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What is the basic mechanism for the emission of beta^(-) or beta^(+) particles in a nuclide? Give an example by writing explicitly a decay process for B’ emission. Is (a) the energy of the emitted beta^(-) particles continuous or discrete, (b) the daughter nucleus obtained through beta-decay an isotope or an isobar of the parent nucleus ? |
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Answer» Solution :The basic mechanism for the emission of a B’ particle in a nuclide is the transformation of a neutron into a proton and a Bf particle is ejected out of nucleus. As a result, the atomic number of nuclide increases by one. As an example consider the following REACTION `" "_(15)^(32)PTO" "_(16)^(32)S+" "_(-1)^(0)e+bar(nu)` or `" "_(27)^(60)Coto" "_(28)^(60)Ni+" "_(-1)^(0)e+nu` Similarly at the time of `beta^(+)` decay a proton inside a nucleus transforms into neutron along with the emission of a `beta^(+)` particle. An example is `" "_(11)^(22)Nato " "_(10)^(22)Ne + " "_(-1)^(0)e + nu` As a result of `beta` emission the atomic number of nucleus decreases by one. (a) The energy of the emitted B-particle is continuous. (b) The daughter nucleus obtained through B-decay is an ISOBAR of the parent nucleus because mass number of both parent and daughter nuclei are same. |
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