 Now, next question the work function of Czm is 2.14 electron volt. So, what is given the value of 5? 2.14 electron volt is the work function. You have to find the threshold frequency first find out the threshold frequency how much it is. In this chapter in this chapter almost everybody knows how to solve it is all about calculation. How much? 5 into 10 to the power 14 hertz. See in this chapter you have only one equation. So, whenever in doubt first write down this equation h mu minus 5 is k and then start from here. For threshold frequency k is 0 just put k equal to 0 you get mu threshold 5 by h. But 5 is given in electron volt you have to convert in joules that into charge of electron you have to do find SI unit is joules. So, this is equal to 2.14 into 1.6 into 10 to the power minus 19 joules 5 into 10 to the power 14. Next part of this question you have to find the incident light wavelength for what wavelength lend up the stopping potential is 0.4, 0.6 volt. For what wavelength of incident light stopping potential is 0.6 volt. Lot of calculation is there in this chapter do not get afraid by it. Because that is what it is tested in this chapter. There is nothing conceptually over here. But one thing that I can say here is that many times I have seen students they put the values in between what is asked wavelength get the expression for the wavelength and then put the values. For example, here what we have h mu minus 5 is what? Kinting edge is what e to stopping potential fine. So, h into C by lambda is equal to e v naught plus work function. So, lambda is what h into C divided by e v naught plus work function fine. Now, you put the values h is what 6.63 10 is power minus 34 speed of light 3 into 10 is power 8 divided by e times v naught. So, this is what 5 is also e times 2.14 and e times v naught is this. So, this plus this is what 2.74. So, this is 2.74 into 1.6 into 10 is power minus 19. This is e times v naught and which 5 you have to even multiply charge of electron. So, this is approximately 4. So, 12.2 you can say divided by 2.74 into what 10 is power minus 7 fine. Now, you have done the correct thing right. So, how much is this 4.1 4.5 I guess the answer is 4.5. So, that is 12.2. No, you cannot do it this is also 2.74, but you know it is I mean less than 4.5. Yeah, slightly less than 4.5. So, you can just do it like this 12.1. I mean we have taken approximate value just into this divided by that I have taken exact 4 and this multiplied by that is 12 that I have written 12.2 fine. So, these are just elementary levels. Let us do few more. Threshold frequency is 3.3 into 10 to the power 14 hertz. What is it 3.3 10 to the power 14 hertz. This is threshold frequency. The incident frequency is 8.2 into 10 to the power 14 hertz. Find out cut off voltage or stop in potential for this situation. What is the formula? H mu minus phi is k max and k max is what? E times may not. So, stopping potential is H times mu minus phi divided by E and what is phi? Phi is H into mu threshold, is not it? When mu is mu threshold, conductivity is 0. See H mu minus phi at mu equal to mu threshold k is 0. So, phi will be equal to H into mu threshold fine. So, this will be equal to H times mu minus mu threshold divided by E. H is 6.63 10 to the power minus 34 divided by charge of electron 1.6 into 10 to the power minus 19 into mu minus mu threshold. Mu minus mu threshold is what 8.2 minus 3.3. What is that? 4.9 into 10 to the power 14. So, this is 4. So, it will like 1.9 volts. See if a particle has a momentum and it strikes a surface, will it create a force? Will it create a force because there is a change in momentum and rate of change in momentum is force and experimentally it is observed also that when light is present on a metal plate, it applies a force experimentally this proven. So, let us try to solve a question that assume that n photons per second hit a plate normally fine. 50 percent gets absorbed let us skew it little bit, 70 percent gets absorbed and 30 percent gets reflected back. Find out the force the metal plate will feel ok it is done. Now, what is the momentum of a photon? Sorry lambda is given I am sorry wavelength of the photon is given. Now, you will be able to do it. So, the initial momentum of the photon is what? H by lambda fine. If it get absorbed what is the final momentum? 0 it is absorbed ok. If it get reflected back what is the final momentum? Negative of H by lambda. I am getting it it get reflected back fine. So, if it is absorbed change in momentum is what? H by lambda only and if it is getting reflected change in momentum is 2 H by lambda fine. So, n photons per second are hitting 0.7 times n change in momentum is this and 0.3 times n change in momentum is that fine. So, total change in momentum per unit time this is how much 0.7 n H by lambda plus 0.3 n into 2 H by lambda fine. This is how much 0.7 plus 0.6 1.3 n H by lambda is it the force? Mechanics is getting mixed up the mechanics is good. So, in a photoelectric experiment it is found that the potential decreases from 1.85 volt to 0.82 volt. This is stopping potential. Stopping potential changes from 1.85 to 0.82. If you change the wavelength from 300 nanometer to 400 nanometer calculate the value of Planck's constant from this data find out H. First get the expression do not put the values assume this to be v 1, this to be v 2, this is lambda 1 and that to be lambda 2 of solubility H c by lambda 1 minus phi is e times v naught. Let us say e times v 1 and H c by lambda 2 minus c phi will not change metal is same. This is e times v 2 and now it is simple right just have to subtract phi will go away will be able to find the value of H fine.