 Hello guys, good morning all of you. Welcome to the session. Today we are going to solve some problems based on the chapter atomic structure. Okay, the session going to help you in your coming exams like KVPY, JEE, all of the no comparative exams, right? So do watch it. Yeah, so we'll start with the questions. Okay, so if you want to try the question on your own, you can pause the video anytime and then you can try on your own, right? Okay, so let us start with the session. So let's see the first question here. Okay, so let us discuss these two questions. The first one you see, the other force alpha particle scattering experiment eventually led to the conclusion and that conclusion is, okay, mass and energy are related. Electrons occupy space around the nucleus, electrons are buried deep into the nucleus. The point of impact with matter can be precisely determined. Okay, so it is the other four, right? So point of impact with matter can be precisely determined. It is not at all given by another four, right? Electrons are buried deep into the nucleus, which is not true. If it is true, then the atoms are not stable, right? So that's why C and D are wrong. Mass and energy are related is the Einstein equation. E equals to mc is square not given by other force. Option B is electron occupy space around the nucleus. Okay, so option D is correct for this particular question. Okay, so we are left with only option B. Okay, the other four experiment was what? If you try to remember this, it is the alpha particle scattering experiment. Okay, we have a gold foil and when the light strikes, sorry, the alpha particle strikes at the gold surface, which is this, suppose, suppose this is the gold foil we have, if you remember this, another four is scattering experiment. Okay, so we have a gold foil and all this gold foil, the alpha particle strikes. Okay, there was source of alpha particle here and this strikes on this surface, right? So what we observe, some of the alpha particle passes through the gold foil without any deflection, right? This is what we observe. Without any deflection, the alpha particle passes through. Okay, some of the alpha particle strikes and it deviates with a very small angle and moves like this. It comes over here and then it deviates like this. This was alpha particle scattering experiment. So some of the alpha particles deviates with a very small angle. Just a quick revision of it if you do not memorize it. Okay, and a very few of them retrace its path, right? It strikes and comes back. Okay, so you see there are spaces which are vacant and then only it is possible that some of the alpha particle passes through without any deviation, right? So that's why we can say there are spaces available vacant within the atom and that's what it means. The electron occupies space around the nucleus. So we have nucleus supposed like this. It is given in other way, but it means the same thing. Suppose this is the atom we have, right? Atom we have, we have nucleus here and there are electrons present in the orbital, right? So within this, in between the electron and nucleus, there are spaces, there are space available, right? These are the space we have, which is vacant available around the nucleus in an atom, right? That is what the second option we have, that space is available around the nucleus in an atom and this is the orbit in which the electron is present, right? Hence option B is correct, okay? Yeah. So this is for question number 11. Now the question number 12 we have, which one of the following sets of quantum number represents impossible arrangement, okay? So impossible arrangement, we know the, you know, values. N is the principal quantum number. L value is, can be anything 0 to N minus 1, okay? N can be anything from 1 to infinity. Only 0 is not possible for N. M value is lies in this range minus L2 plus L and Ls is either plus half or minus half. That is what we know, correct? Now N can be anything. So this is, this is all or possible. Ms is nothing but S, it's been quantum number. This is also possible, right? Plus half or minus half. Now, you see this L and M, first of all, L and N cannot be equal. That's the one thing. The two things that you have to keep in mind that N must not be equals to L and ML can be equal, right? L, or we can say M should not be greater than L. This is a relation we have. Easy one. You can easily do this. So 3 plus half possible. So 3 to 0 to 2, L is possible minus 2 to 2 plus 2, M is also possible. 4, 0, 0 plus half possible. 3, 2, you see this ML belongs to minus 2 to 2 plus 2, right? So this cannot be greater or equal to 2 or M cannot be less than equal to, cannot be less than minus L. That is what the true relation we have. Not greater than L, not less than minus L, including the sign, okay? So minus 2 to 2 plus 2 minus 3 is not possible for L, right? So this option C, you see, it is definitely wrong. 5, 3, 0, minus half, which is correct. So option C is wrong for this one. Hence the answer for the second question is option C. 12th one is option C, okay? I hope you understood these two questions. Let us move to the another question. The ratio of the energy of a photon, the ratio of the energy of a photon, 2000 angstrom, wavelength radiation to that of 4000 angstrom, okay? So we need to find out the ratio of the energy associated with these two wavelengths, okay? Easy one. You see, the answer for this question would be, we can easily solve this. We know the energy associated with the wavelength lambda is equals to Hc by lambda. So what we can conclude from this, the energy and wavelength are inversely proportional. Energy and wavelength are inversely proportional. So what we can write, even by E2 is equals to lambda 2 by lambda 1, okay? Lambda 2 by lambda 1. So this is equals to lambda 2 is 4000 and lambda 1 is 2000, okay? The ratio of the energy of a photon, this, so we have this equation. So when you solve this, you'll get 2. So the ratio is even by E2 is to 1. The answer is option D. Now, the next one, the triad of the nuclei, that is isotonic, okay? So what is isotonic first of all? Isotonic are the atoms which has, which has equal number of neutrons, equal number of neutrons. And how do we calculate neutrons? Suppose we have an atom X, right? Atom X, we have mass number A and atomic number Z. So the number of neutrons is A minus C. That is what the formula we have. Let us use this formula here. So for carbon 14 C6, the number of neutron is 8 for 15 and 7. The number of neutron is 8 for 17 F9. The number of neutron is 8. So we are getting A as the correct option here, okay? Isotonic equal number of neutrons, okay? If you look at the other, 12 minus 6, 6, 14 minus 7, 7, not possible. 14 minus 6, 8, 14 minus 7, 7, not possible. 14 minus 6, 8, 14 minus 7, 7. Hence D is also not possible, right? So option A is correct in this one. Isotonic is equal number of neutrons, okay? So the wavelength of a spectral line for an electronic transition is inversely related to, inversely related to the options are given, okay? The wavelength of a spectral line for an electronic transition. Okay. So wavelength, we know, is lambda, okay? And lambda relation is, suppose we have, we can, you know, approach this question this way. When electronic transition is there, right? So electron moves from an orbit which has some energy to the another orbit which also has some energy, electronic transition like this, initial and final. So if this EI, this delta E, if this EI is greater than, in terms of energy, it is greater than EF, then in this transition, the energy comes out, energy comes out in the form of, in the form of radiations and radiations has wavelength lambda, suppose, wavelength lambda. So delta E is the energy that comes out, that is, that has wavelength lambda. So delta E is equals to Hc by lambda and we have here EI minus EF. Remember, if you solve this, you'll get 1 by lambda is R, R into 1 by, 1 by NF square minus 1 by Ni square. This is what the relation we have, R is the Reedberg constant, okay? Let us see the option. The number of electrons undergoing a transition, no, the nuclear charge of the atom, no. A difference in the energy of the energy levels involved in the transition, the velocity of the electron undergoing the transition. Obviously, you see delta E is equals to Hc by lambda and delta, this lambda is inversely proportional to delta E, right? That's what we have, the difference in the energy of the energy level involved in the transition, correct? That's why question number 15, the correct answer is what? You see this relation here. From this, we can write lambda is inversely proportional to delta E and that is what we have the option, okay? So answer for this is C, okay? Next one. All these questions are C. They have asked in the exam, okay? These exams they have asked. Orbital in which the oboe principle is violated. What is oboe principle? Oboe principle, we also call it as N plus L, right? And we use this tool for the filling of electrons in the orbital. We know the lower energy orbital fills first and then the electron goes into the higher energy orbital. The orbital which has lower value of N plus L has lower energy, correct? Means the lower energy orbital fills first, right? So in the question, it is given 2S and 2P. So 2S and 2P, if you calculate the N plus L value, for 2S it is 2 plus 0, it is 2 and for 2P it is 2 plus 1, that is 3. It means that 2S orbital must be fully occupied first and then the electron goes into 2P orbital, right? So here's 2S is fully occupied and then the electron goes into 2P, right? So this option is not correct. Here you see the 2S orbital is not occupied and the 2P orbital has electron. Hence option B does not follow oboe principle. Answer would be option B. Here also it is correct. Here also it is correct. Okay, so the question is which of this orbital diagram does not obey Hans rule, right? So Hans rule is regarding the pairing of electron and it says the pairing of electron in the orbital belongs to the same sub-shell, in the orbital belongs to the same sub-shell is not done unless all the orbitals are singly occupied. Hence all the orbitals must have 1, 1 electron present into it. If you see this 2P, 1, 1 and 1 should be there. So this orbital is paired here, so this one does not follow Hans rule. Which one of the following relates to the photon as wave motion and as stream of particles, okay? As wave motion and stream of particles. So this is what this is the dual nature, correct? Yes. So you see this following relates to the photon both as wave motion as well as a stream of particle. Obviously the wave plus particle nature, particle nature if you have, this we call it as the dual behavior, right? This we call it as a dual behavior, right? And this we started in, this we started in Planck's quantum theory. If you remember this Planck's quantum theory and then we'll discuss the photoelectric effect, right? Where we come to know here the particle nature of this, we know this wave like, you know the photons has particles, right? There are particles present into this, the stream of light has particles present to this. And since it is a wave also, there are the wave particles, we have photons for light, okay? And quanta for any other radiations, correct? So this is even observed by Planck and we called as Planck quantum theory. And he says what the energy associated with any wave of frequency nu is directly proportional to the frequency of that particular wave. E directly proportional to nu. Then if you remove this proportionality constant, E becomes h nu. And this is the energy associated with one photon which is present in a radiation of frequency nu, right? And that's why we say that the energy present in this photon is quantized, quantized, okay? It is available in packets and in discrete manner, right? That's why this shows both it has particle nature, plus it is wave also because photon is a particle of wave. That's why it consists of both properties or both relations, that is, wave as well as stream of particles. So answer for this is option D is equals to h nu, okay? Nodes, okay? Question number 22 you see. Nodes are what? Nodes are the surface, nodes are the surface or plane where the probability of finding an electron is zero, right? We know there are two types of nodes we have, right? Spherical node or we also call it as radial and non-spherical, non-spherical, which we also call it as angular node, right? Now the formula for this we know n minus l minus 1 and this one is l. So for 3p orbital you see the number of spherical node if you count here, 3 minus p value is 1, 1 it is 1 and there's only one angular node, right? So we have one spherical node, one angular node. Let's see the option now. Two non-spherical nodes, not correct. Two spherical nodes, not correct. One spherical and one non-spherical. This is correct. One spherical, two non-spherical, this is wrong, okay? This is the answer for this question option C. I hope you get it. Let's try these questions now, okay? 23rd one, it is talking about the orbital angular momentum. So what is orbital angular momentum is equals to l into l plus 1, h pi 2 pi. This is the formula we have. Now we know the orbital, the value of this orbital, l value for each orbital, right? Okay? So we have 2s orbital. So we know l is equals to 0 for s-subtial or p it is 1, d it is 2, f it is 3. 0 will substitute here. So the answer orbital angular momentum is coming out to be 0 for this one. Question number 24, the first use of quantum theory to explain that structure of atom was made by, okay? So Planck, Bohr, Eisenberg and Einstein, okay? The first use of quantum theory to explain the structure of atom was made by the answer is what? It is Bohr, right? So Bohr theory was based on the postulates of quantum theory because it contains the wave characteristics also, right? Based on the, this thing, the Planck's equation. That's why the answer for this 24th one is b. Some postulates of Bohr's was based on the quantum theory, okay? Next, 25th one. For a d-electron, the orbital angular momentum, same question we know, l is equals to 2 for d-subtial. So orbital angular momentum is equals to l into l plus 1 root over of it, h by 2 pi. l is equals to 2u substitute, so 3 plus 2, 6 root 6 h by 2 pi. Remember one more thing, h by 2 pi, we can also represent this as this symbol. So both are correct here. So answer for this question is this. So these are the questions that, you know, that most commonly they ask in this particular chapter. We'll discuss, we'll have another session also on the same chapter atomic structure and we'll see the different types of question that has been asked in this particular chapter, okay? Thank you so much.