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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, and m(" "_(11)^(23)Na) = 22.089770 u |
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Answer» Solution :Equation for `beta`-decay is `" "_(10)^(23)Ne (to) " "_(11)^23)Na + " "_(-1)^(0)e+ barnu +Q` Here, Q = energy released in `beta`-decay = maximum K.E. of the electron emitted and is given by `Q = [ m_(N) (" "_(10^(23)Ne) -{m_(N) (" "_(11)^(23)Na) +m_(e) }] U xx 931.5 MeV = [ {m(" "_(10)^(23)Ne) - 10m_(e) } - {m(" "_(11)^(23)Na) - 11m_(e) +m_(e)}]u xx 931.5 MeV = [ m(" "_(10)^(23)Ne)-m(" "_(11)^(23)Na)]u xx 931.5 MeV = (22.994466 u - 22.989770 u) xx 931.5 MeV = 0.004696 xx931.5 MeV = 4.374 MeV` |
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