? One photon will emit here. So, in this case here we can state in this situation for every step one photon is emitted. If it bypasses one step then one photon will be reduced.

So, if there are n layers then maximum number of possible photons will be how many? Maximum possible spectral lines. Now, we do not use the term spectral line because whenever a spectrum is made it is always made by gas and not by any atom. We will write maximum possible photons. So I have removed maximum possible photons emitted by one atom.

So here what will be the answer for one atom? It will be n minus 1. If it is in 5th state, then 5 to 4, 4 to 3, 3 to 2, 2 to 1, 4 will be emitted. If it is in n equals to 2, then 1 n equals to 3, 2 is there.

So if it is in nth state, then n minus 1 maximum possible number of photons which can be emitted if it comes down to n equals to 1 at every step. If we include every step, then it will be maximum. Apart from this, the number of photons will always be less than n-1 and more than 1 because at least one photon is not released. Let's move on to further points.

So, Pooji, please clear it. Thank you, Pooji. So, as to our next point, we will talk about Hydrogen Spectrum and Spectral Series.

We talk about spectrum, so, all types of radiations that come out of hydrogen gas, that means, if hydrogen gas is put in a discharge tube and all the atoms are excited at high energy levels, then when they de-excite, all the radiation that comes out, The combinations that we discussed, as much radiation as possible can be released from the electronic transition of hydrogen gas, they should be passed through a spectrometer and a spectrum plot is made, that is, spectral lines should be arranged in a low wavelength to high wavelength or in the order of high energy to low energy. So, the spectral lines that are found in a photographic film are called hydrogen spectrum. And these spectral lines, i.e. the energy of photons emitted by hydrogen gas, they have a specific order.

From that order, the energy of photons or spectral lines are classified into different groups, which we call spectral series. So, these are the two things we have to discuss quickly, hydrogen spectrum and spectral series. Let's first see the picture of hydrogen spectrum. Every line that you see here is depicting a specific wavelength or a specific energy radiation. If you look carefully, the line with the highest energy or lowest wavelength is on the left-most side because where the wavelength is lowest, you can say energy will be very high.

And as we increase the wavelength, the energies will be relatively low. Group of lines are visible, these group of lines are called spectral series. If we want to see the corresponding transition of these spectral series, we can see in energy level diagram. You can see the second diagram below.

In this diagram, the transition from any energy level from infinite to n equals to 1, which are considered as high energy transitions. So, in high energy transition, correspondingly, Lyman series is called. That means, if an electron jumps from anywhere from any energy level to n equals to 1, then the radiation emitted is and all the radiation is in a group as Lyman series. In this way, from anywhere from any infinity to n equal to 3, when electron jumps to n equal to 2, so this radiation we call as Balmer series. Similarly for n equal to 3, 4, 5, Lyman-Balmer, Parson, Brackett, Fund series, all these spectral series are obtained which are simply arranged in a group of energy or spectral lines.

So these spectral series can be obtained and for every spectral series, John Spector series ki baat. So, in every spectral series there is a first line and a last line. First line is of spectral series corresponding to lowest energy level.

When electron transit from immediate energy level above that, then the radiation that comes out is called first line. If we talk about Lyman series, then in case of Lyman series, the first line is corresponding to transition 2 to 1. And last line is corresponding to transition infinity to 1. Similarly, if we talk about Balmer series, in case of Balmer series, first line will be corresponding to 3 to 2 and last will be corresponding to infinity to 2. And so on, we can write all series lines. So, if we want to get wavelength of spectral series, if we want to get energy, then already in previous slide, I have told you, either you will get it from Redberg formula, or the direct relation we discussed, corresponding to energy.

We can calculate wavelength 12431 by energy in electron volt. Either method you can use. So, this is also an important relation because many questions are directly asked in the name of spectral series. For first and last line, corresponding wavelength or energy comes out. So, directly you can handle these.

So, let's move on to the next point. Poji, please clear it. Thank you, Poji. So one more point which is very crucial for GE advance aspirants, GE main may be up to now one or two times questions have been asked on this.

Neat I don't think I can't recall any question asked on this concept, but it is important for GE advance, it is also important for GE main. And as GE main questions have been asked in NEET many times, so from that point of view it is useful for all students, that is effect of nuclear mass on Bohr model. The Bohr model that we discussed was on the basis of a nucleus and electron revolving around it. This was the concept we considered in an orbit radius Rn and electron velocity Vn. And we discussed a lot of electron properties, that is, its velocity, radius, angular speed, time, current, we took out all those parameters.

But when we talk about practical consideration, then the case we are talking about is not practically like that. This is considered because nuclear mass is considered to be very heavy compared to electron. So we consider nucleus as rest and electron is considered to be rotating around it.

But what happens practically is that since there is no external force in this system, then this system will always revolve about its center of mass. about revolve. Practically it is not possible that in absence of external force and trauma mass acceleration. And if the nucleus is fixed and the electron is rotating, even though the electron is very lightweight and the center of mass is outside the nucleus, it will be moving in a circular path which is not practically possible.

So the actual situation is that if this is a nucleus, this is an electron, center of mass is located somewhere here, so the nucleus revolves about the center of mass and the electron revolves about this. So, the situation is always like this. The position at which the nucleus will be, the electron will always be on its opposite side such that the center of mass will remain at rest. And this is a kind of binary star system which we have talked about in gravitation. In case of a binary star, we always use reduced mass.

So practically, if we want to take the mass of the nucleus into consideration or account it, then all the properties that we had taken out for the electron in Bohr model, All those properties will change. In which properties electron's mass is expressed. Like in radius of nth orbit we talked about mass is expressed in denominator. Radius of nth orbit was given as This was n square h square by 4 pi square k z e square m. This was the mass of electron.

Whereas in case of expression of velocity it was independent of mass. This means that if the effect of mass is not on velocity, then the velocity will remain same. But when we talk about radius, this radius is taken out by considering nucleus at rest and electron moving around it.

And if nucleus is also moving, electron is also moving, such that center of mass is at rest, then in this case, what we have to do first is to transform it in this inertial frame. For that, we use the reduced mass concept. If we consider nucleus at rest, So mass of electron will be replaced with the reduced mass of electron given as M E M N by M E plus M N. This reduced mass derivation we already discussed in concept videos. Now you want to see in concept videos or in the book of Physics Galaxy Volume 4, its full derivation is given.

But rather than that, I want you to keep in mind as a checklist revision point. If ever the mass of nucleus is given in front of us in questions, which have come in J-Advance. So, in that case, we will replace mass of electron with reduced mass of electron. To replace this means that we have analyzed nucleus by considering it on rest.

So, the difference will be that here as the radius is there, radius of 1th orbit will be given as n h square h square by 4 pi square k z e square. Here we will use mu e that is m e m n and in numerator it is m e plus m n. So, in this formula if this was Rn0 then Rn will be given as Rn0 multiplied by, here mE was already there so this will be taken as mE plus mN divided by mN. This is the only correction you need to introduce if you need to account nuclear mass in calculation of any of the electron property. In calculation of any of the electron property, if you want to account the mass of nucleus then you have to account this correction.

Sometimes mass will come in the numerator like in the expression of energy. So what you have to do is multiply mn by me plus mn in the numerator. So you will get the difference. Now where do we need such cases? Once a question was asked in Olympiad or something was asked in G-Advance.

So you need to be careful. Like I take a basic question. We have to calculate the difference of hydrogen and deuterium ionization energy.

So, there will be no difference between Z and N in hydrogen and deuterium. The difference is in the mass of the nucleus. So, there we will have to make a correction in the expression of energy. Once we take the mass of hydrogen, once we take the mass of deuterium. And the difference in the values will be the difference of hydrogen and deuterium ionization energy.

It will not come out directly. So, such cases are often required. You need to be careful.

Let's move ahead. Poji, please clear it. Thank you, Pooji.

So, the next point is the use of Bohr model in different potential energy systems. The Bohr model we have studied is in the electric field of a nucleus, in electrical potential energy, an electron is revolving around it. And we have discussed the entire analysis related to its properties and transitions in this chapter of atomic structure. Many times, such questions arise that instead of a nucleus, is a system given in which electricity electron or a charged particle is revolving and its potential energy expression is given. And we are told to use Bohr model to analyze that situation.

I would like to handle the situation in the form of an example. I just explain it with the help of an illustration. given in a system a charge has potential energy given as Here, we have given potential energy as u is equal to k e logarithm of r. There is already a question which has earlier been asked in J-Advanced. So, for explanation, this is one of the best examples.

But, anything can happen. We can take a sample here that potential energy as a function of distance f r is given. Anything can happen.

Potential energy is given for a charged particle. E is the charge on the charged particle. and we are told to use Bohr model find radius of nth orbit for this charge. That means if we use Bohr model, then Bohr's second postulate will be applicable which states cannot revolve in every orbit of any radius.

Here quantization of angular momentum i.e. mvr equals to nh over 2 pi will be applicable. So, here it is given that in this system a charge has potential energy given as this. We can also say more precisely a revolving charge. i.e. a picture is clear that we are talking about an orbital motion.

Potential energy has a dependency on radius. So, from here we can get this. If we go for the solution.

We can force on charge here will be given as minus of du by dr. Potential energy already negative hogi because it is in revolving stage i.e. attractive forces influence me. So, force ka jo magnitude aayega if you differentiate it this gives ke by r. So, this must be balancing mv square by r for the circular motion.

Boor ki first postulate modify hogi. Differentiate karenge natural log of r ka derivative is 1 by r. So, this gives us m v n square is equal to equals to k e and by p two any second postulate we use we have m v n r n is equals to n h over two pi. So, in this case here the radius of n th orbit we are getting is n h over two pi m v n you can write as root of k e by m. So, this gives us this m also gets cancelled load So this gives nH over 2 pi root of mke.

This is the radius of nth orbit. Same question can be asked, what will be the energy of charge in nth orbit? So we will get kinetic energy plus potential energy.

So in any given case, the consideration will be only this much. That you will differentiate the energy given and get the force. Equated to mv square by r.

This will be the modified equation. This is the modified equation of P1. You can say that the modified equation of the first postulate will come in front of us. So you need to be careful for the second postulate remains same. And you can analyze and calculate the things that have been asked in the question.

So just be careful for such type of questions. It has been asked many times. Let's move ahead. Pooji, please clear it.

Thank you, Pooji. So, we have reached the last point of this checklist, that is atomic collisions. This point I have already spoken earlier when I told you about excitation, that when any atom is supplied with energy, there are two ways for the excitation of an electron.

One is electromagnetic radiation, i.e. giving energy through photons, about which we had talked that if supplied energy is equal to the exact difference of two energy levels, then the electron gets excite and goes to higher energy level. The second method was by collision. That's what I am going to discuss in this topic, atomic collision.

When energy is supplied by collision, then what happens? I will take some elementary cases which are important for your JEE main and NEET. These things are also useful in JEE Advanced.

But if you want to go on detailed analysis, so I would prefer that you go to Physics Galaxy Volume 4, in the first chapter, the last topic is given. With many examples, including the cases which have already been asked in JEE Advanced, you can review from there also. So if we talk practically, there is an N1 orbit and an N2 orbit and an electron exists in N1. If we give it energy from the outside with photons, we had said that it will be excited if the supplied energy is equal to the exact difference of N1 and N2.

Otherwise, suppose an electron is coming from outside and finally it is coming and colliding with this electron. So, now we will discuss this. If the electron energy is supplied by the collision, then depending upon the impact parameter, the amount of supplied energy can be different. If this collision is considered perfectly inelastic, then in that case, inelastic means that all the energy is lost here.

Because electron is such a small elementary particle, we don't worry much about its momentum. That means this electron can supply its entire energy to this electron. Or If it comes from some other direction, i.e. the impact parameter varies a little, so what happens in that case?

In that case, the supplied energy will be less than the maximum energy it contains. Means, there will be a little less energy supply, any fraction of that energy can also be supplied. That is what you need to account for atomic collision.

When this electron is coming and its kinetic energy is given to us is E0. So, we will assume that, this is a note you need to keep in mind, I will take an example also just to explain this specific note. In above collision, the supplied energy to atomic electron, this is the atomic electron, this is the external electron which is projected onto this particle.

So, supplied energy to atomic electron can be Any fraction of E0, depending on impact parameter, that's not necessarily written over here. It is not necessary to write. But we will assume that any fraction of supplied energy can be absorbed. If it comes from above, from below, how much energy will it give in collision? Because collision is a kind of elastic phenomenon in which supplied energy can be any fraction of the maximum possible energy loss by the incoming particle.

Let's talk about an example. We can see that in hydrogen, say if it is n equals to 2 and this is n equals to 4 and 2 has an electron. Or let's take 1. 1 has an electron. If we talk about 1 to 4. So 1 to 4, the energy difference is delta E14 is 12.75 electron volt. That means if we supply energy from photon, then that energy should be exactly 12.75.

then only it will get excited or else it will not get excited. But, if we assume that an electron is coming from here and electron is coming and colliding to this electron and this electron's energy is 12.9 electron volts, say for example. So, if this was a photon, then it would not absorb electron energy because it is not equal to exact.

And the difference of 1 to 5 is more than 12.9 and electron cannot stay in the middle. That means to absorb photon has to be of energy exactly 12.75 electron volt for this transition to occur. But if electron is coming, then we just talked that it can supply any fraction of its energy in collision. And 12.75 is less than this.

This means it can give any energy between 0 to 12.9. It is not necessary that it will give. If there are many collisions, then 12.75 will be transferred in some collision. And what will happen?

This electron will transit. That means at least one possible case is there. So we can say that if energy is being supplied through collision, in case of the colliding particle is also an electron, then the incoming kinetic energy of that electron can absorb less energy than that.

Now you will say that 12.9 is more than 10.2, then it is absolutely correct. This means when an electron hits a wall, it may cause the electron to collapse. in n equals to 2, it can be in n equals to 3 or 4, there are possibilities in all three.

So, there is a possibility in 4, but there is also a possibility in 2 or 3 because they are also less than this. So, depending upon the impact parameter of collision, i.e. the electron collides with it head-on or at which point, depending on the internal mechanism of collision, it can be excited in 2, 3 or 4. So, that's what you need to keep in mind. This is the basis factor in atomic collision. How do we get questions in GE Advanced?

Instead of the coming electron, there can be an atom. then we will not be able to ignore its momentum and we need to account that when there is collision in both atoms, then maximum possible energy loss i.e. the coefficient of restitution should be 0. So, any fraction of maximum possible energy loss can be absorbed by electron for the purpose of transition. Here I am not analyzing it numerically, I have explained it.

You can see it numerically in PG Concept videos, in the book, in Atomic Collision. There is a complete analysis and many illustrations. You can follow the same. So here I have taken up almost all the concepts related to atomic structure which are in J main, NEET and even in G advance.

These are the only related questions. So if you are clear about all the things, then it is very very sure that this topic's question will not be wrong in J main, NEET. So I prefer you all to do your concept checklist properly.

Review the previous year questions thoroughly. Make a checklist of all the things. Review the exams with the students who have them.

Review them 2-3 times. Review them slowly. But make the checklist at your concept level. Make a copy of all the points.

And map the checklist that I am covering here. Wish you all the best.

Transcript for:Key Concepts in Modern Atomic Physics

Transcript for:
Key Concepts in Modern Atomic Physics