 Alright friends, so I am Dheeraj from Centrum Academy. So we are live again and we are taking up a question on nuclei chapter. All right, so this this is a different kind of question which is based on some, you know, scenario as such fine So let us see what this question is about. So as you can see On your screen, there is a huge question. All right So what we'll do we'll we'll solve it in a step-by-step manner. So there are multiple parts to this question Fine, so we'll take one by one all of these So there is a town which has a population of one million. So there is a town fine Which is a population of 10 this power six people All right, the average electric power needed per person is 300 watt So every person requirement is 300 watt Fine to the requirement of town is what? So, you know, it must be about the town which is asked later on. So we are just You know readily finding out some of the obvious parameters like per person energy consumption is given So I'll find out the Consumption by the entire town. So that would be what 300 into 10 is for six. So town will require three into 10 this power eight watt Okay, the reactor is designed to supply power to this town The efficiency with which thermal power is converted into electrical power is 25% fine. So thermal power If it is 100 then 25 will be converted into electrical power fine So if the requirement of town is this much electrical energy So how much thermal energy is required to create this much electrical energy? Assume that to be x fine. So 25% of x Which is x into 0.25 should be equal to 3 into 10 ratio power 8 watts Fine So x will be equal to what 3 divided by 0.25 which is 12 into 10 ratio power 8 joules per Second fine. So this much amount of heat energy Must be supplied to create that much amount of power in the town Okay, so these are some of the initial analysis we did now. Let us see what is asked This I'll write. Yes. This is the thermal energy Required Okay now part a of the question assuming 200 Mev of Thermal energy to come from each fission event Find the number of events that should take place every day now we know that one fission will give 200 Mev Okay, if you convert it into joules you'll get here as 200 into 10 ratio power 6 Into charge of electron This much joules you'll get Right. So one fission gives this much joule and the requirement is 12 into 10 is for eight joules every second Fine. So every second How many reactions should take place? Let's call it that as n and that will be equal to 1.2 into 10 ratio power 9 which is x Okay that divided by This energy, okay, that is equal to 2 into 1.6 Into 10 ratio power This is 8 and that is minus 11. Okay, so this many Events Should happen per second Okay, now how many events should happen per day? that will be equal to and into 24 hours 60 minutes into 60 seconds Right, so this is the answer for part a this many events per second So per day if I have to find out I have to multiply that with number of seconds one day has Okay, so like this you can do part a of the question All right. Let us see what is part B In part B assuming that the fission to take place largely through uranium 235 at what rate? Will the amount of uranium 235 decrease? Okay, we know that every second this many Fission reaction should happen fine and one fission reaction one fission will consume one nucleus of Uranium 235 Fine, so how many nucleases will get consumed of Uranium 235 it will be equal to this only number of events per second Will be equal to number of nucleus that should get consumed every second Isn't it now if you want to find out what is the mass rate of consumption? all you have to do is Divide this n with the Avogadro number Fine, if you divide it with Avogadro number, you'll get number of moles that should get consumed every second This divided by 6.023 Into 10 raised to power 23 fine, so this is number of moles of Uranium 235 that should get consumed every second fine now what is a consumption rate in terms of you know mass you can multiply this with the molecular mass or the you know Molar mass sorry, so if you multiply this with 235 You will get grams per second that should get consumes Gram per second that should get consume Okay, so if you want to find in terms of kg per second you just divided with thousand Fine, so like this you can do the B part of the question Okay, now it says that you should also express in kg per day So this is gram per second So I'm sure you will be able to convert gram per second in kg per day. All right So we can move to the C part of the question assuming that uranium enriched to 3% of 235 Uranium that will be used you need to find how much uranium is needed per month All right, so in B part we have got how many kgs of uranium 235 is required per day fine Let's call that as why so this much uranium 235 suppose is required per day Fine, so for 30 days What is the requirement of uranium 235 that is 30 into y Okay, this much uranium 235 is required per month kg per month. All right now This much uranium 235 is required, but uranium doesn't exist in pure 235 it is written that Only 3% of naturally occurring uranium has uranium 235. Okay, so if that is a case We can say that if x kg of Uranium naturally occurring uranium is required then this into 3% which is point 0.03 Will be equal to 13 to y Okay, so if you solve this equation you'll get the value of x and x is the num x is the amount of naturally occurring uranium that you require To conduct all this Okay, so I hope you have learned something today in case you have any doubts feel free to get in touch with us And we'll be you know clearing your doubts Hello friends. This is Dheeraj. So Today also I have taken up a question and this question is from radioactivity chapter All right, so let's see what this question is about so you can see a question Which is there in your screen a charge capacitor of capacitance C is Discharged through a resistance R. Okay, so whenever you you know here such thing immediately That formula comes in your mind right that charge at any point in time should be equal to q naught e to the power minus t by rc, right? So this is a formula for discharging of the capacitor. All right, then a radioactive sample decays with an average life Tau so average life is tau lambda is equal to what 1 by tau Right because tau is equal to 1 by lambda. All right find the value of r for which the ratio of electrostatic field energy Stored in the capacitor to the activity of the radioactive sample remains constant in time now Energy as a function of time in the capacitor. How will we find out? energy is q square by 2c Isn't it and charge at any point in time is given by equation number one. All right, so I can use equation number one over here and Write the sound as this will be q naught square e to the power minus 2t by rc Okay, this is q square that divide by 2c. All right and ratio of this The energy To the activity should be constant now activity is what activity of a radioactive sample is dn by dt Right and dn by dt is if I just talk about magnitude of the activity is lambda times n Okay, and n as a function of time is given as n naught e to the power e to the power minus lambda t Okay, so this is what we have and now if I take the ratio the ratio should be independent of time Okay, so if I take ratio of energy With activity, let's say activity just dn by dt if I write it like this What I'll get is q naught square by 2c fine lambda n naught e to the power minus 2t by rc multiplied by e to the power lambda t fine, so basically this entire thing This thing is a constant, right? So let let that be a constant k. All right. This is k into e to the power minus of 2 by rc minus lambda times t Okay, now if this ratio is Independent of time then what should happen the ratio is independent of time then coefficient of this time should be equal to zero Right, so 2 divided by rc Minus lambda should be equal to what? Zero, okay, we need to find what we need to find the value of r now the value of r from this You'll get it by solving this equation. So 2 by rc 2 divided by rc is equal to lambda fine, so r is equal to what r is equal to 2 by C times lambda where C is the capacitance So this is how you solve this particular question So I hope you have learned something today in case you have any doubts Please feel free to get in touch with us. Okay. Thank you