 Hello everyone. In the first lecture, we discussed about the historical aspects of the phenomena of radioactivity and also how the different types of radioisotopes are there, whether they are occurring naturally or they are produced artificially and how they find these applications in different areas. Today, we will go into the fundamentals of radioactive decay and as usual, I will just briefly mention that at the end of this lecture, you should be able to answer these questions. For example, what is radioactive decay law? What is half-life and how to measure it from the experimental data? What is mean life of a radioisotope? What are the units of radioactivity? How does the activity of a daughter nucleide depend upon that of the parent activity and how does the radioactive decay in the case of branching? So, we will discuss these details in the details of this lecture. So, let us discuss first the radioactive decay law. We have just taken an example of a radioisotope A which is undergoing some decay. It could be alpha or beta or gamma to B. So, the A is called the parent and B is called the daughter. So, henceforth, once we say parent means the parent isotope which is decaying and whatever is found we call it daughter. Now, the radioactive decay law is like chemical kinetics in chemistry. In chemistry, you have the order of a reaction like if A going to B, you may say it is a first order reaction. Radioactive decay also follows the first order rate law. In the first order rate law, the rate of decay that is if n is the number of atoms of the reactant at any point of time, then minus dn by dt is the rate of decay of the atoms per unit time is proportional to the number of atoms at that point of time. And if you remove the proportionality constant, this lambda becomes a proportional constant. So, the rate equation for radioactive decay becomes minus dn by dt equal to lambda n. So, this is a first order rate equation. And if you want to find a solution of this equation, you integrate this equation. So, if it is an indefinite integral, so minus dn you arrange the variables a minus dn by dn by n equal to n integration dt plus a integration constant c. So, now, you can put the limits here that is the solution of this will become ln n equal to lambda t plus c. So, now, how do you find the c? When t equal to 0, n equal to n 0. So, you put this condition minus ln n 0 equal to c because t is 0. So, you get minus ln n equal to lambda t plus ln c ln n 0 rather it will be minus ln n 0. So, it will become minus ln n upon n 0 equal to lambda t or you can say n upon n 0 equal to e raised to minus lambda t. This is what is the expression for the decay of a radioisopropane. So, n is the number of atoms decaying with time and the number of atoms decay exponentially with time. n 0 is the initial number of atoms. So, the exponential decay of radioactivity is the signature of first order rate law that governs the radioactive decay law. So, this I have shown here as a function of time you plot the number of atoms then activity follows exponential decay. Now, from this graph you can if you want to find out the term called half-life. So, what is half-life? Half-life is the time when the number of atoms have become half. So, suppose initially there were n 0 atoms the time when the atoms left are n 0 by 2 will call is the half-life. So, you can just draw on this graph find out the time when the active number of atoms have become half. So, from the exponential data it may be difficult graphical to find out the half-life, but you can we will discuss very soon how to make it more accurate. So, when n equal to n 0 by 2 t equal to t half and so you can substitute the value of t in this equation and find out what is the value of t half. So, t half will become ln 2 by lambda. How do you do that? You can see here that is n upon n 0 equal to e raised to minus lambda t. So, take the logarithm ln n by n 0 equal to minus lambda t and so lambda. So, it becomes so when n by n 0 is 1 by 2 lambda t equal to ln n 0 by 2 n 0 by n and n is nothing but n 0 by 2. So, it becomes ln 2 and lambda 3 becomes t half. So, you can see here lambda becomes equal to ln 2 by t half or t half equal to ln 2 by lambda and so it is ln 2 is natural of 2 is 0.693 upon lambda. So, this is the relationship between half-life and decay constant. So, if you find out the half-life from graphically from this exponential decay you can find out the lambda or later on I will show you that if you find the lambda then you can find out t half from a another type of plot which I have shown this one. Now, most of the time we don't determine the number of atoms what we determine the activity. So, the activity is nothing but the number of atoms decaying per unit time. So, minus dn by dt is nothing but activity atoms decaying per second and it is given by n lambda that you see from here. So, activity is nothing but number of atoms into lambda. So, activity becomes equal to a 0 e raise to minus lambda t. So, you can if you plot activity functions as a function of time this also will follow the exponential decay because lambda is a constant. So, before we proceed further since we when we do measurements we don't measure the activity directly we measure the counts our detector system gives you counts and so if you say counts per second we have to convert into activity there are some factors called detection efficiency and so on. So, the units of activity like activity we will call as minus dn by dt number of atoms decaying per second equal to n lambda. So, this is called disintegrations per second and to honor the discoverer of the phenomenon of radioactivity we have a unit of 1 bacharyl is equal to 1 disintegration per second. Also there is another unit of radioactivity called curie in the name of modern curie and 1 curie is 3.7 10 power 10 bacharyls. So, you can see the curie is a much bigger unit compared to bacharyl. So, how do how does this number come about this 3.7 10 power 10 bacharyl actually is the activity due to 1 gram of 226 radium radium was discovered by madam curie and therefore, the activity of 1 gram of 226 radium has been defined as 1 curie. Now, you can find out the in terms of bacharyls what will be equal to 1 curie the half life of 226 radium is 1600 years. So, you put in the equation activity equal to n lambda number of atoms in 1 gram of 226 radium will be abrogator number 6.023 10 power 23 atoms in 226 gram this is the n in 1 gram into 0.693 upon t half that is lambda. So, 600 years into 365 days per year into 24 hours per day into 3600 seconds per hour. So, the units get cancelled and so, you have seconds. So, you have atoms per second and so, if you do the calculation it will come to 3.7 10 power 10 atoms per second or bacharyls. So, the curie is a very very large unit you can see the 1 gram of 226 radium is very difficult to handle. So, normally we do not handle curie level of activity we handle in fact, even less than micro curie at level of activity. So, there are other units called milli curie 10 power minus 3 curie 3.7 10 power 7 bacharyls micro curie 3 power minus 6 curie 3.7 10 power 4 bacharyls sometimes even we can have even nano curie also. So, for activity handling in the laboratory people use bacharyls 1 bacharyl 10 bacharyl, but if you are using for industrial applications like a radiation source or radiation you may be handling and that is your curies. Now, let us go a little further I was mentioning about determining the half-life from the decay data using the activity. So, activity is a a 0 e raise to 1 lambda t where a 0 is the initial activity. So, just a little bit of manipulation if you see if I take the logarithm on both sides that ln a is ln a 0 minus lambda t and now see this becomes a straight line of ln a versus t. So, you see here what I have plotted here is ln a on the y axis and time on the x axis and it becomes a straight line. So, it becomes easy to understand and handle lots. So, the slope of this line is lambda and from the activity data also you can now find out the half-life using a using a paper which I will call as semi log paper because on the y axis I have logarithmic unit on the x axis we have the linear unit. And so, when we do when we plot activity actually we will be plotting the count rate. So, the activity of radioastrophies disintegration per second, but when we have the experimental data we will have the count rate in detector and the count rate can be counts per unit time it could be counts per second it could be counts per minute and so on. So, that I will discuss now in more detail and it would be very interesting if you understand to follow this semi log paper what I have shown here is a semi logarithmic paper on the x axis we have the time let us say it is hours. So, 1, 2, 3, 4 hours. So, this is the larger graduations you can see there are the bold lines on the x axis you will find vertical parallel to y axis they are 1, this is in time in linear scale and on the y axis you see say the scale already made in logarithmic units. So, it is cycles you have 10 here 100 here 1000 here so it is called 3 cycle paper. You can plot the data on 3 orders of magnitude 10 to 100 100 to 1000 1000 to 10000. So, now when you plot the at the counts on this scale you do not have to convert the counts into log because the scale is adjusted like that. So, you see here when I say 100 next point is 200 300 400 so on when you have 1000 here then you have 2000 3000 and so on. So, if you understand how to use a semi log paper activity data handling is very very simple. So, what I have plotted here the decay of a redo isotope on a semi log paper and as I discussed in the previous slide also even here also you can see the logarithm of the counts versus time will be a straight line. Now, instead of ln natural log you plot here the log based to the power 10 and so logarithmic plot of activity will have a straight line behavior and now from here suppose initial count was 1000 when counts become 500 so this is 500 at this time and you can read on the x axis the time is 2 hours. So, straight away from the semi log paper you can find out the half life in a much simpler way. Once you find out the half life you can find out the decay constant lambda equal to 0.693 upon t. Now, there is another quantity called mean life. So, the what is mean life? Mean life is the average time a radio-nucleide survives half life is when half the atoms have decayed or half the atoms have survived. So, from the radioactive decay law n equal to n0 this one lambda t let us find out how many atoms survive or what is the time required for the survival of radio-nucleide on the average. So, if you find if you have a function and you want to find out the average how do you find out? So, you want to find out the mean value of t. So, you can take the t dn integrate over 0 to n0 divided by the total number of atoms that is integration of dn over 0 to n0. So, now you can write here now dn equal to the derivative of n here. So, dn will be n0 lambda e raised to the power lambda t upon n0. So, you can see now the variable has changed from n to t and so the limits of will become 0 to infinity instead of 0 to n0 because the radioactive decay that exponents of decay know that the curve will meet the x axis at infinite time and therefore, the time can go from 0 to infinity upon the integral of this will become n0. So, what you have here is now n0 will cancel. So, you will have lambda t e raised to minus lambda t dt and if you solve this integral you will find it will become 1 by lambda. You can do it at your home and find out whether it becomes equal to 1 by lambda. Now, there is one more way of deriving the expression n equal to n0 divided by lambda t based on statistics. In statistics we say lambda the decay constant is a probability of decay of an atom in unit time. So, if you that is the definition of lambda then what is the probability of survival in unit time 1 minus lambda lambda at the probability of decay the probability of survival is 1 minus lambda. You have a unit time interval delta t let us say in delta t what is the probability of survival of an atom it will be 1 minus lambda delta t. And now you want to find out what is the probability of survival till time t. Then this time t you can define in terms of n delta t that means, you divide this time interval into a large number of delta t intervals n intervals. So, it becomes 1 minus lambda delta t to the power n you just multiply the probabilities over the n intervals. And now when delta t becomes very small or n becomes very large. So, n tending to infinity 1 minus it is my lambda t upon n to the power n. So, this is actually you will see limit n tends to infinity 1 minus e raise to minus lambda t upon n to the power n will be it is not e raise to 1 minus lambda t upon n it will become e raise to minus lambda t. And so, what you have here is the probability of survival till time t becomes e raise to minus lambda t when n becomes very large. So, the fraction if there are n 0 atoms in the beginning what is the fraction of atoms that survive till time t n by n 0 equal to e raise to minus lambda t. So, you have n equal to n 0 e raise to minus lambda t. So, that is how from the fundamental definition of lambda as a probability of decay of an atom per unit time also you could derive the expression for radioactive decay. Now, there will be many situations when you will encounter two independently decaying radioisotopes. So, you have a 1 and a 2 are two isotopes decaying independently. They are independent of each other. So, a 1 is decaying with its own half life, a 2 decaying with own half life, but it is a mixture of activity. So, when you plot such a activity on a similar paper. So, I have now seen l and a essentially I am plotting on a similar paper. So, you see here the total activity will be the red line. Now, from how to resolve this such a data on a similar paper when the half life of the two isotopes are different. So, much later time when the short lived isotope has decayed you can extrapolate the linear portion to 0 time and what you get is the activity of the short lived isotope a long lived isotope and when you subtract this activity of this the long lived isotope from the total activity you get this data that is the pure activity of short lived isotope. So, extrapolation to 0 time gives you initial activity of the two isotopes and from the slope of these two you can find out their half lives. So, you can using similar papers you can resolve the mixture of two activities into individual isotopes half lives and their initial activity. There is another case of branching decay that means the same data isotope decays by two modes. So, a going to b and c with different decay constants lambda 1 and lambda 2. Now, in such a case so, the decay of the parent we can say minus dna by dt equal to na lambda a which is nothing but the the growth of a b and c separately. Now, this you can write d and b by dt as lambda 1 a and d and lambda this is lambda 2 a. So, this the decay to b and c can be written as now lambda 1 a these are two branches of decay of a. So, they this is lambda 1 a is first part lambda 2 is second part you take na as separate become lambda 1 plus lambda 2 into na. So, this is becomes the decay constant for the radio isotope a is the sum of the decay constant for the two branches. It is like you know if you have a tank and it has got two outlets to empty it and you open both the taps then the tank will decay empty and depending upon the diameter of the two taps you know the flow will be different the time taken to empty will be different. So, it is similar to that. So, this equation lambda equal to lambda 1 plus lambda 2 you can erase lambda 1 by t half 0.693 upon t half is for the parent equal to 0.693 upon t half 1 for the lambda 1 and 0.693 upon t half 2 for lambda 2. So, you can cancel the 0.690 that is the l n 2. So, 1 by t is 1 by t 1 plus 1 by t 2 or t is equal to in the half life of the parent equal to t 1 t 2 upon t 1 plus t 2 that is t 1 t 2 are called the partial half lives for decay into v and c. So, the relationship between the half partial half lives for the two branches to the parent half life is the way. So, when you measure the activity of either v or c the half life that you will get that of the parent. Because if you measure the rate between one tap in a tank or another tank the tank is getting empty anyway it decaying by both the taps. So, the half life that you will get while measuring either v or c will be the average half life of the radioisotope. So, this is a important like I will give you an example where this is the isotope copper 64. The copper 64 I will tell you later on in the in terms of the masses know it can decay by both beta plus as well as beta minus. So, in beta plus decay it becomes nickel 64 and beta minus decay becomes zinc 64. Now, the average half life. So, when you do the measurement of half life of copper 64 you get 12.7 hours. But when you calculate the partial half life. So, you calculate the half life of this copper 64 for beta plus decay or beta minus decay. I am writing here beta plus comma electron capture that we will discuss in the radioactive decay the nuclear decay data. But right now we will assume that this is the one channel beta plus e c is one channel. So, this is called electron capture when beta plus decay does not happen then there is a competing mode of decay to beta plus decay that is called electron capture we will discuss this more details in the nuclear decay. So, now decay constant for copper 64 is 0.693 upon t half and that t half is 12.7 hours. So, you can write is equal to 0.0564 hour inverse. Now, how do we get the partial half lives for decay into nickel 64 and zinc 64 we set up two equations lambda 1 plus lambda 2 is equal to 0.0546 that is the lambda and lambda 1 by lambda 2. So, here the branching intensities are given 61 percent of the time nickel 64 is formed 39 percent of the time zinc 64 is formed. So, I can write lambda 1 upon lambda 2 it is 0.61 upon 0.39 and that is 1.564. So, I have two equations lambda 1 plus lambda 2 equal to 0.0546 lambda 1 by lambda 2 0.61 by 1.564. So, I can solve these two equations to find out lambda 1 and lambda 2. So, lambda 2 is equal to 0.0213 hour inverse lambda 1 equal to 0.0333 hour inverse and now you can convert this lambda into half life they are called as the partial half lives. You see here the partial half life for decay to nickel 64 is 20.8 hours the partial half life for decay to zinc 64 is 32 hours and the average half life is 12 hours. So, you just see here the partial half lives are more than the average half life because it is decaying by both roots. So, it is less than both the half lives. So, what we discussed today was the radioactive decay and different types of decay is like the decay of a mixture of radioisotopes or the branching decay of a radioisotope into different modes like beta plus, beta minus and so on. And so, in the next lecture I will discuss the radioactive decay and the growth and decay of the daughter products. So, I will stop here. Thank you very much.