Complete the decay reaction `._10Ne^(23) to?+._-1e^0+?` Also, find the maximum KE of electrons emitted during this decay. Given mass of `._10Ne^(23)=22.994465 u`. mass of `._11Na^(23)=22.989768u`.
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Applying conservation of charge number and mass number, we can write the decay reaction as `._10Ne^(23) to ._11Na^(23)+._-1e^0+barv` Neglecting mass of `e^-`, mass defect, `Deltam=m(._10Ne^(23))-m(._11Na^(23))` `=22.994456-22.989768` `Deltam=0.004697u` Max K.E. of electron =Q value of the reaction =`0.004697xx931MeV` `=4.372MeV`
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