 Next question, this came in 1998 for 8 marks, nuclei of radioactive element A are produced at a constant rate alpha. A is produced constant rate alpha. The element has a decay constant of lambda. It has a decay constant of lambda. At t equal to 0, there are n naught nuclei and n naught nuclei are there. You need to find out number of nucleases after time t. So this nuclei is not only disintegrating, but also it is getting produced. Sir, this is just n naught e to the power minus lambda t minus alpha into t. The alpha will be increasing or decreasing the nucleus? Sir, then this alpha t minus n naught e to the power minus lambda t. Since alpha is the production rate. But as alpha produces, those nucleus will also get decayed right with time. So it is not a straight forward. You cannot separate these two processes. Okay sir. You need to solve a differential equation. Okay, should I do it? Yes, I will do it. Okay. Is this what you get? See, suppose it is only getting created. So dA dt will be equal to what? Alpha. And since it is getting decayed, dn by dt will also be equal to minus lambda times n. Fine. So this is prediction and this is decay. So total rate will be the summation of these plates which will be equal to alpha minus lambda n. So this is the differential equation you have which you have to solve. dn divided by alpha minus lambda n is equal to dt. Fine. And this is what you have to integrate from n naught to n. That integration part you will do right? Yes sir. Ramchand and Kondanya, are you following? Yes sir. Yes sir. Yes sir. Yes sir, I am following. Sure. Okay. Now write down if, today we are doing only problem solving siddhant as many of you have exams in the school. We are using this opportunity to discuss as many questions as possible. So alpha is equal to 2 n naught lambda. If alpha is 2 n naught lambda, this will be part of the same question. Okay. Calculate the number of nuclei of A, how many nuclei of A after half life of A? Okay. And as t tends to infinity, what is the number of nucleus of A tends towards? I hope this is what the expression you might have got. Are you getting this expression? Yes sir. Okay. Once you integrate this differential equation, you will get that expression. Okay. And then alpha is 2 n naught lambda. Substitute it and get the value of number of nucleus after t half. t half should be log 2 by lambda. t half is log 2 by lambda. Substitute that. What we got then, sir? 2 n naught into n lambda minus 1. It is actually 3 by 2 n naught. Okay. And when t tends to infinity, n becomes equal to 2 times n naught. How it becomes 2 times n naught? Because this term will go to 0 as t tends to infinity. So this goes to 0 and this is a multiplication with 0. So only alpha by lambda remains and alpha is 2 n naught lambda. So that is straight forward. Okay. So I hope you are appreciating that even though modern physics is simple, but you can't take it for granted. Alright. Fine. So let us take these ones. This came in 2003 for two marks. A radioactive sample emits n beta particles. It emits n beta particles in two seconds. Okay. And it emits 0.75 n beta particles. n is a number. Okay. n times beta. Sorry. n beta particles. Okay. Then 0.75 n beta particles. It will emit in next two seconds. Are you getting my point here? Yes, sir. What is the mean life of the sample? As in t average, that is how much? Which is 1 by lambda as a formula. Anyone got the answer? No? Sir, 2 by lambda 4 by 3. Yes. Correct. See n beta particles in first two seconds. So you can say that initially its rate is proportional to n. Okay. And then I can say n by 2. Then dn by dt after two seconds, 0.75 n by 2. Fine. So if you take the ratio, you are basically getting a ratio of n1 is to n2. This is proportional to number of nucleus initially. This is number of nucleus finally. So when you divide it, you will get log of 1 divided by 0.75 is equal to lambda into t. And t is what? Two seconds. So lambda is equal to 1 by t times log of 3 by 4. So you can get 1 by lambda, which is the average life, that is t divided by log 3 by 4. Okay. This is 4 by 3 CD error. Okay. Okay, sir. Fine. Let us move to next question. Sir, can we do some from atoms also? Atoms. Okay. Just one more question, then we will take a few from atoms. Okay. This I wanted to take up because this is a unique type of question. It came in 2001 for five months. Understand the question. Okay. A radioactive nucleus X. You can also write with me. X decays into Y with a decay constant of lambda X. Decay constant is lambda X. Y further decays into stable nucleus Z. Stable nucleus means it will not decay. Z will not decay with lambda Y as decay constant. Okay. Initially, there are only X nuclei. At t equal to 0, only X nuclei is there. Okay. And the number is n naught, which is 10 ratio power 20. I mean, let us find out everything in terms of variables only. So, number of X nuclei initially were n naught. Okay. You need to first write down the rate equations for X, Y and Z. Rate equation means what? dx by dt, what it is and dz by dt. Rate equations for X, Y and Z. And then you need to prove that number of nucleases Y as a function of time is given as n naught lambda X divided by lambda X minus lambda Y e to the power minus lambda Y t minus e to the power minus lambda X t. Okay. Do this first. Are you guys getting it? Just a minute. This is exactly like sequence of reactions in chemical kinetics. Oh, yeah. Chemical kinetics. So, you have derived something like this in chemical kinetics. Yeah. So, that expression you have to prove is exactly the same. Oh, so you know this already. So, rate at which the X is changing. First of all, it should be negative rate. X number is going down. This is equal to minus lambda X and X. Okay. And the rate at which Y is changing. See, when X is changing, it will increase Y. Fine. So, minus lambda X and X is decrease in X, but it is increase in Y. So, it will be lambda X and X minus lambda Y and Y. Okay. And can you tell me what is d and Z by dt? How much that is equal to? See, Z is only created. It is not decaying. And that is created because Y is disintegrating. This is lambda Y and Y only. Fine. Now, I need to find NY as a function of time. So, how will I integrate this particular expression? Because here I have one variable, two variable and time as a three variable. So, I need to have at max three variables. Right? Yeah, you can use nX is equal to nY is equal to negative. What? So, you can use the disintegration equation, right? n naught is equal to n. n is equal to n naught e to the power minus k. Right. So, lambda naught e to the power minus lambda X into t. Fine. So, once you write it like this, what we have done is that we have written nX variable as a function of time. So, now we have reduced and we have eliminated the nX variable. So, we have now only three variables. This we can integrate to get the value of nY. Okay? Any idea how should we integrate this? How should we go about integrating this? dny. So, you can multiply it with e to the power lambda or 2 or lambda Y into t. Okay. So, then you will get it, then you will get it as a differentiation, like as a derivative of the, of the, of Y. So, then like that you can. So, wait, lambda Y and N Y are all constants, right? Lambda Y is constant, N Y is not a constant. Okay. So, once you integrate this, you will get this particular expression. Now, there is a second part in which it is asked at what time N Y is maximum. So, once you get N Y, just take a derivative of N Y and equal to 0. So, you will get time at which N Y will be maximum. And then at that time, what is the value of X, Z and Y nucleases, right? So, that's how you do those kind of questions. Okay. So, we'll take a small break, we'll meet after 10 minutes. Is it fine? Yeah, stop. So, right now it is 6.13. So, we will meet at 6.23.