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Relative orbital angular moment in nuclear reactions

  1. Jan 5, 2016 #1
    Hi folks,
    i have to calculate the angular Spin and Parity JP of 17O as a result of the shooting of 16O with Deuterons. So the reaction equation should be:
    16O + ²H -> 17O + 1H
    The only further Information given is that the captured neutron has positive parity and an orbital angular momentum number of ln=2, thus the possible nuclear spin of 17O is either 5/2 or 3/2, where the d5/2-state is aligned with a lower energy.
    The way to solve this problem should be to use the conservation of the total angular moment and the parity taking into account that both the initial and the final reaction products have got a relative orbital angular moment li and lf.
    I don't really understand how to calculate li. Once you know li, lf should be a result of the conservation of partity and total angular moment.
    I tried to calculate lf for li=0 and got as possible results for JP(17O)=5/2 lf∈{2,4} and for JP(17O)=3/2 lf∈{0,2}. Now i dont see how to eliminate one of the possible JP∈{3/2, 5/2} because both possibilities dont violate neither the conservation of spin nor the conservation of parity.
    So to make a long story short: can please somebody help me to understand the relative orbital angular moment? :confused:
  2. jcsd
  3. Jan 10, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
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