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6. Atomic and Nuclear Physics Chapter 6.6 Nuclear Physics Scattering experiments and distance of closest approach In scattering experiments such as Rutherford’s, simple considerations can be used to calculate the distance of closest approach of the incoming particle to the target. Consider an alpha particle shot head-on toward a stationary nucleus of charge Q=Ze v=0 v d 2e Ze Scattering experiments and distance of closest approach Initially, the total energy of the system consists of the kinetic energy of the alpha particle. v 2e Ze The initial distance from the nucleus is so big that Ep does not exist. v=0 At the point of closest approach, a distance d from the centre of the nucleus, the alpha particle stops and is about to turn back. The total energy now is the electrical potential energy of the alpha particle and the nucleus, given by: Qq (2e)(Ze) 2Ze 2 Ek k k d d d d Scattering experiments and distance of closest approach Assuming that the nucleus does not recoil, its kinetic energy is zero. Then, by conservation of energy: 2 2Ze 2Ze Ek k d k d Ek 2 Assuming a kinetic energy for the alpha particle equal to 2.0MeV directed at a gold nucleus (Z=79) gives d = 1.1x10-13m. This is outside the range of the nuclear force, which means that the particle is simply repelled by the electrical force. Scattering experiments and distance of closest approach But the distance of closest approach depends on the kinetic energy of the alpha particle. The bigger the Ek, the smaller the distance of closest approach. The smallest it can get is of the same order of the radius of the nucleus. By doing experiments of this kind, physicists have estimated the nuclear radii. It is found that the nucleus radius R depends on mass number through: 1 3 15 R 1.2 A 10 m The mass spectrometer The existence of isotopes can be demonstrated using a mass spectrometer. Ion source S1 - + S2 photographic plate B into the page The mass spectrometer The mass spectrometer In a mass spectrometer single ionized ions of an element (charge e) are made to move through a pair of slits (S1) which collimates the beam. The ions enter a region of magnetic and electric fields at right angles to each other. The B is directed into the page in the region shaded grey. Ion source The positive ions are deflected to the left by the electric field and to the right by the magnetic field. By choosing a suitable value for the magnetic field, the ions can continue through undeflected if the magnetic and electric forces are equal. S1 - + S2 photographic plate B into the page The mass spectrometer eV=evB That is, E v B If the forces are the same, then Thus, only ions with this specific velocity will be able to go through the second slit S2. The selected ions enter then a second region of magnetic field and are thus deflected into a circular path, hitting a photographic plate where they are recorded. mv R eB The radius of the circular path is given by If the beam contains atoms of equal mass, all atoms will hit the plate at the same point. If, however, isotopes are present, the heavier atoms will follow a longer radius circle and will hit the plate further to the right. Measurement of the radius of each isotope's path will allow the determination of its mass. Beta decay and the neutrino The beta decay process originates from a decay of a neutron inside an atomic nucleus: 1 0 n p e e 1 1 0 1 0 0 The neutron decays into a proton (the Z of the nucleus increases by 1), an electron and an antineutrino. A free neutron (i.e., outside the nucleus) decays into a proton according to the equation above with a half-life of 11 minutes A related decay is that of positron emission, in which a proton inside the nucleus turns itself into a neutron accompanied by the emission of a positron (electron’s antiparticle) and a neutrino. 1 1 p n e e 1 0 0 1 0 0 Beta decay and the neutrino 1 0 n p e e 1 1 0 1 0 0 1 1 p n e e 1 0 0 1 0 0 Unlike a free neutron, a free proton cannot decay into a neutron since the rest energy of a neutron is larger than that of a proton. Inside the nucleus the reaction is, however, possible because binding energy is used to make up for the difference. These reactions must be understood as the disappearance of one particle and the creation of three particles on the right-hand side of the decay equation and NOT the splitting of a particle to form another 3 particles. Beta decay and the neutrino The electron antineutrino went undetected until 1953 but its existence was predicted on theoretical grounds. If we consider he neutron decay The mass of the neutron is bigger than the masses of the proton and electron together by 1 0 n p e e 1 1 0 1 0 0 1.008665u – (1.007276+0.0005486)u = 0.00084u This mass corresponds to an energy of 0.00084 x 931.5MeV = 0.783MeV This is the available energy in the decay, which will show up as the kinetic energy of the products Beta decay and the neutrino If only the electron and the proton are produced, than the electron being lighter of the two will carry most of this energy away as kinetic energy. Thus, we should observe electrons with kinetic energies of about 0.783MeV. In experiments, however, the electron has a range of energies from zero up to 0.783 MeV. So, where is the missing energy? Wolfgang Pauli and Enrico Fermi hypothesized the existence of a third particle in the products of a beta decay in 1933. This third very light particle would carry the remainder of the available energy. Fermi coined the word neutrino, the Italian word for “little neutral one” Electron capture A process related to beta decay is electron capture, in which a proton inside the nucleus captures an electron and turns into a neutron and a neutrino: 1 1 p e n e 0 1 1 0 0 0 The creation of a neutron star rests on this process, in which the huge pressure inside the sat drives electrons into protons in the nuclei of the star, turning them into neutrons. Nuclear energy levels The nucleus, like the atom, exists in discrete energy levels. The main evidence for this fact is that in alpha and gamma decays, the energies of alpha particles and gamma ray photons are discrete. E/MeV Like atoms, we can have diagrams for nuclear energy levels 8.17 7.60 211 83 The energy of the particle emitted equals the energy level difference. 211 84 Po Bi 0 207 82 Pb The radioactive decay law The decay law states that the number of nuclei that will decay per second is proportional to the number of atoms present that have not yet decayed, dN N dt Here is a constant, known as the decay constant. Its physical meaning is that it represents the probability of decay per unit time. If the number of nuclei originally present (at t=0) is N0, by integrating the previous equations it can be seen that the number of nuclei of the decaying element present at time t is N N0et The radioactive decay law As expected the number of nuclei of the decaying element is decreasing exponentially as time goes on. If after a certain time t (lets call it t1/2), the number of decaying nuclei is reduced by half N=N0/2. So, N N 0e t N0 1 t1 / 2 N 0e e t1 / 2 2 2 Using the logarithms: 1 1 t1 / 2 ln ln e ln t1/ 2 2 2 ln 1 ln 2 t1/ 2 ln 2 t1/ 2 t1/ 2 ln 2 The radioactive decay law t1/ 2 ln 2 t1/ 2 0.693 This is the relationship between the decay constant and the half-life. This also means that we can have an equivalent formula for the decay equation: 1 N N0 2 t T1 / 2 The number of decays per second is called activity where A0=N0. So, A N0et Why the decay constant is the probability of decay per unit time Since dN N dt we know that in a short time interval dt the number of nuclei that will decay is dN=Ndt. The probability that any one nucleus will decay within the time interval dt is thus: probabilit y dN dt N and so the probability of decay per unit time is equal to the decay constant: probabilit y dt