The nucleus ""_(10)^(23)Ne decays by beta^(-) emission. Write down the beta-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that: m (""_(10)^(23) Ne) = 22.994466 u m (""_(11)^(23)Na) = 22.989770 u.

The nucleus ""_(10)^(23)Ne decays by beta^(-) emission. Write down the

1.	The nucleus ""_(10)^(23)Ne decays by beta^(-) emission. Write down the beta-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that: m (""_(10)^(23) Ne) = 22.994466 u m (""_(11)^(23)Na) = 22.989770 u.
Answer» SOLUTION :The equation is `""_(10)^(23)Ne to ""_(11)^(23)Na + beta + BARUPSILON +Q` `Q = [m_n(""_(10)^(23)Ne) - m_(N) ""_11^23Na - m_e]c^2` (Neglect the rest mass of `barupsilon)` `= {{m_n ""_(10^(23)Ne + 10 m_e} - {m_n(""_11^23Na) + 11m_e}]c^2` `= [m ""_(10)^(23)Ne - m ""_(11)^(23)Na]c^2` `=[22.994466 - 22.989770]931` `= 4.374 MeV`.

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The nucleus ""_(10)^(23)Ne decays by beta^(-) emission. Write down the beta-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that: m (""_(10)^(23) Ne) = 22.994466 u m (""_(11)^(23)Na) = 22.989770 u.

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