Exercise 3 The measured spin of deuteron is found to be I-1. Since the neutron a
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Exercise 3 The measured spin of deuteron is found to be I-1. Since the neutron and proton spins can be either parallel or antiparallel, show that there are four possible ways, presented in the lecture, to couple Sa, Sp, and& to get a total 1-1. Exercise Show that in a nucleon-nucleon scattering problem, if the incident energy is far below 20 MeV, then the assumption that the relative orbital angular momentum is equal to zero (1-0) is justified. Exercise 5 a. What is the effect of parity and time reversal operators on individual spins, S, S, b. Show that under parity and time reversal the terms such as: S,, S,, and AS,+BS, are excluded from being in the spin-dependent potential term. c. Show that terms like: Si, S22, and Si·s, are allowed in the spin-dependent potential term. d. Show that the term V.(r)1 . s fulfils the symmetry requirement (parity and time reversal) Dr. F. Aksouh, 2012Explanation / Answer
Solution 3: Experimentally the measured spin of deuteron is I=1
So let us look at how can we get I=1. I hope you already know that I,the Nuclear Spin, actually represents the Total Angular Momentum of the deuteron nucleus. This in turn is a combination of the Orbital Angular Momentum,L, and Spin Angular Momentum,S of the nucleus.
I = S + L .............(1) The bold letters are for vectors.
where S = sn + sp
and sn=sp=1/2.
So S can either be 0 (=1/2 - 1/2) or 1 (= 1/2 + 1/2). If S is 0 the spins of the proton and neutron are anti-parallel and so cancel each other, if S is 1 the spins are parallel and add to each other. Please refer standard textbook for further information.
1) So from (1) we can easily see if S=0, to give I=1 we need to have L=1. This is one possible combination.
Similarly, if S=0 and L=0 then I=0
Again, if S=0 and L=2 then I=2. So if we increase the values of L and put S=0 we will get higher values of I, but we discard them all as we need I=1. So with S=0 , i.e with snand sp antiparallel to each other, we need L=1 or the p-subshell to get I=1.I hope this is clear.
2) If S=1 and L=0 we get I=1, i.e. if sn and sp are parallel to each other we need the nucleus to be in L=0 or the s-subshell to get I=1. This is another possibility.
3) If S=1 and L=1 then according to the rules of coupling, I=0,1,2.
i.e. I can have values from modulus of (L-S )to modulus of (L+S) in steps of 1.
Modulus of (L-S)=0 and modulus of (L+S)=2 and to move from 0 to 2 we have 1 in between. So what we do is keep on adding 1 to the extreme small value untill the extreme big value is attained. So I= 0, then 0+1=1 and 1+1=2.This is how we got the three values of I.So this is another possibility with S=1 or snand sp both parallel to each other and L=1 or in the p subshell.
4) If S=1 and L=2 then again, according to the rules of coupling I= 1,2,3. I hope you understand the coupling result, i.e modulus of (L-S) or 1 to modulus of (L+S) or 3 in steps of 1. So we have 2 in between. If I go ahead and put L=3 with S=1 then I will have I=2,3,4. I really hope you understand how we got the values of I. If not, please refer standard textbook and study coupling schemes.
But we will discard this result because we need I=1 and not 2,3 or 4.So the last possibility is with S=1 or both sn and sp parallel to each other and L=2 or d subshell.
These are the four required possibilities.
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