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Math Help - emission question

  1. #1
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    emission question

    if 13 Al 27 isotope is stable which emission can 13 Al 29 undergo
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  2. #2
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    For future reference you don't need to give us the atomic number since it's implied in the element name. In this case every Al atom has 13 protons - if it didn't it wouldn't be Al!

    Back on topic, ^{29}\text{Al} needs to lose two atomic mass units (amu) without changing atomic mass number. It's a weird one since to lose 2 amu without changing atomic number implies neutron emission twice which would be ^{29}\text{Al} \rightarrow ^{28}\text{Al} + n \rightarrow ^{27}\text{Al} + n


    Yet google and wolfram say beta minus decay: ^{29}\text{Al} \rightarrow ^{29}\text{Si} + e^- + \bar{\nu_e} and ^{29}\text{Si} is stable.


    I would be inclined to believe google if this is a real world problem since neutron emission usually occurs when there are more than two excess neutrons compared to the stable isotope
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