Tài liệu Chapter XXIII Nuclear Physics pdf

GENERAL PHYSICS III Optics & Quantum Physics Chapter XXIII Nuclear Physics §1 The structure and properties of nuclei §2 Nuclear binding energy and nuclear force §3 Nuclear stability and radioactivity It is well known for you that every atom contains as its center a nucleus that is: positively charged much smaller in size than the atom carrying almost the total mass of the atom This chapter provides deeper knowledge about NUCLEI §1 The structure and properties of nuclei: 1.1 The structure of nuclei: • Nucleus consist of two sorts of particles: protons (p) and neutrons (n) • The mass of nuleons: (the common name: nucleons) mp = 1.007276xu; mn = 1.008665xu; (Note: me = 0.000548580 x u) where 1u = 1.66053873x10-27 kg → the atomic mass units In nuclear physics, the masses of particles are usually expressed in the energy units – MeV (It is based on the formula E = mc ) Then, 1u = 931 MeV  mp = 938.2 MeV; mn = 939.5 MeV; (me = 0.511 MeV) 1.2 The spin and magnetic moment: • Protons and neutrons have the same spin s = ½ • The magnetic moment:  = +2.49  ;  = -1.91  p N n N e    05079   J / T 10 27  : the nuclear magneton N N 2m p (It is 1386 times smaller than the Bohr magneton) • An application of the proton spin: Magnetic Resonance Imaging In order to image tissue of various types, Magnetic Resonance Imaging detects the small difference in the numbers of “up” and “down” hydrogen proton spins generated when the object studied is placed in a magnetic field Magnetic resonance imaging (MRI) depends on the absorption of electromagnetic radiation by the nuclear spin of the hydrogen atoms in our bodies (in water molecules) The nucleus is a proton with spin ½, so in a magnetic field B there are only two possible spin directions with definite energy The energy difference between these states is  B, with  = 1.41 x 10-26 J /Tesla E=2 p p B B=0   pB E   B Example: A typical magnitude of magnetic field is Tesla Then   pB = x (1.41 x 10-26 J/T) x T = 2.82 x 10-26 J E   = 2.82 x 10-26 J x eV/ 1.6 x 10-19 J = 1.76 x 10-7 eV  the frequency f of photons that can be absorbed by this energy f =  / h E difference: = 2.82 x 10-26 J / 6.626 x 10-34 J.sec = 42 MHz Radio waves These radio-band signals depends on the microscopic properties of the object and give the image of the object 1.3 Mass and radius of nuclei: • We have for a nucleus: The mass number = the total number of protons and neutrons The number of neutrons A = Z + N The atomic number = the ordinal numeral of the element in the Periodic Table = the charge number = the number of protons Denote a nucleus by the symbol: XA Z ( 1H1 , 16 2He , 8O ,… ) • The isotopes of an element: the nuclei that have the same Z (that is belong to the same element), but different A Examples: The isotopes of hydrogen: 1H , H2 (deuterium), 1H3 (tritium) The isotopes of carbon: 6C12 (carbon-12), 6C13 (carbon-13); etc It is known today about 2500 different nuclei (including different isotopes) 300 nuclei are stable, the rest nuclei are unstable (radioactive) • The radius of nuclei: The rather exact results of measuments give r 1  15 A / m 1 A / fermi 10 (1fermi 10 15 m) §2 Nuclear binding energy and nuclear force: 2.1 Nuclear binding energy: The component paricles (protons, neutrons) of a nucleus bind strongly together inside it The evidence for this is that we must spend an amount of energy (to bombard the nucleus by other particles) to separate a nucleus to individual protons and neutrons Thus we have for the nuclear binding energy EB the following equation: EB = { [Z.mp + (A-Z).mn] – M }.c2 The total mass of separate nucleons > The mass of nucleus The difference [the total mass of separate nucleons - the mass of nucleus] is called the mass defect of the nucleus • Note: The last equation can be written in terms of atomic masses: EB = { [Z.mH + (A-Z).mn] – Matom }.c2 This equation is more convenient, because there are usually in Tables not nuclear masses, but the atomic masses • Example: For 2He4: M atom = 4.0026 u = 3728.0 MeV EB = [ x 938.7 + x 939.5 ] – 3728.0 = 28.4 MeV the binding energy EB The binding energy per nucleon: It is the ratio the number of nucleons A An important observation is that the binding energy per nucleon is nearly the same for almost nuclei ! The dependence of EB /A on A for all nuclei is as follows The graph represents the dependence of the binding energy per nucleon (EB/A) as a function of mass number A Remarks: • The maximum of EB/A is for the values A ~ 50 ÷ 60 (Cr ÷ Zn) It is about 8.7 – 8.8 MeV • Light nuclei (with few nucleons) have small EB/A (for example, EB/A ~ 1.1 MeV for 1H (deuterium); EB/A ~ 7.1 MeV for 2He4) • EB/A for heavy nuclei decreases as increasing A ( EB/A ~ 7.5 MeV for the isotopes of uranium) This dependence of EB /A on A leads to important consequences Two possibilities of giving off energy: Nuclear Fission: Division of a heavy nucleus into two fragments of comparable mass Nuclear Fusion: Synthesis of some light nuclei to form a larger nucleus Example: • Division of a nucleus with A = 240 into two pieces of A = 120 (fission): (EB) f = 2x8.5x120 MeV (EB) i = 7.5x240 MeV = 2.040 MeV = 1.800 MeV So after division the binding energy of the system increases about 240 MeV Since binding energy is negative → the total energy of the system decreases → the remainder (240 MeV) converts into the kinetic energy of the fragments The division of isotopes of Uranium, Plutonium,… gives large sources of energy – called the atomic energy • Synthesis of two nuclei H2 into the nucleus 2He4 (fusion): (EB) f = 7.1x4 MeV (EB) i = 2x1.1x2 MeV = 28.4 MeV = 4.4 MeV The remainder about 24 MeV is given off This type of energy is called the fusion energy • Comparision: With the same mass of material the synthesis reaction give off much more energy than the division reaction 2.2 Nuclear force: The large binding energy of nucleons inside nucleus is the evidence of the strong interaction between nucleons  there exists forces between them, and this kind of forces is called nuclear forces Some charisteristics of nuclear forces: They must be attractive, and much stronger the repulsive electrical forces between positive-charged protons They are short-range forces: The range is of the order 10-15 m They become negligible at distances d > 2.10-15 m At d < 10 -15 m the attractive forces are replaced by the repulsive They not depend on the eletrical charge of nucleons The binding forces have the same magnitude for two protons, or two neutrons, or proton and neutron This is an important feature and called the charge-independence of nuclear forces They depend on the mutual orientation of nucleons The binding forces favour paires of protons or neutrons with opposite spins They have the property of saturation: It means that one nucleon cannot interact simultaneously with all other nucleons in the nucleus, but only with those few in its vicinity (otherwise the binding energy per nucleon will not be nearly constant) To give a theoretical basis for these characteristics of nuclear forces one introduced models of nuclei There is no completely perfect model, every model has successfulness and disvantage §3 Nuclear stability and radioactivity: Among 2500 known nuclei, there are about 300 stable The others are unstable, and they decay to convert into other nuclei with emitting particles and electromagnetic radiation This kind of processes is called radioactivity We consider the following processes: (i)  -decay; (ii)  -decay; 3.1 Alpha decay: In this process a nucleus emits the nuclei 2He4 ( -particles): Example: XA → Z U238 → 92 YA-4 + 2He4 Z-2 Th234 + 90 He4 • Alpha decay is usually accompained by  -rays (electromagnetic radiation) • The velocity of emmited a-particles is about 10 m/s, the kinetic energy ~ few MeV • When  -particles travel through matter, they ionize molecules of matter, loss their energy and stop The race in normal air: several centimeters, in solid: several tens of micrometers (  -particles cannot pass through an usual paper sheet) • Alpha decay is the tunnel effect of  -particles through a potential-energy barrier as shown in the picture 3.2 Beta decays: + There are types of  -decays: , , and electron capture 3.2.1  decay: ~ X A Z Y A 1 e0   Z  ~ • A typical  process: n  p   e  anti-neutrino  A free neutron is  -radioactive particle • Another example: ~ Th 234 Z  Pa 234 1e0   90  (thorium) (protactinium) • The energy of emitted electrons in this decay is not definte, but has a continuous spectrum from to Emax • The speeds of electrons are up to 0.9995 c  the electron’s motion is relativistic • Anti-neutrino: It is the anti-particle of neutrino Both neutrino and its anti-particle have the rest mass It is neutral  the electric charge of the system is conserved Its spin is ½  the total spin of the system is conserved It shares the given off energy in the  decay process with the electron The given off energy is distributed between the electron and the anti-neutrino in different ways, holding the energy conservation + 3.2.2  decay: • The scheme: • Example: X Z Y 1 e    A Z N 6 C 1e    13 A 13 neutrino pozitron • Note: Since the mass of proton is smaller than that of neutron, p  n    is impossible for free protons e  the process (But it is possible for protons inside nuclei by taking energy from other nucleons) 3.2.3 Electron capture (K-capture): • It is the process in which the nucleus absorbs one of the electrons in K-shell of the atom (or in L-, M-shells, but with small probabilities) In the result, one of protons converts to neutron: p   n  e  and the corresponding transformation of the nucleus is X A 1e Z Y A   Z  • The place of the captured electron will be filled by an electron from other shells  this process is accompaned by emitting electromagnetic waves in the band of X-rays • Example: K 40 1e0 18 Ar 40   19  3.3 Decay rate and half-lives: Suppose that we have an amount of radioactive matter Due to radioactive processes (decays), the initial nuclei transform into other nuclei  the number of the initial matter decreases in time • Denote by N(t) the number of nuclei (and also atoms) at the moment t During an interval dt the number of nuclei (atoms) decreases dN : dN  Ndt  ln N     t const N ( t ) N e t where N0 = N(t=0) is the number of nuclei of the initial matter at t=0;  is a constant which characterizes the radioactivity of the matter It is different for different elements  called the decay constant is • The half-life T 1/2 : By definition, it is the time required for the number of radioactive nuclei to decrease to one-half the original number N0 • We can derive the relation between the half-life and the decay constant: N0  N e T1 / 2 T1 / ln 693     • Unites for T1/2 → s, for  1/s  → • The half-lives are very different for different elements For the known elements the range of half-lives is from 3x10 -7seconds to 5x1015 years Summary Any nucleus consists of the component particles: protons (p) and neutrons (n) Proton has the electrical charge +1, neutron has no charge Spin of proton and neutron is ½ The nucleus ZX A consists of Z protons and (A-Z) neutrons The binding energy of the nucleons inside a nucleus is determined by the mass defect of the nucleus (the difference between the total mass of separate nucleons and the mass of nucleus) The dependence of the binding energy per nucleon on the mass number A gives predictions about characteristics of nuclear forces between nucleons inside nucleus It allows also to explain two ways of having energy: fission and fusion Radioactivity: The most of nuclei are unstable under decay processes After decay processes the nucleus transforms into other nucleus The rate of radioactive process is described by the equation: N ( t ) N e  t  The half-life of an radioactive element T1/2: It is the required time to decrease to one-half the initial amount of radioactive matter  .. .Chapter XXIII Nuclear Physics §1 The structure and properties of nuclei §2 Nuclear binding energy and nuclear force §3 Nuclear stability and radioactivity... measuments give r 1  15 A / m 1 A / fermi 10 (1fermi 10 15 m) §2 Nuclear binding energy and nuclear force: 2.1 Nuclear binding energy: The component paricles (protons, neutrons) of a nucleus... 1.008665xu; (Note: me = 0.000548580 x u) where 1u = 1.66053873x10-27 kg → the atomic mass units In nuclear physics, the masses of particles are usually expressed in the energy units – MeV (It is based