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i have to calculate the angular Spin and Parity J^{P}of^{17}O as a result of the shooting of^{16}O with Deuterons. So the reaction equation should be:

^{16}O + ²H ->^{17}O +^{1}H

The only further Information given is that the captured neutron has positive parity and an orbital angular momentum number of l_{n}=2, thus the possible nuclear spin of^{17}O is either 5/2 or 3/2, where the d_{5/2}-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 l_{i}and l_{f}.

I don't really understand how to calculate l_{i}. Once you know l_{i}, l_{f}should be a result of the conservation of partity and total angular moment.

I tried to calculate l_{f}for l_{i}=0 and got as possible results for J^{P}(^{17}O)=5/2 l_{f}∈{2,4} and for J^{P}(^{17}O)=3/2 l_{f}∈{0,2}. Now i dont see how to eliminate one of the possible J^{P}∈{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?

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# Relative orbital angular moment in nuclear reactions

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