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The nucleus ""_(10)^(23)Ne decays by beta^(-)– mission. 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 :`""_(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`. | |