Relative orbital angular moment in nuclear reactions

[Rajees]

Hi folks,
i have to calculate the angular Spin and Parity J^P 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 l_n=2, thus the possible nuclear spin of ¹⁷O 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(¹⁷O)=5/2 l_f∈{2,4} and for J^P(¹⁷O)=3/2 l_f∈{0,2}. Now i don't see how to eliminate one of the possible J^P∈{3/2, 5/2} because both possibilities don't 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?

 

[eljose79]

Hello,

To calculate the angular Spin and Parity JP of 17O in this reaction, we need to use the conservation laws of total angular momentum and parity. Let's break down the steps to solve this problem:

1. Determine the initial relative orbital angular momentum li:
In this reaction, the initial state of 16O and ²H have a relative angular momentum of l=2, as stated in the forum post. This means that li=2.

2. Determine the final relative orbital angular momentum lf:
The final state of 17O and 1H also have a relative orbital angular momentum, which we can denote as lf. To determine lf, we need to use the conservation laws of total angular momentum and parity. This means that the sum of the initial relative orbital angular momentum and the relative orbital angular momentum of the captured neutron (ln=2) must be equal to the sum of the final relative orbital angular momentum and the relative orbital angular momentum of the proton (lp=0). Mathematically, this can be written as:
li + ln = lf + lp
Substituting the values, we get:
2 + 2 = lf + 0
lf = 4
Therefore, lf=4.

3. Determine the possible values of JP:
Now that we have determined lf, we can use the formula for calculating the possible values of JP:
JP = |li - lf|, |li - lf| + 1, ..., li + lf
Substituting the values, we get:
JP = |2 - 4|, |2 - 4| + 1, ..., 2 + 4
JP = 2, 3, 4, 5, 6
Therefore, the possible values of JP for 17O in this reaction are 2, 3, 4, 5, or 6.

4. Determine the correct value of JP:
To determine the correct value of JP, we need to use the information given in the forum post that the d5/2-state is aligned with a lower energy. This means that the lower energy state should have a lower value of JP. In this case, the state with JP=3/2 is the lower energy state, and hence is the correct value of JP for 17O in this reaction.

So, to summarize, the angular Spin and Parity JP of 17O in this reaction is 3/