 So we will proceed further from here. So the relation we have now is ln n by n naught is equal to minus lambda times t. So here there is one very popular term which is half life. Half life simply means amount of time that is required to decrease the number of nucleus to half of its original value. So half life means time taken for n naught to go to n naught by 2. So n is n naught by 2, so this is log of n naught by 2 divided by n naught. This is equal to minus lambda times t. So log of half which is equal to minus of log 2 is equal to minus lambda t half. This is half life, it is called t half, but t half is equal to log of 2, natural log of 2 divided by lambda and natural log 2, the value for that is 0.693. So write down, so this is the half life. This you should remember because at times the value of lambda is not directly given but t half is given. So from there on you can calculate the value of lambda and then you can use it for further analysis. This lambda has a name, this is called decay constant. This is called decay constant. It depends only on the type of nucleus or what is a nucleus, which nucleus we are talking about whether it is uranium, plutonium, which nucleus it is, it depends on that for radium it will be something else, for some other nucleus it will be something else, like that. And also have you done chemical kinetics? So this is first order reaction, the first order where rate of reaction is directly proportional to number of particles, number of reactants. But then the difference between the chemical reaction and this nuclear first order reaction is this that in chemical reaction the rate also depends on external conditions like pressure, temperature, isn't it? But then in chemical reactions the rate is independent of, the rate is independent of external conditions, write this down. It doesn't matter what is the external, it doesn't matter what is the external condition, the rate doesn't depend on the external conditions. Because the forces that are getting involved here, they are nuclear forces which are tremendous. So temperature and pressure will hardly affect these type of forces. So the rate doesn't depend on it, is it clear? So this is T half, let us solve a particular question, write down. The half life of uranium 238, uranium 238 half life is 4.5 into 10 raise to power 9 years. So you can see that it needs tremendous amount of time to get consumed. So it's not very easy to get rid of a radioactive substance once it starts disintegrating. So it will just last for years and years, that is why we need to be very, very careful with the radioactive substance. So the half life is this and you need to find out activity of 1 gram of uranium 238. Activity is nothing but a measure of what is dn by dt. The SI unit of dn by dt is disintegrations per second. Disintegrations per second or it is also called as becquerel. Because becquerel was the first scientist who discovered this phenomenon. It is bq, becquerel. So the SI unit of dn by dt is this. Now you know, in this chapter you will be again and again converting years into seconds because second is the SI unit end of the day. So it is a good idea to remember how many seconds is one year. So one year, write down one year is 3.15 roughly this into 10 raise to power 7 seconds. Anyone got the answer? So calculated. No. This is what you have done. And dn by dt, mod of that is lambda times n. So lambda is this 0.693 into n. n is what? How will you find n? n is number of nucleus, right? How will you determine how many nucleases are there? It is 1 gram of uranium 238, right? So number of moles is 1 divided by 238. This into Avogadro number. Number of nucleus is equal to number of atoms only, isn't it? You have done this, Aditya? n is number, it is not mass. So you will get roughly 1.23 into 10 raise to power 4 disintegrations per second or you can also write it as becquerel, bq, okay? Now here, you might have heard about Madame Curie when we discuss radioactivity, few scientists named like becquerel and Madame Curie that comes in our mind. So Madame Curie has also done a lot of research in it. So there is a unit of disintegration in the name of Curie, okay? So Madame Curie has done research with respect to radium and other elements. So she has introduced a unit Curie for disintegration per second, which is related with SI unit with this. This is 1Ci, which is 1 Curie, write it down, 1 Curie is equal to 3.7 into 10 raise to power 10 becquerel, okay? So you have to remember this because at times, you know, the units will be given for disintegration in terms of Curie, it is 3.7 into 10 raise to power 10 becquerel, all right? Should I raise this? So now there is one more important concept in radioactivity, which is average life, okay? For example, recently there was an article in Times of India where the average life of Indians is found out to be around 68 or 69, around that, okay? So how they calculate that average life, what they do, they find out out of a sample, let us say there is a sample of 1,000 people, okay? So if 10 people, they live for let us say 30 years, okay, which is very less, and you know, 500, they live only for 50 years, okay? And then 490, I am just, you know, randomly assuming, 490, they live for 80 years and nobody lives beyond 80 years. So how will you find the average life in this case? What you will do? 10 plus 500 times 50 plus 80 times 490 divided by 1000, exactly, so you take weighted average. So weighted average will make sure that you are adding up all the ages. And taking the average, this is the average life, this divided by 1000, okay? Similarly since nucleus is disintegrating and let us say that the life of the last nucleus is infinite, can I find the average life of the nucleus, okay? So we have, let us say, Tn by dt, which is given as, you know, minus lambda times n, which is what? Minus lambda times n naught e to the power minus lambda t, okay? And suppose I am only interested in the magnitude, as in the number of nucleus that are disintegrating, in next dt second will be what? Lambda n naught e to the power minus lambda t dt, isn't it? This is equation number one. So I can say that this dn nucleus will get disintegrated in next dt second. Is it clear, all of you? Yes, sir. So what is the life of these dn nucleus? How much is the life? These dn nucleus is where they present at t equal to 0, yes or no? They were there or not? Tell me one by one, where these dn elements were present at t equal to 0, yes or no? They were there, right, t equal to 0, all the nucleus were alive, they were there, okay? And they were there till t equal to t seconds, okay? But then in next dt second, they are gone. So what will be their life from 0 to t, so t seconds, are you getting it? Yes, sir. So these dn nucleus were present till t seconds and then they are disappearing in next dt second. So the life of dn nucleus is t, okay? Now I am finding the life of the nucleus in terms of variable t. So what I will do in order to find the average life, let us say tau average, I have to literally add up, I have to sum it up, t with multiplication of number of nucleus that will go away in next dt seconds, okay? And divide it with total number of nucleus similar to this, are you getting it? Yes, sir. Yes, sir. Fine. So since we are talking about the continuous function and we are taking differential element, the summation will get replaced by integration, so this will be an integration of this. And lambda into n naught is a constant, so it will come out of integral, this will become t times e to the power minus lambda t dt, this divided by n naught, so this n naught will be gone, getting it. So this will be the integral and the limits of time will be what? t will be integrated from where to where, from 0 to 0 to infinity, there will be few nucleus as whose time of disintegration is infinite, are you with me on this? Yes, sir. Okay, now can you find the value of this integral, quickly find out, you have learned integration by parts, right? Find out the value of this integral, this is the formula for integration by parts, lambda square or 1 by lambda square? Oh, yes, sir, 1. So remember this, the average life is 1 by lambda, okay? This formula is also quite useful because at times in the numerical, they talk about average life, okay? So half life is 0.693 divided by lambda and average life is 1 by lambda, okay?