 Hello and good morning, my friends. I'm Atul Kedia. I'm a fourth year undergraduate student at IT Bombay. And I'm going to discuss some quantum mechanics problems with you. So let's begin with the first problem. This is the problem we are encountering right now. I'll give you a couple of minutes to attempt it, give it a thought, and then we'll discuss how to go about with the solution. So just to make sure that you are able to read the question, I'll just read it out loud. Show that the Bohr condition, Bohr's condition of quantization of angular momentum leads to the condition of formation of standing waves of electrons along the circumference of the Bohr model of hydrogen atom. So the Bohr's quantization condition comes from one of his postulates that the energy difference between transitions of electrons from one state to the other has to be an integral multiple of h. So let's go about with the problem. The condition is that the angular momentum has to be an integral multiple of the reduced Planck's constant. So taking this beyond, it'll be the angular momentum for an electron around an atom at a radius. Let's say this is a nucleus of the atom. And the electron is revolving at a distance of, let's say, around r, around the nucleus. So the angular momentum would be given by p into r. Precisely speaking, it will be r cross p. But then the radius and the direction of the momentum would be perpendicular to each other. So the cross product would equivalently result into r into p with some direction. So r into p will be equal to n into h over 2 pi. Now we know from De Broglie's hypothesis that the momentum and the wavelength of an electron are interrelated. The relation is given by the wavelength of a massive particle that is given by h over the momentum. So now we can make a substitution of the Planck's constant over here by lambda into the momentum. And then we get a very unique, very interesting result. The result comes out this way. p r into p is equal to n lambda into times p over 2 pi. Here, I've changed from reduced Planck's constant to the Planck's constant divided by 2 pi. Here, I have substituted the Planck's constant by lambda times p. And then, so from the Bohr's quantization condition, we have arrived at this relation. Here, we simply cancel out the momentum of the particle. And so this gives 2 pi into radius is equal to n times the wavelength of the electron. So this is a very profound implication. This, in one way, says that the electron has to have an integral multiple of the wavelength of the electron has to be. And the multiple of an electron's wavelength should be equal to the circumference of the electron's revolution circuit. So in way, you can see it this way. The relation was 2 pi into radius would be equal to n times lambda, where the radius is the radius of revolution. And the wavelength is the de Broglie wavelength of an electron. So now, assuming, picturing it in this way, that there's a nucleus in the center, the electron is revolving at a distance of r. From here, circularly. Now, this is the condition says that the circumference of this should be equal to n times the wavelength, which means, which implies that if the electrons wavelength, if the electrons, the waves crest starts over here, let's say, then the wavelength comes out. And since it's a multiple of the wavelength has to be equal to the circumference of the circle, there's a constructive interference between the wave that is coming back from the other side of the electron. So you can see this way. It comes out this way. And then, eventually, it will have a constructive interference. So you can think of it this way. That if this n were equal to 1, let's say, that means the circumference is exactly equal to the wavelength. So the length covered overall in the circumference is equal to the wavelength. And so after one revolution, the electron would have covered exactly one wavelength amount of distance. So there would have been such an interference. This being the crest over here, this being the trough, this being the crest, this being the trough. So this is one complete revolution. Similarly, if this were not true, then we would have observed, if we would have started from one point, come around, we would have reached with a phase difference, which would have led to a destructive interference amongst the two waves. The original going wave, the original wave that was leaving from here, and the wave that is coming from the other side. I hope that is clear. I can make another illustration of this. Starting from here, here, here, here. If we say one wavelength amount of length is covered in this much, let's say, if we take 2 pi r is equal to 4 times the wavelength, n being 4. Then this much circumference covers one wavelength, so like this. Then remaining covers another wavelength amount of distance. Next one covers another wavelength amount of distance. And the fourth one, again, covers the same amount. Had this multiple been n by 2 or something, there would have been a destructive interference after coming back over here. So in that case, the electron would not have been possible. It would have not been stable at all. So this is a very profound implication that the electron has to have a wavelength that is somewhat related to the circumference of the radius at which it is rotating. We can go to the next problem now. So I'll give you a couple of minutes to attempt this problem again. The dispersion relation for a lattice wave propagating in one-dimensional chain of atoms of mass m bound together by force constant beta is given by the following equation. Omega is given by omega 0 times sine of k a by 2, where omega 0 is equal to the square root of 4 beta by m. Here a is the distance between atoms and beta is the expression given. Show that in the long wavelength limit, the medium is not in dispersive. And then find the phase and group velocities at k equals pi by a. So the dispersion relation given here is this. Omega equals omega 0 times sine of k a by over 2, where omega 0 is given to be under root of 4 beta by m. So in the long wavelength limit, when the wavelength is extremely long, let's say wavelength is tending to infinity, we could simply say that k is related to wavelength by this relation as we know. In the long wavelength limit, k would be tending to 0. k would be a very small quantity. And so we can say that the dispersion relation could be simply written as omega is equal to omega 0. We can make a sine a small angle approximation, which is sine of let's say k a over 2 minus of k a over 2, the whole cube, 1 by 3 factorial plus higher order terms. In the small angle approximation, we could simply say that small k approximation. We could simply say that the higher order terms are negligible as compared to the first order term. So omega can simply be written as omega equals omega 0 times k a over 2. It is an approximation. So now we have a simpler dispersion relation for the system given. For this, if we try to calculate the group and the phase velocities, the phase velocity is given by, as Professor Shivprasad must have covered in class, that the phase velocity is given by v phase is equal to omega over k. And v group is given by del omega over del k. So in this case, the phase velocity comes out to be exactly equal to the group velocity. And both of them are equal to omega 0 a over 2. So now this is a very interesting result that we have obtained. Here we are getting the phase velocity and the group velocity are equal. So this, in turn, implies that any disturbance that is created in the wave would be propagating with a constant speed and would not be varying over time or space. These velocities are dependent absolutely on constants, omega 0 a and omega 0 n a. So we know that the wave and the wave propagates absolutely at a constant speed. This can also be seen as this thing. Any disturbance that is created on the wave, let's say this is the wave over here, the group velocity would be given this way. So I'm making a picture of how the wave actually looks. The wave would actually be a product of two factors. The phase velocity factor and the group velocity factor. Together, both of them would give us the total wave. The group velocity would individually give the bigger, would be comparable to the wave packet that we think about in quantum mechanics. The wave packet of a photon or any particle is what the group is supposed to be, whereas the velocity of phase is actually the actual space perturbations or actual space oscillations and velocities, so to say. So this can be thought of as the velocity of group would be as these, whereas the phases would be individual particles in between which would be oscillating with some frequency, which is probably different from the group phase, group frequency. But the velocities of both of them would be the same. All of these particles would be moving together. All of them would be moving together. And these internal crests and troughs would be maintained as time passes. So after some time, we can just expect the entire wave to be traveling as it is the same picture and moving together along the time direction, time or space, either of those. So the second part asks us to find the group and phase velocities at k equals pi by a. So at k equals pi by a, the original dispersion relation was firstly omega 0 into sine of k a over 2. The phase velocity would be omega over k, which would be equal to omega 0 sine of k a over 2 over k. And this would be the phase velocity. And the group velocity would be del omega over del k, which is equal to omega 0 into derivative of sine k a over 2 with respect to k. That would be a over 2 cos of k a over 2. OK, so these evaluated at k equals pi by a. k equals pi by a, sine of k equals pi by a would be 1. So this would be omega 0 over k. And the group velocity, on the other hand, would be 0, because since cos of pi by 2 would be 0. All right, so like I said, in the earlier case, we got both the phase velocity and the group velocity as a constant. The constant was equal to, if you don't remember, it was equal to omega 0 a over 2 in the previous case, vg and v phase in the earlier case. But now, we are getting both of them are firstly not equal in this, when we don't do the approximation. And when we do a substitution of k equals pi by a, we get a difference in both of them. So this also implies that one of the wave packet does not move at all, because the phase velocity, the group velocity is given as 0. Whereas the phase velocity is non-zero, that means the internal waves, the waves in between each group would be moving and propagating through space and time. Whereas in the earlier case, we saw that both of them moved at the same speed. And so the entire wave moved together in space. All right, I think we can go to the next problem now, third problem. I'll give you a few minutes to give it a thought. So the question was, find the group and phase velocities of the matter wave associated with a free particle under the assumption the frequency is defined using the kinetic energy or the total relativistic energy. So we can keep the relativistic energy case hold on for a time, time being. We have to find the phase and group velocities for when the energy is related by the kinetic energy only of a particle. So the group and the phase velocities, when the particle's energy is related by only its kinetic energy. The particle's energy, let's try to write the dispersion relation firstly, is that the particle's energy E is given just by the kinetic energy, which is equal to the momentum squared over 2 times the mass. All right, the momentum given by the de Broglie hypothesis is related to the wavelength. The momentum is related to the wavelength by p is given by h over lambda, where lambda is the particle's de Broglie wavelength. We can find the relation between p and the wave vector k through this, because lambda is just equal to 2 pi over k. So p would be equal to h over hk over 2 pi, or simply h cross times k. So we've got a relation between. So if we substitute p into the energy's expression, where E is equal to p square over 2m, we would get a relation between the energy and k. That will be h cross squared k squared over 2 times the mass. Now we know that we can attribute this energy to the wave nature of the particle with a particular wavelength, the wavelength and frequency. The frequency of which would be given by the relation energy is equal to h cross times omega, or which is simply equal to h into nu. So this in turn reduces to omega is equal to h cross omega is equal to h cross squared k squared over 2m, which is equal to omega is equal to h cross k squared over 2 times the mass. Now this is a simple relation between omega and k, and we can just simply find the group and the phase velocities using this. The group velocity as we did before would be given by the derivative of omega with respect to k. And the phase velocity would be given by just by dividing omega by k. That would be h cross k over 2m. And group velocity would be h cross k over m since the derivative would take out the two. All right, so this is one relation that we have found out using the energy being related just by the kinetic energy over here. And then in putting the momentum as h cross into k, and the energy as h cross into omega, we found out the omega versus k relationship. And then we found out the phase and the group velocities. Now if we take this, if we try to attempt the same problem using the relativistic energies, we would get a slightly different result. Let's try this out. The second part of this question says this. Then if the energy is related by the total relativistic energy, then how does the phase and the group velocity turn out? So the relativistic case. So probably you might be knowing from before, although it has not been covered in class yet. But in the relativistic case, the energy is given by the famous Einstein's equation e squared is equal to p squared c squared plus m not squared c raised to 4, where m not is the rest mass of the particle. And the momentum, p is the momentum of the particle. c over here is just the speed of light. And this is the total energy. So it's not a linear relation anymore. It's the energy squared is related by the momentum squared and the mass squared in some way. There's a restriction this way. And the entire problem changes because of this relation. So now going beyond substituting p is equal to h cross time k. We get energy is equal to, we reduce the square root sign, h cross square k squared c squared plus m not squared c raised to 4. All right. Now we equate this to h cross into omega and solve it beyond. So h cross into omega is equal to squared, h cross squared k squared c squared plus m not squared c raised to 4 raised to half. And now just finding the phase and the group velocities would be simple. We have a omega and k relationship. You can just divide omega by k and find the phase velocity. And we can differentiate omega with respect to k and find the group velocity. So omega by k, which is equal to the phase velocity, is equal to this not so good looking expression, some k relationship plus some constant, some other constant over k. And the group velocity would be the derivative of this with respect to k. So differentiating this with respect to k would give us firstly 1 over h cross from over here. And so half of this 2k into h cross square c squared over this entire bracket. This entire bracket comes over here on differentiating. So again, now we see the expression has changed entirely. It is dependent on k over square root of k squared plus square root of something like k squared, some constants c not plus some c 1 over here also c not and some c 1. These relationships aren't exactly as we found out in the previous case, which were just linear. The phase velocity and the group velocity were linearly related to k earlier. But in this case, we see there's a difference. So that's how relativity changes the entire situation. The phase and group velocities change entirely due to this. All right, I think we can go to the next problem. It's a very interesting problem. So I'll read out a lot of the question once. A wave packet is constructed by superposing waves, their wavelengths varying continuously in the following way, y of x comma t, which is the wave, it is given by an integral of a of k, an amplitude, times cos of kx minus omega t times dk, where k is the wave vector. And now there are some conditions given on A. A k is just a constant between some range of k from k equals minus delta k by 2 to k not plus delta k by 2. And A is equal to 0 otherwise. Now we have to sketch the approximate by the actual wave and estimate the uncertainty in the position that we obtain from this wave given. So we can continue with the problem now. I'll go about with this. So firstly, we are given A of k is a constant A in the range of k between k not minus delta k over 2 to k not plus delta k over 2 and 0 otherwise. So I'll just make a plot to keep this in mind. So if this is k versus A of k, it would be 0 all throughout. And then there will be a box type relation and then 0 again. So this is a constant over here. A is a constant equals A naught. And the range is given by the condition given over here. This is k not minus delta k over 2 and this is k not plus delta k over 2. So now evaluating the actual wave given as the relation given over here, the displacements along the y direction are given as y of xk as the integral. Now we can reduce this integral by substituting the values of A in the ranges given. So A is equal to, so from the range 0 to k not minus delta k over 2, A will be just equal to 0. So the integral would be 0 times cos of kx minus omega t times dk plus in the range of k not minus delta k over 2 to k not plus delta k over 2, it will be just a constant times cos of something times dk. Plus the third term would be exactly as this term except that the integral would be from k not, I'll just write it down. The integral would be from k not plus delta k over 2 till infinity. And again the coefficient would be 0 times cos of something times dk. So again these two integrals would be 0 obviously because the coefficients are given as 0. We only have to evaluate this one integral over here in the range from k equals minus a k not minus delta k over 2 to k not plus delta k over 2. So the displacements y of x comma t would be equal to A times the integral of k not minus delta k over 2 to k not plus delta k over 2 of cos of kx minus omega t times dk. All right. So let's go ahead and evaluate this integral. The integral of cos would be just sine, sine divided by the coefficients. So we can write that down. Y of x comma t would be evaluated as A times, some A times. The integral would evaluate as a sine of kx minus omega t, sine of kx minus omega t over x within the range of k being from k not minus delta k over 2 to k not plus delta k over 2. All right. So now substituting, we make the substitutions and do some simplifications. This would be, we can just remove the x over here and group it with rA and this would turn out to be sine of k not plus delta k over 2 times x minus omega t minus the whole thing subtracted by sine of slightly tedious k not minus delta k over 2 times x minus omega t. So now we make some trigonometric simplifications. This is sine of A minus sine of B. We can write it as cos of A plus B by 2 into sine of A minus B by 2. 2 times of sine of, it will be 2 times cos of A plus B by 2 and sine of A, this minus this by 2. So the relation simplifies to 2 times A over x times cos of sum of these 2 divided by 2 which would be k not x minus omega t. This is a simpler relation and multiplied by sine of the difference of these 2 divided by 2. So we can write it as this, delta k into x minus omega t. This would be divided by 2 and this would remain as it is, alright. So we have simplified our original integral from our original integral of y of x, t equal to integral of A k into cos of k x minus omega t, we have arrived at this relationship between the wave displacements y of x, t and these 2 trigonometric terms. Now getting back to the question, what is the question, the question asked is that sketch this, we will do the sketching in a while. Before that we will just notice that one of these in the limit that k not is much greater than k delta k. This one would have, this sine function would have a very slow oscillation frequency since delta k is a very small quantity as compared to k not, I will just write it down, k not is given if in the limit k not is much greater than delta k. The oscillations with respect to sine function, the sine function would be much slower since much slower in space coordinate and so in like we saw in the earlier problem, these could be said as the group velocity, group frequency. This would correspond to the group frequency whereas these would correspond to individual particles oscillating within each group. So, you can see it this way, since delta k is a smaller quantity, let us say if this were the overall delta k factor oscillation, this is the sine delta k times x over 2 oscillation. Within each, within this much of length, there will be a large number of oscillations of the cos factor. So, I will just write it down at this as this. So, firstly if we say that at x equals 0, we can firstly ignore the omega t factor for a while and discuss this only at a constant time, time equals 0, t equals 0 let us say. So, at t equals 0, the sine factor, sine of kx minus delta k into x over 2 would be oscillating in this fashion again continuing and at x equals 0, it would be 0 and at x equals, x equals delta pi over 2 delta k, it would again be 0, ok. So, where we know delta k is a very small quantity, hence this length is a very large quantity. But within this length, each of these oscillations would happen. So, as for at any given x value, these would be oscillating since these are relatively larger quantities and we would observe oscillations like this as we had seen earlier in the previous problem. Again, this highlights on the group and the phase velocities and now, so the envelope would actually look like this, if we take it beyond it would be, it would look something like this where the number of oscillations within the, within each group would be different from how many are there which would be dependent entirely on the k naught and delta k naught how I have made right now, alright. So, the problem also asks us to verify the uncertainty in the position given by the envelopes, envelopes wavelength. So, firstly we notice that the envelopes, envelopes wavelength would be given by when this factor sign of 2 delta k x over 2 would be, would change by 2 pi, so that would, would make a delta k into the wavelength, let us say x prime over 2 would equal to 2 pi. So, that will be equal to, the wavelength would be equal to x prime is equal to 4 pi over delta k alright. So, to calculate the uncertainty in delta x, delta x, the uncertainty in delta x has to be taken as the difference between 1 maxima and 1 minima within the group velocity, group wave. So, that would correspond to a phase shift of just pi instead of pi by 2 and so we will get delta x would be equal to, so this can be taken as the wavelength delta x would be equal to 2 pi over delta k alright. So, now we have found a relationship between the delta x and the delta k, the relationship was delta x is equal to 2 pi over delta k. Now, we make some simplifications, now taking delta k to the other side delta k times delta x would be equal to 2 pi and noticing that p is given by k times h cross, so multiplying both sides by h cross and substituting k h cross into delta k as delta p. So, I will just write it down, multiplying both by h cross would give us h cross delta k delta x is equal to 2 pi times h cross alright. So, now this can be written as delta p since h cross is just a constant and so delta p would be given by h cross times delta k, I will just write it down delta p would be just h cross times delta k and so this is delta p times delta x is equal to 2 pi times h cross which is equal to which and now we have found out the uncertainty principle. We have obtained at the uncertainty principle through just by assuming that the oscillations of a particle are given by some sine functions and so this is a very profound result that we have obtained right now alright. I think we can move on to the next problem, so the information given in the problem is that the average value of momentum p x is 0 and the momentum in x direction is given as 0 and so the uncertainty the standard deviation in the momentum is just given by the square root of the average value the expectation value of p x squared. Now we are requested to obtain firstly the minimum kinetic energy that the proton and electron would have if they were confined to a nucleus of approximately approximate diameter of 10 to the minus 14 and we have to see that this argument results in the electron not being it turns out to be that this energy is very unphysical. The electron could not have this much amount of energy staying within the nucleus, so I think we will go about with the problem. So the delta, so we are given that delta problem 5 the expectation value of momentum or the average value is given as 0 and the standard deviation in the momentum is given by square root of the average value of p of x times squared. Had this not been 0 we would have had another term over here written as p x the average of p x the whole squared, but since it is given 0 we can just ignore this alright. So firstly we are to evaluate the kinetic energy using this. So the kinetic energy would be given by p squared over 2 m. So we are given that the electron or the proton are confined to a diameter of 10 to the minus 14. So we can say that delta x is equal to 10 to the minus 14 of meters alright. Using the uncertainty relation that delta x times delta p is equal to let us say anything h cross or h bar it does not really matter we will just be working by with the order of magnitudes this entire problem can be just needs us to do an order of magnitude analysis it does not need really to give a precise answer. So we can just take it as h cross usually the exact relation would be h cross over 2 but then this is this would work absolutely. So delta p would be given by h cross over delta x we make the substitution over here for the kinetic energy of the nucleus in the of the particle in the nucleus and it would be kinetic energy is equal to the h cross squared over 2 times mass times delta x the whole squared alright. So now keeping the mass as a variable. So in the part in the question it is asked that we can we have to solve this for electrons as well as protons we can keep we can substitute at the end of each for the electron and a proton separately and see how the results are for now we will just make the substitution this is equal to the h cross would be given by 10 to the minus 34 the whole squared over 2 times the mass we can ignore the 2 also but let us just keep it over here times delta x squared delta x squared would be minus 14 the whole squared alright. So this would turn out to be of the order 10 to the minus 40 over 2 times the mass. If we try to evaluate this in this is in Joules so if we try to evaluate this convert this into electron volts we will have to multiply and divide by electrons and the net result would be something like divide by electrons so it will be 10 to the minus 10 to the positive 119 electron volts divide by 1.6 I will just give you a couple of minutes to attempt this I think I am getting the number somewhere wrong I was hoping to get a positive and extremely positive number oh ok alright alright I think I have gotten it so yeah this is the relation right now it is the kinetic energy is given by this I have converted so till over here the kinetic energies were in Joules I will just write it down in Joules and over here it is in electron volts so I will just copy the relation in another sheet and explain kinetic energy is equal to 10 to the minus 40 over 2 times the mass multiplied by 10 raise to 19 over 1 over 1.6 electron volts alright. So simplifying this a little bit gives 10 to the minus 21 times some we can ignore this 3.2 times mass so now substituting for the mass of electron and proton separately for electron it would be the mass of the electron is around 9 into 10 to the minus 31 in SI units. So, this would this would be this would give us 10 to the minus 21 over some constant some this ok we can ignore the constants and write it this way 9 into 10 to the minus 31 and similarly this is for electron ok electron and similarly for the proton the same analysis would give us this Ke would be equal to 10 to the minus 21 over 3.2 into the mass which is equal to 10 to the minus 27 in this case and to some into some constant we can ignore that it is an approximation. So, this gives us so alright both of them are in electron volts this one gives us a total of this is approximately 10 to the 10 in order whereas, this one is over here of the order of 10 to the 6. So, kinetic energy of electron is about this much and of a proton is about 10 to the 6 whereas, often electron is of the order of 10 to the 10 electron volts which is which can be said as a almost to giga electron volts this is almost like a 10 giga electron volts this is extremely large a normal electrons the electrons rest mass is given by is about 0.5 mega electron volts whereas, this is in giga giga electron volts this is extremely large and unphysical for an electron to have whereas, for a proton on the other hand the same result gives us something like 10 to the 10 electron volts which is equal to a mega electron volt which is very physical and a protons I will write this down mass of an electron a protons mass is given is about 900 or 930 mega electron volts. So, this is very physical as compared to the kinetic energy of the proton that we have observed is very physical it makes a lot of sense for a proton to have such a kinetic energy and for electron to not have such a kinetic energy since the electron cannot be restricted in such a small area also another analysis of the momentum we could do using the uncertainty relation and the delta x we were given the delta x that was given was 10 to the minus 14 since I am running short of time I will just I will do a small analysis of the momentum of the particles of each of the particles in within that nucleus given this delta x we can find the delta v the uncertainty in the velocity would be given by h by h or h cross by delta x into m the mass alright. So, this would be of the order of 10 to the minus 34 over delta x is given as a 10 to the minus 14. So, we can simply write this as ok right we will write it down as 10 to the minus 14 and mass would be 10 to the minus for an electron it would be 31 and for a proton it would be minus 27. So, if we reduce this for an electron we get delta v for an electron would be about 10 to the 11 if we reduce this to this and same the same thing for a proton would be around 10 to the 7 since this turns to 10 to the minus 27 the entire thing the entire relation reduces to 10 to the 7. Now, from relativity in hindsight we know that the speed of an electron speed of any particle cannot be more than the speed of light over here we are observing that the speed of electron is coming the uncertainty in the in the electrons velocity or in some way the electrons speed is coming to be more than the speed of light which is entirely unphysical. Hence, we can easily say that the electron is cannot be bound within this much delta x which is to say the it cannot have been bound within a nucleus. So, moving on there are two other parts to this problem. So, the next part is pretty simple we can go ahead the next part is just says that in a square well potential ok. So, we have to find the ground state energy of a particle in an infinite square well potential again this is an approximate problem we are not we are not dealing with exact values the delta x over here can be taken simply as delta x is equal to for an infinite potential delta x can simply be taken as the length of the potential itself. Again, so now the energy ground state energy delta k the ground state energy or the energy so to say would be p squared over 2 m which would be again we can substitute p as h cross over L delta p is equal to h cross over delta x which is equal to delta h cross over the length. So, now we get h cross squared over 2 m L squared. If I am sure that you have already done this problem of a particle in an infinite well now we are seeing that the relation over here is gives that we are we have obtained is not exactly not entirely equal to what we have expected we would have expected the actual answer would have been pi times h cross squared over 2 m L squared for a ground state for the ground state energy whereas, we are observe we have calculated as h cross squared over 2 m L squared. Now, this is to say we are we are dealing with just approximations we have taken delta x as just equal to L had more information information being given about the nature of uncertainty in the position we would have had a better estimation of what the energy would have been and hence we could have actually had obtained the correct relation of pi squared h cross squared over 2 m L squared. Alright, so we can go to the next problem the next part but next part is next part is slightly simple. So, I would leave that for you people to attempt that since we are running out of time we will attempt the sixth problem now. So, I will simply explain the problem what the problem states the problem states that in actual physical atoms the excited atoms usually the electrons in higher state not their ground state they usually emit some photons and come back to the ground state. Now, we are given over here that the lifetime of at which for which the electron stays in the highest state is of the is of around 10 to the minus 8 seconds. Now, we are expected to find the uncertainty in the energy in the B part and the uncertainty in the frequency using this information the information given is again the that the life the lifetime of an electron at a highest energy state is equal to 10 to the minus 8 seconds. So, delta t is given to be of 10 to the minus 8 seconds. Alright, we know the uncertainty principle for relating the uncertainty in energy and the uncertainty in the time as delta E times delta t would be equal to h cross simply to say h cross I will ignore any factors do or anything. So, the second part just simply tells us to evaluate this value of uncertainty in the energy that can be simply done by making the substitutions. In the first part rather we are asked to find the frequency the uncertainty in the frequency. So, we will evaluate we will go ahead and evaluate that that is the uncertainty in the energy would be the uncertainty h cross h into the uncertainty in the frequency times the delta t would be equal to h cross. Now, cancelling the h cross and h we would get delta nu delta t is equal to 1 over 2 pi. So, delta nu is simply equal to 1 over 2 2 pi times delta t this is equal to this is equal to 10 to the 8 over 2 pi. So, we can just simply say that it is around 10 to the 7 sorry this is 10 to the 7 hertz the uncertainty in the frequency is coming out to be 10 to the 7 hertz. Now, so this is the uncertainty we have observed. Now, we are asked to find out the uncertainty the fractional uncertainty the delta of nu by nu for the radiation of a wavelength given as 5000 angstroms. So, we are supposed to find delta nu over nu given lambda is equal to 5000 angstroms. Now, we can simply we can simply do the calculations and find the frequency frequency would be equal to C over the lambda which would turn out to be 3 over 5 times 10 to the 10 to the 8 plus 10 to the minus 7. So, that will be 10 to the 15. So, this is the frequency of light and so the fractional uncertainty in the particles frequency would be simply 10 to the minus 10 to the 7 over this factor where 10 to the 15 we can ignore the 3 by 5. So, delta nu over nu is given by this. So, we see that it is a very small factor it is a of the order of 10 to the minus 8 hertz it is of the order of 10 to the minus 8. So, so we can simply say that the so when we observe the spectrum of an electron excited of an electron de-exciting from an excited state we can simply say that the electron the uncertainty in the frequency is very less, but still if we resolve the frequency to of the order of 10 to the 8 then we could see the uncertainty we could see a width within the spectrum power spectrum observed over there. Alright. So, I think this is the last problem now we can go to some questions if anyone has doubts questions and regarding anything in quantum mechanics. Center 1 2 5 6 go ahead for your question. How group of those are generated when single particle is under motion. Okay your question is how a group is formed when we are dealing only with a single particle so to say right. Yeah obviously. Yeah alright so it is an interesting question. So firstly so the whole the entire discussion about quantum mechanics relies on the fact that a particle is not just one particle but there is one particle we can just we can say that it's a particle associated with a wave with it alright. So, there's a wave so to say for each particle if there's an interference of one particle with another particle we can say that the there's a interference of two waves and then there's a there's a there's this difference that comes in the phase and the group velocities. So you're right we are not the the phase and the group velocities does not come due to one particle but the due to the interference of multiple particles let's say two particles and then there's a so that will lead to some differences in the frequencies and which would be which would give us the a difference in the phase and the group velocities. Another question is there sir. Yeah. Then according to Dave Wembley's hypothesis lambda is equal to H by P. Yeah. Where P stands for L in the way. Yeah. Why didn't we use group or group velocity in that occasion. Okay so alright so I think Professor Shivprasad could answer this much better I'll give it an attempt this okay so the de Broglie's hypothesis is about one particle when de Broglie says that lambda is equal to H by the momentum the he said it is said only for one particle the so let's say one electron over there so we can when we make a substitution to find the lambda it's for one particle now the group and the phase velocities that you said are for multiple like I said for when two particles interfere there's a group there's a phase and a group velocity so hence there's a difference the the de Broglie's hypothesis does not apply to that an interfered particle if had we a better understanding of how the interfered particle can be thought of as one particle together then we would have said one part as one particle alone then we could have said that we could have applied the de Broglie's hypothesis to find one wavelength but since that is not true and we and I'm pretty sure that we wouldn't be finding that because okay so firstly because we as we saw the wave changes the waves amplitude firstly changes and its internal frequencies also internal wave also changes so there's a difference the this is not exactly a normal wave which which was assumed in the de Broglie's hypothesis so there's a difference I think Professor Shifrasad could explain this much better diagram the wave what you had drawn it looks like coming from a single particle that's I have an elementary doubt in that actually in the wave packet that I made to a crest and a trough and within within there are electr there are oscillations are increasing and then decreasing are you talking about that yes that that is my doubt yeah so that is about that is when more than one particle when more than one particles are there together so again again the k relationship that we saw so in problem number four we saw that the displacements were given as an integral over k all right so that means not only one k value are available there are multiple k values available for in the in the analysis whereas according to de Broglie's hypothesis there'll be only one part one k value all right so there's a difference so over here there are multiple particles whereas then the de Broglie's hypothesis deals with only with one particle any one particle center 1176 go ahead from karnataka sir in the second problem okay while solving the second problem we got vg is equal to zero yeah yeah that means particle will be a drift yes yes I think that in the contradict result I think okay let's just look at the problem so the dispersion relation was a relation was given as a omega over some sine function so and we were we found out the case in the group velocities for k equals pi by a so yeah I think it is possible that for this one k value k equals pi by a the phase the group velocity comes out to be zero but it in general that need not be the case so I would say that the particle as such is not moving but in in other cases of k values there the particle would be moving and we would get a non-zero group velocity and in the same problem yeah in the next part we got v phase greater than vg is that possible sir yeah v phase is equal to omega by k yeah will be h cross k divided by 2m and vg is equal to do b omega divided by dk is h cross k divided by m so that is so yeah vg is twice the v phase yes that's correct there's a there's a difference between the two so the phase velocity was given as h cross k by 2m and the group velocity was given as h cross k by m so physically we could think of this as this the group as such is moving with a slower rate but the oscillations within are moving much faster so let's say initially the when the group is over here the oscillations were within them but as time passes the oscillations move move beyond and they move into the next next group so the oscillations keep shifting from one group to the next one and travel beyond that's how it is next one three five go ahead from Tamil Nadu the first problem which we have solved that circumference of the circle is equal to an integral multiple of wavelength yes right yes which is applicable only for the hydrogen atom or any other atoms this would apply to any atom which has only one electron so in in the in physics we call it like as hydrogen like atoms these would be atoms that have only one electron we could say it is a helium plus ion or a lithium two plus ion so any atom that would have only one electron so firstly the deeper the boss boss model ball model applies only to atoms that are hydrogen like which have basic which basically means which has only one electron associated to the nucleus so there's one nucleus which could have any number of protons or neutrons but there's there should be only one electron associated with it so it could be lithium two plus ion or a helium plus ion or anything beyond that or a hydrogen atom so this relation this result that we obtained is true for all of them the phase velocity of the wave is greater than the velocity of light then what about the physical nature of the de Broglie wave the phase velocity of a particle of the particle is greater than the velocity if it turns out to be greater than velocity of light you're asking what would be the nature of de Broglie I mean how would de Broglie's okay so like I said earlier phase and the group velocities come as come into the picture when we are talking about multiple particles interacting together all right so in that case again applying the de Broglie's hypothesis does not make sense because that applies only to a single particle case and anyways if the phase velocities come out to be larger than speed of light then that's not entirely a problem because those aren't physical matter moving with the speed of something greater than light this actual physical particles are only the groups that are formed the groups are the actual physical particles that are moving that turns out to be larger than the speed of light then that is a problem not if the phase velocity comes out to be larger than the problem larger than the speed of light if you take the de Broglie relation lambda is equal to actually P for a particle of having a mass m moving at certain velocity and it will have only a unique momentum then the corresponding wavelength also should be unique can we say that the de Broglie wave is monochromatic no because an assumption that you made was the momentum is unique so like like we know there's an uncertainty in the momentum and the position of a particle if we okay let's take an example if we take the electrons wavelength to be a constant particles wavelength to be a constant one value that would mean the particle has one momentum and we can we can picture it as a wave moving in free space continuous wave moving from from minus infinity to plus infinity moving in free space this would be of one wavelength and the de Broglie's hypothesis would give us one momentum associated with it but then with this with this condition we have lost the uncertainty the certainty in the position of the position of the particle so when we when we say that the particle is a wave when we take a wave which is which is spread it out around the whole space we are saying that the uncertainty in the position is infinite like we can find the electron from minus infinity to plus infinity anywhere but then so so we have lost the uncertainty the the uncertainty principle has shown it shown its role over there now if if you if we make the particle a little bit constricted to some small range of space and it is finite texture yes it is not from minus infinity to flux plus infinity wave packet is a finite size yes that's what i'm saying so if we make it a finite sized particle that's what i was i was about to say that if we make it a finite size particle we would have to include more terms in the in the wavelength condition so so okay if we want to distribute if you if you want to find a Fourier transform of or the or the spectrum of or the dispersion relation for this particle which is localized with respect to the frequencies it would be it would have a lot of terms not just one term like like in the previous case so then again the uncertainty has shown its role and now we don't have a distinct we don't have a distinct wavelength associated with this this and and the momentum so again this is interrelated as you may see like that you see you see the uncertainty in the wavelength the which is somehow related to to the momentum by the de Broglie's hypothesis and we know that the momentum and we see that there's an uncertainty in the momentum because there's an uncertainty in the lambda in the wavelength so that gives us the uncertainty in principle back again the uncertainty in the x in the and the multiply the uncertainty in the momentum is a constant is greater than greater than a constant at any time in the fifth problem we solved today we proved that the delta v in the order of 10 to 11 and delta vp is in order of 10 to 7 and based on that how can we say that electron could not exist within the nucleus can you comment on this yeah so as I said then the so we said we found delta p delta vp to be of the order of 10 to the 11 and the other one was for so the uncertainty in the electrons velocity was of the order of more than speed of light and that of the proton was less than speed of light now in some way the uncertainty of the law of a particle gives us reflects about the particle's actual velocity so once we are saying that the uncertainty is greater than c then we can clearly say that this is not this is unphysical relativity implies that the absolute speed of speed of any particle could be equal to speed of light so as soon as in any argument if we see something which is something exceeding the speed of light we simply say that that's unphysical that's that's impossible so yeah so thank you very much it was interesting solving these problems today with you thank you and bye bye