Lecture 3 Radioactivity
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1 Objectives In this lecture you will learn the following We shall begin with a general discussion on the nucleus. Learn about some characteristics of nucleons. Understand some concepts on stability of a nucleus. In particular, we shall discuss Neutron/Proton ratio for a nucleus to be stable. Move on to quantify radioactivity and associated concepts. Understand the radioactive law and learn the concept of half life. Finally digest the analysis of radioactive chains and observe the general behaviour. Analyse a typical three element chain and identify its characteristics.
2 Review of Nuclear Physics Most of the concepts in nuclear engineering can be understood by considering only protons, neutrons and electrons as described by Rutherford's model. Just as chemical energy is released by the rearrangement of electrons, nuclear reactions can be understood by considering rearrangement of protons and neutrons. Chemical Energy Nuclear Energy Rearrangement of Electrons Rearrangement of Protons and Nuetrons Mass of Proton and Neutron are approximately equal to 1/N Avogadro in grams. Every element can be represented by In the above Z - No of Protons or Charge Number. A - No of Protons + No. of Neutrons or Mass Number. Isotope have Same Z, but different A e,g.,,. Typically the Radius of Nucleus = m. Radius of Atom = m. Volume ratio = From above we understand that there is enormous hollow space. This implies that we need a large number of neutrons for collisions to occur.
3 Stability Of Nucleus In nature all elements are not stable. For example, Oxygen is stable, while Uranium is not. Interestingly, Neutron/Proton ratio influences stability. There is an optimum ratio needed for a nucleus to be stable and this ratio changes with the mass of the nucleus. The figure shown below indicates the zone where stable and unstable isotopes lie. The ratio of N/P varies from The stable band is shown by a thick line, while the unstable one is shaded. The unstable ones emit radiation, transform themselves and move towards stability. The process of transformation of unstable nucleus by spontaneous emission of radiation is called radioactivity.
4 The fundamental radioactive law is that the probability of nuclear disintegration rate is constant. Consider the following: Population > N Time > dt Disintegration > dn = Constant = (say) The above equation is a first order Ordinary Differential Equation (ODE). If the initial value is. Separating the variables and integration leads to This implies that the population decays exponentially as shown in the figure. Thus, technically speaking a radioactive isotope has infinite life.
5 Activity Every disintergration results in emission of radiation dn/dt = Rate of emission of Radiation. This is termed as Activity and is denoted by α. α is numerically equal to λn (from radioactive law). SI unit of α is Becquerel denoted by Bq. 1 Bq = 1 disintegration per second (dps) or s -1. An older unit is often used and is called Curie. 1 Curie = 3.7 Χ10 10 Bq. It represents activity of 1g of Radium.
6 Half Life As pointed out earlier, the life of a radioactive isotope is technically infinite. Hence a half life is defined for every radioactive isotope We can now rewrite the radioactive decay law as In 5 halflives, In 10 halflives Thus in about 10 half lifes the population is reduced by about 1000 times.
7 Analysis of Decay Chains-I Fission Products are Radioactive A B C D.. Practical chains are long. A three isotope chain has all the characteristics of any long chain. Consider A B C (Stable) The Population Balance Equation can be written as Rate of change = Production Rate Destruction Rate. Let N A, N B and N e be the population of A, B and C respectively. The conservation of species A, B & C can be written as,, The solution of first equation with the initial condition leads to the solution With N A = N A0 at t = 0 Similarly the second equation can be written as Using as Integral factor, we can write, Integration of the above with the initial condition, N B = N B0 leads to
8 Since The following figure shows the variation of population of A, B and C for the data shown inside the figure.
9 Special Cases-I For λ B > λ A After some time, the first term goes to zero and the term 2 in the parenthesis of second term also becomes negligible. This leads to This is called transient equilibrium. Here though both N A and N B varies with time, the ratio of them remains a constant. It may be observed in the last slide that for t > 3, N B /N A = 1.
10 Special Cases-II For λ B >> λ A The above condition is called secular equilibrium. It implies that irrespective of their initial activities they quickly attain the same activity. This can be appreciated by the figure shown below, where for t > 0.3, the ratio of the activity reaches the value 1.05.
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