 So, now we will talk about wave nature of particles, when I say particle I do not mean a very very small particle can be a extended object also, particle means that I will not consider the rotation of it, I will just consider translation, translation as in velocity, So, D Broglie has studied the wave nature of particle systematically and he has proposed a model in which he said that the wavelength of any particle is h divided by magnitude of linear momentum, fine. How much is the linear momentum? Mass into velocity, so magnitude is this, the mass into velocity if you do you get the magnitude of linear momentum, fine. So, the wavelength will be h by p, this is what D Broglie has proposed. Let us do a small numerical and find out how does that affect our day to day life. Suppose there is a mass of 1 kg that moves with the velocity of 1 meter per second or let us say 10 meter per second, which is very high velocity, fine out the wavelength, got it? How much? Wavelength will come out to be 6 point into 10 square minus 33 meters, you are not able to calculate this. How much? Minus 35. H by p, h is 6 point 63, 10 square minus 34 divided by 10, yeah 35, do not make such errors. So, wavelength is very very less, that is why you feel that it is moving in a straight line, you never feel the wave of it, but then it is wave also. When you will be able to observe a greater wavelength, when the momentum is very less, as in very very less and when the momentum will be very very less, when the mass is very less, what comes into mind when I talk about the smallest, very very small particle electron, right? So, with electron the wavelength comes out to be, you know something which you can measure, this you cannot even measure, there is no instrument which can measure up to this much accuracy for a wavelength, fine? So, electron is a particle also for whom the wave nature is very prominent, ok? You might have already learned in chemistry that sp3 hybridized orbital, what is that? When two wave functions of electron meet, the standing wave get created and different kinds of standing wave get created, sp3, sp3, d2, sp2, fine? So, when the standing wave get created, standing wave will have node and antinode, antinode is where the maximum chance electron exist, electron density is maximum, fine? So, we have already studied in greater detail in chemistry about the wave nature of the electron, but we never highlighted it, ok? But here, let us try to find out wavelength for an electron which is moving with 10 raise power 5 meter per second, find out, this is the velocity of electron, find is wavelength, how much is the wavelength? It is 1 by 3, 3.3, we need 2 power, you still trust yourself, there is no much 10 raise power, this is 3 and 8, minus 3, there is approximately 9 and this approximately is 6.6, let us see 7, 9, 6, 3, 30, 20, 28, we will get 10, 0 7.4 nanometers. This is more or less like the wavelength which we usually deal with. So electrons dual nature or electrons wave nature comes out very prominently. You can observe it. Other bigger masses wave nature you will not be able to observe. So that is why the center of focus when we talk about wave nature of particle is not electrons. Those photoelectrons that are coming out in EM wave radiation, those electrons will also have a wavelength. You find out their linear momentum and then h by linear momentum will be their wavelength. Find out. Suppose kinetic energy is k for the electron that are coming out in photoelectro commission. Find out their wavelengths. Just find out in terms of h m and k. If k is a kinetic energy, how much is the momentum? This is half mass of electron into v square. So mass of electron into v square is 2 into k. So m square v square is 2 m into k. So m into v is what? Under root of 2 mass of electron into kinetic energy. So h divided by under root 2 m into k. This is the wavelength of the electron in that photoelectro commission. Now suppose these electrons gains some kinetic energy. What is the best way for electron to gain the kinetic energy? Accelerate against the potential difference. If potential difference is v, the kinetic energy becomes e times v. So if this electron gets accelerated by a potential difference of v, kinetic energy becomes e times v where e is a charge of the electron. Now when you substitute this over here, I mean this k is because of the photoelectro commission. This k is because of the acceleration, accelerating potential. But k remains k only. So wavelength formula will become h divided by under root of 2 mass of electron e into v. So when you substitute all the values, see here you can see that h is a constant 2 m e. Everything is constant except v. v can be changed. So when you put the values, you will get 1.22 divided by under root of v nanometers. So if I tell you accelerating potential is v, suppose 100 volt, you should be able to directly find the answer. Or else you have to substitute a lot of values and solve it. If you remember this, it will be better. But this is nanometers, this is not meters. So this is 1.227, sorry, 3 digits. Electron, alpha particle and a proton have same kinetic energy. Which one of these will have shortest D Brueglie wavelength? This wavelength which you have found out, it is D Brueglie wavelength. Which one of these? Shortest. Are you trying to solve it theoretically? That is not how you should solve it. It is not a theoretical question. Get the expression and then compare. Yeah, from expression if you can compare. Lambda is h by root over k. 2 m k. So see k is same for all of them, h is same to whoever's mass is largest, will have shortest wavelength. Which one of them? Alpha. So alpha will have shortest wavelength. Next question. A particle is moving 3 times as fast as the electron. Velocity of particle is 3 times the velocity of electron. The ratio of D Brueglie wavelength of the particle to that of electron. So lambda of particle divided by lambda of electron is given as 1.813 into 10 raise to power minus 4. Find out the mass of particle. Answer is 1.6 into 10 raise to power minus 27. Lambda is what h by mass of particle velocity of particle divided by h by mass of electron velocity of electron. This is 1.813 into 10 raise to power minus 4. We have h gone. So this is V e by V p divided by m e equals to this. So V e by V p is one-third and mass of electron is known. So you will be able to find the mass of particle. So mass of particle comes out to be 1.67 10 raise to power minus 27 kg. So this could be a proton or a neutron. Next, what is a D Brueglie wavelength associated with an electron which has accelerated through a potential difference of 100 volt. This is direct formula substitution. We have a formula for electron that gets accelerated by a potential difference of V. How much is that? 1.27 divided by under root V nanometers. And V is 100. So 0.128 nanometers. We will not increase number of decimal places. We will round off because you cannot increase accuracy in that calculation. It remains 3 decimal points. So let us keep it 3 decimal points. Fine. Any doubt till now? So the wave nature of particle is straightforward. Okay. Now one important thing, I mean not with respect to problem solving, but when this understanding or getting developed that particle can behave like a wave, there was no experiment to prove it as such. It was not visible. It was just theory. So Davidson and Germer, first time demonstrated using an experiment, Davidson and Germer experiment that electron can show a wave nature. Okay. It was very weird observation, but this is how it is. Okay. The experiment goes like this. See more or less the experiment is this only. In your book, what they have done? They have elaborated electron gun. They have a whole set up of how electron gun is made, but that is not our focus. Our focus is what happens after electron comes out. Okay. So I have drawn only that much. So what is that? This is to galvanometer. Fine. So what happens is the stream, a stream of electron will go like this, hit the nickel target and then reflect off. And why it gets reflect off? Because this nickel target will have small, small lattice structure will be there. Fine. So when electron goes, it can go inside and then come outside due to reflection and then go like this. Okay. Or it can come like this and reflect off like that. Or it can just get absorbed or it will just go away. It will pass through like gold foil experiment. Okay. So there is this scattering which angle it will go to. It is not determined. As if there are so many electrons will get scattered away at different different angles. Fine. So basically you have made an arc of a circle keeping this as a center. So this wherever this galvanometer goes, its distance from this point will be same. Fine. So if you assume that there is equal probability for electron to get scattered anywhere, so this galvanometer will catch equal number of electrons wherever it will go. Okay. So current the galvanometer reads should be same every time. But what happens is completely weird. You get suppose maximum current here and then you get minimum current. Then you get again maximum. So like that you get maximum, minimum, maximum, minimum, maximum, minimum kind of observation. And then you recall such thing happens in interference. Fine. So what is happening is this wave that directly get reflected from here and this wave that goes slightly inside and comes out, these two wave will interfere and there will be constructive or destructive interference. And now if you treat wave, sorry, electron as a particle, it is like two electrons meeting and destroying each other. Gone. And two electron meeting creating four electrons. That's absurd if you think electron is a particle. But if you think it like a wave, then it makes sense. One wave can meet the other wave and both wave will disappear. Fine. And one wave will meet another wave and intensity becomes four times and it will become double intensity becomes four times. So this kind of experiment has shown that interference or diffraction electrons undergo interference and diffraction if electron comes out from the nickel damper like this. This proves that electron can be here like a wave also. Any doubt till now from this chapter, we are done this chapter. One more small thing that we can talk about although not part of the chapter is this. Suppose the electron has to revolve in an orbital radius of r. This is nucleus, radius is r. Electron is moving like that. Fine. And we have assumed that electron is a wave. So how it will travel? Travel like a wave. It goes like this. Fine. So suppose it goes on like that, then when it meets, it has to join in phase. I am getting my point. If this point do not end up here, then a new wave will get created. So every time a new wave get created, there is no repetition of the path. So if electron has to repeat its path again and again, it has to follow like this. So if you take this point and stretch it like that, total length will be 2 pi r. 2 pi r. Total length will be 2 pi r. And one more thing is from here to here, one wavelength, there to there, one more, one more, one more. Like this, you see that entire thing can be divided in integer times of the wavelength. It is integer times wavelength, total length. 2 pi r should be equal to integer times wavelength. Getting it? This thing, there should be, this is what? One wavelength. This is again one more wavelength. One more wavelength. So like that, you can divide this entire thing into integer times the wavelength. Now according to D Brugli, what is the wavelength? How much it is? H by P, right? P is what? M into V. So from here, you will get M V r is equal to N H by 2 pi. This is what we call? This is both. So D Brugli has developed his concept independent of what both was doing, but both of them coincides when it comes to the idea of being angular momentum quantized. Angular momentum will be quantized only when you treat electron as a wave, right? Any doubt in the chapter? We are done with the first chapter of modern physics. Now you can solve questions at any level yourself. Nothing.