 Let us try to solve a numerical in this experiment, I am talking about in this experiment what is the distance of closest approach to the nucleus for 7.7 MeV alpha particle before it momentarily comes to rest and reverses its direction. The atomic number of the gold is given as 79, the atomic number of gold is given as 79, the hint is you have to use conservation of energy, that see you have a nucleus the closest approach happens when it is a head on collision by other way it will just reflect and go away. So, if the alpha particle starts to move towards the nucleus, how much charge it has 2 times the charge of the electron how much charge nucleus has 79 times charge of the electron it has 79 protons in it fine initially this has a kinetic energy of this much and potentially between this and that is 0. Then suppose it comes very close this is the distance of closest approach I can say that this is the distance r the potential between this charge and that charge you can write down. So, basically all you have to do is sum of potential energy and kinetic energy initially and equate it to potential energy and kinetic energy finally. What is the final kinetic energy 0 it comes to rest fine initial potential energy is what 0 they are infinitely separated from each other that is 0 and initial kinetic energy is 7.7 MeV. So, this is 7.7 mega is 10 is power 6 electron volt as in you have to multiply charge of the electron this is q 1 plus k 1 this should be equal to what u 2 plus k 2 u 2 is what k times q 1 q 2 by r plus 0 k is 1 by 4 pi epsilon naught get it quickly what is r k is 9 to transform 9 q 1 is 2 e q 2 is 79 e this into 1.6 into standards for minus 19 divided by r that is equal to 7.7 10 is power 6 into charge of the electron. So, this e is go there will be one more e q is coming into e square. So, when you simplify you will get the value of r as 3 into 10 square minus 14 which is actually 30 centimeter or 30 centimeter, but in actual case the radius is 6 the radius of the gold foil this thing is 6. So, the alpha particle never touches the gold nucleus it comes that far and then moves away using consumption of energy you can actually find out if you know the radius of the gold nucleus what energy is required by the alpha particle touch the gold that also you can find out fine. So, this is all about the atoms in general in general now we are going to talk about electrons going forward our study is electrons the behavior of electrons why not the nucleus because there is separate chapter on the nucleus the next chapter of model physics is nuclei only. So, our focus right now is electrons right down electron orbits. So, now the situation is pretty clear that at the center there is this possibly charged particle just called nucleus and around it the electron will revolve fine. So, the picture looks like this there is this nucleus fine and the electron will revolve around it right I am talking about right now hydrogen atom not hydrogen molecule single. So, there is one electron that is moving let us say the other speed is V it moves in a circular orbit fine. Now, if it if anything has to move a circular orbit what is its acceleration it is uniform circular motion speed is constant. So, acceleration is there will be an acceleration towards the center okay. So, there will be a radial or centripetal acceleration of the value V square by r okay where r is the radius. So, if there is an acceleration I mean this thing you know right if anything as you move in a circle its radius should be V square by r that should be known okay. Now, if that is the radial acceleration does it mean there should be a force also if there is an acceleration there should be a force how much mass is the acceleration. So, there should be a force along the direction of acceleration which is radial that force should be equal to mass into acceleration. So, m V square by r so, there should be a force that creates this mass into acceleration what is the force here what it is electrostatic force force of attraction fine like for example when earth revolves around the sun what is the centripetal force gravitational force there should be a centripetal force if something is moving in a circle fine. So, this is equal to 1 by 4 pi epsilon naught q 1 q 2 is e square only because the nucleus also has the same charge as that of electrons this is e square by r this is clear okay this comes by force balancing this is actually r square. So, one of the r will cancel fine in the electrostatic force divided by r square comes. So, you will get m V square is equal to what 1 by 4 pi epsilon naught e square by r. Now, what is kinetic energy of the electron this is what this is mass of electron only right. So, kinetic energy of the electron will be how much half into mass of electron into V square which is equal to what half of 1 by 4 pi epsilon naught e square by r fine this is kinetic energy what about potential energy how much is the potential energy between electron and the nucleus how much electrostatic potential energy you do not know you forgot electrostatic completely potential energy is 1 by 4 pi epsilon naught we just used it now distance of closest approach q 1 q 2 by distance between the two charges in this case what will be it 1 by 4 pi epsilon naught into e square minus e square one is positive charge other is negative charge fine. So, this is potential energy let us call it P e because kinetic energy is k e is better to call it P e. So, this is potential energy now what is total energy k e plus P e right. So, if you add these two 1 and 2 you get a total energy fine that will come out to be equal to minus of half is this thing clear total energy is a negative quantity it is negative. Now, if electron leaves a nucleus what is the energy 0 it goes to infinity energy 0 right now it is negative will it leave the nucleus it will not because its energy is already less than what it would have been had it leaves if it leaves energy will increase and every object wants to have as less energy as possible it is already less. So, why it will leave the nucleus getting a point fine. So, how much energy you have to give for it to leave the nucleus this much energy if you give added it becomes 0 fine. So, this energy can be referred as binding energy this much energy is holding the electron with the nucleus and now you can also create a formula for yourself that total energy is equal to minus of kinetic energy is equal to half of potential energy fine. So, if I just tell you one of the energies the other two you should be able to get it immediately fine any doubt on this anything no this thing you have done in chemistry also right down it is found that 13.6 electron cold energy is required to separate hydrogen atom into proton and an electron. It is like taking away electron from the hydrogen you need this much energy calculate the radius of the atom and velocity of the electron did you get the answer lot of calculation is there everybody knows how to solve but difficult to calculate that is why that is much smaller than nucleus itself this is a total energy fine. So, if you have to give this much energy for electron to leave. So, if you add this much how much it should become this should become 0. Minion energy required for this to leave the nucleus when it leaves energy 0 getting it. So, one electron we get cancelled from here. So, from here you will be able to calculate what is R ok the answer for R is 5.3 into 10 is power minus 11 did you get anything I got 5.3 you got 5.3 I got 5.3 power is a easier thing to get anyway how you get velocity see kinetic energy is how much kinetic energy is negative of this negative total energy. So, half times 1 by 4 pi epsilon naught e square by R once you get the value of R you can substitute it here, but good thing is the value of this is known how much this is 30.6 e this is this entire thing is it not this plus this if it is 0 and this should be equal to that this should be equal to half mv square mass of electron into velocity square mass of electron is 9.1 that is how much 31 fine. So, from there you get the velocity