The antielectron is often represented by the symbol e+, but in beta decay, it is written as β+ to indicate that the antielectron was emitted during a nuclear decay. Antielectrons are the antimatter counterpart of electrons that are almost identical, have the same mass, spin, etc., but have a positive charge and an electron family number of -1. When a positron hits an electron, mutual annihilation occurs, in which the entire mass of the anti-electron-electron pair is converted into pure photonic energy. (The reaction, e+ + e− → γ + γ, preserves the number of electron families as well as all other conserved quantities.) If we know that a nuclide [latex]_Z^Atext{X}_N[/latex] decays β+, then its equation β+decay (b) The following decay is mediated by the weak nuclear force: The lambda particle then decays by the weak nuclear interaction according to As in the previous example, we must first find Δm, the mass difference between the parent nucleus and the decay products, using the masses specified in Appendix A. Then the energy emitted is calculated as before with E = (Δm)c2. The initial mass is only that of the parent nucleus, and the final mass is that of the daughter nucleus and the electron produced during decay. The neutrino is massless or almost massless. However, since the masses given in Appendix A apply to neutral atoms, the daughter nucleus has one more electron than the parent nucleus, and thus the extra mass of electrons equal to the β– is contained in the atomic mass of Ni. Thus, Δm = m(60Co) − m(60Ni). The neutrons released during fission with an average energy of 2 MeV in a reactor undergo on average many collisions (elastic or inelastic) before being absorbed.
During the scattering reaction, a fraction of the kinetic energy of the neutron is transferred to the nucleus. It is possible to derive the following equation for target or moderator nucleus mass (M), incident neutron energy (egg) and scattered neutron energy (Es) using the laws of conservation of momentum and energy and the analogy of billiard ball collisions for elastic scattering. When a proton turns into a neutron, which particle should be released? Consider an example of element charge and baryon number and lepton conservation in a weak interaction process (where the weak nuclear force intervenes). Disintegration: The process by which an atomic nucleus of an unstable atom loses mass and energy by emitting ionizing particles The conservation of the number of leptons states that the sum of the lepton numbers must be equal before and after the interaction. There are three different lepton numbers: the electron-lepton number Le,Le,The muon-lepton number Lμ,Lμ and the tau-lepton number Lτ.Lτ. For each interaction, each of these quantities must be kept separately. For electrons and electron neutrinos, Le=1; The=1; for their antiparticles, Le=−1; The=−1; all other particles have Le=0.Le=0. Similarly Lμ=1Lμ=1 for muons and muon neutrinos, Lμ=−1Lμ=−1 for their antiparticles and Lμ=0Lμ=0 for all other particles. Finally, Lτ=1,−1Lτ=1,−1 or 0, depending on whether we have a tau or tau neutrino, their antiparticles, or another particle.
Lepton number conservation ensures that the number of electrons and positrons in the universe remains relatively constant. (Note: The total number of leptons is, as far as we know, conserved in nature. However, observations have shown variations in the number of leptons in the family (e.g., Le) in a phenomenon called neutrino oscillations.) Some nuclides are stable and apparently live forever. Unstable nuclides decay (i.e. they are radioactive) and eventually produce a stable nuclide after many decays. We call the original nuclide parents and its decay products girls. Some radioactive nuclides decay in a single step to form a stable nucleus. Example: 60Co60Co size 12{» lSup { size 8{«60»} } «Co»} {} is unstable and decays directly to 60Ni60Ni size 12{» lSup { size 8{«60»} } } «Ni»} {}, which is stable. Others, such as 238U238U size 12{» lSup { size 8{«238»} } U} {}, decay into another unstable nuclide, resulting in a decay series in which each subsequent nuclide decays until a stable nuclide is finally created.
The decay series from 238U238U size 12{» lSup { size 8{«238″} } U} {} is particularly interesting because it produces the radioactive isotopes 226Ra226Ra size 12{» lSup { size 8{«226»} } «Ra»} {} and 210Po210Po size 12{» lSup { size 8{«210»} } «Po»} {} that the Curies first discovered (see [link]). Radon is also produced (222Rn222Rn size 12{» lSup { size 8{«222»} } «Rn»} {} in the series), a natural hazard increasingly recognized. Since radon is a noble gas, it emanates from materials such as soil, which even contain traces of 238U238U size 12{» lSup { size 8{«238″} } U} {} and can be inhaled. The decay of radon and its progeny causes internal damage. The decay series 238U238U size 12{» lSup { size 8{«238″} } U} {} ends with 206Pb206Pb size 12{» lSup { size 8{«206»} } } «Pb»} {}, a stable isotope of lead. The decay equation has already been used for 239Pu239Pu size 12{» lSup { size 8{«239»} } «Pu»} {}; it is β−β− size 12{β rSup { size 8{ – {}} } } {} decay produces 137Ba137Ba size 12{» lSup { size 8{«137»} } «Ba»} {}. The parent nuclide is an important reactor waste product and has a chemistry similar to potassium and sodium, resulting in its concentration in your cells when ingested. This process is mediated by weak interaction.
The neutron is transformed into a proton by the emission of a virtual W− boson. At the quark level, the W emission transforms a down quark into an up quark and a neutron (one up quark and two down quarks) into a proton (two up quarks and one down quark). The virtual boson W− then decays into an electron and an antineutrino. Figure 5. The β+ decay is the emission of a positron that eventually finds an electron to annihilate it, characteristically producing gammas in opposite directions. If a nuclide [latex]_Z^Atext{X}_N[/latex] decays β−, then its decay equation β− is [latex]text{X}_Nrightarrowtext{Y}_{N-1}+beta^{-}+bar{nu}_e[/latex] (β− decay), For reaction (a), the net number of baryons of the two reactants is 0+1=10+1=1 and the net number of baryons of the four products is 0+1+0+0=1.0+1+0+0=1. Since the net baryon numbers of the reactants and products are the same, this reaction is permissible on the basis of the law of conservation of the number of baryons. In the analysis of nuclear reactions, we apply the many laws of conservation. Nuclear reactions are subject to classical conservation laws for charge, momentum, angular momentum, and energy (including resting energies). Other conservation laws not expected by classical physics are electric charge, lepton number and baryon number. Some of these laws are respected in all circumstances, others are not.
We have accepted the conservation of energy and momentum. In all the examples given, we assume that the number of protons and the number of neutrons are preserved separately. We will find circumstances and conditions in which this rule is not true. If we look at non-relativistic nuclear reactions, this is essentially true. However, we will find that these principles need to be extended when we consider relativistic nuclear energies or those with weak interactions. This process saves load, energy and momentum. However, this does not happen because it violates the law of conservation of the number of baryons. This law requires that the total number of baryons in a reaction before and after the reaction be equal.
To determine the total number of baryons, each elementary particle is assigned a number of baryons B. The number of baryons has the value B=+1B=+1 for baryons, –1 for antibaryons and 0 for all other particles. To return to the above case (decay of the neutron into an electron-sediment pair), the neutron has a value B = + 1, B = + 1, while the electron and positron each have a value of 0. Thus, decay does not occur, since the total number of baryons goes from 1 to 0. The proton-antiproton collision process, where Y is the nuclide that a proton has more than X (see Figure 4). So if you know that a particular nuclide decays β-, you can find the daughter nucleus by first finding Z for the parent and then determining which element has the atomic number Z + 1. In the example of the β decay of 60Co given above, we see that Z = 27 for Co and Z = 28 Ni. It is as if one of the neutrons in the parent nucleus decays into proton, electron and neutrino. In fact, neutrons outside nuclei do just that – they live on average only a few minutes and β decay in the following way: β−β− size 12{β rSup { size 8{ – {}} } } {} decay of 3H3H size 12{» lSup { size 8{3} } H} {} (tritium), a manufactured isotope of hydrogen used in some digital clock screens and produced primarily for use in hydrogen bombs. Calculate the energy released in the β+β+ size 12{β rSup { size 8{+{}} } } {} {} of 22Na22Na, whose equation is given in the text. The masses of 22Na22Na and 22Ne22Ne size 12{» lSup { size 8{«22»} } «Ne»} {} are 21.994434 and 21.991383 u. If you look at the periodic table of elements, you will find that Th has Z = 90, two less than U, Z = 92.