The nucleus decays byemission. Write down thedecay equation and determine the maximum kinetic energy of the electrons emitted. Given that: = 22.994466 u = 22.989770 u.

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In emission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus. emission of the nucleus is given as: It is given that: Atomic mass of = 22.994466 u Atomic mass of = 22.989770 u Mass of an electron, me = 0.000548 u Q-value of the given reaction...
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Inemission, the number of protons increases by 1, and one electron and an antineutrino are emitted from the parent nucleus. emission of the nucleusis given as: It is given that: Atomic mass of= 22.994466 u Atomic mass of= 22.989770 u Mass of an electron,me= 0.000548 u Q-value of the given reaction is given as: There are 10 electrons inand 11 electrons in. Hence, the mass of the electron is cancelled in theQ-value equation. The daughter nucleus is too heavy as compared toand. Hence, it carries negligible energy. The kinetic energy of the antineutrino is nearly zero. Hence, the maximum kinetic energy of the emitted electrons is almost equal to theQ-value, i.e., 4.374 MeV. read less
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