Nuclear Physics MSCPH511 Past Paper PDF 2022

MSCPH 511 M. Sc. III SEMESTER NUCLEAR PHYSICS DEPARTMENT OF PHYSICS SCHOOL OF SCIENCE UTTARAKHAND OPEN UNIVERSITY HALDWANI NUCLEAR PHYSICS MSCPH511 DEPARTMENT OF PHYSICS SCHOOL OF SCIENCES UTTARAKHAND OPEN UNIVERSITY Phone...

MSCPH 511 M. Sc. III SEMESTER NUCLEAR PHYSICS DEPARTMENT OF PHYSICS SCHOOL OF SCIENCE UTTARAKHAND OPEN UNIVERSITY HALDWANI NUCLEAR PHYSICS MSCPH511 DEPARTMENT OF PHYSICS SCHOOL OF SCIENCES UTTARAKHAND OPEN UNIVERSITY Phone No. 05946-261122, 261123 Toll free No. 18001804025 Fax No. 05946-264232, E. mail [email protected] htpp://uou.ac.in  Board of Studies Prof. P. D. Pant Prof. S.R. Jha, Director School of Sciences School of Sciences, I.G.N.O.U., Maidan Uttarakhand Open University, Haldwani Garhi, New Delhi Prof. P. S. Bisht, Prof. R. C. Shrivastava, SSJ Campus, Kumaun University, Almora. Professor and Head, Department of Physics, Dr. Kamal Devlal CBSH, G.B.P.U.A.&T. Pantnagar, India Department of Physics School of Sciences, Uttarakhand Open University Department of Physics (School of Sciences) Dr. Kamal Devlal, Assistant Professor & Programme Coordinator Dr. Vishal Kumar Sharma, Assistant Professor Dr. Gauri, Assistant Professor Dr. Meenakshi Rana, Assistant Professor (AC) Dr. Rajesh Mathpal, Assistant Professor(AC) Unit Writing and Editing Editing Writing Dr. Vishal Kumar Sharma Dr. Vishal Kumar Sharma Assistant Professor Assistant Professor Department of Physics Department of Physics School of Sciences, Uttarakhand Open University School of Sciences, Uttarakhand Open University Course Title and Code : Nuclear Physics (MSCPH511) ISBN : Copyright : Uttarakhand Open University Edition : 2022 Published By : Uttarakhand Open University, Haldwani, Nainital- 263139 Printed By :  Contents Course 11: Nuclear Physics Course code:MSCPH511 Credit: 3 Unit Block and Unit Title Page Number Number BLOCK 1: Nuclear Properties and Nuclear Models 1 Nuclear Properties 1-29 2 Nuclear Binding 30-55 3 Nuclear forces I 56-82 4 Nuclear Forces II 83-101 5 Nuclear Models 102-129 BLOCK 2: Radioactivity 6 Radioactivity 130-157 7 Alpha Decay 158-173 8 Beta Decay 174-193 9 Gamma Decay 194-215 BLOCK 3: Nuclear Reactions 10 Nuclear Reactions 216-262 11 Fission and Fusion 263-318  1 NUCLEAR PHYSICS MSCPH511 UNIT 1 NUCLEAR PROPERTIES Structure of the Unit 1.1 Introduction 1.2 Objectives 1.3 Introduction to Nuclear Terminology 1.4 Discovery of Neutron 1.5 Rutherford Scattering and Nuclear Size Estimation 1.5.1 Nuclear Radius 1.5.2 Measurement of Nuclear Radius 1.6 Angular Momentum 1.7 Nuclear Statistics 1.8 Parity and Symmetry 1.9 Magnetic Dipole Moment 1.10 Electric Quadrupole Moment 1.11 Summary 1.12 Glossary 1.13 References 1.14 Suggested Readings 1.15 Terminal Questions 1 UTTARAKHAND OPEN UNIVERSITY HALDWANI  2 NUCLEAR PHYSICS MSCPH511 1.1 INTRODUCTION In this unit, we will study the basic concepts of the nucleus, constituents of the nucleus and some basic nuclear properties like nuclear size, nuclear mass, nuclear charge, etc. Before discussing the unit in detail we must have some knowledge about the atomic structure. Now question arises, what is the basic building block of all the matters? The answer is an “atom”and it is the smallest amount of matter that retains all the properties of an element and atomic theory also suggests that it is composed of smaller particles that no longer have the same properties as the overall element and consists of two main components which will be discussed in this unit. Now, when we look into the origins of nuclear physics, this can be traced back to atomic structure investigations, which began in 1896 with Henry Becquerel's discovery of radioactivity. It's helpful to review how the nuclear atom came to be in order to grasp the concept of the nucleus. The majority of scientists in the early nineteenth century agreed that chemical elements are made up of atoms, but they had no idea about atomic structure. The fact that all atoms contain negatively charged electrons was one of the first indications. So, in order for an atom to remain neutral, it must possess positively charged matter of some kind. But how did it come to be like that way? Rutherford discovered this when he used a thin gold foil to conduct his famous alpha-scattering experiment. The alpha particles were supposed to pass through the foil with hardly any deflection. This is based on the Thomson model, which assumes that the electric charge inside an atom is uniformly distributed throughout its volume. However, they were surprised to find that while the majority of the alpha particles did not deviate much, a few were distributed over quite wide angles. Some have even become scattered. This is when the nuclear atom notion was formed, in which the atom is claimed to be made up of a tiny nucleus carrying all of the positive charge and approximately all of its mass, with the electrons separated by some distance away. 2 UTTARAKHAND OPEN UNIVERSITY HALDWANI  3 NUCLEAR PHYSICS MSCPH511 1.2 OBJECTIVES This unit introduces the basic concepts regarding nucleus and their basic properties. After studying this unit, you should be able to- Describe the basic terminology used in nuclear physics like atomic number, atomic mass, isobars, isotopes etc. Define fundamental properties of the nucleus such as nuclear size, nuclear mass and nuclear charge Know about Nuclear spin, magnetic moment and electric quadrupole moment. 1.3 INTRODUCTION TO NUCLEAR TERMINOLOGY Probing the fundamental particles and their interactions, identifying and interpreting the features of nuclei, and making technological improvements that benefit society are the three main components of nuclear physics. In nuclear physics a nuclear species is characterized by the total amount of positive charge contained by the nucleus and also by its total number of mass units. The value of the net nuclear charge is equal to+𝑍𝑒 , where 𝑍 is the atomic number and e denotes the magnitude of the electronic charge. The fundamental positively charged particle inside the nucleus is the proton,which is also the nucleus of the simplest atom, Hydrogen. A nucleus with the atomic number Z almost contains Z protons, and on the other hand an electrically neutral atom must therefore contain Z negatively charged electrons. As we know that the mass of the electrons is negligible as compared to the proton mass ( 𝑚𝑃 ≅ 2000 𝑚𝑒 ),thereforethe mass of the electron can often be ignored while we have the discussions about the mass of an atom. The mass number of a nuclear species is denoted by the symbol 𝐴 , which is the integer nearest to the ratio between the nuclear mass and the fundamental mass unit. 3 UTTARAKHAND OPEN UNIVERSITY HALDWANI  4 NUCLEAR PHYSICS MSCPH511 In general, Ais greater than Z, nearly for all the nuclei and in most cases, it increases by a factor of two or more as compared to Z. Thus, there are more massive components in the nucleus. Before 1932, it was supposed that the nucleus contained 𝐴protons, in order to provide theproper mass, along with 𝐴 − 𝑍 nuclear electrons to give a net positive charge of𝑍𝑒. One question also arises here: Does electron exist inside the nucleus? The answer is provided with certain facts that are not satisfied for the electrons to remain inside the nucleus. 1. The nuclear electrons would have to be held to the protons by a strong force, maybe even stronger than the Coulomb force. Despite this, there is no evidence for a strong force between protons and atomic electrons. 2. If we try to confine the electrons in a region of a small space as small as a nucleus (∆𝑥~10−14 𝑚) the uncertainty principle would require that these electrons have a momentum distribution with a range ∆𝑝~ ħ⁄∆𝑥 = 20𝑀𝑒𝑉/𝑐. Electrons which emits from the nucleus in radioactive β decay have energies value generally less than 1 𝑀𝑒𝑉; never we have seen a decay of electrons with20 𝑀𝑒𝑉 energies. Thus the existence of 20 𝑀𝑒𝑉 electrons inthe nucleus isnot confirmed by this observation. 3. The total intrinsic angular momentum (spin) of nuclei for which the value of 𝐴 − 𝑍is odd would disagree with the observed values if 𝐴protons and 𝐴 − 𝑍electrons were present in the nucleus. For example, consider the nucleus of deuterium with 𝐴 = 2,Z= 1), which according to the proton-electron hypothesis would contain 2 protons and 1 electron. The proton and electron both have intrinsic spin angular 1 momentum2and from the quantum mechanical rules for adding spins of particles would 1 3 1 require that these three spins of 2 combines to give a total of value either2 or 2.But the observed spin of the deuterium nucleus is 1. 4. Magnetic dipole moments in nuclei containing unpaired electrons should be far higher than those measured.If a single electron were present in a deuterium nucleus, we would anticipate the nucleus to have a magnetic dipole moment about the same size as an 4 UTTARAKHAND OPEN UNIVERSITY HALDWANI  5 NUCLEAR PHYSICS MSCPH511 electron, however the observed value of the magnetic moment for the deuterium nucleus 1 is 2000 of the value of the magnetic moment of one electron. All sorts of above reasons were eliminated by the discovery of neutron in 1932 by Chadwick. The neutron in general are electrically neutral and it has a mass almost equal to that of proton mass.Thus a nucleus with 𝑍protons and 𝐴 − 𝑍 neutrons has the proper total mass and charge, without the need to introduce nuclear electrons. In general, a nuclear species or nuclide is represented by 𝐴𝑍𝑋, where X represents the chemical symbol with the number of neutrons 𝐴 − 𝑍, where A is the atomic mass and Z represents the atomic number which have the same value as that of protons. For example: 11𝐻 , 42𝐻𝑒, 63𝐿𝑖 , 168𝑂, 56 26𝐹𝑒. The nucleon family consists of two members: neutrons and protons. We use the term nucleons when we only want to talk about nuclear particles without specifying whether they're protons or neutrons. As a result, a nucleus with mass number A has A nucleons. Nuclides with the same proton number but different neutron numbers are called isotopes; for example, the element chlorine has two isotopes that are stable against radioactive decay,35Cl and 37Cl. It also has many other unstable isotopes that are artificially produced in nuclear reactions; these are the radioactive isotopes (or radioisotopes) of Cl. A nuclide with the same N but different𝑍; these are called isotones. The stable isotones with N = 1are 2Hand3He. Such Nuclides which have the same mass number A are known as isobars; thus, stable3He and radioactive 3H are isobars. 1.4 DISCOVERY OF NEUTRON The neutron was not found until 1932, when James Chadwick calculated the mass of this neutral particle using scattering data. Since the time of Rutherford, scientists have known that the atomic mass number A of nuclei is somewhat more than double that of most atoms' atomic number Z, and that the nucleus contains almost all of the atom's mass. Protons and electrons were thought to be the fundamental particles around 1930, however this required that a number 5 UTTARAKHAND OPEN UNIVERSITY HALDWANI  6 NUCLEAR PHYSICS MSCPH511 of electrons be bound in the nucleus to partially negate the charge of A protons. However, by this time, the uncertainty principle and "particle-in-a-box" confinement calculations had established that there just wasn't enough energy available to confine electrons in the nucleus. By putting the particle's De-Broglie wavelength equal to that dimension, a rough scale of the energy required for confinement to that dimension can be produced. If we assume a hydrogen atom has a diameter of 0.2 nm, the equivalent confinement energy is roughly 38 eV, which is the correct order of magnitude for atomic electrons. However, it takes around 250 MeV of energy to confine an electron to a nuclear dimension of about 5 fermis. The maximum confinement energy available from the nucleus's electrical attraction is given by 𝑍𝑘𝑒 2 79(1.44 𝑀𝑒𝑉. 𝑓𝑚) = ≈ 23 𝑀𝑒𝑉 ≪ 250 𝑀𝑒𝑉 … … ….. (1) 𝑟 5𝑓𝑚 So, from above result there are no electrons in the nucleus. Bothe and Becker observed in 1930 that bombarding beryllium with alpha particles from a radioactive source created neutral radiation that was penetrating but non-ionizing, which was an experimental breakthrough. They assumed it was gamma rays, but Curie and Joliot demonstrated that when this radiation was used to bombard a paraffin target, it expelled protons with an energy of roughly 5.3 MeV. As can be seen through momentum and energy analysis, that wasn't the case with gamma rays: Fig. 01: Curie and Joliot experiment 6 UTTARAKHAND OPEN UNIVERSITY HALDWANI  7 NUCLEAR PHYSICS MSCPH511 This analysis is similar to that of a head-on elastic collision in which a small particle collides with a considerably larger one. The required energy for the gamma ray explanation was significantly higher than any energy measured in the nucleus, thus the neutral radiation must be some form of neutral particle. If the neutral particle had a mass comparable to that of the proton, the 5.3 MeV energy of the expelled protons could be easily explained. This would only require 5.3 MeV from the neutral particle in head-on collisions, which is within the range of observed nuclear particle emissions. By bombarding targets other than hydrogen, such as nitrogen, oxygen, helium, and argon, Chadwick was able to demonstrate that the neutral particle could not be a photon. Not only were these interactions incompatible with photon emission on energy grounds, but their cross-section was orders of magnitude larger than that of photon Compton scattering. Chadwick was left with the problem of estimating the mass of the neutral particle. He decided to bombard boron with alpha particles and study how the neutral particles interacted with nitrogen. Figure 02: Discovery of neutron The masses of boron and nitrogen were well known; thus, these specific targets were chosen. When energy conservation is applied to the combined interactions, the following formulas result. 1 1 1 𝑚𝛼 𝑣𝛼2 + 𝑚𝛼 𝑐 2 + 𝑚𝐵 𝑐 2 = 𝑚𝑁 𝑣𝑁2 + 𝑚𝑁 𝑐 2 + 2 𝑚𝑛 𝑣𝑛2 + 𝑚𝑛 𝑐 2………….(2) 2 2 Solving for the mass energy of the neutron gives UTTARAKHAND OPEN UNIVERSITY 7 HALDWANI  8 NUCLEAR PHYSICS MSCPH511 1 𝑚𝛼 𝑣𝛼2 + 𝑚𝛼 𝑐 2 + 𝑚𝐵 𝑐 2 − 𝑚𝑁 𝑐 2 2 𝑚𝑛 𝑐 ≈ 2 … … … … … (3) 𝑣2 1 + 𝑛2 2𝑐 The speed of the neutron is the last remaining unknown on the right hand side of the equation. Chadwick blasted hydrogen atoms with his created neutrons, assuming that the neutron mass was near to that of the proton, in order to determine the speed of the protons following the impacts. He then used the above energy expression to generate a neutron mass of 9381.8 MeV by setting the neutron speed equal to those proton speeds. Chadwick obtained the initial value for the neutron mass, which matched the current accepted estimate of 939.57 MeV by using a consistent set of measurements. 1.5 RUTHERFORD SCATTERING AND NUCLEAR SIZE ESTIMATION A Preliminary idea about the nuclear model was first suggested by Rutherford in 1910.The diagram of Rutherford’s experimental set up is shown in the figure 03. Obviously the apparatus has to be closed in vacuum. A thin film of gold foil is made to strike by a beam of alpha particles. The scattered alpha particles were detected by a fluorescent screen-telescope arrangement. Rutherford observed that many of the alpha particles went straight through the foil or they were deflected by a very small angle. But a few alpha particles were deflected by a very large angle. Further some were even got scattered back. This observation was not expected on the basis of idea of atomic structure at that time: that the positive charge is uniformly smeared over the entire volume of the atom. 8 UTTARAKHAND OPEN UNIVERSITY HALDWANI  9 NUCLEAR PHYSICS MSCPH511 Figure 03: Rutherford’s gold foil experiment For explanation of these deflections, Rutherford assumed that the positive part of the atom was concentrated in a very small volume at the center of the atom. This core, which is surrounded by the cloud of electrons, which makes the entire atom electrically neutral, is now called as nucleus. This nuclear model of the atom accounts for the scattering of alpha particles at large angles in the following way. An alpha particle approaching the center of the atom experiences an increasingly large Coulomb repulsion. Since the atom is mostly empty space , most of the alpha particles do not go sufficiently close to the nucleus to get sufficiently deflected and so they pass through the foil without any deviation. However, an alpha particle that passes close to the nucleus is subjected to a very large Coulomb repulsive force exerted by the massive positive core and is deflected at a large angle in a single encounter. 1.5.1 Nuclear Radius From the above discussion we have a basic idea about the nucleus. Now we shall discuss about the estimation of Nuclear Radius by considering Rutherford’s α-Scattering experiment using classical approach. Let us consider that the gold nucleus in the scattering experiment has a radius R. An α-particle trajectory can be specified by its impact parameter b as sown in the figure 9 UTTARAKHAND OPEN UNIVERSITY HALDWANI  10 NUCLEAR PHYSICS MSCPH511 Figure 04: Deflection of an alpha particle by a gold nucleus For the less coloumb reflection i.e for b is greater than R the deflection of alpha paeticle is also less. Also when b is less than R, the alpha particle goes through the nuclear positive charge distribution and so the nuclear charge above the trajectory and the one below the trajectory work in opposite directions and hence again the deflection produced in the alpha particle trajectory is small. When the alpha particle just almost touches the nucleus (which is considered to be a positive charge), this is the case when coulomb repulsion becomes maximum and correspondingly for b~R, the deflection of alpha particle also becomes maximum. For this qualitative discussion, let us consider a trajectory corresponding to small deflection. As the particle approaches near the gold nucleus, it is slowed by the Coulomb repulsion and is speeded up on its way out i.e away from the nucleus as shown in the figure below Fig.05: Particle trajectory with impact parameter b and small deflection θ 10 UTTARAKHAND OPEN UNIVERSITY HALDWANI  11 NUCLEAR PHYSICS MSCPH511 The Coulomb repulsion in the region close to the scattering gold nucleus is given by the formula 1 (2𝑒)(𝑍𝑒) 𝐹= ……………..(4) 4𝜋𝜖0 𝑏2 Where 2e represents the charge on α particle. Z represents the atomic number and for gold (Z=79). Now this repulsive force can be approximately considered to be operating in the direction perpendicular to the direction of incidence, over a distance b. If we consider the particle’s velocity to be v then the time during which the force acts is given by 𝑏 ∆𝑡 = … … … …. (5) 𝑣 By this force there is a momentum ∆𝑝 in the direction perpendicular to the incident direction. By Newton’s law Δ𝑃 𝐹= Δ𝑡 1 2𝑍𝑒 2 𝑏 ∆𝑝 = 𝐹∆𝑡 = ………………….(6) 4𝜋𝜖0 𝑏 2 𝑣 From fig.05 the deflection θ is ∆𝑝 1 2𝑍𝑒 2 ⁄𝑏𝑣 𝜃~ = ………………(7) 𝑝 4𝜋𝜖0 𝑚𝑣 Where m denotes the mass of the α-particle. So, we have 1 2𝑍𝑒 2 1 𝑏 = ……………(8) 4𝜋𝜖0 𝑚𝑣 2 𝜃 Above equation shows the approximate relation between deflections and impact parameter. UTTARAKHAND OPEN UNIVERSITY 11 HALDWANI  12 NUCLEAR PHYSICS MSCPH511 Rutherford in his experiment observed the maximum deflections were of the order of 1 radian. This doesn’t mean that large deflections were not observed by Rutherford. Occasionally, deflections close to π radians were observed by him. In above equation if we put the values corresponding to maximum deflections i.e θ = 1 and b = R, we have 1 2𝑍𝑒 2 𝑅~ … … … … … ….. (9) 4𝜋𝜖0 𝑚𝑣 2 If we put other values Z= 79 for gold, mass of α particle 𝑚 = 6 × 10−27 𝑘𝑔 𝑣 ~ 107 m/s for speed of α particle 𝑒 = 1.6 × 10−19 𝑐𝑜𝑢𝑙𝑜𝑚𝑏𝑠 We obtain the radius has order of 10-14m. This distance is extremely small as compared to the atomic radius value 10-10 m, which is smaller by a factor of 104. Since protons are positively charged and the gravitational force is negligible inside the nucleus, the query then arranges us to how all these protons state together in a nucleus of size having value 10-14m. There is certainly a nuclear interaction acting between protons and neutrons in bracket nucleons inside the nucleus over dispenses of about 10-14m that is strong enough to counteract the massive column repulsion between the closely packed protons. Because the nuclear interaction's attraction force has no effect on the additional nuclear atomic structure, it must be a short-range force, effective only across distances of 10-14m. This new force must, by definition, be more dependent on distance than an inverse square force. With very high energy particles generated by accelerators, the magnitude of the region where this nuclear short-range force is considerable (relative to the Coulomb force) may be examined. When such particles are fired at thin foils, the distribution of scattering angles reveals that the radius R of the nucleus is 12 UTTARAKHAND OPEN UNIVERSITY HALDWANI  13 NUCLEAR PHYSICS MSCPH511 proportional to the cube root of the scattering nuclei's mass number, i.e. R = RoA1/3 , where Ro has the value of 1.07 x 10-15 m. 1.5.2 Measurement of Nuclear Radius In the previous section we have estimated the nuclear size by the use of Rutherford’s α- scattering experiment. There are many experimental methods for measuring nuclear size. In this section we shall study about the experimental method of measuring the nuclear radius which was first employed by R. Hofstadter and his colleagues by the important bearing that nuclear size has our understanding about the nuclear force. This method is based on measurement of nuclear charge distribution and it assumes that charge distribution and mass distribution within nuclei are essentially the same. By analyzing scattering of high-energy electrons from various nuclei we have obtained the charge distribution in nuclei. Incident beams of high energy Thin target Beam collector θ electrons from accelerator Detector for scattered electrons Fig.06: Experimental Setup for nuclear radius measurement In this experiment electrons are chosen as bombarding particles and its interaction within the nucleus is well known eletro-magnetic interaction. As shown in the fig.06 the elastically scattered electrons are observed by a detector as a function of angles. It is easy to see that the de Broglie wavelength of 200 MeV electrons is about 10 -14 m (by using the relation𝜆 = ℎ⁄𝑝 ) which is the size of the nucleus. Thus, to get information about 13 UTTARAKHAND OPEN UNIVERSITY HALDWANI  14 NUCLEAR PHYSICS MSCPH511 nuclear charge distribution, the incident electron beam energy has to be the order of 200 MeV. The assumption that the nuclear charge is uniformly spread over a spherical volume and not a point charge leads one to expect diffraction effects. Portions of electron waves incident on different parts of the nucleus will be scattered in a particular direction, with phase differences resulting in constructor for destructive interference at some angle. Fig.07 is taken from the work of Hofstadter and his collaborators and shows the angular distribution of elastically scattered 200 MeV electrons from nuclei assumed to have spherical uniform charge distribution (having uniform density up to radius r as shown in fig.08. The actual nuclear charge distribution is not exactly given by a step function, but is shown by dotted line. One can clearly see in fig.07 that experiments do show diffraction maxima and minima as expected from a uniform charge distribution. For comparison, the expected result (which is not obtained experimentally) from a point charge assumption is also shown in fig.07. For theoretical calculations Hofstadter and his colleagues assumed a charge density of this form 𝜌0 𝜌(𝑟) = ……………….(10) 1+𝑒 (𝑟−𝑅)⁄𝑏 𝜌0 Where 𝜌0 stands for density at the nuclear center, R is the radius at which 𝜌 falls to 2 and 𝑏 measures how rapidly 𝜌 falls to zero at the nuclear surface. 14 UTTARAKHAND OPEN UNIVERSITY HALDWANI  15 NUCLEAR PHYSICS MSCPH511 Fig.07: Shows experimental angular distribution of scattered electrons from three types of nuclei Fig. 08: step function shows here the uniform charge distribution It is clear from the distribution relation given by equation 4 posseses a property that for 𝑅 ≫ 𝑏, 𝜌 is close to 𝜌0 , until (R-r) is a few times b, at which point 𝜌 decreases to very small values in a distance determined by b, but not by R. If we put the following values, then this charge distribution agrees with the experimental data involving many nuclei UTTARAKHAND OPEN UNIVERSITY 15 HALDWANI  16 NUCLEAR PHYSICS MSCPH511 𝑛𝑢𝑐𝑙𝑒𝑜𝑛𝑠 𝜌0 ~ 1.65 × 1044 , 𝑎~0.55 𝑓𝑒𝑟𝑚𝑖 𝑚3 and 𝑅~1.07𝐴1⁄3 fermi ………………..(11) Where 1 Fermi = 1 F = 10−15 m. If we vary the values of R and b, then we obtain the charge distribution which represents the observed angular distribution shown in Fig. 07. The fig. 09 shown below represents that the charge density is essentially the same for almost all nuclei( except lighter nuclei H and He) decreasing slowly for the increasing values of A. There exists a general relationship for the nuclear radius, which has been experimentally verified, as given by equation 11, i.e. 𝑅 = 𝑅0 𝐴1⁄3 Fig. 09: Nuclear charge distribution found by using high-energy electrons as probes in some nuclei 16 UTTARAKHAND OPEN UNIVERSITY HALDWANI  17 NUCLEAR PHYSICS MSCPH511 Other methods are also there for obtaining the nuclear radius. An important method probes the extent of the nuclear force rather than the nuclear charge. We have the same general relationship 𝑅 = 𝑅0 𝐴1⁄3. Example : 01 : Determine the radii of O16 and Pb206 nucleus.(given that R0= 1.4fm) Solution :Using the relation R=R0A 1⁄3 and substituting the values of A in the following formula, we have R(O16) = 1.41.4 × (16)1⁄3 = 3.33 fm. R(O16) = 1.4 × (208)1⁄3 = 3.29 fm. Example : 01 : Determine the stable nucleus that has a radius one third that of Os189. Solution : By Using the relation R=R0A 1⁄3 1 𝑅 𝐴 1⁄3 𝐴 1⁄3 = = ( ) = ( ) 3 𝑅𝑂𝑠 𝐴𝑂𝑠 189 189 𝐴 = =7 27 The element with the mass number (A=7) is Lithium(Li7). 1.6 ANGULAR MOMENTUM The angular momentum of an isolated system is known to be conserved according to elementary quantum mechanics. Since the nucleus is an isolated system, its angular momentum is a constant quantity. Most commonly, the total nuclear angular momentum J is denoted by symbol I corresponding to nuclear spin term. So, the total angular momentum J of a nucleon is the vector sum of its orbital and spin angular momenta values 17 UTTARAKHAND OPEN UNIVERSITY HALDWANI  18 NUCLEAR PHYSICS MSCPH511 J = L+S …………………….(13) where L is orbital angular momentum and S is spin angular momentum. The magnitude of the total angular momentum of a nucleon is given by J= j ( j + 1) ……………….(14) J can take on the half integral values, 1/2, 3/2, 5/2,……… The magnitude of the spin angular momentum is given by S = s ( s + 1) ……………………….(15) Where S = ½ is the spin angular momentum quantum number and the magnitude of the orbital angular momentum is given by L = l ( l + 1) ………………………….(16) L can take up integral values 0,1,2,3, Quantum mechanical considerations show that the total orbital and spin angular moments of the nucleus is given by PI2 = I ( I + 1) 2 PL2 = L ( L + 1) 2 ………………….(17) PS2 = S ( S + 1) 2 It is determined during measurement which part of the angular momentum is larger along the direction of the applied electric or magnetic field. These have the magnitudes I, L, and S, respectively, for the three examples indicated above. 18 UTTARAKHAND OPEN UNIVERSITY HALDWANI  19 NUCLEAR PHYSICS MSCPH511 The nuclear spin is always zero (I=0) for even Z and even N nuclei, according to measurements of the ground state spin of nuclei. This demonstrates that the nucleons inside the nucleus have a tendency to pair off into nucleon-like pairs when their angular momenta are equal and oppositely aligned. The measured values of the ground state spins of the nuclei are small integers or half odd integers; the highest measured value is 9/2, which is small when compared to the sum of the absolute values of Ii and Si of all the individual nucleons contained in the nucleus. This is an important point to keep in mind. This is consistent with what was said earlier about pair creation inside the nuclei. The majority of nucleons of either kind appear to assemble into pairs of zero spin and orbital angular momentum protons and neutrons, giving the core itself zero total angular momentum. The nuclear spin, which is thus determined by the few remaining nucleons outside the core is therefore a small number, integral or half odd integral. 1.7 NUCLEAR STATISTICS Nuclear statistics can be Either Bose-Einstein statistics or Fermi-Dirac statistics, provide a description of a system with many particles, such as the nucleus, in terms of quantum mechanics. Thus, it is possible to divide all fundamental particles into two types Bose-Einstein Statistics: The wave function of a system of two identical subatomic particles is either symmetric or anti-symmetric in the exchange of the two particles' coordinates in the case of a subatomic particle subject to the laws of quantum mechanics. If the sign of the wave function is not altered by such an interchange, we have a symmetric wave function, and for that Bose-Einstein Statistics holds. Bosons, which are defined as particles with integral spin or zero spin, are subject to Bose-Einstein statistics. Photon, meson, and deuteron are what they are. Even mass number nuclei are subject to Bose-Einstein statistics. Fermi-Dirac Statistics: But when the exchange results in a change in the wave function's sign, we get an asymmetric wave function, and the associated statistics are 19 UTTARAKHAND OPEN UNIVERSITY HALDWANI  20 NUCLEAR PHYSICS MSCPH511 known as Fermi-Dirac statistics. Fermions are all particles with half integral spin, which all correspond to Fermi-Dirac statistics. Protons, neutrons, and electrons also comes into this category. All odd mass number (A) nuclei correspond to F.D. statistics. No two fermions can exist in the same quantum state because all fermions obey the Pauli exclusion principle. On the other hand, Boson particles do not follow to the exclusion principle and a number of Bosons can exist in the same quantum state. 1.8 PARITY AND SYMMETRY Another very important property that particles possess is Parity.The particle's parity is a characteristic of the wave function representing its quantum mechanical state. Positive parity refers to a wave function that represents a single particle that does not undergo a sign change upon reflection through the origin, whereas negative parity refers to a wave function that undergoes a sign change. So we have  ( x, y, z ) =  ( − x, − y, − z ) =  ( x, y, z ) , for positive parity...........(18)  ( x, y, z ) =  ( − x, − y, − z ) = − ( x, y, z ) , for negetive parity...........(19) A wave function that describes multiple particles can be expressed as the sum of the wave functions of each individual particle or as a composite of those wave functions. Since the parities of the individual particle wave functions are a product, the parity of the entire system is evidently determined by this. The mass density and charge density of nuclei are symmetric no matter what the parity is since they are always equal. Symmetry is an important concept in atomic and nuclear physics. Schrodinger's equation for two identical, non-interacting particles travelling in the same potential has a straightforward solution that is  AB =  A (1)  B ( 2 )....................(20) where  A and  B stand for the positions of particles 1 and 2, and are two solutions to the same one-body wave equation. However, the wave function  AB is not an acceptable function for two UTTARAKHAND OPEN UNIVERSITY20 HALDWANI  21 NUCLEAR PHYSICS MSCPH511 identical particles because since it assumes that we can label the particles and distinguish between those in state A and those in state B. This difficulty can be avoided by writing the wave function in either one or the other of the two following forms, both of which are solutions of Schrodinger's equation if Eqn. 20 is a solution 1 S =   A (1)  B ( 2 ) +  A ( 2 )  B (1) ........(21) 2 1 A =   A (1)  B ( 2 ) −  A ( 2 )  B (1) ........(22) 2 When labels 1 and 2 are switched in both equations, the probability density is symmetric. As a result, both are valid solutions to the two-body problem. Because I and 2 can be switched around,  S does not change and it remains symmetric , while  A is anti-symmetric and sign changes. 1.9 MAGNETIC DIPOLE MOMENT Now in this unit we shall study about the magnetic dipole moment µ which is associated with a current loop of area A and having a current I µ = IA …………………..(23) Fig. 10: Magnetic dipole with circular loop of area A 21 UTTARAKHAND OPEN UNIVERSITY HALDWANI  22 NUCLEAR PHYSICS MSCPH511 A spinless particle revolving with with electric charge e generates a magnetic dipole moment which is equal to the number of revolutions per second times the charge times the area enclosed by the orbit. Now for this circular loop with area A(=π r2) ,we get the value of magnetic dipole moment by substituting the values of I and A  ev  2 evr =  r =..................(24)  2 r  2 where v is the velocity of the particle and r is the radius of the orbit. The orbital angular momentum of this particle is mvr. The ratio of the magnetic dipole moment to the angular momentum is called the gyromagnetic ratio which is given by  e g= =..................(25) p1 2m Naturally, this relationship will hold for the two vectors' components as well as for p, in any direction, such as the direction of a magnetic field. We know from quantum physics that the mechanical moment p's component along a specified direction (the z-direction) is m h, which can be either a positive or negative integer or zero. Consequently, we obtain for the magnetic dipole moment's z-component e  z =   m1....................................(26)  2m  This expression yields the correct result for the z-component of the magnetic dipole moment that is due to orbital motion of electrons in an atom. It is believed to give the correct result also for the orbital motion of protons in a nucleus. The mass in the denominator of Eqn. 26 should then be the proton mass mp.. In contrary to magnetic dipole moments, we can introduce the nuclear magneton defined by e N = = 5.0505 10−27 Jm2 wb −1................(27) 2m p 22 UTTARAKHAND OPEN UNIVERSITY HALDWANI  23 NUCLEAR PHYSICS MSCPH511  N shows its analogy with the Bohr magneton which is given by the formula 𝑒ħ which is the unit of 2𝑚 𝑒 atomic magnetic moment.  N is much smaller than  B , being only 1/1836 part of the later. The measured values of the magnetic moments of the electron, proton and the neutron are given below e = −1.001145358 B  p = 2.7927  N n = −1.9131 N.............................(28) In the above equation 28 negative sign results from because of the direction of the angular momentum vector and the magnetic dipole moment vector are opposite to each other. The total magnetic dipole moment for a nucleus is the sum of the moments associated with the spins of the protons and neutrons and the moments associated with the orbital motion of the proton. Since the z-component of the dipole moment is the only component which can be observed through its interaction with a magnetic field. 1.10 ELECTRIC QUADRUPOLE MOMENT The nuclear electric quadrupole moment is a parameter which describes the effective shape of the ellipsoid of nuclear charge distribution. A non-zero quadrupole moment Q indicates that the charge distribution is not spherically symmetric. By convention, the value of Q is taken to be positive if the ellipsoid is prolate and negative if it is oblate. First we have the value of electric potential V of any distribution of electric charges at a distance R in the Z-direction given by equation 29 1 1  V=  4 0  R  R 1 R 1 ( )  dV + 2   zdV + 3   3z 2 − r 2 dV +....... ................(29) 23 UTTARAKHAND OPEN UNIVERSITY HALDWANI  24 NUCLEAR PHYSICS MSCPH511 where,  is the charge density. The integration is performed over the region containing the electric charge. In the above expression we have a rapidly convergent series and as the value of R increases, only the first two or three terms become important in this series. The first term corresponds to the net charge and for larger R this is the only important term. The integral in the second term is called the dipole moment and the third term is called the quadrupole moment. It turns out that a nucleus must have a zero electric dipole moment and so the lowest order contribution to V due to a point charge arises from the quadrupole moment. Two examples of non-zero electric quadrupole moments are shown in Fig. 11 Fig. 11. Electric Quadrupole Moment The four-charge (quadrupole) system in the left part of the Fig. 11 has net charge and dipole moment zero and so the entire electric field is produced by the electric quadrupole moment. An ellipsoidal charge distribution shown on the right leads to a highly deformed nucleus; it has a quadrupole moment; however, its dipole moment is zero. The charge density at point r(x, y, z) is Ze ( r ) given by where Ze is nuclear charge. The quadrupole moment Q is given by ( ) Q = Z 3 z 2 − r 2  (r )d 3 r ( ) = Z  r 2 3cos 2  − 1  ( r ) d 3 r...............(30) 24 UTTARAKHAND OPEN UNIVERSITY HALDWANI  25 NUCLEAR PHYSICS MSCPH511 For a spherically symmetric charge distribution  ( r ) , the quadrupole moment vanishes, for a prolate nucleus, the charge is concentrated along the z-axis and so Q is positive. Q is negative for an oblate nucleus as shown in Fig. 12 Fig. 12 Oblate and Prolate shapes From Eq. 30, it is clear that Q has same dimensions as that of area and is given in terms of m² or barns (10-24 cm³) or F2 (fermi²). In a quantum-mechanical definition of the quadrupole moment, the charge density  replaced by the probability density and the expression is summed over all the protons. In principle, these calculations are simple and directly yield the quadrupole moment referred to the z-axis. In classical or semiclassical calculations, one should consider the fact that the nuclear symmetry axis is not, in general, the space-fixed z-axis to which measured quadrupole moments are referred. Let us assume that the J-axis can be regarded as a symmetry axis, or at least that the time average of the charge distribution has a rotational symmetry about the J-axis. It can be UTTARAKHAND OPEN UNIVERSITY 25 HALDWANI  26 NUCLEAR PHYSICS MSCPH511 defined as a quadrupole moment Q given by Eqn.30 with the z-direction along the J-vector. The angle è between this vector and the space fixed z-axis in the state for which mJ = J is given by J J cos  = =......(31) J ( J + 1) ( J + 1) It can be shown by simple classical calculations that the relationship between Qj and the observed quadrupole moment Q, which relates to the space-fixed z-axis in the state mJ =J, is Q= ( J − 12 ) Q......(32) ( J + 1) J This shows that a nucleus with J=½ has a zero-quadrupole moment with reference to a space- fixed axis. The same is true for a nucleus with J=0 for which there are no constraints on the orientation, and therefore the time-averaged charge distribution is symmetric. An important aspect of quadrupole moment measurement is that it makes it possible to determine nuclear deformation. Knowledge of nuclear deformation is important for studying collective behavior of nuclear particles. 1.11 SUMMARY In this unit basic properties of Nucleus have been discussed in the simpler manner. In this unit Rutherford’s scattering and nuclear size estimation have also been explained to understand the nuclear radius. Learners will also be able to solve the numerical based on nuclear size determination. The relation between mass and binding energy are discussed. The nuclear wave electrical properties like nuclear statistics, parity and symmetry have also been explained in this unit. The concepts of magnetic dipole moment and electrical Quadrupole moment have also been explained for the nucleus with suitable examples. 26 UTTARAKHAND OPEN UNIVERSITY HALDWANI  27 NUCLEAR PHYSICS MSCPH511 1.12 GLOSSARY Nucleus the central region of an atom where the majority of the mass is concentrated. Isotopes atoms of the same element that have different numbers of neutrons but the same number of protons and electrons. Isobars are elements that have the same number of nucleons (sum of protons and neutrons). Mirror Nuclei atomic nucleus that contains a number of protons and a number of neutrons that are mutually interchanged in comparison with another nucleus. Scattering a change in the direction of motion of a particle because of a collision with another particle Angular Momentum The property of any rotating object given by moment of inertia times angular velocity. Parity property important in the quantum-mechanical description of a physical System Magnetic dipole moment the measure of the objects tendency to align with the magnetic field. Electric Qudrupole moment a parameter which describes the effective shape of the ellipsoid of nuclear charge distribution 1.13 REFERENCES 1. B.L.Cohen, Concepts of Nuclear Physics; Tata McGraw-Hill Publisher 2. Kenneth S.Krane, Introductory Nuclear Physics, Wiley, New York, 2008. 3. S.B.Patel, Introductory Nuclear Physics, New Age International Publishers. UTTARAKHAND OPEN UNIVERSITY27 HALDWANI  28 NUCLEAR PHYSICS MSCPH511 4. S N Ghoshal, Nuclear Physics, S.Chand Publication 5. H.M.Agrawal, Nuclear Physics, Prentice Hall of India 6. A.Beiser, Concepts of Modern Physics, McGraw-Hill, New York, 2003. 7. H.A.Enge, Introduction to Nuclear Physics, Addison-Wesley, London,1966. 8. M.L.Pandya and R.P.S.Yadav, Elements of Nuclear Physics, Kedar Nath Ram Nath, Delhi. 1.14 SUGGESTED READINGS 1. Blin-Stoyle Nuclear and Particle Physics, Chapman and Hall 2. I.Kaplan, Nuclear Physics, Narosa ,2002 3. S.S.M. Wong.,Introductory Nuclear Physics, Prentice Hall of India, New Delhi, 2005 4. B.R.Martin, Nuclear and Particle Physics, Wiley, New York, 2006. 5. J.B.Garg, Nuclear Physics: Basic Concepts, Macmillan, New Delhi, 2011. 6. V.Devnathan, Nuclear Physics, Narosa, New Delhi, 2006. 1.15 TERMINAL QUESTIONS 1. Explain the term nuclear radius. 2. Explain the term nuclear magnetic dipole moment. 3. Define the term statistics and parity for the nucleus. 4. Show that a nucleus a zero electric dipole moment. 28 UTTARAKHAND OPEN UNIVERSITY HALDWANI  29 NUCLEAR PHYSICS MSCPH511 5. Show that nuclear density is constant for all nuclei. 6. Mention various methods for determining the size of the nucleus and describe any one in detail. 7. Explain the term electric quadrupole moment and find out an expression for quadrupole moment. 8. Estimate the ratios of the major to the minor axes of 51Sb123. The quadrupole moment is -1.2b. Take R= 1.5A1/3fm. 9. Explain in brief the different properties associated with the nucleus. 10..Discuss one method for the determination of the size of the nucleus. 11.. Explain the relationship between the nuclear mass with nuclear size. 29 UTTARAKHAND OPEN UNIVERSITY HALDWANI  30 NUCLEAR PHYSICS MSCPH511 UNIT 2 NUCLEAR STABILITY Structure of the Unit 2.1 Introduction 2.2 Objectives 2.3 Binding Energy 2.3.1 Mass Defect 2.3.2 Binding Energy per Nucleon 2.3.3 Packing Fraction 2.4 Nuclear Reaction 2.5 Types of Nuclear Reactions 2.6 Mass Balance and Energy in Nuclear Reaction 2.7 Q-Value of Equation 2.8 Solution of the Q-Value of Equation 2.8.1 Exoergic Reactions 2.8.2 Endoergic Reactions 2.9 Glossary 2.10 Summary 2.11 References 2.12 Suggested Readings 2.13 Terminal Questions 30 UTTARAKHAND OPEN UNIVERSITY HALDWANI  31 NUCLEAR PHYSICS MSCPH511 2.1 INTRODUCTION In the previous unit we have obtained the basic information of the nucleus and its properties. Now, in this unit we shall study about the binding energy which is responsible for the stability of the nucleus and we know that lighter nuclei are more stable than heavier nuclei in which the number of neutrons are in excess as compared to the number of protons. In this consequence we will also study about mass defect and packing fraction. Our current understanding of the nuclear structure is mostly based on studies in which a chosen nucleus is blasted with various particles, such as protons, neutrons, and deuterons. When these particles are close enough to the target nuclei to interact, either elastic or inelastic scattering may occur, one or more completely distinct particles may be ejected from the nucleus, or the incident particle may be trapped and generate a gamma ray. After the bombardment, a nuclear reaction is said to have occurred when the mass number and/or atomic number of the target nuclei changes. 2.2 OBJECTIVES After studying the unit the learners will be able to understand Binding energy, mass defect and packing fraction Nuclear Reaction and its types Q-value and its solution for nuclear reaction Endoergic and Exoergic nuclear reactions Solve the numerical problems of finding the Q-value 31 UTTARAKHAND OPEN UNIVERSITY HALDWANI  32 NUCLEAR PHYSICS MSCPH511 2.3 BINDING ENERGY 2.3.1 Mass Defect Now in this unit we shall study about mass defect which is a very important aspect while we discuss binding energy responsible for nuclear stability. Mass of an atom is concentrated in the central part called nucleus, which is made up of neutrons and protons. It has been observed that the mass of a nucleus is always less than the sum of the individual masses of the protons and the neutrons, which constitute it. This difference is a measure of the nuclear binding energy which holds the nucleus together, is known as the mass defect Δ𝑚. If 𝑚(𝑍, 𝑁) is the mass of the nucleus of an atom consisting of 𝑍 protons and 𝑁 neutrons, the mass defect Δ𝑚 is given as Δ𝑚 = 𝑍𝑚𝑝 + 𝑁𝑚𝑛 − 𝑚(𝑍, 𝑁) Or Δ𝑚 = 𝑍𝑚𝑝 + (𝐴 − 𝑍)𝑚𝑛 − 𝑚(𝑍, 𝑁)………………(1) where 𝑚𝑝 and 𝑚𝑛 are the masses of a proton and a neutron. It is convenient to talk in terms of the atomic masses, therefore, adding and subtracting the mass of 𝑍 atomic electrons on the RHS of the above equation, we get Δ𝑚 = 𝑍(𝑚𝑝 + 𝑚𝑒 ) + (𝐴 − 𝑍)𝑚𝑛 − 𝑚(𝑍, 𝑁) − 𝑍𝑚𝑒 Δ𝑚 = 𝑍𝑀𝐻 + (𝐴 − 𝑍)𝑚𝑛 − 𝑀(𝑍, 𝑁) … … … … … … … … … …. (2) where 𝑚𝑒 is the mass of one electron, 𝑀(𝑍, 𝑁) is the atomic mass and 𝑀𝐻 is the mas of neutral hydrogen atom. This missing mass may be regarded as the mass, which would be converted into energy, if a particular atom is to be formed from the requisite number of electrons, protons and neutrons. 32 UTTARAKHAND OPEN UNIVERSITY HALDWANI  33 NUCLEAR PHYSICS MSCPH511 This is also the amount of energy required to break up the atom into its constituents. Therefore, mass defect is a measure of the binding energy of an atom/nucleus. Example 1: Calculate the mass defect of deuterium 12 𝐻, which is an isotope of hydrogen known as heavy hydrogen Solution: Let us would expect that its mass should be equal to the mass of one neutron in its nucleus. Thus, one atom, i.e. 𝑚𝑛 + 𝑀𝐻 = 1.008665 + 1.007825𝑢 Therefore, the expected mass of deuterium is 2.016490𝑢. However, the measured mass of deuterium, is found to be 2.014102𝑢. Therefore, for deuterium, the mass defect (Δ𝑚) = 2.016490 − 2.014102𝑢 Δ𝑚 = 0.002388𝑢. 2.3.2 Binding Energy per Nucleon According to the proton-neutron model of the nucleus, it is obvious that the particles that make up the stable nucleus are kept together by potent attraction forces, and that effort is required to break them apart. In other words, the nucleus needs energy to be given in order to be divided into its component parts. We use the well-known Einstein mass-energy connection E=mc2, where E and m are the energy and mass of the particle and c is the speed of light in a vacuum, to investigate what shape this energy can take. Simply said, this shows that energy and mass are different representations of the same entity.. Therefore, we should anticipate that the mass of the nucleus as a whole will be less than the sum of the masses of its constituent parts. In fact, a significant deal of experimentation has led to this observation. 33 UTTARAKHAND OPEN UNIVERSITY HALDWANI  34 NUCLEAR PHYSICS MSCPH511 The difference between the energy of the constituent particles and the energy of the entire nucleus is thus used to determine the binding energy of the nucleus. Take the example of a nucleus zMA, where the binding energy is given by 𝐵 = [𝑍𝑀𝑃 + 𝑁𝑀𝑁 − 𝑧𝑀 𝐴 ]𝑐 2 …………………………..(3) where 𝑀𝑃 = Mass of the Proton, 𝑍 = Number of Protons, 𝑀𝑁 = Mass of the Neutron, 𝑁 = Number of Neutrons = (𝐴 − 𝑍) 𝑧𝑀 𝐴 = Measured mass of the neutral atom, [also written as 𝑀(𝑍, 𝐴) ] The above expression is, now-a-days, generally expressed as 𝐵 = [𝑍𝑀𝐻 + 𝑁𝑀𝑁 − 𝑧𝑀1 ]𝑐 2 where 𝑀𝐻 represents the mass of the neutral hydrogen atom. Since there are A nucleons in the nucleus, the binding energy per nucleon is also given as 𝐵𝑒 𝑐2 = [𝑍𝑀𝐻 + 𝑁𝑀𝑁 − 𝑧 𝑀 ]……………………….………(4) 𝐴 𝐴 When binding energy fraction 𝐵/𝐴 is plotted against 𝐴, the curve similar to Fig.1 is obtained. We find from this curve that 𝐵/𝐴 almost remains constant between 𝐴 = 30 and 𝐴 = 100 and decreases for small and large values of 𝐴. The coulomb repulsion between the protons, which obviously makes the nuclei less and less stable, is what causes the reduction for large A. Because only a small number of other nucleons are attracted to each particular nucleon in light nuclei, their separation distances are greater, which again affects stability. The drop in B/A for small A is a surface effect. It is obvious that 34 UTTARAKHAND OPEN UNIVERSITY HALDWANI  35 NUCLEAR PHYSICS MSCPH511 the particles at the surface are less tightly bonded than those inside. stronger ties than those in the interior across. The fraction of constituents at the nuclear surface increasewith the size of the nucleus. This effect reads B/A for a low A. The nuclides with an even number of protons and neutrons have greater B/A values than the nearby odd mass nuclides, according to an analysis of the aforementioned curve. Since there are no remaining unpaired neutrons in such nuclides, the even number of protons can couple off all the nucleons, explaining the even-even nucleon's better stability and natural abundance.. This low mass number nuclides, the stability rule is N = Z, 𝐻 3 , Li2 BH 2 , N14 are examples of odd-odd examples of odd-odd nuclides which are most stable. Example 1: Masses of helium nucleus, proton and neutron are 4.0026u, 1.007895u and 1.008665u. Find the energy required to knock out nucleons from the helium nucleus. Solution: We have the equation 4 2 He → 2𝑝 + 2𝑛 Mass of two protons and 2 neutrons is 35 UTTARAKHAND OPEN UNIVERSITY HALDWANI  36 NUCLEAR PHYSICS MSCPH511 = 2 × (1.007895 + 1.008665) = 4.03312𝑢 Therefore, mass defect is given as Δ𝑚 = 4.03312 − 4.0026 = 0.03052𝑢 And the equivalent energy is = (0.03052𝑢) × 931.49 = 28.42 MeV 2.3.3 Packing Fraction We have seen that atomic masses are not whole numbers. This divergence of the masses of the nuclides from whole number was studied by a number of workers like Aston and is expressed in terms of packing fraction. Packing fraction is defined as Atomic mass - Mass number 𝑓 = Mass number 𝑧𝑀 𝐴 − 𝐴 = … … … … … … … …. (5) 𝐴 where 𝑀 𝐴 is the actual mass of a nuclide on the physical atomic mass scale and A mass number. The quantity 𝑧𝑀 𝐴 − 𝐴 is known as Mass-defect'. If packing fraction is represented as 𝑓, then 𝑧𝑀 𝐴 = 𝐴(1 + 𝑓)………………….……………..(6) When 𝑓 is plotted against 𝐴, we get the following curve of the shape as shown in fig.2 36 UTTARAKHAND OPEN UNIVERSITY HALDWANI  37 NUCLEAR PHYSICS MSCPH511 Fig 2: Plot of packing fraction(f) versus mass number(A) The packing fraction for all the elements fall on this curve except He4, C14, and O16.For 𝑂16 the value of packing fraction is zero. As 𝐴 increases, the falls of the packing fraction becomes negative, passes through a flat minimum rises gradually becoming positive again at values of 𝐴 above 180. A a positive packing fraction has an atomic mass 𝑀 greater than its mass nude whilh This means that the loss of mass due to binding energy requirements, whera, nucleons combine to form a nucleus, is less than that for oxygen. The packing fractions for light elements like oxygen, carbon, helium which do not lie on the frac. have positive values but much smaller (≈ 5 × 10−1 or less) than other light curve like nitrogen, lithium and hydrogen (≈ 80 × 10−4 for hydrogen ) light elements of the former type (O16 , C12 , He4 ) have more stable nuclei’s that latter. 37 UTTARAKHAND OPEN UNIVERSITY HALDWANI  38 NUCLEAR PHYSICS MSCPH511 Example 1: The value of fB can be estimated as follows: For deuteron (H2), since Z=1, N=1, ( ) B.E H 2 = M H + M n − M d = (1.007825 + 1.008665 − 2.0414102 )  931.5 = 2.224 MeV ( ) fB H 2 = 2.224 2 = 1.112 MeV per nucleon For alpha-particle (He4), Z=2, N=2, ( ) B.E He 4 = ( 2 1.007825 + 2 1.008665 − 4.002603)  931.5 = 28.3MeV ( f B He 4 = ) 28.3 4 = 7.075 MeV per nucleon For (O16),since Z=8, N=8, ( ) B.E O16 = ( 8 1.007825 + 8 1.008665 − 15.994915 )  931.5 = 127.62 MeV 127.62 fB = = 7.98 MeV per nucleon 16 The binding energy fractions of the different nuclei represent the relative strengths of their binding. Thus H2 is very weakly bound, compared to He4 or O16. 2.4 NUCLEAR REACTION Typically, an equation representing a nuclear reaction may be written as 38 UTTARAKHAND OPEN UNIVERSITY HALDWANI  39 NUCLEAR PHYSICS MSCPH511 x+ X =Y + y......(7) In words this would read like: when an incident projectile x hits the target nucleus X, a nuclear reaction takes place and as a result there is a new nucleus Y and an outgoing particle The above reaction can also be written in short form as X (x, y) Y. In 1919, Lord Rutherford discovered that protons are created when nitrogen is attacked with polonium particles, and that these protons are capable of piercing 28 cm of air. Rutherford deserves credit for piercing the "inaccessible armour" of typical non-radioactive nuclei and causing a transmutation because this was the first nuclear reaction to be set off un a laboratory. You might think of the nuclear reaction as 14 7 N ( , p ) 178 O......(8) In 1930 Cockcroft and Walton used artificially accelerated protons and produced the following nuclear reaction. 7 3 Li + p → 24 He +  7 3 Li ( p,  ) 24 He......(9) Thousands of nuclear reactions have previously been examined by various laboratories throughout the world, and the number is constantly increasing. In this chapter, we'll look at the scenario when the target nucleus is at rest before to the collision and all of the particles can be considered non-relativistically. Finding out when a specific nuclear reaction becomes energetically feasible is what we're interested in learning. By using the equations of momentum and energy conservation, we may determine this. 2.5 TYPES OF NUCLEAR RECATION Nuclear reactions are classified on the basis of the projectile used, the particle detected and the residual nucleus. 39 UTTARAKHAND OPEN UNIVERSITY HALDWANI  40 NUCLEAR PHYSICS MSCPH511 i. Scattering: In the scattering reaction, the projectile and the detected (outgoing) particle are the same. The scattering is elastic when the residual nucleus is left in the ground state. When the residual nucleus is left in an excited state, the scattering is called inelastic. ii. Pickup reactions: When the projectile gains nucleons from the target, the nuclear reaction is referred to as pickup reaction, e.g. 16 8 ( O 12 H , 13H ) 15 8 O iii. Stripping reactions: In this type, the projectile loses nucleons to the target nucleus, e.g. 16 O 8 ( 4 2 He, 12 H ) 18 9 F In a pickup or stripping reaction it is assumed that the nucleon involved in the process enters or leaves (the shell of) the target nucleus without disturbing the other nucleons. These reactions are therefore referred to as direct reactions. In contrast, we have the following type compound nuclear reactions. Of iv. Compound nuclear reactions: Here the projectile and target form a compound nucleus which has a typical life span of 10−16 sec. When this time is compared with a typical nuclear time, i.e., the time taken by the projectile to traverse the target nucleus ( 10−22 sec.) as in the case of direct reactions, we can conclude that the decay compound nucleus does not depend on the way it was formed. of A This situation is often described as: the compound nucleus does not "remember" how it was formed. Usually, the same compound nucleus is given rise to by a number of nuclear reactions. This compound nucleus can decay in a number of ways or channels. This is 64 illustrated by taking for example, the compound nucleus 30 Zn formed in an excited state ( ) * 64 30 Zn by two different methods and then decays as given below: 40 UTTARAKHAND OPEN UNIVERSITY HALDWANI  41 NUCLEAR PHYSICS MSCPH511 ( ) * p + 2963Cu → 64 30 Zn → 3063Zn + 01n → 3062 Zn + 01n + 01n → 62 29 Zn + 01n + p ( ) * + 60 20 Ni → 64 30 Zn → 63 30 Zn + 01n → 62 30 Zn + 01n + 01n → 62 29 Cu + 01n + p.....(9) These reactions were experimentally studied by S.N. Ghoshal [Phys. Rev. 80939 (1950)] Figure 4 depicts how a compound nucleus in an excited state can have different modes of decay, giving rise to different residual nuclei and detected particles. Fig 4: Modes of decay in a compound nucleus in an excited state 41 UTTARAKHAND OPEN UNIVERSITY HALDWANI  42 NUCLEAR PHYSICS MSCPH511 The compound nucleus model has been successfully applied for targets of A>10 and projectile energies up to about 15 MeV. For projectile energies between 15 MeV and 50 MeV the direct reaction model and the optical model are found to be successful. In this unit, our main emphasis is on understanding energies of nuclear reactions. 2.6 THE BALANCE OF MASS AND ENERGY IN NUCLEAR REACTIONS Consider the nuclear reaction, x+ X =Y + y Since the total mass and energy is conserved, we have (E x ) ( ) ( + mx c2 + M X c2 = EY + MY c2 + Ey + my c 2 )......(10) Where, E x = kinetic energy of the projectile mx c 2 = rest energy of the projectile And similarly, EY and M Y c 2 , E y , my c 2 and M X c 2 The target nucleus X is assumed to be at rest. The Q value is expressed as, Q = EY + Ey − Ex......(11) i.e., it is the change in the total kinetic energy. 42 UTTARAKHAND OPEN UNIVERSITY HALDWANI  43 NUCLEAR PHYSICS MSCPH511 This change in the total kinetic energy in a nuclear reaction is clearly the nuclear disintegration energy. From Eq. (10) and Eq. (11) becomes, Q = EY + E y − Ex = ( M X + mx ) − ( M Y + m y )  c 2.......(12) Thus, we see that Q is also change in the total rest mass. We have an exoergic nuclear reaction if Q is positive. The nuclear reaction is an endoergic reaction if the Q value is negative. From Eq. (12) it is clear that to determine the Q value or nuclear disintegration energy for a nuclear reaction, we must know the kinetic energy of the particles. E y , the kinetic energy of the residual nucleus, is usually small and hard to measure. In the following section, we will see how E y can be eliminated and how the Q equation can be set up. 2.7 THE Q EQUATION The analytical relationship between the kinetic energy of the projectile and outgoing particle and the nuclear disintegration energy Q is called as the Q equation. To understand the dynamics of two body nuclear reactions in laboratory coordinate system, refer to Fig.5 43 UTTARAKHAND OPEN UNIVERSITY HALDWANI  44 NUCLEAR PHYSICS MSCPH511 Fig. 5: Two body nuclear reaction in lab system Conservation of mass-energy gives Q = EY + Ey − Ex As remarked before, E y is small and hard to measure and is therefore eliminated. Conservation of linear momentum along the direction of projectile x gives, 2mx Ex = 2my Ey cos  + 2M Y EY cos ......(13) Angles  and  are coplanar as linear momentum perpendicular to the  plane is equal to zero. Therefore, conservation of linear momentum normal to projectile direction (in the  ,  plane) gives, 0 = 2MY EY sin  − 2my Ey sin .............(14) From these equations E y and  are eliminated. Squaring and adding Eqs. (13) and (14), M Y EY = mx Ex + my Ey − 2 mx my Ex Ey cos  44 UTTARAKHAND OPEN UNIVERSITY HALDWANI  45 NUCLEAR PHYSICS MSCPH511 But EY = Q − Ey + Ex mx my 2 mx m y E x E y Q = E y − Ex + Ex + Ey − cos  MY MY MY  my   mx  2 mx m y E x E y =E y  1 +  − Ex 1 − − cos ......(15)  M Y   M Y  M Y This is the standard form of the Q equation. It is interesting to note that one can also obtain Eq.15 by solving the momentum triangle and substituting for E y. This is indicated below: Fig 6: Momentum triangle, using the notation of fig. 05 We see that, PY2 = px2 + p y2 − 2 px p y cos ......(16) 45 UTTARAKHAND OPEN UNIVERSITY HALDWANI  46 NUCLEAR PHYSICS MSCPH511 2MY EY = 2mx Ex + 2my Ey − 2.2 mx my Ex Ey cos  giving, mx my 2 mx my Ex E y EY = Ex + Ey − cos ......(17) MY MY MY From Eq. (6) EY = Q − Ey − Ex Substituting this in Eq (12) we get,  my   mx  2 mx my Ex E y Q =E y 1 +  − Ex  1 − − cos   MY   MY  MY Which is the Q Equation (10). The kinetic energy E x , E y and  all are measured in laboratory system. Since the Q equation is based on mass-energy conservation in a nuclear reaction, it holds for all types of reactions. The exact masses can be replaced by the corresponding mass numbers, in many applications without significant error. For very accurate calculations, the neutral atomic masses are used. Let us see how we can use the isotopic masses (neutral atomic masses) to obtain Q value in alpha and beta decay reactions. For convenience, let the element symbol represent the isotopic mass. We recall that Q = ( mx + M X ) − ( my + MY ) where the masses are the nuclear masses and Q is in mass units. i. Consider the reaction U→ 238 92 Th + 24 He 234 90 Q = (U − 92e ) − (Th − 90e ) + ( He − 2e ) 46 UTTARAKHAND OPEN UNIVERSITY HALDWANI  47 NUCLEAR PHYSICS MSCPH511 Where e stands for electronic mass. i.e., Q = (U − 92e ) − (Th + He − 92e )  = U − (Th + He )  ∴ Isotopic masses can be used to evaluate Q. ii. Consider the reaction, C → 147 N + −10e + 00 14 6 Q = ( C − 6e ) − ( N − 7e ) + e  = ( C − 6e ) − ( N − 6e ) =C−N iii. Consider the reaction, 63 30 Zn → 2963Cu + +10e + 00 Q = ( Zn − 30e ) − ( Cu − 29e ) + e  = ( Zn − 30e ) − ( Cu − 30e + 2e ) = Zn − ( Cu + 2e ) Thus, we see that the isotopic masses can be used to compute Q value in a positron decay reaction, provided two electron masses are included with that of the product particle. 47 UTTARAKHAND OPEN UNIVERSITY HALDWANI  48 NUCLEAR PHYSICS MSCPH511 2.8 SOLUTION OF THE Q EQUATION We are often interested in E y , the energy of the detected particle and its variation with E x , the energy of the bombarding projectile, for a fixed Q. For this purpose, the Q equation (Eq. 10) can be regarded as a quadratic in Ey. Then its general solution can be conveniently put in the form, Ey =  2 + w......(18) Where, mx m y E x = cos ......(19) my + M Y And, M Y Q + E x ( M Y − mx ) w=......(20) my + M Y When Ey is real and positive the reaction is energetically possible. Emission of m y becomes energetically impossible when Q value is negative, ( M Y − mx ) is negative (i.e., a heavy projectile) and a large angle of observation  , making cos negative. Equation (13), for a various energy E x of the following particle, tells us about the types of nuclear reactions which can occur. Let us now consider energies of exoergic and endoergic reactions. 2.8.1 Exoergic Reactions: Here Q > 0. a. Very low energy projectiles: e.g., thermal neutrons. 48 UTTARAKHAND OPEN UNIVERSITY HALDWANI  49 NUCLEAR PHYSICS MSCPH511 We can take E x  0 MY Q v = 0 and w = my + M Y And so, MY Ey = Q......(21) my + M Y i.e., E y is same for all angles  An example of such reaction is, C ( n,  ) 49 Be 12 6 b. Finite energy projectiles: For most of the reactions, mx M Y and so, w  0 for all E x. Thus, it can be seen from Eq. (13) that E y is single valued for all E x , and is given by, Ey = v + v2 + w......(22) An example of such a reaction is, 10 5 B ( , p ) 136 C and Q = +4.0 MeV c. Double values of Ey: In a number of reactions E y is no single valued. For example, in the reaction, 15 7 N ( d , n ) 168 O, Q  10 MeV 16 If the observed particle m y is chosen to be the residual nucleus O , then, my mass number , AY = 16 and M Y − AY = 16 (for neutron). Equation (15) gives, 49 UTTARAKHAND OPEN UNIVERSITY HALDWANI  50 NUCLEAR PHYSICS MSCPH511 Q − Ex w= 17 Which is negative for all deuteron bombarding energies greater than, Ex = Q = 10 MeV Thus, for Ex > 10 MeV , we have two real positive values of E y at  = 0. That is, in the forward direction, there are two monoenergic groups of 16O nuclei. Physically this implies that in the center of mass system, these two groups are projected forward and backward. 2.8.2 Endoergic Reactions: Q < 0. For every nuclear reaction with a positive Q value, the inverse reaction has a negative Q value of the same magnitude. Thus, for example, 168 O ( n, d ) 157 N has Q = −10 MeV , or 136 C ( p,  ) 105 B has Q = −4 MeV a) Very low energy projectiles: Ex  0, Q  0 v 2 + w2  0 and so... Ey is imaginary, implying that no reaction occurs. ( E x is insufficient to start the reaction.) b) Threshold energy ( Ex )thresh : In an endoergic reaction, the energy -Q is needed to excite the compound nucleus sufficiently so that it will break up. The bombarding particle must supply this energy in the form of kinetic energy. However, not all of that kinetic energy is available for excitation because some is used to give momentum to the compound nucleus which, is distributed among the products of the reaction. Thus, the bombarding particle (projectile) must supply some energy in addition energy in addition to - Q so that the energy -Q becomes available for the excitation of the compound nucleus. UTTARAKHAND OPEN UNIVERSITY 50 HALDWANI  51 NUCLEAR PHYSICS MSCPH511 The smallest value of projectile energy (bombarding energy) at which an endoergic reaction can take place is called the threshold energy for that reaction. From Eq. 13, it can be seen that the reaction first becomes possible when E x is large enough to make, v 2 + w2 = 0 E x has its minimum possible value at  = 0 , which is the ( Ex )thresh. This condition gives: mx my Ex M Y Q + Ex ( M Y − mx ) + =0 ( my + M Y ) my + M Y 2 Which gives,  m + MX  ( Ex )thresh = −Q  x ......(23)  MX  In obtaining Eq. (18), we have made use of the fact that the value of Q is much less than the c2 masses of particles in a nuclear reaction, and so my  mx + M X − MY......(24) Equation (18) can be also obtained in a straightforward way by considering M c and Vc as mass and velocity of the compound nucleus. Then we have, mx vx = M cVc mx Or Vc = vx Mc 51 UTTARAKHAND OPEN UNIVERSITY HALDWANI  52 NUCLEAR PHYSICS MSCPH511 Now, the part of kinetic energy of the projectile needed for the excitation of the compound nucleus is, 1 1 Q= mx vx2 − M cVc2 2 2 2 1 1 m  = mx vx2 − M c  x  vx2 2 2  Mc  1  m  = mx vx2  1 − x  2  Mc  But M c = mx + M X 1  Mx  Therefore, −Q = mx vx2   2  mx + M X  The threshold energy is then, 1  m + MX  ( Ex )thresh = mx vx2 = −Q  x  2  MX  Which is Eq.23. 2.9 GLOSSARY Kinetic Energy - the energy possessed by a body due to its motion. Magnetic Moment - a vector quantity that provides a measure of the torque exerted on a magnetic system (as a bar magnet or dipole) when placed in a magnetic field. 52 UTTARAKHAND OPEN UNIVERSITY HALDWANI  53 NUCLEAR PHYSICS MSCPH511 Mass Defect - the difference between the mass of an atomic nucleus and the sum of the masses of its constituent particles. MeV (Mega Electron Volt) – the energy possessed by a particle with one electronic charge in passing through a potential difference of one million volts. Nuclear Binding Energy – the minimum energy required to separate an atomic nucleus into it constituent particles i.e., protons and neutrons. Nuclear magneton - a unit of magnetic moment, used to measure proton spin and approximately equal to 1/1836 times Bohr magneton. Q value - the Q value for a reaction is the amount of energy absorbed or released during the nuclear reaction. Exoergic(exothermic)reaction - a nuclear reaction that releases energy and have positive Q- value. Endoergic (or endothermic)reaction – a nuclear reaction that requires an input of energy to take place with negative Q-value. 2.10 SUMMARY In this unit, we have explained various important terms like binding energy, mass defect, binding energy per nucleon and packing fraction with suitable examples. In this consequence we have defined the nuclear reaction and also its various types has also been explained. In this unit we have also learnt that to determine the Q value or nuclear disintegration energy for a nuclear reaction, we must know the kinetic energy of the particles. We have an exoergic nuclear reaction if Q is positive. The nuclear reaction is an endoergic reaction if the Q value is negative. The analytical relationship between the kinetic energy of the projectile and outgoing particle and the nuclear disintegration energy Q is called as the Q equation has also been explained 53 UTTARAKHAND OPEN UNIVERSITY HALDWANI  54 NUCLEAR PHYSICS MSCPH511 mathematically. In addition to this the solution of the Q equation has also been established in this unit for better understanding of the topic. 2.11 REFERENCES 1 Nuclear Physics - Problem-based Approach including MATLAB, Hari M. Agrawal. PHI Learning, Dehi 2 Modem Physics, Stephen T. Thorthon - Cengage Learning India, New Delhi 3 S.B.Patel, Introductory Nuclear Physics, New Age International Publishers. 4 Kenneth S.Krane, Introductory Nuclear Physics, Wiley, New York, 2008. 5 B.R.Martin, Nuclear and Particle Physics, Wiley, New York, 2006. 2.12 SUGGESTED READINGS 1 B.L.Cohen, Concepts of Nuclear Physics; Tata McGraw-Hill Publisher 2 S N Ghoshal, Nuclear Physics, S.Chand Publication 3 H.A.Enge, Introduction to Nuclear Physics, Addison-Wesley, London,1966. 4 M.L.Pandya and R.P.S.Yadav, Elements of Nuclear Physics, Kedar Nath Ram Nath, Delhi. 5 I.Kaplan, Nuclear Physics, Narosa ,2002 6 S.S.M. Wong.,Introductory Nuclear Physics, Prentice Hall of India, New Delhi, 2005 2.12 TERMINAL QUESTIONS 1. What do you meant by nuclear binding energy. 2. Explain the terms mass defect and packing fraction. 54 UTTARAKHAND OPEN UNIVERSITY HALDWANI  55 NUCLEAR PHYSICS MSCPH511 3. Plot the general shape of the binding energy curve (Binding energy per nucleon versus mass number A. Explain fission and fusion on the basis of this plot. 4. What do you mean by Q value of a nuclear reaction? 5. Write down the expression for Q value in the class of α decay. 56 6. Find the Binding Energy per nucleon for Fe.(Given that mass of Fe atom= 55.93493 amu). 7. Find the B/A for deuterium atom. Mass of deuterium atom(1H2) is 2.01402 amu. 8. Find the B/A for 7Li. Mass of lithium atom is 7.016004 amu. 9.The B/A for Deuterium and Helium are 1.1 MeV and 7 MeV respectively. What would be the energy released when two deutrons are fused together to form a He nucleus. 55 UTTARAKHAND OPEN UNIVERSITY HALDWANI  56 NUCLEAR PHYSICS MSCPH511 UNIT 3 NUCLEAR FORCES I Structure of the Unit 3.1 Introduction 3.2 Objectives 3.3 Basic Understanding of Nuclear Forces 3.4 Basic properties of deuteron 3.4.1 Binding Energy 3.4.2 Size 3.4.3 Spin of Deuteron 3.4.4 Magnetic Dipole Moment 3.4.5 Quadrupole Moments 3.5 Existence of excited states of deuteron 3.6 n-p scattering at low energies with specific square well potential 3.6.1 What is Scattering 3.6.2 Neutron – Proton Scattering at Low Energies 3.7 Results of Low Energy n- p and p-p Scattering 3.8 Glossary 3.9 Summary 3.10 References 3.11 Suggested Readings 3.12 Terminal Questions 56 UTTARAKHAND OPEN UNIVERSITY HALDWANI  57 NUCLEAR PHYSICS MSCPH511 3.1 INTRODUCTION In the previous unit we have studied the basic information about binding energy which refers to the stability of nucleons. Now in this unit we shall study about the forces that binds the nucleon together i.e. nuclear force.in this consequence we known that large number of nuclei, available in nature, are stable. Now the natural question is what bounds the nucleons together in the nucleus to make it stable? It is further known that most of the nuclei consist of more than one proton – the positive charged particle; it implies that there exists in the nucleus forces which are strong enough to overpower the coulomb repulsion and hold the nucleons together. The forces which hold the nucleons together are commonly called nuclear forces and are short range forces as it is evident from the fact that binding energy is proportional to the number of constituent nucleons. 3.2 OBJECTIVES The main aim of studying this unit is to understand the nuclear forces which is responsible for binding the nucleons. For understanding the nuclear force, we will study the Basic properties of deuteron viz, its binding energy, its size, spin, magnetic and quadrupole moments etc. After going through this unit learners should be able to: Understand the various properties of deuteron. Analyze the existence of excited states of de