Okay, so I need to figure out which partial wave isn't allowed when a two-nucleon system in a 1S0 state emits a particle X with intrinsic parity -1. The question mentions Pauli statistics and some quantum number conditions. Let me start by breaking down the problem step by step.

First, the initial state is a 1S0 state. The 1S0 notation in term symbols is (2S+1)LJ. For 1S0, S is 0 (since 2S+1 =1 implies S=0), L is 0 (because it's 1S0, L=0). So the initial state has spin 0 and orbital angular momentum 0. The total angular momentum J is L + S = 0 + 0 = 0. So the initial state has J=0.

Now, the system emits a particle X with intrinsic parity -1. The problem states that the final NN state must satisfy T(NN) = S(NN) + L(NN) + 1 mod 2. Wait, T is the isospin, S is spin, L is orbital angular momentum. So the equation is T = (S + L) + 1 mod 2. But I need to check what each of these represents.

But before that, maybe I should think about conservation laws. The process is a decay, so angular momentum and parity must be conserved. The initial state has total angular momentum J=0. The emitted particle X has some angular momentum L_X. The final state is two nucleons (so their orbital angular momentum L_NN) and the particle X with angular momentum L_X. But wait, the term symbol for the final state would be something like (2S+1)L_NN for the nucleons and (intrinsic parity for X) * (parity of their orbital angular momentum). So the total parity should be the product of the initial parity, the intrinsic parity of X, and the parity from the orbital angular momentum of X.

Wait, but the parity of the initial state is the parity of the 1S0 state. For nucleons, the parity of the 1S0 state is (-1)^L, where L is the orbital angular momentum. Since L=0, the parity is 1. So initial parity is 1. The emitted particle X has intrinsic parity -1. The parity of the emitted particle's orbital angular momentum is (-1)^L_X. So total parity of the final state is 1 * (-1) * (-1)^L_X = (-1)^{L_X +1}. 

But the initial parity is 1, so the final state's parity must also be 1. Therefore, (-1)^{L_X +1} = 1. That implies that (-1)^{L_X} = -1, so L_X must be odd. Because (-1)^(odd) = -1, so (-1)^{odd +1} = (-1)^even = 1. So L_X must be odd. So the emitted particle X must have an odd orbital angular momentum. The possible partial waves (orbital angular momentum states) for X are s (L=0, even), p (L=1, odd), d (L=2, even), f (L=3, odd), etc. So the allowed partial waves for X are p, f, etc., which are odd. So any even partial wave (s, d, g,...) would not conserve parity. Wait, but the question is asking which partial wave is not permitted. So if the allowed are odd, then even ones are forbidden. So in the choices, the even partial waves (s, d) are forbidden. Wait, but let's check the options.

Looking at the options:

A. 3D3 + f → D is L=2 (even), so f would be L=3 (odd). But wait, the notation here is a bit confusing. The term symbol for the emitted particle is given as, for example, 3D3. Wait, but the problem statement says "the lowercase letter denotes the angular momentum state of X". Wait, the problem says "the lowercase letter denotes the angular momentum state of X." Wait, maybe that's a typo. Let me check the original problem again.

Wait, the problem says: "Make use of the conservation of both angular momentum and parity. Note that the term symbol (e.g. 1S0) is in general defined (2S+1)L(J), where S is the total spin, L the total orbital angular momentum written in spectroscopic notation, and J the total angular momentum."

Wait, but when the problem refers to the "angular momentum state of X", perhaps it's referring to the orbital angular momentum quantum number. So the term symbols given in the options are for the final states. For example, option A is 3D3 + f. Wait, but 3D3 is a term symbol for the nucleon system. Wait, maybe I misinterpret. Let me re-read the problem.

The question says: "the final NN state has T(NN) = 0. Which of the following partial waves is not permitted, where the lowercase letter denotes the angular momentum state of X?"

Wait, perhaps the "partial waves" refer to the orbital angular momentum state of X. So each option is a combination of the final nucleon state (the term symbol) and the orbital angular momentum (the lowercase letter). For example, option A is 3D3 + f, where 'f' is the orbital angular momentum of X. But I'm a bit confused here.

Wait, let me parse the problem again. The question is: the final NN state has T(NN) = 0. Which partial wave (orbital angular momentum state of X) is not permitted. The allowed partial waves would be those that conserve angular momentum and parity.

So the process is NN (initial) → NN (final) + X. So the initial angular momentum is J_initial = 0. The final angular momentum must also be J_final = 0. The angular momentum of the final NN state (T(NN)) plus the angular momentum of X (L_X) must equal the initial angular momentum. But wait, angular momentum is a vector, so it's more complex. However, the problem states that T(NN) = S(NN) + L(NN) + 1 mod 2. Wait, that's the condition given. Let me parse that.

The problem says: Pauli statistics require that the isospin, angular momentum, and spin quantum numbers of the final NN state satisfy T(NN) = S(NN) + L(NN) + 1 (mod 2). So T(NN) is the isospin of the final NN state. S(NN) is the spin of the final NN state. L(NN) is the orbital angular momentum of the final NN state. The equation is modulo 2. So T = (S + L) + 1 mod 2.

Wait, but I thought that in Pauli's exclusion principle for identical fermions, the total wavefunction must be antisymmetric. The wavefunction is a product of the spatial part, spin part, and isospin part. So the combined symmetry must be antisymmetric. For two fermions, the spatial part, spin part, and isospin part must combine such that their product is antisymmetric. Since nucleons are fermions, their overall wavefunction must be anti-symmetric under exchange. So if the spatial part is symmetric (even L), then the spin-isospin part must be antisymmetric. If the spatial part is antisymmetric (odd L), then spin-isospin part must be symmetric.

But maybe that's a different approach. The problem gives a specific condition: T(NN) = S(NN) + L(NN) + 1 mod 2. So maybe we can use that.

Given that the initial state has T(NN) = 0. Wait, the initial state is 1S0, which is a singlet state (S=0) and L=0. So the initial NN state has S=0, L=0. What is T_initial? The initial state is a 1S0 state, which is for two nucleons. But the 1S0 term symbol here is for the NN system. The term symbol is (2S+1)LJ. For 1S0, 2S+1=1 implies S=0. L=0. So the spin is 0, orbital angular momentum 0. The isospin T_initial is 0? Wait, no. Wait, the isospin of the NN system is determined by their isospin states. But in the 1S0 state, the spin is 0, so the NN system must have total spin S=0. But the isospin T is a different quantum number. For two nucleons, the possible isospin states are T=0 (singlet) and T=1 (triplet). The 1S0 state refers to the spin state, not the isospin. So T_initial is not necessarily zero. Wait, but the problem states that the final state has T(NN) = 0. So the final NN state has isospin T=0. So the initial state's isospin must be the same as the final state's, because isospin is conserved in strong interactions (assuming X is a boson, which it is here since it's a bound system emitting a particle; but if X is a fermion like a pion, then interactions would conserve isospin). Wait, but the problem mentions that the system emits a particle X of intrinsic parity -1. A pion has isospin 1, but if X is a nucleon or something else, but the problem doesn't specify. But the problem states that the final NN state has T=0. So the initial state's isospin must be T_initial. If the initial state is 1S0, which refers to spin 0, but the isospin could be anything. Wait, but the problem says T(NN) = 0 for the final state. So maybe the initial state's isospin must be T_initial = T_final. But if X is a scalar (spin 0, isospin 0), then the isospin of the NN system remains the same. So T_initial = T_final. Therefore, the initial state's isospin must be T_initial = 0. But I'm not sure. Let me think again.

Alternatively, maybe the problem is focusing on the condition T(NN) = S(NN) + L(NN) + 1 mod 2. So for the final state, T = S + L + 1 mod 2. Given that the problem states that the final state has T(NN) = 0. So substituting into the equation: 0 = S + L + 1 mod 2. Therefore, S + L ≡ 1 mod 2. So the sum of the spin and orbital angular momentum of the NN system must be odd. So either S is even and L is odd, or S is odd and L is even.

Additionally, we have conservation of angular momentum. The initial state has total angular momentum J_initial = 0. The final state must have total angular momentum J_final = J_initial = 0. The angular momentum of the final state is the sum of the NN's angular momentum L_NN and the orbital angular momentum L_X of X. But wait, the total angular momentum J_final is the vector sum of L_NN (orbital angular momentum of the NN system) and L_X (orbital angular momentum of the emitted particle X). However, since the problem states that the NN system's L(NN) is part of the final state's angular momentum, perhaps the total orbital angular momentum of the system (including X) must add up to J_initial = 0.

But this is getting complicated. Let me try to approach it step by step.

First, initial state: NN in 1S0. So J_initial = 0. The emitted particle X has L_X (orbital angular momentum). The final NN state has L_NN (orbital angular momentum of the NN system) and spin S_NN. The total angular momentum of the final state is the sum of L_NN + L_X + S_NN. But since angular momentum is a vector, the total J_final must be equal to J_initial, which is 0. However, the problem gives a condition that T(NN) = S(NN) + L(NN) + 1 mod 2. So maybe the problem is considering the quantum numbers modulo 2, not the actual vectors. So perhaps the condition is that the sum of S and L of the NN system (S(NN) + L(NN)) must be odd (since T(NN) = 0 = (S + L) + 1 mod 2 → S + L ≡ 1 mod 2).

So the sum of the spin and orbital angular momentum of the NN system in the final state must be odd. So either S is even and L is odd, or S is odd and L is even.

Additionally, the total angular momentum of the final state must be J = 0. So the sum of the NN's orbital angular momentum (L_NN), the emitted particle's orbital angular momentum (L_X), and the NN's spin (S_NN) must result in J=0. Since angular momentum is a vector, this is a vector sum, but since we don't know the directions, perhaps we can reason in terms of possible quantum numbers. However, this might be too vague. Maybe the problem assumes that the spins and orbital angular momenta are such that their sum is zero.

But perhaps a better approach is to consider parity conservation. The initial state has parity 1 (since 1S0 has parity (-1)^L = (-1)^0 = 1). The emitted particle X has parity -1. The parity of the final state is the product of the parity of the NN system and the parity of X's orbital angular momentum. The parity of the NN system is (-1)^{L_NN}, and the parity of X's orbital angular momentum is (-1)^{L_X}. So total parity of the final state is [(-1)^{L_NN}] * (-1) * [(-1)^{L_X}] = (-1)^{L_NN + L_X +1}.

This must equal the initial parity, which is 1. Therefore:

(-1)^{L_NN + L_X +1} = 1 → (-1)^{L_NN + L_X} = -1 → L_NN + L_X must be odd.

So L_NN + L_X ≡ 1 mod 2.

But from the problem's condition, T(NN) = S(NN) + L(NN) + 1 mod 2, and T(NN) = 0. So:

0 = S(NN) + L_NN + 1 mod 2 → S(NN) + L_NN ≡ 1 mod 2.

So we have two conditions:

1. S(NN) + L_NN ≡ 1 mod 2
2. L_NN + L_X ≡ 1 mod 2

We need to find the possible values of L_X and L_NN that satisfy both.

Let me denote:

Equation 1: S + L_NN ≡ 1 mod 2
Equation 2: L_NN + L_X ≡ 1 mod 2

From equation 1, S + L_NN is odd. So S and L_NN must be of opposite parity.

From equation 2, L_NN + L_X is odd. So L_NN and L_X must be of opposite parity.

Therefore, combining both equations, S and L_X must have the same parity.

Because:

From equation 1: S ≡ 1 - L_NN mod 2
From equation 2: L_X ≡ 1 - L_NN mod 2

Therefore, S ≡ L_X mod 2. So the spin of the NN system and the orbital angular momentum of X must have the same parity.

Now, the problem asks which partial wave (orbital angular momentum state of X) is not permitted. The partial wave here refers to L_X. So we need to find which L_X (even or odd) is not allowed based on the above conditions.

But the problem gives options where the partial wave is given as a letter (like p, s, etc.), which I think corresponds to L_X. For example, 's' is L=0, 'p' is L=1, 'd' is L=2, 'f' is L=3, etc. So even L_X would be s, d, g,... and odd L_X would be p, f, j,...

From the parity condition, L_X must be such that L_X ≡ 1 - L_NN mod 2. But L_NN can be either even or odd. However, the problem gives that the final NN state has T(NN)=0. So T(NN) = 0 = (S + L_NN) + 1 mod 2. So S + L_NN ≡ 1 mod 2. Therefore, S and L_NN have opposite parity.

But the spin S of the NN system can be 0 or 1 (for two nucleons). Wait, no. The spin of the NN system can be 0 (singlet) or 1 (triplet). So S can be 0 or 1. Let's consider each case.

Case 1: S = 0 (singlet state)
Then from equation 1: 0 + L_NN ≡ 1 mod 2 → L_NN ≡ 1 mod 2 → L_NN is odd (1,3,5,...)
From equation 2: L_NN + L_X ≡ 1 mod 2 → since L_NN is odd, L_X must be even (0,2,4,...)
Therefore, L_X must be even.

Case 2: S = 1 (triplet state)
Then from equation 1: 1 + L_NN ≡ 1 mod 2 → L_NN ≡ 0 mod 2 → L_NN is even (0,2,4,...)
From equation 2: L_NN even + L_X ≡ 1 mod 2 → L_X must be odd (1,3,5,...)

So depending on whether the NN system's spin is 0 or 1, L_X must be even or odd. But the problem states that the final NN state has T(NN)=0. However, the isospin T of the NN system is determined by their initial state. Wait, the initial state's isospin is not given. The problem says the initial state is 1S0, which refers to the spatial and spin state, but not the isospin. So the initial isospin T_initial could be anything. But the problem states that the final NN state has T(NN)=0. Assuming the strong interaction conserves isospin (which it does for particles like pions), then T_initial = T_final = 0. Therefore, the initial state's isospin must be T_initial = 0. However, the initial state's spin is 0 (since 1S0), but isospin is a separate quantum number. So if the initial isospin is T_initial = 0, then the final state's T(NN) must also be 0. Therefore, the NN system's isospin T must be 0. But how does that affect the possible S and L of the NN system?

Wait, the isospin condition is given as T(NN) = S(NN) + L(NN) + 1 mod 2. Since T(NN) = 0, we have 0 = S + L + 1 mod 2 → S + L ≡ 1 mod 2. So regardless of the initial isospin, the final state must satisfy S + L ≡ 1 mod 2. But the problem states that T(NN) is 0 for the final state, so the condition applies.

So going back, the possible L_X depends on the parity of L_NN, which in turn depends on S. But since the initial state's isospin is T_initial = 0 (assuming conservation), but I'm not sure if that affects the spin or orbital angular momentum. Maybe I need to focus on the conditions derived earlier.

From the parity condition, L_X must be even if L_NN is odd, and odd if L_NN is even. But L_NN can be anything, as long as S + L_NN is odd. However, the problem gives us the final state's T(NN)=0, so S + L_NN ≡1 mod 2. Therefore, the parity of L_NN is opposite to that of S.

But the problem is asking which partial wave (L_X) is not allowed. So if L_X must be even or odd based on the conditions, then the partial wave with the opposite parity would be disallowed.

Wait, but which partial wave is not allowed? Let me look at the options again.

The options are:

A. 3D3 + f → NN state 3D3, X orbital angular momentum f (L=3, odd)
B. 1S0 + p → NN state 1S0, X p (L=1, odd)
C. 7D1 + p → NN state 7D1 (L=3, odd?), X p (L=1, odd)
Wait, but 3D3 is (2S+1)LJ notation. So 3D3 implies L=2 (since D is L=2), so the NN state is L=2 (even). Then the X's orbital angular momentum is f, which is L=3 (odd). So L_X=3 (odd).

Similarly, option A: NN state 3D3 (L=2, even) + X p (L=1, odd). So L_X=1 (odd). Then from the parity condition, L_NN + L_X ≡1 mod 2. L_NN=2 (even) + L_X=1 (odd) → even + odd = odd ≡1 mod 2. So condition satisfied. Then the parity equation: L_NN + L_X = 2 +1=3 → (-1)^3 = -1, initial parity is 1. So final parity would be 1 * (-1) * (-1)^{L_X} → Wait, initial parity is 1. The parity of the NN system is (-1)^{L_NN} = (-1)^2 = 1. The parity of X's orbital angular momentum is (-1)^{L_X} = (-1)^1 = -1. So total parity: 1 * (-1) * (-1)^1 = 1 * (-1) * (-1) = 1. Which matches the initial parity. So condition satisfied.

But the problem is asking which partial wave is not allowed. If all the options satisfy the conditions, then perhaps there's another condition I'm missing. Wait, but perhaps the isospin condition also affects the possible L_X.

Wait, the problem says that the initial state is 1S0, which refers to spin 0. The isospin of the initial state could be anything, but the problem states that the final state has T(NN)=0. However, if the initial state's isospin is T_initial, then the final state's T must be the same. But the problem doesn't specify the initial isospin. However, the problem says that the system emits a particle X of intrinsic parity -1. If X is a boson (like a pion), then its isospin could be 0 or 1. If it's a fermion, like a nucleon (isospin 1/2), but X's intrinsic parity is -1. Wait, but the problem just says X has intrinsic parity -1. So maybe X is a pseudoscalar particle, like a pion (which has T=0). So if X has T=0, then the final NN state's T must equal the initial NN state's T. But the problem states that the final NN state's T is 0. Therefore, the initial NN state's T must also be 0. But the initial state's 1S0 term refers to spin 0, but isospin could be anything. However, if X has T=0, then the initial state's T must equal the final state's T. Since the final state's T is 0, the initial state's T must be 0. Therefore, the initial NN state must have isospin T=0. But the initial state's 1S0 term could have any isospin, but perhaps the problem implies that T_initial = T_final = 0. Therefore, the initial NN state must have T=0. So the initial NN state's isospin must be 0. Therefore, the initial NN state is a singlet in isospin. So the initial NN state's spin is 0 (since it's 1S0) and isospin is 0. Then, when they emit X (T=0), the final NN state must also have T=0. So the condition T(NN) = S(NN) + L(NN) + 1 mod 2 still holds, with T=0. Therefore, S(NN) + L(NN) ≡1 mod 2.

But the problem is about the orbital angular momentum of X. Let's consider each option again.

Option A: 3D3 + f. The NN state is 3D3, which is L=2 (even). X has L=3 (f is L=3, odd). So L_NN=2 (even), L_X=3 (odd). Then L_NN + L_X = 2 + 3 =5 (odd), which satisfies the parity condition. Also, S + L_NN ≡1 mod 2. Since the NN state is 3D3, which is (2S+1)LJ. 3 implies 2S+1=3 → S=1. So S=1 (triplet state). Then S=1, L_NN=2. 1 + 2 =3 ≡1 mod 2. So condition satisfied. Therefore, this is allowed.

Option B: 1S0 + p. NN state is 1S0 (L=0, even). X has L=1 (p, odd). So L_NN=0 (even), L_X=1 (odd). L_NN + L_X=0+1=1 (odd). Parity condition satisfied. S of NN state is 0 (from 1S0). S=0, L_NN=0. 0 +0=0 ≡0 mod 2. Wait, but earlier condition is S + L_NN ≡1 mod 2. So 0 +0=0≡0≠1 mod 2. Therefore, this violates the condition. Therefore, this option is not allowed. Wait, but the NN state is 1S0, which has S=0. So S=0, L_NN=0. Then S + L_NN=0≡0 mod 2. Which does not satisfy S + L_NN ≡1 mod 2. Therefore, this option violates the condition. Therefore, this partial wave is not allowed.

Wait, but the problem states that the final NN state has T(NN)=0. So if the option B leads to S + L_NN=0, which would make T(NN)=0 +1=1 mod 2=1≠0. Therefore, this is not allowed. Therefore, option B is not permitted.

But let me check the other options.

Option C: 7D1 + p. The NN state is 7D1, which is (2S+1)LJ. 7 implies 2S+1=7 → S=3 (which is not possible for nucleons, since S can only be 0 or 1). Wait, hold on. For nucleon-nucleon system, the spin S can only be 0 or 1. So S=3 is impossible. Therefore, the NN state 7D1 is not possible. Therefore, this option is invalid. So option C is not a valid term symbol, but the problem is asking which partial wave is not permitted. However, the question says "the final NN state has T(NN) = 0". If the NN state is 7D1, which is impossible, then this option is invalid. So perhaps the answer is C? But the problem is about the orbital angular momentum of X, not the NN state. Wait, but the problem states "the final NN state has T(NN) = S(NN) + L(NN) + 1 (mod 2)". So the NN state must have S(NN) + L(NN) +1 ≡0 mod 2. So if the NN state is 7D1, which is invalid (since S=3 is impossible), then this option is invalid. Therefore, option C is not permitted. However, the problem is asking which partial wave is not permitted. The partial wave refers to X's orbital angular momentum. But if the NN state is impossible, then the entire option is invalid. Therefore, option C is invalid. But the problem is about the orbital angular momentum of X, not the NN state. So perhaps the answer is C, but I need to check.

Option D: 3P0 + s. The NN state is 3P0, which is L=1 (odd). X has L=0 (s, even). So L_NN=1 (odd), L_X=0 (even). Then L_NN + L_X=1 +0=1 (odd). Parity condition satisfied. S of NN state is 0 (from 3P0). So S=0, L_NN=1. S + L_NN=0+1=1≡1 mod 2. Which satisfies the condition. Therefore, this option is allowed.

So among the options, options B and C are not allowed. Option B violates the S + L_NN condition, and option C has an impossible NN state. However, the problem asks which partial wave is not permitted. The partial wave is X's orbital angular momentum. The options are:

A. f (L=3)
B. p (L=1)
C. p (L=1)
D. s (L=0)

Options B and C have L_X=1 (odd), which would require L_NN even. But for the NN state 1S0, S=0, L_NN=0. So S + L_NN=0+0=0≡0 mod 2. Which doesn't satisfy the condition. Therefore, option B is not allowed. Also, option C's NN state is impossible. Therefore, both B and C are not allowed. But the problem is asking for the partial wave (L_X) that is not permitted. However, the options might be considering only the parity condition, not the NN state's validity. For example, if the NN state is 1S0 (S=0), then S + L_NN must be odd (1 mod 2). Therefore, L_NN must be odd. But the NN state in option B is 1S0 (L=0, even). So L_NN=0 (even). Then S + L_NN=0 +0=0≡0 mod 2. Which violates the condition. Therefore, option B is not allowed. Similarly, option C's NN state has L=1 (odd), S=3 (invalid), so the NN state is impossible. Therefore, option C is invalid. However, the problem is about the partial wave of X, not the NN state. But the question says "the final NN state has T(NN) = 0". So the NN state must have T(NN)=0, which requires S + L_NN +1 ≡0 mod 2. Therefore, the NN state must have S + L_NN ≡1 mod 2. If the NN state's S is determined by the term symbol, then for term symbols like 1S0 (S=0), 3P0 (S=0), 7D1 (invalid), etc., the condition must hold. Therefore, option B's NN state is 1S0 (S=0), L_NN=0. Then S + L_NN=0 +0=0≡0 mod 2, which doesn't satisfy the condition. Therefore, option B is not allowed. Similarly, option C's NN state is 7D1 (S=3, impossible), so it's invalid. But since the problem is asking for the partial wave (L_X) that's not permitted, regardless of the NN state's validity, but the NN state's condition must be satisfied. Therefore, option B is forbidden because it violates the parity and angular momentum conditions, and option C's NN state is invalid. However, the problem is giving options where the NN state terms are given. Therefore, the correct answer is either B or C. But the problem states that the NN state has T(NN)=0. So if the NN state is 1S0 (option B), then S=0, L_NN=0. Then S + L_NN=0≡0 mod 2, which violates the condition. Therefore, option B is not permitted. Similarly, option C's NN state has L=1, but S=3 (invalid), so it's impossible. However, the problem is asking which partial wave is not permitted, so the answer is B or C. But the problem might be focusing on parity and angular momentum conservation. Let's think again.

The problem is about the partial wave (orbital angular momentum of X) being not permitted. The conditions are:

1. L_NN + L_X ≡1 mod 2 (parity condition)
2. S + L_NN ≡1 mod 2 (from T(NN)=0)

But the NN state's term symbol gives S and L_NN. Therefore, for each option, we need to check whether S + L_NN ≡1 mod 2.

Option A: NN state 3D3 (S=1, L=2). L_NN=2. Then S + L_NN=1 +2=3≡1 mod 2. So condition satisfied. Also, L_X=3 (odd). L_NN=2 (even). L_NN + L_X=2+3=5≡1 mod 2. Condition satisfied. Therefore, allowed.

Option B: NN state 1S0 (S=0, L=0). L_NN=0. S + L_NN=0+0=0≡0 mod 2. Not equal to 1. Therefore, condition not satisfied. Not allowed.

Option C: NN state 7D1 (S=3, L=1). But S=3 is impossible for nucleons. Therefore, invalid. So this option is invalid. However, if we ignore the NN state's validity (assuming it's allowed for some reason), then L_NN=1. S=3. S + L_NN=3 +1=4≡0 mod 2. Again, condition not satisfied. Therefore, even if the NN state were valid, this would not be allowed.

Option D: NN state 3P0 (S=0, L=1). L_NN=1. S + L_NN=0 +1=1≡1 mod 2. Condition satisfied. L_X=0 (even). L_NN=1 (odd) + L_X=0 (even)=1. Condition satisfied. Therefore, allowed.

Therefore, options B and C are not allowed. However, the problem is asking which partial wave is not permitted. The partial wave refers to X's orbital angular momentum. The options are:

A. f (L=3)
B. p (L=1)
C. p (L=1)
D. s (L=0)

Options B and C have L_X=1 (odd). For these, L_NN must be even. The NN state's S + L_NN must be odd. For option B, NN state 1S0 (S=0, L=0). So S + L_NN=0+0=0≡0 mod 2. Not allowed. For option C, NN state 7D1 (S=3, invalid). Even if S=3, L_NN=1. S + L_NN=3+1=4≡0 mod 2. Not allowed. Therefore, options B and C are invalid. However, the problem might be considering only the parity condition, not the NN state's validity. Alternatively, the problem might be considering that the NN state's term symbol is valid (which it isn't in the case of C), but the problem is asking about the partial wave. Alternatively, perhaps the problem is considering that the NN state must satisfy S + L_NN ≡1 mod 2, and the partial wave must be such that L_NN + L_X ≡1 mod 2. Therefore, if the NN state's L_NN is even, L_X must be odd. If L_NN is odd, L_X must be even. But the NN state's L_NN can be determined from the term symbol. For example, in option A, NN state is 3D3 (L=2, even). So L_NN is even, so L_X must be odd. Therefore, allowed. Option B: NN state 1S0 (L=0, even). L_X must be odd. But option B has L_X=1 (odd), but the condition S + L_NN=0 +0=0≡0 mod 2. Therefore, not allowed. Option C: NN state 7D1 (L=1, odd). L_X must be even. But option C has L_X=1 (odd), which is not allowed. Therefore, both B and C are invalid. However, the problem is asking which partial wave is not permitted. Since the answer must be one of the options, and the problem mentions that the final NN state has T(NN)=0, which requires S + L_NN ≡1 mod 2. Therefore, the NN state's S and L_NN must satisfy this. So for each option:

Option A: S=1, L_NN=2 → 1+2=3≡1 mod 2. Allowed.
Option B: S=0, L_NN=0 → 0+0=0≡0 mod 2. Not allowed.
Option C: S=3 (invalid), L_NN=1 → 3+1=4≡0 mod 2. Not allowed.
Option D: S=0, L_NN=1 → 0+1=1≡1 mod 2. Allowed.

Therefore, options B and C are invalid. However, the problem is asking which partial wave is not permitted. The partial wave is L_X, which must be even or odd based on the NN state's L_NN. But the answer needs to be one of the options. Since the problem states that the final NN state has T(NN)=0, the correct answer must be the one where the NN state's L_NN and S do not satisfy the condition. Therefore, the answer is option B or C. However, the problem provides four options, and only one is correct. Let's check the parity condition again.

The parity of the final state must be 1 (initial parity). The parity of the final state is (-1)^{L_NN + L_X} * (-1)^{L_X} = (-1)^{L_NN + 2L_X} = (-1)^{L_NN}.

Because 2L_X is even, so (-1)^{L_NN + 2L_X} = (-1)^{L_NN}.

This must equal the initial parity, which is 1. Therefore, (-1)^{L_NN} =1 → L_NN must be even. Therefore, the NN state's L_NN must be even. Therefore, the NN state must have even L. Let's check the options:

Option A: 3D3 → L=2 (even). Allowed.
Option B: 1S0 → L=0 (even). Allowed.
Option C: 7D1 → L=1 (odd). Not allowed.
Option D: 3P0 → L=1 (odd). Not allowed.

Therefore, options C and D have odd L_NN. Therefore, their parity would be (-1)^{odd} = -1, which would violate the parity condition. Therefore, options C and D are invalid. However, the problem states that the NN state has T(NN)=0. For options C and D, the NN state's L_NN is odd. Therefore, S + L_NN must be odd. For option C: S=3 (invalid), L_NN=1. 3+1=4≡0 mod 2. Not allowed. For option D: S=0, L_NN=1. 0+1=1≡1 mod 2. Allowed. So option D is allowed. Therefore, the only invalid options are C and potentially others.

But the problem is asking which partial wave is not permitted. The partial wave refers to X's orbital angular momentum. The parity condition requires that L_NN is even. Therefore, if the NN state's L_NN is even, then the partial wave (L_X) can be even or odd, but must satisfy L_NN + L_X ≡1 mod 2. Since L_NN is even, L_X must be odd. Therefore, the partial wave must be odd. Therefore, the partial wave (L_X) must be odd. The options are:

A. f (L=3, odd)
B. p (L=1, odd)
C. p (L=1, odd)
D. s (L=0, even)

Therefore, options A, B, C have L_X odd, which would require L_NN even. However, option D has L_X even, which would require L_NN odd. But the parity condition requires L_NN even. Therefore, option D is not allowed. Wait, no. The parity condition is (-1)^{L_NN} =1 → L_NN even. Therefore, L_X can be even or odd, but L_NN + L_X ≡1 mod 2. Since L_NN is even, L_X must be odd. Therefore, the partial wave must be odd. Therefore, options B, C, and A are odd. Option D's L_X is even, which would require L_NN odd. But parity condition requires L_NN even. Therefore, option D is not allowed. However, the problem states that the NN state has T(NN)=0, which requires S + L_NN ≡1 mod 2. If the NN state is 3P0 (option D), S=0, L_NN=1. Then S + L_NN=1≡1 mod 2. Therefore, condition satisfied. But the parity condition requires L_NN even. Therefore, option D's L_NN=1 (odd) violates the parity condition. Therefore, option D is not allowed. Therefore, the answer is option C? Wait, no. The problem is asking which partial wave is not permitted. The options are:

A. 3D3 + f → L_X=3 (odd)
B. 1S0 + p → L_X=1 (odd)
C. 7D1 + p → L_X=1 (odd)
D. 3P0 + s → L_X=0 (even)

But the parity condition requires L_X odd. Therefore, option D's L_X=0 (even) is not allowed. However, the problem states that the final NN state has T(NN)=0. For option D, NN state is 3P0 (S=0, L=1). Then S + L=0+1=1≡1 mod 2. Condition satisfied. But parity condition requires L_NN even, which is violated here. Therefore, option D is not allowed. Therefore, the answer is D. But wait, the problem is asking which partial wave is not permitted. The partial wave is L_X. If the parity condition requires L_X odd, then options A, B, C are allowed (odd), and D is not. Therefore, the answer is D. However, the problem also gives the option C's NN state as 7D1 + p, which is invalid because S=3 is impossible.

But the problem is about the partial wave (L_X) being not permitted. The parity condition requires L_X odd, so the not allowed partial waves are even. Only option D has even L_X. Therefore, the answer is D. However, the problem might be considering the NN state's validity. For example, if the NN state is 3P0 (option D), which has L=1 (odd), then L_NN=1. Then S + L_NN=0 +1=1≡1 mod 2. Condition satisfied. But parity condition requires L_NN even. Therefore, L_NN=1 is invalid, so option D is not allowed. Therefore, the answer is D. But I need to check again.

Alternatively, the problem might not consider the NN state's validity and only focus on the parity condition. The parity condition requires L_X odd. Therefore, the only allowed partial waves are options A, B, C (odd), and D is not allowed. Therefore, the answer is D. But the problem states that the NN state has T(NN)=0. Therefore, the NN state's L_NN must be even. Therefore, options with L_NN even are A and B. For these, L_X must be odd. Therefore, options A and B are allowed. Option C's NN state has L_NN=1 (odd), which violates the parity condition. Therefore, option C is not allowed. However, the problem is asking which partial wave is not permitted. The partial wave refers to X's L. The parity condition requires L_X odd. Therefore, options B and C have L_X odd (allowed), option D has L_X even (not allowed). Therefore, the answer is D.

Wait, but the problem's options are:

A. 1S0 -> L_X=3 (odd)
B. 1S0 -> L_X=1 (odd)
C. 7D1 -> L_X=1 (odd)
D. 3P0 -> L_X=0 (even)

Since the parity condition requires L_X odd, the not allowed partial wave is D. Therefore, the answer is D. However, the problem is about the final NN state having T(NN)=0. So the NN state's L_NN must be even. Therefore, the allowed NN states are those with even L_NN. Options A and B have even L_NN. For these, L_X must be odd. Therefore, options A and B are allowed. Options C and D have odd L_NN. Therefore, their parity conditions are violated. However, the problem is asking which partial wave is not permitted. The partial wave is L_X, which must be odd. Therefore, options A, B, C have odd L_X (allowed), D has even L_X (not allowed). Therefore, the answer is D.
</think>