Nuclear and Particle Physics Module 1 By Dr. Manjula Sharma Asstt. Prof. (Physics) Deptt. of Physical Sciences and Languages COBS, CSKHPKV, Palampur, HP-176062 Mail Id: [email protected] Contact no: 7018006656 DISCOVERY OF NUCLEUS Ernest Rutherford discovered the nucleus of the atom in 1911.In 1911, Rutherford, Marsden and Geiger discovered the dense atomic nucleus by bombarding a thin gold sheet with the alpha particles emitted by radium. Rutherford and his students then counted the number of sparks produced by these alpha particles on a zinc sulphate screen. From this observation, they concluded that almost all the atomic matter was concentrated in a tiny volume situated at the atome center, the atomic nucleus. Constituents of Nuclei 1) Proton electron theory :- Before the discovery of neutron by Chadwick in 1932, protons and electron were considered as the building blocks of the nucleus. According to this theory:- a) Electrons are emitted by radioactive nuclei in beta – decay process. b) From mass spectrum data it was found that masses of different nuclei were whole number multiple of hydrogen and this hydrogen is named as proton. c) Since, atom is whole as neutral so it was considered that number of electrons and protons are same. FAILURES :- 1) WAVE MECHANICAL CONSIDERATION :- According to Heisenberg uncertainity principle Total momentum of electron inside the nucleus must be = 0.527 * 10-20Jsm-1 After this if rest mass energy is calculated then it is 10MeV , but we know rest mass energy is 3 to 4 MeV. So electons do not lie inside nucleus. 2) NUCLEAR SPIN :- According to this theory nitrogen have half integral spin but experimentally it is integral. 3) NUCLEAR MAGNETIC MOMENT :- Magnetic moment of electrons is about 2000 times is more than protons so that is why electrons do not lie inside nucleus. 4) WAVELENGTH :- Wavelength of electron is larger than nuclear dimensions so electrons do not lie inside nucleus. 2) PROTON NEUTRON THEORY :- Electron proton theory failed due to properties of free electron . Further it was predicted that electron is bound with proton and it forms a new particle called NEUTRON. It was discovered by James Chadwick. Facts in support of it :- 1) Wave mechanical consideration :- According to this momentum of proton = 0.527 * 10-20 Jsm-1 So total energy of proton is 940 MeV and experimentally it is 938 MeV. 2) Nuclear spin :- According to this spin will be integral or half integral is A(mass no.) is even or odd. 3) Nuclear magnetic moment :- Magnetic moment of neutron is same in magnitude but opposite in direction. 4) Finite size :- Since both protons and neutrons have same size so it is not difficult for nucleus to have neutrons in it. Properties :- 1) Nuclear size :- Experimentally, V ∝ A where, V = Volume of nucleus and R = Radius of nucleus 1/3 -15 R = R0A where R0 varies from 1.2 to 1.6 fermi (1 F = 10 ). 2) Nuclear shapes :- Quadrupole moment (Q) measures the departure of nucleus from its spherical symmetry . Q<0 OBLATE SHAPE , Q>0 PROLATE SHAPE AND Q =0 SPHERICALLY ELLIPSOIDAL . 3) Nuclear density :- 4) Nuclear mass :- 5) Nuclear charge :- 6) Nuclear Wave mechanical properties :- (a) Parity P = +1; Parity is even or positive; P = -1; Parity is odd or negative. Parity is also related to orbital quantum number as, P = (-1)l (b) Statistics In quantum mechanical treatment, a composite system with a group of particles such as a nucleus is described by either Bose- Einstein (B-E) or the Fermi-Dirac (F-D) Statistics. Particles obeying F-D statistics are called Frmions (e.g. protons, neutrons, electrons) and the particles obeying B-E statistics are called Bosons ( e. g. photons, π-mesons). 7) Angular momentum of Nucleus (I) :- 8) Magnetic moment of Nucleus (μI) :- Associated with each nuclear spin is a magnetic moment which is associated with the angular momentum of the nucleus. It is common practice to express these magnetic moments in terms of the nuclear spin in a manner parallel to the treatment of the magnetic moments of electron spin and electron orbital angular momentum. For the electron spin and orbital cases, the magnetic moments are expressed in terms of a unit called a Bohr magneton which arises naturally in the treatment of quantized angular momentum. 9) Electric moment of Nucleus :- 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. The quantity Q0 is the classical form of the calculation represents the departure from spherical symmetry in the rest frame of the nucleus. The expression for Q is the quantum mechanical form which takes takes into account the nuclear spin I and the projection K in the z-direction. Mass Defect The forces that bind nucleons together in an atomic nucleus are much greater than those that bind an electron to an atom through electrostatic attraction. This is evident by the relative sizes of the atomic nucleus and the atom (10−15and 10−10 m, respectively). The energy required to pry a nucleon from the nucleus is therefore much larger than that required to remove (or ionize) an electron in an atom. In general, all nuclear changes involve large amounts of energy per particle undergoing the reaction. Packing Fraction The term packing fraction (f) is the mass defect per elementary particle contained in the nucleus where, ΔM = mass defect M = Actual mass of an element in a. m. u. A = Mass no. i. e. number of protons plus neutrons in the nucleus. Binding Energy The energy equivalent to mass defect (△m) in a nucleus is the energy needed to hold or bind the nucleons together inside the nucleus. This energy, called binding energy of nucleus. It is the measure of the stability of the nucleus. Clearly binding energy of the nucleus will be equal to the total energy released when the nucleons combine to form the nucleus. Thus binding energy is also defined as the energy required to break up a nucleus into its constituent protons and neutrons and to keep them at rest at infinite distances from each other. Therefore work equal to the binding energy must be done to split the nucleus completely into its constituent nucleons. Binding energy is expressed as : B = △mc2 2 B = [ZMP+ (A-Z)Mn – Mnucleus]c So mass defect appears as binding energy. Binding energy per nucleon:- The average energy required to extract one nucleon from the nucleus is called binding energy per nucleon. Binding energy determines the stability. If binding energy per nucleon is small, the nucleus is less stable whereas binding energy per nucleon is more then nucleus is more stable. Therefore heavy nuclei are less stable then lighter nuclei. If we plot a graph between the binding energy per nucleon and mass no. the graph is shown as: Binding energy per nucleon curve From the curve the conclusions are drawn: 1 2 3 1)Binding energy per nucleon of light nuclei like 1H , 1H , 1H etc. is low. 2)For mass number ranging from 2 to 20 there are sharply defined peaks corresponding to 4 8 12 16 20 2He ,4Be , 6C , 8O , 10Ne . These peaks indicate that these nuclei are most stable then the other nuclei in their neighbourhood. 3)It has a broad maximum to value of 8.5 Mev/nucleon in the mass, ranging from 40 to 140. It 56 has a peak value of 8.8 Mev/nucleon for 26Fe . It is the most stable nucleus, since the highest maximum energy required to pull a nucleon away from it. 4)For mass number more than 56, the binding energy per nucleon decreases slowly and is least 234 value for 92U is 7.6 Mev. 5)Binding energy per nucleon for lighter and heavier nuclei are less and they are less stable than middle one. In order to attain stability the heavier nucleon may split into lighter nuclei(by process of nuclear fission) and lighter nuclei may unite together to form heavier nucleus (by nuclear fusion).In both the processes large amount of energy is released. 6) For the nuclides having A<28 there exists cyclic recurrence of peaks. These peaks corresponds to those nuclei whose mass are multiple of four and contain equal number of protons and neutrons. 4 8 12 16 20 So the binding energy of 2He , 4Be , 6C , 8O , 10Ne are greater than those of their neighbours. Thus these nuclei are more stable . further it is noted that similar but less prominent peaks are observed at the values of Z and N equal to 20,28,50,82 and 126. These numbers are called magic numbers. Average binding energy for light nuclei (A<28) Nuclear Stability Certain Nuclei/isotopes are more stable than others. Their stability is determined by the ratio of the number of neutrons to the number of protons in the nucleus. At low atomic masses, the stable ratio is approximately 1:1. At about an atomic mass number of 20 this starts to increase until it is around 1.5:1 for the very heavy elements. This is due to the fact that with higher numbers of protons more neutrons are needed due to the repulsion of the protons from electrostatics. 1:1 is stable only below Z=20 and that after that the stable nuclei become neutron rich. This ratio is not exact but represents a "band of stability" around which unstable isotopes cluster.