 What else should I send to pick the 30th one of the previous chapter? 30. Yeah, sir All right, right Okay, so here is another question parallel beam of light of wave length 100 nanometer who all have joined now other than Ramcharan, Kondanya and Amov anybody else joined in? No, sir So parallel beam of light of see today, I guess many are not coming since tomorrow is your Physics exam in school. So we'll focus on problem solving only. I'll not be taking up New concepts today. So we'll just take up the questions and see how we can analyze those Not only from these chapters, we'll be taking other chapters also if time permits So a parallel beam of light of wavelength 100 nanometer Passes through a sample of atomic hydrogen in the ground state fine Now assume that when a photon supplies some of energy some of its energy to a hydrogen atom so This is a photon of wavelength this much. So energy of This photon is at C by lambda Okay, so part of this energy is utilized by the hydrogen atom To excite itself and rest of the energy is another photon Okay, moving in the same direction as the incident photon. So you can imagine a scenario like this that you have A glass tube Inside this glass tube, you have atomic hydrogen then a parallel beam is coming like this of Wavelength 100 nanometer Okay, I move you did this question Okay, neglecting the light emitted by the excited Hydrogen atom in the direction of incident beam what wavelengths may be observed in the transmitted beam so What it says that when hydrogen atom get excited? From n equal to 1 it could go to n equal to 2 3 4 like that. Okay, so when hydrogen atom gets excited it will Get de-excited also. So whatever wavelength gets emitted because of the de-excitation of hydrogen atom That doesn't come in the direction of parallel beam. So parallel beam Direction is this x-axis. So when hydrogen atom gets de-excited probably those wavelengths get emitted perpendicular to the direction of the Incident ray fine So we need to find what we need to find what wavelength is observed in the transmitted beam So this wavelength you need to find so this is lambda 1 Let's say lambda incident. So lambda transmitted is what? Can you tell me how you will analyze this situation? before solving It is given that some part of the energy is Used by the hydrogen atom to excite itself and rest of the energy is rejected You have to find out wavelengths that are observed in the transmitted beam any ideas. How will you do this? Okay, see here First we will try to find out energy of the photon Which is how much in terms of electron volt it will be at C by e lambda Can you quickly find out how many electron volt the energy of the photon is? This is incident photon 12.4 to electron volts Sure This will be 6.6 into 3 It will be 12.4 electron volts here. Yeah now see this Photon excites the hydrogen atom from its ground state Right at the ground state the energy is minus 13.6 electron volt. This is n equal to 1 so after excite man it will go to n equal to 2 Which is 3.4 electron volt or it could go to You guys remember these values, right? Yes, sir minus 1.9 0.8 correct so This is n equal to 4 minus 0.85 electron volt now it is given that all the hydrogen atoms were in the ground state so This gap is what? This gap is 10.2 electron volt Okay, so out of 12.4 electron volt it will absorb 10.2 electron volt. So how much will be remaining? 2.2 electron volt Fine, so when you equate this to the photon of this much energy in terms of electron volts You'll be getting wavelength of the transmitted beam because it will absorb 10.2 electron volt out of 12.4 and reject 2.2 electron volt are you getting this Yes, so this is one wavelength and then it could go to n equal to 3 probably How much is this energy gap from n equal to 1 to n equal to 3? 12.1 12.1 so it will go to n equal to 3 also so balance of the energy which is 0.3 electron volt This will also emit a photon on Lambda transmitted to a wavelength now. What is the difference in energy between 1 and 4? How much it is? 12.7 12.7 so it will not be able to go to n equal to 4 yes or no What if What if the n equal to 2 excited Electron absorbs further it can jump up are you getting it? Yes, right, but then we are ignoring that scenario because I think in the question also that scenario is not taken up So we'll assume that everything is happening from ground state onwards. So there will be two wavelengths that will be getting Transmitted is this thing clear yes now B part is Radiation detector is placed near to the gas to detect the radiation coming perpendicular to the incident beam We need to find out the wavelength of radiation that may be detected by the detector I think Naman joined from YouTube. Naman if you have any doubts, please WhatsApp me With the pic of your doubt. I can take it up here itself Huh, so now there's a detector which is With detects the radiation coming perpendicular direction. So beam will go along x axis perpendicular to beam The photons will come from where? Which photons will come perpendicular to beam these electrons which are excited when they get de-excited Getting it these will emit photons These photons will come out perpendicular direction getting it. It is written that beam will go parallel and The wavelengths that are emitted because of de-excitation once it get excited Will be coming perpendicular to the direction of the incident fine So That's it know that out Fine, so let me Give you your questions. So initially we'll take up questions from Etsy Verma, then I'll open up I know past year Photoelectric effect we haven't done yet, right for Kormangala. We haven't done it at least Yes, we haven't done For Indra Nagar we did What are the chapter is left other than items for Kormangala? So we haven't done communication systems Okay, that's a small Yeah, so other than that other than that we have done Magnetism and matter Okay, that's all See magnetism and matter I have taken a class already Fine, so that was also online only so exactly the way I take the class So if it is possible we can focus our class timings on problem solving and you can watch that Anyways, like it's totally up to you. We can discuss that later on also Fine, so let us Do this particular thing Which particular question you guys are in 23rd you said right? Yes, sir. Okay We'll be taking up Try to do 28 Meanwhile, I'll get a book on past year J questions So I haven't done semiconductors. Okay, but we have revised so many class 11 chapters. So I guess That's how you guys are ahead. I was wondering how many Should I do it Just be 270 others See this this question doesn't require any calculation as such Okay, so I guess some of you are Doing a lot of calculations, but that's a natural tendency, you know, but then at times you have to Look at the question very carefully here. You see that, you know sample gives five into 10 is for nine gamma raised per second initially and It drops to 2.5 into 10 is for nine per second finally So there is a drop of half in terms of activity Now this is the activity of Iron or activity of cobalt Tell me Who emits gamma? Who is emitting? So this is the activity of fe right? But then the half life of gamma radiation is Extremely small at as if it is negligible Right. So immediately when fe gets formed It goes to ground state by emitting gamma radiation Okay, so the activity of the iron Okay is Proportional to how many excited iron nucleus get formed Yes or no cobalt will give you Iron in its excited state then excited state of iron get converted into the ground state of iron by emitting gamma. So this Gamma radiation number is proportional to how many excited iron nucleases are there fine and and The number of excited Iron Nucleases that are getting formed is equal to number of cobalt nucleases that are decreasing Yes, right. So rate of formation of iron is equal to rate of decay of cobalt Yeah, and Rate of decay of cobalt is proportional to number of cobalt So it is exactly half life From five it goes to 2.5 Are you getting what I'm trying to say here? Yes, sir See the number of iron excited state getting formed per second is equal to number of cobalt that are getting decayed per second and number of cobalt decayed per second is Proposal to number of cobalt and number of Decay per second they are given so you can understand it takes only half life