Slide 1 : Chapter Nuclear Physics Gaurav Arora 1 Nuclear Physics
Slide 2 : Introduction u The branch of physics which deals with the study of nucleus and nucleus related phenomena is called Nuclear Physics. u A nucleus is made up of elementary particles called neutrons and protons. A neutron is a neutral particle carrying no charge. A proton carries a unit positive charge. Neutron is slightly heavier than proton. Mass of proton (mp) = 1.6726 × 10–27 kg = 1.007825 a.m.u.Mass of neutron (mn) = 1.6749 × 10–27 kg = 1.008665 a.m.u. Protons and neutrons are together referred as nucleon. Atomic Number of an element is the number of protons present inside the nucleus of an atom of the element. It is represented by Z. Mass Number of an element is the total number of protons and neutrons present inside the atomic nucleus of the element. It is represented by A. In an atom we haveNumber of protons = Z Number of electrons = ZNumber of nucleons = A Number of neutrons = A – Z Gaurav Arora 2 Nuclear Physics
Slide 3 : Properties of Nucleus 1. Charge : The charge of nucleus is given by QNucleus = Zewhere Z is atomic number 2. Size : The radius ‘r’ of the nucleus depends upon the atomic mass A of the element. It is given by r = R0A1/3where R0 = 1.2 × 10–15 m = 1.2 Fermi NOTE : (i) Density of nucleus is constant for all isotopes.(ii) Density of nucleus is very high while that of the atom is very small. This is due to the fact that the atom contains a lot of empty space in it. Gaurav Arora 3 Nuclear Physics
Slide 4 : 4. Nuclear quantum states : Like electrons, nucleus is also capable of existing in various discrete energy levels known as nuclear quantum states. When nucleus makes transition from one states to another, the difference of energy of the two levels is emitted in the form of electromagnetic radiation which always lie in g-region, irrespective of nucleus. 5. Nuclear spin : Just like the orbital electrons of the atom, the nucleons of the nucleus revolve in various orbits. This orbital motion give angular momentum. Nucleons like electrons also have an intrinsic angular momentum called spin angular momentum. Hence, the nucleus as a whole possesses angular momentum due to nucleons. Nuclear spin is the angular momentum arising out of the spin as well as the orbital motion of the nucleons. Gaurav Arora 4 Nuclear Physics
Slide 5 : Isotopes and Isobars Gaurav Arora 5 Nuclear Physics
Slide 6 : Nuclear Forces Nuclear force is much stronger than gravitational and electromagnetic forces and is operative when nucleons are at a separation of a femtometre Nuclear force is responsible for stability. Unlike gravitational or electromagnetic force, nuclear force is not represented by a simple formula. Various-characteristics of nuclear forces are given below : (i) Nuclear forces are attractive in nature. (ii) Nuclear forces are charge independent. Nature of force remains the same whether we consider force between two protons, between two neutrons or between a proton and a neutron. (iii) These are short range forces. (iv) Nuclear forces are spin dependent. Nucleons having parallel spin are more strongly bound to each other than those having anti-parallel spin. Gaurav Arora 6 Nuclear Physics
Slide 7 : (vi) The nuclear forces show saturation properties i.e., each nucleon interacts with its immediate neighbours only, rather than with all the other nucleons in the nucleus. (v) When distance between nucleons becomes less than 0.5 Fermi, the nuclear forces become strongly repulsive. Yukawa Theory of Nuclear Forces According to this theory a nucleon consists of a core surrounded by a cloud of mesons, which may be charged or neutral. The mesons constantly get exchanged back and forth between two neighbouring nucleons. In this process the two nucleons remain bound to each other. Gaurav Arora 7 Nuclear Physics
Slide 8 : The very shot range of strong nuclear forces plays an important role in stability of nucleus. For a nucleus to be stable, the electrostatic repulsion between protons must be balanced by the attraction between nucleons due to strong nuclear force. Inside a nucleus one protons repell all other protons. Since the electrostatic force has a long range of action. But a proton or a neutron due to strong nuclear force attracts only its nearest neighbours, thus to in a large sized nucleus to counter balance the repulsive force more number of neutrons are required to maintain the stability of nucleus. In the graph almost all points representing stable nuclei fall above the straight line N = Z, reflecting that the number of neutrons are greater than number of protons. Gaurav Arora 8 Nuclear Physics
Slide 9 : In very heavy nuclei as number of protons are large, there comes a point when balance of repulsive and attractive forces cannot be achieved by an increased number of neutrons. For heavy nuclei, the size is also large and due to the limited range of strong nuclear force, extra neutrons in the nucleus can not balance the long range electric repulsion of extra protons. Gaurav Arora 9 Nuclear Physics
Slide 10 : In a stable nucleus, because of strong nuclear force of attraction, the nucleons are held tightly together in a small volume. As system is stable we can relatively say that the total potential energy of system is negative and to separate all the nucleus from each other some energy must be supplied to break the nucleus. 2mH + 2mn = 2 (1.007825) + 2(1.008665) Gaurav Arora 10 Nuclear Physics
Slide 11 : Some amount of mass from independent nucleons is converted into energy and released when nucleons bounded with each other to form a stable nucleus. The binding energy of above nucleus X can be given as DEBE = Dmc2 Gaurav Arora 11 Nuclear Physics
Slide 12 : The mass defect per nucleon i.e., Dm/A = P, is called the packing fraction of the nucleus. Mass Energy Equivalence Generally nuclear masses are given in atomic mass unit where 1 amu = 1.656 × 10–27 kg If in a reaction 1 amu mass is converted into energy then using Einstein’s mass energy relationship the amount of energy released is DE´ = Dmc2 Gaurav Arora 12 Nuclear Physics
Slide 13 : 1 amu mass is equivalent to 931.5 MeV energy = (1.656 × 10–27) (3 × 108)2 joule = 931.5 × 106 eV = 931.5 MeV Gaurav Arora 13 Nuclear Physics
Slide 14 : Lets consider an example of 56Fe and 209Bi. Their binding energies are given as DEFe = 492.8 MeV DEBi = 1640 MeV From the above values it seems that to break the nucleus of Bi more energy is required hence it is more stable then that of Fe. But we should not ignore the bigger size of bismuth nuclei. If we find binding energy per nucleon for both of these nuclei, we get Now we can see that removal of one electron from Fc nucleus is more difficult as compared to that from Bi nucleus. So Fe nuclei are more stable then Bi. Thus to judge or compare the stability of different nuclei we see binding energy per nucleon not the nuclear binding energy. Gaurav Arora 14 Nuclear Physics
Slide 15 : Variation of Binding Energy per Nucleon with Mass Number The binding energy per nucleon is a characteristic property of elements. The graph is figure shows the variation of binding energy per nucleon with mass number for all the elements of periodic table. Gaurav Arora 15 Nuclear Physics
Slide 16 : It is observed that nuclei with A > 209 are unstable and hence radioactive. Gaurav Arora 16 Nuclear Physics
Slide 17 : Example:A neutron breaks into a proton and electron. Calculate the energy produced in this reaction in MeV. Mass of an electron = 9 × 10–32 kg. Mass of proton = 1.6725 × 10–27 kg. Mass of neutron = 1.6747 × 10–27 kg. Speed of light = 3 × 108 m/sec. Mass defect of the process is given by Dm = [Mass of neutron – (mass of proton + mass of electron)] = [1.6747 × 10–27 – (1.6725 × 10–27 + 9 × 10–31)] = 0.0013 × 10–27 kg Solution : Gaurav Arora 17 Nuclear Physics
Slide 18 : Nuclear Reactions I + T ¾® C ¾® P + O Incoming Target Compound Product Outgoing Particle nucleus nucleus nucleus radiation When nitrogen is bombarded with a-particle following reaction take place : 2He4 + 7N14 ¾® 9F18 ¾® 8O17 + 1H17N14 gets converted into 8O17 due to bombardment with a-particles. Example : Gaurav Arora 18 Nuclear Physics
Slide 19 : Q-Value of a Nuclear Reaction Let Kei KEp and KEo be the kinetic energies associated with I, P and Q respectively while the target “T” is at rest initially. Q-value of a nuclear reaction is given by Q = KEp + KEo – KEi Let mi, mt, mp and mo, respectively, be the masses of ‘I’, ‘T’ ‘P’ and ‘O’Before reaction, Energy of ‘I’ = mic2 + KEi Energy of ‘T’ = mic2Total energy of the system = mic2 + KEi + mic2 After reaction, Energy of ‘P’ = mpc2 + KEp Energy of ‘O’ = moc2 + KEo Total energy of the system = moc2 + KEp + moc2 + KEo Gaurav Arora 19 Nuclear Physics
Slide 20 : According to the law of conservation of energy mic2 + KEi + mic2 = mpc2 + KEp + moc2 + KEo Q = KEp + KEo - KEi = [(mi + mt) – (mp + mo)]c2 Q = Dmc2where ‘Dm’ is the mass defect between initial and final particles. Case (i) A reaction is said to be exoergic if Q is positive. Q is positive if (mp + mo) < (mi + mt).The part of mass which disappears gets converted into the energy in accordance with Einstein’s mass-energy relation. Case (ii) A reaction is said to be endoergic if Q is negative, Q is negative if (mp + mo) < (mi + mt) i.e., the sum of the masses of product particles is greater than that of reactant particles. For this reaction is proceed, the incoming particle mass posses an energy equivalent to the mass defect. Gaurav Arora 20 Nuclear Physics
Slide 21 : Solution : Gaurav Arora 21 Nuclear Physics
Slide 22 : This shows that the original unstable products must have emitted 4 b particles. Again mass number on right hand side is 2 units less than on left hand side i.e., two neutrons are also produced. Now the reaction can be represented as (ii) Mass defect, Dm = mass of L.H.S. – mass of R.H.S Mass of L.H.S. = (235.044 + 1.009) = 236.053 amu Mass of R.H.S. = (97.905 + 135.917) + 4(0.00055) + 2(1.009) = 235.842 amuThus mass defect is Dm = (236.053 – 235.842) amu = 0.211 amuThe equivalent energy released is DE = Dm × 931.2 MeV = 0.211 × 931.2 MeV = 196.48 MeV Gaurav Arora 22 Nuclear Physics
Slide 23 : A deuterium reaction that occurs in an experimental fusion reactor is in two stages : (i) Two deuterium nuclei fuse together to form a tritium nucleus, with a proton as a by-product written as D(D, p) T. Compute (a) the energy released in each of two stages (b) the energy released in the combined reaction per deuterium, and (c) what percentage of the mass energy of the initial deuterium is released. Example : Gaurav Arora 23 Nuclear Physics
Slide 24 : Given (a) The reaction involved in the process is Solution : Mass defect of the above reaction is Dm = [2(2.014102) – (3.016049 + 1.007825) = 0.00433 amu Energy released in the process is E1 = 0.00433 × 931.2 MeV = 4.033 MeV Gaurav Arora 24 Nuclear Physics
Slide 25 : The reaction involved in the process is Mass defect of above reaction is Dm = [(3.016049 + 2.014102) – (4.002603 + 1.008665)] = 0.01888 amu Energy released in the process is E2 = 0.01888 × 931 MeV = 17.585 MeV Total energy released in both processes = 17.585 + 4.033 + 21.618 MeV Gaurav Arora 25 Nuclear Physics
Slide 26 : (b) Energy released per deuterium atom is (c) % of rest mass of released Gaurav Arora 26 Nuclear Physics