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The nucleus .^(23)Ne deacays by beta-em...

The nucleus `.^(23)Ne` deacays by `beta`-emission into the nucleus `.^(23)Na`. Write down the `beta`-decay equation and determine the maximum kinetic energy of the electrons emitted. Given,`(m(._(11)^(23)Ne) =22.994466 am u` and `m (._(11)^(23)Na =22.989770 am u`. Ignore the mass of antineuttino `(bar(v))`.

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`""_(10)^(23)Ne to ""_(11)^(23)Na + e^(-) + bar(v) + Q , Q = [m_N (""_(10)^(23)Ne) - m_N (""_(11)^(23)Na) - m_(e)] c^2`, where the masses used are masses of nuclei and not of atoms as in Exercise . Using atomic masses `Q = [m (""_(10)^(23)Ne) - m(""_(11)^(23)Na)]c^2`. Note `m_e` has been cancelled. Using given masses, `Q = 4.37 MeV`. As in Exercise , maximum kinetic energy of the electron (max `E_e) = Q = 4.37 MeV`.
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