ISOTOPIC SPIN OF LIGHT NUCLEI

A. Baz’ and Ya. Smorodinskii

Part I. THEORY

Chapter I. ISOTOPIC SPIN OF NUCLEONS

§ 1. Introduction

There are many reasons to believe that the nuclear forces acting between nucleons possess charge invariance. This means that the interactions of the three possible pairs of nucleons—proton–proton, neutron–neutron, and proton–neutron—when they are in identical states (in the sense of the dependence of the wave functions on coordinates and spins), are identical to one another.

However, despite the widespread acceptance that the hypothesis of charge invariance has received in nuclear physics, there is at present no direct proof of its validity.

Indeed, the interaction of two nucleons has been studied directly only in the phenomena of scattering of neutrons by protons and of protons by protons. It is well known, however, that in the investigated energy region (approximately up to 10 MeV) scattering associated with nuclear forces occurs only in states with \(l = 0\) (\(S\)-states). Therefore, from the analysis of these experiments one could only conclude that the neutron–proton and proton–proton interactions are similar only in the \({}^{1}S\) state (let us recall that, by virtue of the Pauli principle, two protons cannot be in the \({}^{3}S\) state). The participation in scattering of states with \(l > 0\) is so insignificant that in the region of low energies (up to \(\sim 10\) MeV) no conclusions can be drawn about the interaction in these states. The region of higher energies, to which many works have been devoted, represents a separate problem, connected, in particular, with the charge invariance of the interaction of nucleons with \(\pi\)-mesons (as well as with other mesons, which possibly play an essential role in the interaction between nucleons). However, even in the region

there is no exact proof of charge invariance at high energies, although the entire body of experiments available in this domain does not contradict the hypothesis.

Therefore, consideration of the structure of the spectra of light nuclei from the point of view of those regularities which follow from the adoption of the hypothesis of charge invariance acquires great importance. As it turns out, precisely the structure of nuclear spectra is at present the most convincing proof of the validity of the hypothesis under consideration, at least in the region of not too high energies.

In this review we set ourselves the task of showing what general properties of spectra are connected with the hypothesis of charge independence of nuclear forces, and what conclusions can be drawn concerning the accuracy of this hypothesis.

The review consists of two parts. In the first part we shall set forth the basic physical ideas on which the theory of isotopic spin is built. In doing so we shall avoid complicated mathematical questions, confining ourselves to not quite rigorous but transparent considerations. Theoretical questions have been treated in the recently published papers of Shapiro1 and Zel’dovich[^2], to which we refer the reader.

The second part contains a description of the levels of light nuclei and their analysis with respect to isotopic spin. A large part of the data on the levels has been taken from two papers on light nuclei by Ajzenberg and Lauritsen[^3] and by Endt and Kluyver[^4].

§ 2. Quantum characteristics of nuclear levels

As is well known, each level of a quantum system (and, in particular, of a nucleus) is characterized by a set of quantum numbers. These numbers are connected with the various symmetry properties of the system; depending on whether the symmetry under consideration is exact or not, one distinguishes exact and approximate quantum numbers. The former—such as the energy of the system and the angular momentum—are strictly conserved in any processes occurring in the system; the latter, generally speaking, may fail to be conserved, and are characterized by the fact that processes associated with their nonconservation have a substantially smaller probability. Examples of the latter may be the orbital and spin quantum numbers in atoms, whose conservation is valid only within the framework of the \(LS\)-coupling scheme. In addition to the exact quantum numbers mentioned above—the energy \(E\) and spin \(I\)—the levels of a system are also characterized by parity \(P\), arising from the invariance of the properties of the system under reflection of all coordinates in the origin.

Let us recall that from the fact that a reflection performed twice is the identity operation it follows that the wave func-

...of the system under a single reflection may either change sign or remain unchanged. In the first case one says that the system is “odd,” in the second, that it is “even.”

These three numbers \(E\), \(I\), and \(P\) exhaust the exact quantum numbers.

Let us now turn to the “inexact” quantum numbers. The introduction of such numbers is connected with specific assumptions about the properties of a system of nucleons. The approximate nature of these assumptions is the source of the inexactness of the quantum numbers.

Thus, the model of nuclear shells in the \(jj\)-coupling scheme leads to the appearance of a whole series of inexact quantum numbers—the angular momenta and parities (or, what is the same thing, the total and orbital angular momenta) of the individual nucleons in the nucleus. The inexactness of such quantum numbers is obvious: it is determined by the neglect of the interaction between the nucleons.

The concept of isotopic spin has the same inexact character.

The origin of this concept is connected with the hypothesis, discussed by us, of charge invariance of nuclear forces. Such a hypothesis reduces to the requirement that the Hamiltonian (or Lagrangian) function of the system remain unchanged when any proton is replaced by a neutron, or, conversely, any neutron by a proton. This means, in particular, that the Hamiltonian function of the system must be symmetric with respect to the simultaneous interchange of the coordinates and spins of any particles.

The condition of invariance of the Hamiltonian of the system with respect to arbitrary permutations leads to the imposition of certain conditions on the wave function of the nucleus. Only in the simplest case of a system of two particles do these conditions reduce to the condition of symmetry or antisymmetry of the wave function. In the general case they are formulated more complicatedly. These conditions can be described in the most compact and transparent way in the form of the theory of isotopic spin, and charge invariance itself—in the form of the law of conservation of isotopic spin.

§ 3. Isotopic spin of nucleons and systems of nucleons

Within the framework of the hypothesis of charge invariance, the neutron and the proton are regarded as two charge states of one and the same particle—the nucleon. It then turns out that such a unification of the neutron and the proton into one particle makes it possible to formulate, in a very compact way, all the consequences following from the hypothesis of charge invariance and, moreover, this device is very convenient in classifying the states of a system of neutrons and protons.

The corresponding apparatus is constructed as follows. In accordance with what was said above, the nucleon must be described by a two-component wave function, which may be written in the form of a column. In this notation the proton and neutron states of the nucleon are represented respectively as

\[ \psi_p=\begin{pmatrix}1\\0\end{pmatrix},\qquad \psi_n=\begin{pmatrix}0\\1\end{pmatrix}. \]

Let us introduce the operator \(\tau_1\), which transforms a neutron into a proton. By definition, \(\tau_1\) must have the property that

\[ \tau_1\psi_n=\psi_p,\qquad \tau_1\psi_p=0. \]

It is easy to see that this operator can be represented in the following form:

\[ \tau_1=\begin{pmatrix}0&1\\0&0\end{pmatrix}. \]

Similarly, the operator \(\tau_2\), possessing the property

\[ \tau_2\psi_p=\psi_n,\qquad \tau_2\psi_n=0, \]

may be written as

\[ \tau_2=\begin{pmatrix}0&0\\1&0\end{pmatrix}. \]

Next we introduce the operators \(\tau_\xi,\ \tau_\eta,\ \tau_\zeta\) according to the following equalities:

\[ \tau_\xi=(\tau_1+\tau_2)= \begin{pmatrix}0&1\\1&0\end{pmatrix}; \qquad \tau_\eta=i(\tau_2-\tau_1)= \begin{pmatrix}0&-i\\ i&0\end{pmatrix}; \qquad \tau_\zeta= \begin{pmatrix}1&0\\0&-1\end{pmatrix}. \]

The operators \(\tau_\xi,\ \tau_\eta,\ \tau_\zeta\) defined in this way coincide with the Pauli matrices known from spin theory and, consequently, possess the same formal properties as the latter. In particular, it is not difficult to verify that the operator \(\frac12\tau_\zeta\) acts on the wave functions of the neutron and proton in the following way:

\[ \frac12\tau_\zeta\psi_p=\frac12\psi_p,\qquad \frac12\tau_\zeta\psi_n=-\frac12\psi_n. \]

All the relations of the theory of isotopic spin are identically analogous to the relations of the nonrelativistic theory of spin. The role of the state with spin projection \(\frac12\) is played by the proton state of the nucleon; to the state with spin projection \(-\frac12\) there corresponds the neutron state; and the role of the operators of the spin projections on the Cartesian coordinate axes is played by the operators \(\frac12\tau_\xi,\ \frac12\tau_\eta,\) and \(\frac12\tau_\zeta\). Hence there arises the name for the operators \(\frac12\tau_\xi,\ \frac12\tau_\eta\) and \(\frac12\tau_\zeta\): projections of the isotopic-spin operator of the nucleon; and the two charge states of the nucleon are regarded as states with different projections of the nucleon isotopic spin \(\frac12\tau\) on the \(\zeta\)-axis.

In order to carry the analogy between ordinary and isotopic spin still further, a three-dimensional isotopic space is formally introduced, in which \(\tau_\xi\), \(\tau_\eta\), and \(\tau_\zeta\) may be regarded as components of the vector operator \(\tau\). The meaning of introducing isotopic space is that the operator \(\frac{1}{2}\tau\) thereby acquires the intuitive meaning of the angular-momentum operator in this space; the nucleon must then be regarded as a particle with isotopic spin \(1/2\), and the problem of classifying the states of a system consisting of neutrons and protons reduces to the familiar problem of classifying the states of identical particles with spin \(1/2\), encountered in the theory of the electron shell of the atom.

Since, within the hypothesis of charge invariance, the interactions \((pp)\), \((pn)\), and \((nn)\) are equal, neutrons and protons in a nucleus differ from one another only because of the Pauli principle, which forbids such states of the system in which two neutrons or two protons are in one and the same quantum state. On the other hand, the Hamiltonian of the system is symmetric with respect to the interchange of the spins and spatial coordinates of any two particles. If, therefore, neutrons and protons are regarded as different states of a single particle—the nucleon—then it is evident that charge invariance leads to the requirement that the Hamiltonian of the nucleon system be invariant with respect to the interchange of all five coordinates \((x y z s_z \tau_\zeta)\) of any two nucleons. It follows directly from this that the wave function of any such system must either remain completely unchanged under interchange of the coordinates of any two nucleons (a symmetric wave function), or must change sign under it (an antisymmetric wave function). However, it is known that neutrons and protons separately obey the Pauli principle, i.e., under interchange of the spatial coordinates and spins of two neutrons or of two protons the wave function of the system changes sign. Since \(\tau_\zeta\) does not change under such an interchange \(\left(\frac{1}{2}\tau_\zeta=-\frac{1}{2}\right.\) for all neutrons and \(\left.\frac{1}{2}\tau_\zeta=\frac{1}{2}\right.\) for all protons), such an interchange may be regarded as an interchange of all five coordinates of two nucleons. But if the wave function changes sign under interchange of all five coordinates of at least one pair of nucleons, then it is easy to show that it must be antisymmetric with respect to interchange of all five coordinates of any pair of nucleons. This statement is usually called the generalized Pauli principle.

Thus, a system consisting of neutrons and protons, under the assumption of charge invariance of nuclear forces, may be regarded as a system of identical particles—nucleons (with a charge degree of freedom), obeying Fermi statistics.

Next there arises the question of the classification of the energy levels of such a system. For this purpose let us introduce the concept of the isotopic spin of a system of nucleons. The isotopic spin of a system of nucleons \(T\) is defined as the sum of the isotopic spins

\[ \mathbf{T}=\sum_{i=1}^{N}\frac{1}{2}\boldsymbol{\tau}^{(i)}, \]

where the summation is over all nucleons. It is clear that the operator \(\mathbf{T}\), thus defined, is a vector in isotopic space and possesses, just as \(\frac{1}{2}\boldsymbol{\tau}\) does, all the properties of angular momentum. In particular, when adding the isotopic spins of nucleons the usual rules for the addition of angular momenta are applicable*), and, for example, in the case of a system of two nucleons, \(T\) may take the values 0 and 1, in the case of three nucleons—\(\frac{1}{2}\) and \(\frac{3}{2}\), in the case of four—0, 1, and 2, and so on. It is easy to see what physical meaning states with different isotopic spins have. For this purpose we note that, according to the definition of \(\mathbf{T}\) as a vector, \(T\) is in any case not smaller than its projection on the axis: \(T\geq |T_\xi|\). But the expression for \(T_\xi\) can be written in the following form:

\[ T_\xi=\sum_{i=1}^{A}\frac{1}{2}\tau_\xi^i=\frac{1}{2}(Z-N), \]

where \(Z\) is the number of protons, \(N\) the number of neutrons, and \(A\) the total number of nucleons, \(A=N+Z\). Thus, \(T_\xi\) is equal to one half of the neutron excess of the nucleus, taken with the opposite sign. (In the literature the old Wigner sign convention is often used, in which \(\frac{1}{2}\tau_\xi\) of the neutron is equal to \(+\frac{1}{2}\). In this normalization the signs are changed in the other formulas as well.)

It follows from this that if the isotopic spin of some state is \(T\), then this state can be realized only in such systems for which \(|Z-N|\leq 2T\). Thus, for example, states with \(T=0\) can be realized only when \(Z=N\), while states with \(T=1\) when \(Z=N\) or \(Z=N\pm1\).

As was noted above, the introduction of isotopic spin makes it possible to study a system of nucleons by the same methods as a system of identical particles with spin \(\frac{1}{2}\). In particular, one can without particular difficulty generalize the methods of constructing the wave function of a system of electrons to the case of systems possessing isotopic spin. We shall not, however, occupy ourselves with this, since

\[ \text{*) In quantum mechanics it is shown that the law of addition of quantum vectors is a simple consequence of the commutation rules.} \]

we shall not be interested in the concrete calculation of the numerical characteristics of the levels, for which the exact form of the wave functions is needed. For the purposes of our review it is sufficient to restrict ourselves to the vector model.

§ 4. Conservation law for isotopic spin

Consider a system of \(Z\) protons and \(N\) neutrons, in some stationary state, and suppose that the hypothesis of charge invariance of the nuclear forces is strictly fulfilled. As was already said above, this assumption means that the properties of the system remain unchanged when any neutron is replaced by a proton, or conversely. Let us now replace one neutron by a proton. A new system is thereby obtained, but its Hamiltonian is exactly equal to the old Hamiltonian. Therefore the stationary states of both systems must be found from the solution of one and the same Schrödinger equation

\[ H\psi = E\psi, \]

and the only difference arises because of the Pauli principle, owing to which some of the states possible in one system will be forbidden in the other. Those states which can be realized both in the one and in the other system will, obviously, possess completely identical properties (they will have the same energy, angular momentum, parity, etc.).

To understand what all this means from the point of view of isotopic spin, let us note the following. When one neutron is replaced by a proton, the projection of the isotopic spin of the system changes by one unit: \(T_\zeta \to T_\zeta + 1\) (\(T_\zeta \to T_\zeta - 1\), if a proton is replaced by a neutron), i.e. such a replacement corresponds to a certain rotation in isotopic space. Consequently, the circumstance that the Hamiltonian of the system remains unchanged under this replacement may be regarded as the invariance of the Hamiltonian of a system of nucleons with respect to rotations in isotopic space. It follows immediately from this that the total isotopic spin of the system must be conserved. The proof in this case is carried out in complete analogy with the proof of the conservation law for the ordinary angular momentum.* Thus we arrive at a very important law: the isotopic spin of a system

* We have given a non-rigorous but intuitive derivation of the conservation law for isotopic spin. It can be derived rigorously as follows: the operators for the transformation of a neutron into a proton and of a proton into a neutron, \(\tau_2\) and \(\tau_1\), and all the more \(\tau_\xi\), under the assumption of charge invariance, commute with the Hamiltonian. Therefore the operators \(\tau_\xi\), \(\tau_\eta\), \(T_\xi\), \(T_\eta\), and \(T_\zeta\), which are linear combinations of \(\tau_1^{(i)}\), \(\tau_2^{(i)}\), commute with the Hamiltonian of the system. Hence it follows at once that \(T\) is conserved.

nucleons, under the assumption of charge invariance of nuclear forces, is an integral of motion.

A consequence of this conservation law is that each stationary (in the quantum-mechanical sense) state of a system of nucleons must possess a definite isotopic spin. This follows directly from the general theorem stating that any two commuting operators must have a common system of eigenfunctions. Applying this theorem to the case in which the commuting operators are the isotopic spin operator \(T\) and the Hamiltonian \(H\) of the system, we obtain the statement formulated above.

Chapter II. SELECTION RULES FOR ISOTOPIC SPIN

§ 5. Introduction

The law of conservation of isotopic spin leads to definite selection rules in various nuclear reactions. Two principal cases may be distinguished:

a) only nucleons take part in the reaction, for which it is known that their total isotopic spin is conserved. In this case we directly appeal to the law of conservation of isotopic spin and obtain that the entire nuclear process must proceed in such a way that at each of its stages the isotopic spin of the system is equal to the initial isotopic spin;

b) not only nucleons but also other particles (\(\beta\)-particles, \(\gamma\)-quanta) take part in the reaction, the emission of which changes the isotopic spin of the system, so that the law of conservation of isotopic spin cannot be applied directly. In this case, however, it turns out that, for certain relations between the isotopic spins of the initial and final states of the nucleon system, the matrix elements corresponding to emission or absorption vanish identically. The conditions under which this occurs depend on the specific form of the interaction and determine the selection rules for isotopic spin in this case.

Let us consider these cases in more detail.

§ 6. Reactions in which only nucleons participate

In the interaction of heavy particles (neutrons, protons, deuterons, etc.) with light nuclei, one may neglect the Coulomb interaction in comparison with the specific nuclear forces. Therefore it may be considered that, in this region of nuclei, the hypothesis of charge invariance of nuclear forces is applicable with sufficient accuracy. Hence a number of interesting consequences follow:

a) Since the isotopic spin of deuterons and \(\alpha\)-particles is equal to zero, in reactions of the type \((dd)\), \((d\alpha)\), \((\alpha d)\), and \((\alpha\alpha)\) the initial and final states of the nucleus must have one and the same isotopic spin. Moreover, if the reaction proceeds through an intermediate nucleus, the latter can be formed only in those states whose isotopic spin is equal to the isotopic spin of the initial nucleus. Thus, reactions of this type are a means for determining the isotopic spins of various states of nuclei.

In the most interesting case, when the initial nucleus has isotopic spin equal to zero—which, for example, occurs for the ground states of almost all light nuclei of the type \(2n\)—we find that in reactions of this type formation of the final nucleus in a state with \(T \ne 0\) is impossible. In addition, in these reactions only those levels of the intermediate nucleus for which \(T=0\) can appear. Thus, for example, in the reaction \(\mathrm{O}^{16}(d\alpha)\mathrm{N}^{14}\) it has been found that, at all deuteron energies, the group of \(\alpha\)-particles corresponding to the excited state of \(\mathrm{N}^{14}\) with energy \(E=2.31\ \mathrm{MeV}\) is absent. Since the ground state of the nucleus \(\mathrm{O}^{16}\) has \(T=0\), and no other selection rules (in angular momentum or parity) can explain this prohibition, it follows at once from the law of conservation of isotopic spin that the level of \(\mathrm{N}^{14}\) with \(E=2.31\ \mathrm{MeV}\) must be assigned isotopic spin \(T \ge 1\). And indeed, from independent considerations it is known\(^5\) that the isotopic spin of this level is \(T=1\).

Similarly, if initially the nucleus had \(T=\frac{1}{2}\), then in reactions \((dd)\), etc., only levels with \(T=\frac{1}{2}\) will appear in the intermediate nucleus, and final nuclei can be formed only in states with \(T=\frac{1}{2}\).

b) If the intermediate nucleus formed in some reaction is in a state with a definite isotopic spin \(T\), then the decay of this state must occur in such a way that the vector sum of the isotopic spins of the particles after the decay is equal to \(T\). Thus, if the intermediate nucleus is in a state with \(T=1\), then emission of an \(\alpha\)-particle or a deuteron is possible only in the case when the nucleus thereby formed is also in a state with \(T=1\). For example, in the reaction \(\mathrm{N}^{13}(p\alpha)\mathrm{C}^{12}\) the excited state of the intermediate nucleus \(\mathrm{O}^{16}\) with excitation energy \(E=12.95\ \mathrm{MeV}\) does not appear at all. At the same time, from energy considerations it is known that in this reaction the nucleus \(\mathrm{C}^{12}\) can be formed only in states with \(T=0\) (the first state with \(T=1\) in \(\mathrm{C}^{12}\) has a very large excitation energy, \(\sim 15\ \mathrm{MeV}\)). Comparing these data, it is easy to conclude that the isotopic spin of the level of \(\mathrm{O}^{16}\) with \(E=12.95\ \mathrm{MeV}\) must be equal to unity (higher \(T\) are impossible for energy reasons).

§ 7. Selection Rules in β-Decay

In calculating β-decay probabilities one has to deal with the calculation of matrix elements of operators that depend on the isotopic spins of nucleons. It turns out that many properties of such matrix elements can be obtained in a general form. For this purpose we use the fact that isotopic-spin operators possess the same formal properties as the operators of the ordinary angular momentum. In particular, they satisfy the same commutation rules as do the components of the angular-momentum vector. On the other hand, it is known that the commutation rules make it possible to obtain a number of general formulas for matrix elements of various combinations of the components of the angular-momentum vector. Since the formulas thereby obtained are based only on the commutation rules, they are fully applicable also to analogous matrix elements of isotopic-spin operators.

We first give a summary of the selection rules obtained in this way for some operators encountered in the calculation of probabilities of β- and γ-transitions.

a) If the operator \(F\) is invariant with respect to rotations in isotopic space (this occurs, for example, when \(F\) does not depend at all on the isotopic spin of the nucleons), then the matrix elements of \(F\) between states with definite isotopic spins \(T, T'\) and their projections \(T_\zeta, T'_\zeta\) satisfy the following conditions:

\[ (TT_\zeta|F|T'T'_\zeta) \ne 0,\quad \text{only if } T=T';\; T_\zeta=T'_\zeta . \]

b) If the operator \(P_\zeta\), under rotations in isotopic space, transforms as the \(\zeta\)-component of a vector (the simplest example of such an operator is the \(T_\zeta\)-operator of the \(\zeta\)-projection of the isotopic spin of the nucleus), then the following selection rules hold: \((TT_\zeta|P_\zeta|T'T'_\zeta)\ne0\), only if \(T-T'=\pm1;\; T_\zeta=T'_\zeta\), or if \(T=T';\; T_\zeta=T'_\zeta\ne0\).

c) If the operator \(P_1\), under rotations in isotopic space, transforms as \(P_1=(T_\xi+iT_\eta)\), then the following selection rules hold for the matrix elements of \(P_1\): \((TT_\zeta|P_1|T'T'_\zeta)\ne0\), only if \(T-T'=0,\ \pm1;\; T_\zeta=T'_\zeta+1\).

d) If the operator \(P_2\), under rotations in isotopic space, transforms as \(P_2=(T_\xi-iT_\eta)\), then \((TT_\zeta|P_2|T'T'_\zeta)\ne0\), only if \(T-T'=0,\ \pm1;\; T_\zeta=T'_\zeta-1\).

Let us now turn to β-decay proper. In β-decay a nucleon passes from one charge state into another (in \(\beta^-\)-decay a neutron passes into a proton, while in \(\beta^+\)-decay—

(proton into neutron). Correspondingly, the operator describing the process of β-decay has the following form:

\[ \mathfrak{M}_{1}=\sum_i B_i(x,y,z,s_z)\,\tau_1^{\,i}\quad(\beta^{-}\text{-decay}), \]

\[ \mathfrak{M}_{2}=\sum_i B_i(x,y,z,s_z)\,\tau_2^{\,i}\quad(\beta^{+}\text{-decay}), \]

where \(\tau_1\) and \(\tau_2\) are, respectively, the operators for the transformation of a neutron into a proton and, conversely, of a proton into a neutron, while the operators \(B_i\) do not depend on the coordinates of the isotopic spin (in the standard notation of the theory of β-decay, \(B_i=1\) for the scalar and vector variants, \(B_i=\sigma_i\) for the tensor and pseudovector variants, and \(B_i=\beta_i\gamma_5\) for the pseudoscalar variant). The summation is carried out over all nucleons of the nucleus.

From the expressions for \(\mathfrak{M}_1\) and \(\mathfrak{M}_2\) it is seen that, under rotations in isotopic space, they transform respectively as \(\tau_\xi \pm i\tau_\eta\). Therefore we may immediately conclude that a β-transition is possible only between such states whose isotopic spins \(T\) and \(T'\) differ by no more than one unit: \(T-T'=0\pm1\). Here \(\Delta T=0\) in the case of the Fermi selection rules, and \(\Delta T=0,\pm1\) in the case of the Gamow—Teller selection rules. Indeed, the Fermi matrix elements (the scalar and vector variants of the theory of β-decay) in the nonrelativistic case reduce to matrix elements of the operator

\[ \sum_i \tau_1^{(i)}=T_\xi+iT_\eta \quad \left( \text{or } \sum_i \tau_2=T_\xi-iT_\eta \text{ in the case of } \beta^+\text{-decay} \right), \]

which commutes with \(T^2\). Therefore, in such a transition \(T\) is conserved.

In the case of the Gamow—Teller matrix elements (the tensor and pseudovector variants), which reduce to matrix elements of the operator

\[ \sum_i \sigma_i\tau_1^{(i)} \quad \left( \text{or } \sum_i \sigma_i\tau_2^{(i)} \right), \]

no additional restrictions arise, and the usual selection rule for operators of this type is fulfilled: \(\Delta T=0,\pm1\).

§ 8. Selection rules for \(\gamma\)-radiation

Analogously to the selection rules for β-decay, one can obtain selection rules for \(\gamma\)-transitions. The transition operator (the Hamiltonian of the interaction of the nucleons with the electromagnetic field) in this case has the form (we neglect the very weak interaction associated with the magnetic moments of the neutron and proton)

\[ H=\sum_i \frac{e}{c}\cdot\frac{1}{2}\left(1+\tau_\zeta^{(i)}\right)\mathbf{v}_i\mathbf{A}(\mathbf{r}_i), \]

where \(\mathbf v_i, \mathbf r_i\) are the velocity and coordinate of the \(i\)-th nucleon, \(\mathbf A\) is the vector potential of the electromagnetic field, and the summation is carried out over all nucleons. In this formula, into which all nucleons enter in a completely symmetric way, the fact that, owing to the absence of charge, the neutron does not interact with the electromagnetic field is automatically taken into account. This is achieved by introducing the operator

\[ \frac{1}{2}\left(1+\tau_\zeta^{(i)}\right), \]

which, when acting on the wave function of a nucleon, is equal to zero or to unity depending on the charge state in which the nucleon is found (zero in the case of a neutron and unity in the case of a proton). The operator \(H\) can be rewritten as the sum of two parts:

\[ H=H_0+H_1, \]

where

\[ H_0=\sum_i \frac{e}{2c}\,\mathbf v_i \mathbf A(\mathbf r_i); \qquad H_1=\sum_i \frac{e}{2c}\,\mathbf v_i \mathbf A(\mathbf r_i)\cdot \tau_\zeta^{(i)}; \]

\(H_0\) does not depend on the isotopic spin of the nucleons and is therefore a scalar in isotopic space. Hence there follow the selection rules for \(\gamma\)-radiation associated with this part of the interaction Hamiltonian: \(\Delta T=0\). The second term, \(H_1\), transforms under rotations in isotopic space as the \(\zeta\)-component of a vector. Therefore, for radiation associated with this part of the interaction operator, the following selection rules hold:

\[ \Delta T=0,\ \pm 1 \quad \text{for } T_\zeta \ne 0, \]

\[ \Delta T=\pm 1 \quad \text{for } T_\zeta=0, \]

i.e., in the case of nuclei with \(T_\zeta=0\) (\(N=Z\)), \(H_1\) can cause transitions only between levels with different isotopic spins. On the other hand, it is evident that \(H_0\) cannot lead to dipole electric transitions (\(E1\)), since in its form this part of the Hamiltonian coincides with the Hamiltonian describing a system of identical particles with charge \(e/2\), and, as is known, such a system cannot emit dipole electric radiation. Hence there follows an important rule: in nuclei with \(N=Z\), dipole transitions between levels with the same isotopic spin are impossible.

This conclusion is fully confirmed by experiment. Indeed, it was found that in nuclei with \(T_\zeta=0\) (for example, \(B^{10}\), \(N^{14}\)) \(E1\)-transitions between levels with identical isotopic spins are forbidden, whereas in neighboring nuclei with \(T_\zeta\ne0\) (\(Be^{10}\), \(C^{14}\)) no prohibition with respect to \(T\) was observed.

It is necessary, however, to note that the prohibitions of \(E1\)-transitions by isotopic spin are not absolutely strict. They reduce the probability of transitions by several orders of magnitude, but do not forbid them completely. There are two reasons for this. First, allowance for spin interactions leads to the possibility of \(E1\)-transitions and, second, each nuclear state contains admixtures of states with other isotopic spins, so that an \(E1\)-transition can occur owing to the presence of admixtures.

All the conclusions of this paragraph are equally applicable both to processes of emission of \(\gamma\)-quanta with the transition of the nucleus to a lower state and to processes of absorption of \(\gamma\)-quanta with the subsequent disintegration of the nucleus. In reactions of the latter type a number of characteristic features arise. Consider, for example, nuclei of the \(4n\) type. In such nuclei the ground state has isotopic spin \(T=0\), while the first state with \(T=1\) is found only at an energy \(\sim 12\text{–}15\) MeV. This leads to the existence of a threshold for the capture of electric dipole quanta, and therefore intense capture of \(\gamma\)-quanta and the subsequent disintegration of the nucleus become possible only at energies greater than \(\sim 15\) MeV.

In some cases the threshold is located still higher. Thus, for example, in the reaction \(\mathrm{C}^{12}(\gamma\alpha)\mathrm{Be}^{8}\) the first excited state of the \(\mathrm{C}^{12}\) nucleus with \(T=1\) is at the energy \(E=15.2\) MeV, and the disintegration of the nucleus from this level into an \(\alpha\)-particle and \(\mathrm{Be}^{8}\), which is in the ground state, is energetically possible. However, in the ground state of \(\mathrm{Be}^{8}\) \(T=0\), and therefore such a disintegration is forbidden by the law of conservation of isotopic spin. A reaction with emission of an \(\alpha\)-particle will be allowed only if the energy of the \(\gamma\)-quanta is sufficient for the formation of \(\mathrm{Be}^{8}\) in an excited state with \(T=1\). This corresponds to a \(\gamma\)-quantum energy \(E\sim 26\) MeV. Thus, in this reaction the threshold energy for \(E1\)-capture reaches the value \(\sim 26\) MeV.

Analogous prohibitions also arise under irradiation by \(\gamma\)-quanta of not very high energy of nuclei\(^7\) of the type \(N=Z+1;\ A=4n+3\). Since in the ground state of such nuclei the isotopic spin is \(T=1/2\), then, on absorbing \(\gamma\)-quanta, the nucleus can pass into states with \(T=1/2\) or \(3/2\). Excited states with \(T=1/2\) may decay in two ways: by emitting either a neutron or a tritium nucleus. Indeed, the nuclei formed in the decay may, according to the law of conservation of isotopic spin, have \(T=0\) or \(1\) (the isotopic spin of tritium is \(T=1/2\)), so that no prohibitions by isotopic spin arise. If, however, the excited state of the nucleus has \(T=3/2\), then the final nucleus may have isotopic spin equal to 1 (or 2), but cannot have isotopic spin equal to zero. Therefore its decay can occur only when the birth of the nucleus in a state

$T=1$ is energetically allowed. This leads to the fact that the emission of tritium proves impossible.

Indeed, after the emission of a neutron there remains an odd-odd nucleus with $N=Z$, and in such nuclei the states with $T=1$ lie very close to the ground state. In the case of emission of a tritium nucleus, however, there must remain an even-even nucleus with $N=Z$, in which all lower-lying states have $T=0$ (the first state with $T=1$ lies approximately at $E\sim 12—15$ MeV), and therefore there is not enough energy to form the residual nucleus in a state with $T=1$. Thus, at not very high excitation energies of the initial nucleus, from levels with $T=3/2$ only neutron emission is possible, while from a level with $T=1/2$ both neutron and tritium emission are possible. In this way it is possible to determine the isotopic spin of excited states of odd-even nuclei of the type $A=4n+3$. It was precisely in this way found that in $\mathrm{Li}^7$ the first excited state with $T=3/2$ lies at $E=9.3$ MeV.

Chapter III. ACCURACY OF ISOTOPIC SPIN

In the preceding chapters we completely neglected those properties of neutrons and protons which, in fact, make it possible to distinguish these particles (mass, charge, magnetic moments). In reality neutrons and protons are not entirely equivalent; in particular, neutrons are not subject to the action of Coulomb forces, whereas in the interaction of two protons it is necessary, in addition to nuclear forces, also to take into account the interaction of their charges. This leads to the fact that the Hamiltonian of a system of nucleons is, generally speaking, not charge-invariant and, consequently, isotopic spin cannot be regarded as an exact quantum number. However, in light nuclei, where the Coulomb interaction is small in comparison with nuclear forces (below we shall establish the criterion of smallness), the charge-noninvariant terms in the Hamiltonian may be regarded as a small addition to the “unperturbed” Hamiltonian, which is charge-invariant, and ordinary perturbation theory may be used to take them into account. In this approximation the total isotopic spin will no longer be conserved, so that the states of a system of nucleons will already be mixtures of states with different isotopic spins; but as long as we may regard the charge-noninvariant terms in the Hamiltonian only as a small addition, only one value of the isotopic spin (the unperturbed state) will play the dominant role in this mixture. The admixtures in this case are small, and the isotopic spin still retains its meaning as an approximate quantum number characterizing the different states of a system of nucleons.

§ 9. Separation of charge-noninvariant terms

The exact Hamiltonian of the nucleus may be written in the following form:

\[ H=V_0+\frac14\sum_i p_i^2 \left(\frac{1+\tau_\zeta^{(i)}}{m_p}+\frac{1-\tau_\zeta^{(i)}}{m_n}\right)+ \]

\[ +\sum_{i>k}\frac{e^2}{4r_{ik}} \left(1+\tau_\zeta^{(i)}\right)\left(1+\tau_\zeta^{(k)}\right), \]

where \(V_0\) is the term describing the nuclear interaction of nucleons (it is assumed to be charge-invariant), while the second and third terms are, respectively, the kinetic energy of the nucleons and the energy of the Coulomb interaction. This expression may be rewritten in the following way, by separating from the kinetic-energy operator its charge-invariant part:

\[ H=V_0+\sum_i\frac{P_i^2}{4} \left(\frac1{m_p}+\frac1{m_n}\right)+ \]

\[ +\frac{m_n-m_p}{m_n}\sum_i\frac{P_i^2}{2m_p}\tau_\zeta^{(i)} +\sum_{i>k}\frac{e^2}{4r_{ik}} \left(1+\tau_\zeta^{(i)}\right)\left(1+\tau_\zeta^{(k)}\right)= \]

\[ =H_0+\frac{m_n-m_p}{m_n}\sum_i\frac{P_i^2}{2m_p}\tau_\zeta^{(i)}+ \]

\[ +\sum_{i>k}\frac{e^2}{4r_{ik}} \left(1+\tau_\zeta^{(i)}\right)\left(1+\tau_\zeta^{(k)}\right) =H_0+v_1+v_2, \]

where \(H_0\) is the charge-invariant part of the Hamiltonian, while \(v_1\) and \(v_2\) are respectively the charge-noninvariant part of the kinetic-energy operator and the Coulomb-energy operator. In the expression obtained, \(H_0\) is the “unperturbed” Hamiltonian, and \(v_1\) and \(v_2\) are additions which lead to the isotopic spin ceasing to be a quantum number. This means that the wave function of the nucleus is a sum of wave functions corresponding to different values of the isotopic spin. We shall assume that the principal role is played by some one value of the isotopic spin, while the remaining functions constitute a small “admixture.” Such an assumption is legitimate, since we are interested in the question of the accuracy of isotopic spin in the region of light nuclei, in which the charge-noninvariant part is a small perturbation.

Thus let us consider a system of states with given angular momentum \(J\), parity \(P\), and isotopic spin \(T\). Denote

the wave functions of these states by

\[ \psi_m \quad (m=0,\,1,\ldots). \]

These states differ in isotopic spin and, possibly, in some other quantum numbers*) and are eigenfunctions of the unperturbed operator \(H_0\). Let the ground state be characterized by the wave function \(\psi_0\). Then the exact wave function of the nucleus has the form

\[ \psi=\psi_0+\sum \alpha_{0m}\psi_m . \]

The coefficients \(\alpha_{0m}\) are found from the well-known formula of perturbation theory. We are interested only in the squares of their moduli:

\[ |\alpha_{0m}|^2= \left| \frac{(\psi_0|v|\psi_m)}{(E_0-E_m)} \right|^2, \]

where the numerator contains the matrix element of the perturbation \(v=v_1+v_2\), calculated with the aid of the wave functions of the unperturbed operator \(H_0\).

It is customary to characterize the degree of “purity” of the state by the “fraction of admixture” (Radicati \(^{8,9}\))

\[ \xi=\sum_{m\ne 0}|\alpha_{0m}|^2 . \]

The calculation of this quantity requires knowledge of the functions \(\psi_m\). It is possible, however, to give a simple estimate of the sum if the difference of energies is replaced by some average difference \((\Delta E)\). Then, replacing the summation over \(m\ne 0\) by summation over all \(m\), we obtain:

\[ \xi=\sum_{m\ne 0} \left| \frac{(\psi_0|v|\psi_m)}{(E_m-E_0)} \right|^2 < \frac{1}{(\Delta E)^2} \sum_m (\psi_0|v|\psi_m)(\psi_0|v|\psi_m)^* = \]

\[ = \frac{1}{(\Delta E)^2} \sum_m (\psi_0|v|\psi_m)(\psi_m|v|\psi_0) = \frac{1}{(\Delta E)^2} (\psi_0|v^2|\psi_0). \]

In the transformations we have used the Hermitian character of the matrix \(v\) and the rules of matrix multiplication.

The matrix element \((\psi_0|v^2|\psi_0)\) is easily estimated from experimental data, since it is simply the mean value of the square—

*) In systems consisting of many particles, specifying \(J, P, T\) does not yet determine the state uniquely. Further classification depends, generally speaking, on the specific properties of the nuclear forces.

ISOTOPIC SPIN OF LIGHT NUCLEI

of the operator \(v = v_1 + v_2\). It is not difficult to see that the term \(v_1\) may be neglected in comparison with \(v_2\). Indeed, from the definition of the operators \(v_1\) and \(v_2\) one readily obtains the following estimates for their mean values:

\[ \overline{v_1} = \frac{m_n-m_p}{m_n}\cdot \overline{\sum_i \frac{P_i^2}{2m_p}\,\xi^{(i)}} \approx \frac{m_n-m_p}{m_n}\cdot \varepsilon \cdot T_z, \]

where \(\varepsilon\) is the kinetic energy of a nucleon in the nucleus \((\varepsilon \sim 8\ \mathrm{Mev})\). It follows that, in order of magnitude, \(\overline{v_1}\sim 0.01\ \mathrm{Mev}\), whereas

\[ \overline{v_2} = \sum_{i>k} \frac{e^2}{4r_{ik}} \left(1+\xi_z^{(i)}\right) \left(1+\xi_z^{(k)}\right) \approx \frac{Z(Z-1)}{2}\cdot 0.5\ \mathrm{Mev}, \]

where \(Z\) is the nuclear charge, and \(0.5\ \mathrm{Mev}\) is the mean energy of the Coulomb interaction of two protons in the nucleus. Thus, we arrive at the conclusion that in calculating \(\xi\) one should take into account only the Coulomb energy of the protons, since the difference in the masses of the proton and neutron leads to a very small effect. We obtain the following estimate:

\[ \xi < \frac{\overline{v_2}^{\,2}}{(\Delta E)^2}. \]

Determination of \(\Delta E\) requires knowledge of the positions of many levels. For orientation, we may replace \(\Delta E\) by the distance to the nearest level with a different isotopic spin, but with the same \(J\) and \(P\). In this way we obtain an upper bound for \(\xi\).

Such an estimate leads to values \(\xi \approx 10^{-3} - 10^{-4}\) for Be and \(\xi \approx 0.1 - 0.5\) for \(\mathrm{O}^{16}\). These values are clearly overestimated. Analysis of experimental data on violations of selection rules leads to smaller quantities\(^{8-12}\). Apparently, the isotopic spin of ground states remains meaningful up to \(Z\sim 20\). For excited states it ceases to be a quantum number considerably earlier.

The general conclusion that strongly excited states do not possess a definite isotopic spin is confirmed by experimental data\(^{13}\). Thus, from an analysis of data on the reactions \(\mathrm{N}^{15}(p\alpha)\mathrm{C}^{12}\) (ground state) and \(\mathrm{N}^{15}(p\gamma)\mathrm{O}^{16}\) (ground state), it was established that the excited state of \(\mathrm{O}^{16}\) with energy \(13.09\ \mathrm{Mev}\) apparently does not possess a definite isotopic spin, but is a mixture of states with \(T=0\) and \(T=1\) (this conclusion is justified by the fact that from this level, with approximately equal probability, decay occurs both to \(\mathrm{C}^{12}+\alpha\) and to \(\mathrm{O}^{16}+\gamma\), while

as, according to the selection rules for isotopic spin, the first decay is possible only if in \(O^{16}\) \(T=0\), whereas the second decay path requires \(T=1\). At the present time one more level is known which, apparently, does not possess a definite isotopic spin—this is the excited state of the \(B^{10}\) nucleus with energy \(7.43\) MeV. As in the first case, such a conclusion is substantiated by the high probability of decay of this state both according to the scheme \(B^{10*}\to \alpha+Li^6\), and according to the scheme \(B^{10*}\to B^{10}+\gamma\).

Chapter IV. SIMILAR LEVELS OF LIGHT NUCLEI

§ 10. Similar states*)

As has already been said more than once above, in light nuclei with \(Z\lesssim 15\text{—}25\) the Coulomb interaction of the protons is small in comparison with the specific nuclear forces. It follows directly from this (see Ch. I, § 5) that when one proton in a nucleus is replaced by a neutron, or conversely, we obtain a new nucleus whose Hamiltonian will differ only slightly from the Hamiltonian of the first nucleus; the only essential difference arises from the fact that some of the states possible in one of these nuclei will be forbidden by the Pauli principle in the other. The same states that are possible in both nuclei have identical properties, namely these states (similar states) have identical angular momenta, parities, isotopic spins, and internal structure; the energy difference between two such states in one nucleus will almost exactly coincide with the energy difference between the corresponding states in the other nucleus, etc. A certain difference between these two nuclei arises because of the difference in Coulomb energy. However, in most cases this leads to the fact that all levels of one nucleus are simply shifted relative to the corresponding levels of the other nucleus, while the energy difference between two corresponding levels changes only slightly (some exceptions to this rule will be discussed below). Let us note here that the Coulomb shift makes it possible to calculate very simply the energy of the electrical interaction of the protons in a nucleus. For this purpose, in two nuclei differing by the replacement of a neutron by a proton, one must find similar levels and compare their energies. The difference between these energies, taking into account the difference in the masses of the neutron and the proton, will give the Coulomb energy per one proton. This is apparently the most direct and simple method for determining the Coulomb energy of light nuclei.

*) See also the work of B. S. Dzhelepov \(^{14}\), as well as the recently published review by B. S. Dzhelepov \(^{26}\).

From the point of view of the properties that follow from the hypothesis of charge invariance, light nuclei are divided into two principal groups—the group of nuclei of type \(2n\): even-even and odd-odd nuclei, and the group of nuclei of type \(2n+1\): odd-even nuclei. The first group is characterized by the fact that, since the nuclei of this group consist of an even number of nucleons, the various states of these nuclei can have only integral values of isotopic spin: \(0, 1, 2, \ldots\). The principal feature of the energy levels of nuclei of this type is that states with \(T=0\) prove to be energetically more favorable than states with \(T=1\), and the latter in turn are energetically more favorable than states with \(T=2\), and so on. Owing to this, there arises the concept of triads of nuclei, i.e., of triples of selected nuclei with different ratios of the numbers of neutrons and protons (for example, \(\mathrm{Be}^{10}\), \(\mathrm{B}^{10}\), and \(\mathrm{C}^{10}\)). Two nuclei of such a triad have \(T_\zeta=+1\) and \(T_\zeta=-1\) (\(\mathrm{C}^{10}\) and \(\mathrm{B}^{10}\), respectively), while the third has \(T_\zeta=0\) (\(\mathrm{B}^{10}\)). Correspondingly, in the first two nuclei only states with \(T=1, 2, \ldots\) are possible, whereas in the third, in addition, states with \(T=0\) are possible. Thus, states with \(T=0\), occurring in nuclei with \(T_\zeta=0\), have no analogues in the other nuclei of the triad, while states with \(T=1, 2\) can occur in all members of the triad, and to each such state in one of the nuclei there correspond similar states in the other members of the triad; moreover, the angular momenta, parities, relative positions, and other characteristics of similar levels are the same in all members of the triad.

Nuclei of type \(2n+1\) consist of an odd number of nucleons; their states can possess only half-integral values of the isotopic spin \(T=\frac{1}{2}, \frac{3}{2}, \ldots\). States with \(T=\frac{1}{2}\) prove to be energetically much more favorable than states with \(T=\frac{3}{2}\), as a result of which all stable nuclei of this type with \(|T_\zeta|=\frac{1}{2}\) have, in the ground state, \(T=\frac{1}{2}\). Thus, in this case nuclei are grouped into pairs of selected nuclei with projections of isotopic spin \(T_\zeta=\frac{1}{2}\) and \(T_\zeta=-\frac{1}{2}\) (the so-called mirror nuclei).

For triads of selected even nuclei and for pairs of selected odd nuclei with \(T_\zeta=\pm\frac{1}{2}\), the general name charge multiplets is used in the literature. Let us note one more term frequently encountered in the literature, namely supermultiplets. This term is used in classifying the states of nuclei under the assumption of pure \(LS\)-coupling. In this case the symmetry of the spatial part of the nuclear wave function is determined by three numbers: the ordinary and isotopic spins of the nucleus, \(S\) and \(T\), and an additional quantum number \(Y\), which characterizes the symmetry of the product of the wave functions of the ordinary and isotopic spins. States of two isobaric nuclei that have identical values of \(S, T\), and \(Y\) are said to belong to one supermultiplet.

§ 11. Methods for determining the isotopic spin of nuclear states

In determining the isotopic spins of various states of light nuclei, the following considerations are usually used:

a) The use of selection rules for isotopic spin makes it possible in many cases to determine the isotopic spin of one or another nuclear state, if the isotopic spin of the initial or final particles is known.

b) In determining the isotopic spins of some nuclei, energetic considerations are often of help. Consider, for example, the $\alpha$-particle. Since it consists of two neutrons and two protons $(T_z = 0)$, the ground state may have $T = 0, 1, 2$. However, it is easy to see that in the ground state the $\alpha$-particle has $T = 0$, since otherwise the existence of a stable isotope $\mathrm{H}^4$ with a binding energy differing only slightly from the binding energy of the $\alpha$-particle (a small Coulomb shift) would be possible. Experimentally, however, it has been found that no such state of $\mathrm{H}^4$ exists. This is the decisive argument in favor of the fact that, in the ground state, the isotopic spin of the $\alpha$-particle is zero.

In the same way it is proved that the mirror nuclei $\mathrm{H}^3$ and $\mathrm{He}^3$ have isotopic spin equal to $1/2$. Indeed, if their isotopic spin were equal to $3/2$, then stable nuclei consisting of three neutrons or three protons would exist, which in fact is not observed. Finally, by exactly the same method one can conclude, from the absence of a stable state in a system of two neutrons, that the isotopic spin of the deuteron is zero.

Analogous considerations often help also in determining the isotopic spins of various states of heavier nuclei. In particular, arguments of this kind are the main evidence in favor of the fact that the ground states of almost all light even-even and odd-odd nuclei have $T = 0$. Let us consider, for illustration, the nucleus $\mathrm{Be}^8$. It is known that the binding energy of the nucleus $\mathrm{Be}^8$ $(T_z = 0)$ exceeds by $\sim 16$ MeV the binding energy of the nucleus $\mathrm{Li}^8$ $(T_z = -1)$, which differs from $\mathrm{Be}^8$ only by the replacement of one proton by a neutron and therefore can occur only in states with $T = 1$. On the other hand, it is clear that the Coulomb energy must be larger in $\mathrm{Be}^8$, which has one more proton. Therefore it is easy to conclude that the first state of $\mathrm{Be}^8$ which can also occur in $\mathrm{Li}^8$ (a state with $T = 1$) has an energy in any case not less than 16 MeV. Hence it follows directly that the ground state and all excited states of the nucleus $\mathrm{Be}^8$ with excitation energy less than $\sim 15$ MeV must have $T = 0$.

b) The isotopic spin of certain states can be determined from the structure of the given state, i.e., from the known state of the nucleons in the given state of the nucleus. For example, it is known that the wave function of the relative motion of the nucleons in the deuteron is a superposition of the states \({}^{3}S_{1}\) and \({}^{3}D_{1}\), which are symmetric with respect to interchange of the spatial coordinates and spins of the nucleons. Since the total wave function must be antisymmetric, these states have isotopic spin equal to zero. Thus, from the known wave function of the deuteron it follows, in agreement with the result obtained earlier, that in the ground state of the deuteron \(T = 0\).

As is known, besides the ground triplet state, the deuteron also has a virtual singlet state \({}^{1}S\), manifested in the scattering of neutrons by protons. The state \({}^{1}S\) has \(T = 1\) and, consequently, an analogous virtual state exists in the system of two neutrons or two protons, as is also observed experimentally.

Structural considerations play a major role in determining the isotopic spins of various states of nuclei because, according to the shell model, nucleons in a nucleus are in states with specified orbital angular momenta. In this case it becomes possible to calculate the angular momenta and isotopic spins of all states of such a system, and therefore knowledge of the angular momentum of one or another nuclear state sometimes makes it possible to draw conclusions also about the isotopic spin of this state (when there is a one-to-one correspondence between the isotopic spin and the angular momentum of the state).

§ 12. Some features in the arrangement of levels

The data on the levels of light nuclei presented in the second part make it possible to establish a number of regularities.

Let us construct graphs on which the energy differences between the lowest levels with \(T = 0, 1, 2, 3\) are plotted for nuclei with even mass number \(A\). The following features of these graphs are immediately apparent: the points corresponding to the differences between the lowest levels with \(T = 1\) and \(T = 0\) (difference (10)) lie, as is seen from Fig. 1, on two smooth curves—one for nuclei of the type \(4n\) (curve \(I\)) and the other for nuclei of the type \(4n + 2\) (curve \(II\)). The first curve in the region \(6 < A < 40\) varies from 17 MeV for \(\mathrm{Be}^{8}\) to 6 MeV for \(\mathrm{A}^{36}\), whereas the second curve in this same region nowhere rises above 3.6 MeV (for \(\mathrm{Li}^{6}\)). Similarly, the points corresponding to the energy differences between the first levels with \(T = 2\) and \(T = 1\) (difference (21)) lie on two smooth curves—one for nuclei of the type \(4n\) and the other for nuclei of the type \(4n + 2\) (Fig. 2).

The first curve \((4n)\) nowhere rises above \(5.5\) MeV \((\mathrm{Si}^{28})\), whereas the second curve \((4n+2)\) in this same range lies entirely in the narrow band \(8.5\)–\(11.0\) MeV. In contrast to

Fig. 1.

the differences (10) and (21) of the energy differences between the first levels with \(T=2\) and \(T=0\), and also with \(T=3\) and \(T=1\), whose isotopic spins differ by 2 (differences (20) and (31)), are smooth functions of the mass number. In this case the nuclei \(4n\) and \(4n+2\) fall on the same curve (curves III in Figs. 1 and 2).

Fig. 2.

If analogous graphs (Fig. 3) are constructed for nuclei with odd \(A\), we find that all points \(\left({}^{3}/_{2}\,{}^{1}/_{2}\right)\) and \(\left({}^{5}/_{2}\,{}^{3}/_{2}\right)\) lie on comparatively smooth curves.

In determining the isotopic spins of nuclei with \(Z<10\), one uses considerations following from shell theory, the results of nuclear reactions, the differences in binding energies of isobaric nuclei, and so on. In the region of nuclei with \(10<Z<25\), one cannot use either nuclear-reaction data, since the conservation law for isotopic spin, appar-

dimensional, there is no longer any place for it, nor on grounds of the shell theory, since at present the scheme of filling the various shells in this region is not yet precisely known. The means for determining isotopic spins here are mainly comparisons of the levels of isobaric nuclei. Therefore the question arises of the reliability of identifying states by isotopic spin for such nuclei; moreover, it is necessary to make sure that for these nuclei isotopic spin still retains its meaning. On this

Fig. 3.

occasion the following can be said. In the region of nuclei with \(Z < 10\), where the validity of introducing isotopic spin as a quantum number is fully confirmed by all experimental material, there exists a clearly expressed dependence of the difference (10), etc., on the mass number \(A\). The fact that for \(Z > 10\) these regularities continue to appear leads to the conclusion that isotopic spin is a characteristic of the states of such, already comparatively heavy, nuclei.

The regularities established above make it possible in a number of cases to draw conclusions about the stability or instability of a given isotope and to indicate approximately its binding energy. Thus, for example, the isotope \(\mathrm{Al}^{30}\), for which \(T_z=-2\), is still unknown. Extrapolating curve \(III\) in Fig. 2, we find that the first state with \(T=2\), which should be the ground state for \(\mathrm{Al}^{30}\), lies approximately \(15\) MeV above the first state with \(T=1\) (the ground state of \(\mathrm{Si}^{30}\)). Taking into account the difference of Coulomb energies in \(\mathrm{Si}^{30}\) and \(\mathrm{Al}^{30}\) (it amounts to \(5\)–\(6\) MeV), we arrive at the conclusion that the binding energy of \(\mathrm{Al}^{30}\) must be \(\sim 9\) MeV smaller than that of \(\mathrm{Si}^{30}\).

Considering the possible decay paths of this isotope, one may conclude that the only energetically possible decay path for \( \mathrm{Al}^{30} \) can be \(\beta\)-decay according to the scheme
\[ \mathrm{Al}^{30}\beta^- \to \mathrm{Si}^{30}. \]
Similarly, it can be shown that the following, as yet unknown isotopes should exist: \( \mathrm{Na}^{26} \) with a binding energy \(\sim 12\) Mev less than that of \( \mathrm{Mg}^{26} \), \( \mathrm{Ne}^{24} \) with a binding energy \(\sim 4\) Mev less than that of \( \mathrm{Na}^{24} \), \( \mathrm{Cl}^{41} \) with a binding energy \(\sim 5\text{--}6\) Mev less than that of \( \mathrm{Ca}^{40} \), etc. In addition, one can approximately estimate the binding energy of certain isotopes whose existence is now known, but whose binding energy has not been established. Thus, the binding energy of \( \mathrm{K}^{44} \) should be \(\sim 5\) Mev less than that of \( \mathrm{Ca}^{44} \), and the binding energy of \( \mathrm{A}^{42} \) should be \(\sim 1\text{--}2\) Mev less than the binding energy of \( \mathrm{K}^{42} \). Verification of these predictions appears very interesting.

Part II. LEVELS OF LIGHT NUCLEI

In this part a brief summary is collected of the levels of light nuclei for \(A \leq 50\). The purpose of the figures in this review is to illustrate the relative position of the levels of isobars and to give some idea of the distribution of the magnitude of isotopic spin in nuclei. The review does not exhaust all the known material*) and is probably not free from some arbitrariness in identification.

The level schemes are arranged so that similar states of neighboring nuclei are superposed. To this end, a correction for the Coulomb energy and for the difference between the proton and neutron masses was introduced into the mass difference of isobars known from tables. The quantity obtained determines the distance between the ground states in the schemes. Because of the uncertainty in the Coulomb energy, such an arrangement is not very precise, and an error of several hundred kev is not improbable. It should be remembered that, in view of what has been said, the arrangement of the levels of neighboring nuclei does not determine the energies of the \(\beta^\pm\)-spectra.

It is clear from the schemes that the experimental data are distributed extremely unevenly among the various nuclei, and that further experiments are necessary for a more detailed analysis of such nuclei.

\( \mathrm{He}^{4},\ \mathrm{Li}^{4} \) and \( \mathrm{H}^{4} \)

\( \mathrm{He}^{4} \) has \(T_\zeta=0\), whence states with \(T=0,1,2\) are possible for it, while for \( \mathrm{Li}^{4} \) and \( \mathrm{H}^{4} \), \(T_\zeta\) is respectively \(+1\) and \(-1\), and, consequently, states with \(T=0\) cannot be realized in them. In \( \mathrm{He}^{4} \), besides the ground state with \(T=0\), there are two more excited states with energy \(\sim 22.5\) Mev and \(\sim 23\) Mev\(^{15}\). The isotopic spin of these states is apparently equal

*) With few exceptions, the data on levels have been borrowed from the two reviews mentioned above\(^{3,4}\).

zero, since they do not appear in the reaction \(T(p\gamma)\mathrm{He}^4\). Hence it follows that the possible states of \(\mathrm{Li}^4\) and \(\mathrm{H}^4\) must lie still higher in energy, and since at energies \(>21\) MeV dissociation into \(T+p\) or \(\mathrm{He}^3+p\) is already energetically possible, stable states in \(\mathrm{H}^4\) and in \(\mathrm{Li}^4\) cannot exist.

\(\mathrm{Li}^5\) and \(\mathrm{He}^5\)

\(\mathrm{Li}^5\) and \(\mathrm{He}^5\) are unstable mirror nuclei with a similar system of levels. All three levels now known possess isotopic spin \(T=\frac12\), since the first two levels of \(\mathrm{He}^5\) appear in the reaction \(\mathrm{He}^4+n\) (we recall that the isotopic spin of the \(\alpha\)-particle is zero, and that of the nucleon is one half), while the level with \(E=16.8\) MeV appears in the reaction \(d+T\).

Fig. 4. The relative positions of the levels of the mirror nuclei in Fig. 4 differ from the experimentally observed ones by the difference in the Coulomb energies of these nuclei and by the correction taking account of the mass difference between the neutron and the proton. With such a displacement, similar levels (in the figures they are connected by dotted lines) of the mirror nuclei coincide.

\(\mathrm{Li}^6,\ \mathrm{He}^6\), and \(\mathrm{Be}^6\)

In \(\mathrm{Li}^6\) (\(T_z=0\)) states with \(T=0,1,\ldots\) are possible, while in \(\mathrm{Be}^6\) (\(T_z=1\)) and \(\mathrm{He}^6\) (\(T_z=-1\)) states with \(T=0\) cannot be realized. Thus in \(\mathrm{He}^6\) all levels of \(\mathrm{Li}^6\) can occur, with the exception of levels with \(T=0\). Hence it follows that the level of \(\mathrm{Li}^6\) corresponding to the ground state of \(\mathrm{He}^6\) must have \(T=1\) (the ground state of \(\mathrm{He}^6\) must have \(T=1\), since if it were equal to 2, there would exist a superheavy isotope of hydrogen, \(\mathrm{H}^6\), which is not found in nature). To find this level, we note that if one excludes from consideration the Coulomb energy and the mass difference between the neutron and the proton, then the difference between the energies of the ground states of \(\mathrm{He}^6\) and \(\mathrm{Li}^6\) should be exactly equal to the energy of the first level with \(T=1\). The average energy of the Coulomb interaction of two protons in light nuclei is \(0.4\)—\(0.5\) MeV, and the mass difference between the neutron and the proton is \(0.78\) MeV. Comparing these figures, it is easy to conclude that the ground state of \(\mathrm{He}^6\) corresponds to a level of \(\mathrm{Li}^6\) with \(E=3.58\) MeV, to which, therefore, \(T=1\) should be assigned. Hence it follows directly that the ground and first excited states of \(\mathrm{Li}^6\) have \(T=0\).

This conclusion is confirmed by the results\(^{16}\) of a study of the reaction \(\mathrm{Be}^9(p\alpha)\mathrm{Li}^6\). Namely, the intermediate state \(\mathrm{B}^{10}\) formed in this reaction, with energy \(E=8.89\) MeV, decays by

scheme \(a + \mathrm{Li}^{6*}\); in this case \(\mathrm{Li}^{6}\) is formed only in the state with energy \(E = 3.58\) MeV, but the ground and first excited states of \(\mathrm{Li}^{6}\) are not formed. Hence it follows that the isotopic spin of the second excited state of \(\mathrm{Li}^{6}\) \((E = 3.58)\) is \(T \ne 0\).

Fig. 5.

The analogue of \(\mathrm{He}^{6}\) should have been the nucleus \(\mathrm{Be}^{6}\) \((T_z = 1)\). However, because of the comparatively large Coulomb energy (approximately \(1.5\) MeV greater than in \(\mathrm{Li}^{6}\)), this nucleus is unstable with respect to decay into \(\mathrm{He}^{4} + 2p\) and therefore is not found in nature.

\[ \mathrm{Li}^{7}\ \text{and}\ \mathrm{Be}^{7} \]

\(\mathrm{Li}^{7}\) and \(\mathrm{Be}^{7}\) are a pair of mirror nuclei with an entirely similar system of levels. Therefore below we shall speak only about the levels of \(\mathrm{Li}^{7}\). The ground state and the first two excited states of \(\mathrm{Li}^{7}\) appear in the reaction \(\mathrm{Li}^{6} + d \to \mathrm{Li}^{7} + p\), and therefore one may conclude that they all have \(T = \tfrac{1}{2}\). The level with \(E = 7.5\) MeV appears in the reaction \(\mathrm{Be}^{9}(d\alpha)\mathrm{Li}^{7}\), together with the first three levels of \(\mathrm{Li}^{7}\), which have \(T = \tfrac{1}{2}\). Therefore this level should also be assigned \(T = \tfrac{1}{2}\). From the analysis of the \((\gamma n)\) and \((\gamma T)\) reactions it follows (see Chap. II) that the levels with \(E = 9.6\) and \(17.5\) MeV have \(T = \tfrac{3}{2}\), while the levels with \(E = 14;\ 12.4;\ 10.8\) MeV should be assigned \(T = \tfrac{1}{2}\). The isotopic spin of the level with \(E = 6.6\) MeV remains unclear. Since this level appears in the reaction \((pp')\), but does not appear in the reaction \(\mathrm{Be}^{9}(d\alpha)\mathrm{Li}^{7}\), one may expect that its isotopic spin is equal to \(\tfrac{3}{2}\). However, to this

Fig. 6.

one should treat this conclusion with caution, since it has not yet been checked in other reactions (for example, \( \mathrm{Li}^7(\alpha\alpha')\mathrm{Li}^{7*} \)).

\( \mathrm{B}^8, \mathrm{Li}^8 \), and \( \mathrm{Be}^8 \) *)

As was already said in the preceding paragraph, it is clear from energy considerations that all levels of \( \mathrm{Be}^8 \) with \(E < 16\) MeV have \(T = 0\). At what energy the first level with \(T = 1\) is located is not precisely known at present; however, one may expect that this level lies at \(E = 17\) MeV. This follows from the results of the reaction \( \mathrm{C}^{12}(\gamma\alpha)\mathrm{Be}^8 \), in which, by virtue of the selection rules for isotopic spin, with high probability only those states of \( \mathrm{Be}^8 \) for which \(T = 1\) can be formed. Investigation of this reaction has shown that \( \mathrm{Be}^8 \) is formed mainly in a state with \(E = 16.7\) MeV\(^{18}\), whence the above assertion follows.

The ground and first excited states of \( \mathrm{Li}^8 \) have \(T = 1\). This follows, for example, from the fact that both of these states appear in the reaction \( \mathrm{Li}^7(\mathrm{d}\mathrm{p})\mathrm{Li}^8 \).

Fig. 7.

The third member of the \( \mathrm{B}^8 \) triad, because of its comparatively large Coulomb energy, has too short a lifetime and has scarcely been investigated; however, from the fact that the binding energies of \( \mathrm{Li}^8 \) and \( \mathrm{B}^8 \), after subtraction of the Coulomb energy and the mass difference of the neutron and proton, coincide almost exactly, it follows that \( \mathrm{B}^8 \) is an analogue of \( \mathrm{Li}^8 \) and therefore in the ground and first excited states has \(T = 1\).

\( \mathrm{B}^9 \) and \( \mathrm{Be}^9 \)

\( \mathrm{Be}^9 \) and \( \mathrm{B}^9 \) are mirror nuclei, as is indicated by the fact that, after subtracting the Coulomb energy and the difference between the neutron and proton masses, the binding energies of these nuclei almost exactly coincide. This fact

*) The level scheme of \( \mathrm{Be}^8 \) is given according to the work of Bonner and Cook \(^{17}\).

is very important, since although the \(B^9\) nucleus is unstable with respect to decay into \(Be+p\) and therefore has been poorly investigated (a small Coulomb shift is sufficient for this; since the ground state of \(Be^9\) lies below the threshold for decay into \(Be^8+n\) by only \(1.6\) MeV), its excited states can appear in certain reactions; and since \(B^9\) and \(Be^9\) are mirror nuclei, we can immediately conclude that all their low-lying levels are similar and, consequently, it is sufficient to study only the excited states of \(Be^9\).

Fig. 8.

The isobaric nucleus

\[ Li^9\left(T_\xi=-\frac{3}{2}\right) \]

has a binding energy \(14.1\) MeV less than \(Be^9\) \((T_\xi=-\frac{1}{2})\). Hence we immediately conclude that all states of \(Be^9\) with \(E<14\) MeV have \(T=\frac{1}{2}\). This is in agreement with all reactions observed in this energy region. From this one may also conclude that in \(Be^9\) there should exist a level with \(T=\frac{3}{2}\) somewhere in the region \(E\sim 15\) MeV (this figure is obtained from the difference of the binding energies of \(Be^9\) and \(Li^9\), to which one must add a correction for the Coulomb interaction of the extra proton in \(Be^9\) and for the neutron–proton mass difference), but it has not yet been found, since in this energy region there are no suitable reactions.

The next levels of \(Be^9\) with \(E=17.2\) and \(17.45\) MeV apparently have \(T=\frac{1}{2}\), since they appear in the reaction \(Li^7+d\).

Fig. 9.

\(Be^{10},\ B^{10},\ C^{10}\)

The nucleus \(B^{10}\) \((T_\xi=0)\) may be in states with \(T=0,1,2,\ldots\), while the nuclei \(Be^{10}\) \((T_\xi=-1)\) and \(C^{10}\) \((T_\xi=1)\)

cannot occur as states with \(T=0\). Comparing the energies of the ground states of the nuclei \(B^{10}\) and \(Be^{10}\), it is easy to conclude that the ground state of \(B^{10}\) has \(T=0\). It follows immediately from this that the states of the nucleus \(B^{10}\) with \(E=0.72;\ 2.15\) and \(3.58\) MeV have \(T=0\), since all of them appear in the reaction \(B^{10}(dd')B^{10*}\). The state with \(E=1.74\) does not appear in this reaction and should be assigned \(T=1\), i.e. this is precisely the state which should occur as the ground state in \(Be^{10}\) and \(C^{10}\). This conclusion is fully confirmed by energy considerations, since the binding energy of \(B^{10}\) (if one takes into account the Coulomb energy and the neutron–proton mass difference) exceeds the binding energy of \(B^{10}\) by just \(\sim 1.7\) MeV.

Applying the law of conservation of isotopic spin to various reactions which proceed through excited states of \(Be^{10}\) with excitation energy \(<10\) MeV, it is easy to obtain that all these states of \(Be^{10}\) have \(T=1\). Corresponding states should also exist in \(B^{10}\). At present it has been established almost reliably\(^{19,20}\) that the level of \(Be^{10}\) with \(E=3.37\) MeV corresponds to the level of \(B^{10}\) with \(E=5.16\) MeV, the level with \(E=5.94\) MeV to the level with \(E=7.19\), the level with \(E=6.24\) to the level \(E=7.48\), and the level with \(E=7.54\) MeV to the level with \(E=8.89\) MeV. In making this comparison, the results of various reactions were used in connection with the selection rules for isotopic spin. Thus, for example, the level of \(B^{10}\) with \(E=8.89\) MeV is assigned \(T=1\), since, first, from this level there is decay into \(Li^6\) in the state with \(E=3.58\) MeV (in this state \(T=1\)) and an \(\alpha\)-particle, and second, from this level there is an intense \(\gamma\)-transition to the ground state of \(B^{10}\) (\(T=0\)), which would not occur if the state of \(B^{10}\) with \(E=8.89\) MeV had \(T=0\).

Fig. 10.

The nucleus \(C^{10}\) has been investigated much less well than \(Be^{10}\); however, from energy considerations it is clear that it is an analogue of \(Be^{10}\) (the difference of the binding energies of \(B^{10}\) and \(C^{10}\), taking into account the Coulomb energy and the neutron–proton mass difference, is, as in the case of \(Be^{10}\), \(\sim 1.7\) MeV). Therefore \(C^{10}\) must have exactly the same level scheme as \(Be^{10}\).

\(B^{11}\) and \(C^{11}\)

These are mirror nuclei with a similar system of levels and \(T_\zeta=\pm \frac{1}{2}\). After subtracting the Coulomb energy and the neutron–proton mass difference, the binding energies of the nuclei \(C^{11}\) and \(B^{11}\) are equal. From

from the reaction \(B^{10}(dp)B^{11}\) it follows that all states of \(B^{11}\) (and, consequently, also of \(C^{11}\)) with excitation energy \(E<9\) MeV have \(T=\frac12\). Thus the absence of any stable isotopes \(N^{11}\) and \(Be^{11}\) becomes clear, since these nuclei have \(|T_\zeta|=\frac32\), and, consequently, the binding energies of these nuclei must be at least by \(\sim 9\) MeV smaller than the binding energies of \(B^{11}\) and \(C^{11}\). With such small binding energies the nuclei \(Be^{11}\) and \(C^{11}\) cannot exist, but decay into \(Be^{10}+n\) and \(C^{10}+p\), respectively.

Fig. 11.

\(B^{12},\ C^{12}\), and \(N^{12}\)

The binding energy of the nucleus \(C^{12}\) (\(T_\zeta=0\)), after subtraction of the Coulomb energy and of the mass difference of the neutron and proton, exceeds the binding energy of the nuclei \(B^{12}\) (\(T_\zeta=-1\)) and \(N^{12}\) (\(T_\zeta=1\)) by approximately 15 MeV. Hence it follows that all excited states of \(C^{12}\) with \(E<15\) MeV must have \(T=0\). This conclusion is confirmed by a whole series of reactions: for example, the radiative-capture reaction \(C^{12}+\gamma\) practically does not occur for \(\gamma\)-quantum energies less than 15 MeV (recall that, since for \(C^{12}\), \(T_\zeta=0\), \(E1\)-transitions can occur only between levels with different isotopic spins). At present the position of the first level of \(C^{12}\) with \(T=1\) is not known precisely; however, there are indications1 that this level has \(E=15.09\) MeV. If this conclusion is correct, then the level of \(C^{12}\) with \(E=15.09\) MeV is analogous to the ground states of \(B^{12}\) and \(N^{12}\). At present there are data[^2] on still another level of \(C^{12}\), namely on a level with \(E=16.07\) MeV. From this level, whose moment is \(2^-\), emission of an \(\alpha\)-particle is possible; however, the corresponding width is very small (\(\Gamma_\alpha\sim 5\) keV, whereas at such energies one should expect widths \(\sim 1\) MeV). This is interpreted as an indication that the isotopic spin of this state of \(C^{12}\) is equal to unity (the fact that emission of an \(\alpha\)-particle is nevertheless possible is ascribed to an admixture of a state with \(T=0\) to this level of \(C^{12}\)). If such an identification is correct, then the level of \(C^{12}\)

with \(E = 16.07\) MeV is the analogue of the first excited level of the nuclei \(B^{12}\) and \(N^{12}\).

The nuclei \(B^{12}\) and \(N^{12}\) have identical ground-state energies (allowing, of course, for the Coulomb energy and the neutron–proton mass difference) and must possess a similar system of levels. It is known concerning their levels that all levels with \(E < 6\) MeV have \(T = 1\). This follows from the fact that all these levels appear in the reactions \(Be^{9}(\alpha p)B^{12}\) or \(B^{11} + n\).

\(C^{13}\) and \(N^{13}\)

The ground-state energies of these nuclei with \(T_z = \pm \tfrac12\) (after subtracting the Coulomb energy and the neutron–proton mass difference) approximately coincide. The spins and parities of several of the first excited states of these nuclei also coincide. This, in fact, is as it should be, since \(C^{13}\) and \(N^{13}\) are mirror nuclei. At first sight this is not consistent with the circumstance that the first excited level in \(N^{13}\) is greatly lowered in comparison with the corresponding level of \(C^{13}\) (2.37 as against 3.08). The reason for this can be understood, since according to the shell model \(C^{13}\) and \(N^{13}\) are a system consisting of a neutron (or proton) above the closed shells of the \(C^{12}\) nucleus. In the ground state this nucleon is in the \(1p_{1/2}\) shell, and the first excited state of these nuclei corresponds to the transition of this nucleon to another shell, probably \(2s_{1/2}\). In such a transition, naturally, the average distance between the outer nucleon and the closed shells changes. This leads to a decrease of the Coulomb energy in the first excited state of the \(N^{13}\) nucleus as compared with the ground state, i.e. to a lowering of this level in \(N^{13}\) in comparison with \(C^{13}\), where this effect is absent. Hence, incidentally, one can estimate the increase of the mean radius of the nucleon in the transition

\[ 1p_{1/2} \to 2s_{1/2}: \quad \frac{\Delta r}{r} \sim 20\%. \]

Fig. 12.

The isotopic spin of all states of \(C^{13}\) (and correspondingly \(N^{13}\)) with energy \(< 9\) MeV is equal to \(\tfrac12\). This follows from the fact that all these levels appear in the reactions \(C^{12}(dp)C^{13}\) or \(B^{10}(\alpha p)C^{13}\). The position of the first level with \(T = \tfrac32\) is at present unknown.

C\(^{14}\), N\(^{14}\), and O\(^{14}\)

The nucleus N\(^{14}\) (\(T_c=0\)) can be in states with \(T=0,1,2,\ldots\), while in the nuclei C\(^{14}\) (\(T_c=-1\)) and O\(^{14}\) (\(T_c=1\)) states with \(T=0\) cannot be realized. On the other hand, it is clear from energy considerations that the ground level of C\(^{14}\) corresponds to the first excited state of N\(^{14}\) with \(E=2.31\) MeV, so that the ground state of N\(^{14}\) should be assigned \(T=0\), and the first excited state \(T=1\) (the ground state of C\(^{14}\) has \(T=1\), as follows from the reaction C\(^{12}\)(Tp)C\(^{14}\)). This conclusion is also confirmed by the fact\(^{3}\) that, first, the state of N\(^{14}\) with \(E=2.31\) MeV does not appear in the reaction of inelastic deuteron scattering on N\(^{14}\): N\(^{14}\)(dd′)N\(^{14*}\); secondly, this state does not appear in the reaction O\(^{16}\)(dα)N\(^{14*}\) (as we shall see below, O\(^{16}\) has \(T=0\), so that this result directly indicates the isotopic spin of the state N\(^{14}\) with \(E=2.31\) MeV, \(T=1\)); and, finally, in favor of such an identification is the fact that the spin of this state of N\(^{14}\), \(I=0+\), is equal to the spin of the ground state of C\(^{14}\).

Fig. 13.

In C\(^{14}\) an excited level with \(E=6.10\) MeV (\(T=1\)) is known. This level apparently corresponds to the level of N\(^{14}\) with \(E=8.06\) MeV. Such a conclusion is drawn from the fact that, first, from this level an \(E1\)-transition to the ground state of N\(^{14}\) is allowed; secondly, the moment of this level is \(1-\), as for the level of C\(^{14}\) with \(E=6.10\) MeV; and, finally, the energy difference between the level of N\(^{14}\) with \(E=8.06\) MeV and the first level with \(T=1\) (\(E=2.31\) MeV) is approximately equal to 6 MeV. The excited levels of N\(^{14}\) with \(2.31<E<8.06\) apparently have \(T=0\), since in C\(^{14}\) no levels below 6.10 MeV have yet been found.

The nucleus O\(^{14}\) has been investigated much more poorly than C\(^{14}\), but from the binding energies of these nuclei it follows that their ground states are similar.

N\(^{13}\) and O\(^{15}\)

As follows from the binding energies of these nuclei and from the similar arrangement of their levels, these are mirror nuclei. From the reaction

From \(N^{14}(dp)N^{15}\) it follows that all levels of \(N^{15}\) (and, consequently, also of \(O^{15}\)) with \(E<9\) MeV have \(T=\tfrac12\). The position of the first level with \(T=\tfrac32\) is not precisely known at present; however, from the fact that the binding energy of the nucleus \(C^{15}(T_\xi=-\tfrac32)\) is 8.8 MeV less than the binding energy of \(N^{15}\), it follows that the level of \(N^{15}\) with \(T=\tfrac32\) must lie somewhere in the region of 11 MeV (in \(N^{15}\) the excess proton, compared with \(C^{15}\),

Fig. 14.

i.e. the Coulomb energy of \(N^{15}\), is \(\sim 3\) MeV greater than in \(C^{15}\), but \(C^{15}\) has an excess neutron, which in turn increases the mass of this nucleus by \(\sim 0.8\) MeV. Hence we obtain the estimate given above: \(8.8+3.0-0.8=11\) MeV).

\(N^{16},\ O^{16},\ F^{16}\)

The nucleus \(O^{16}(T_\xi=0)\) has states with \(T=0,1,2,\ldots\), whereas in the nuclei \(N^{16}\) and \(F^{16}\) states with \(T=0\) cannot be realized. The binding energy of the nucleus \(N^{16}\) is \(\sim 10\) MeV less than that of \(O^{16}\). Therefore the ground state of \(O^{16}\) should be assigned \(T=0\). This agrees with the shell model, according to which in \(O^{16}\) in the ground state all shells are closed (the isotopic spin of a closed shell is zero). The first excited state of \(O^{16}\) with \(T=1\) has energy \(E=12.51\) MeV \(^{23}\). The fact that this state has \(T=1\) follows from the fact that from this level an \(E1\)-transition to the ground state is possible, but decay according to the scheme \(O^{16}\to C^{12}+\alpha\) is forbidden. In addition, the spin of this \(2^{-}\) level coincides with the spin of the ground state of \(N^{16}\). The fact that at lower energies \(O^{16}\) has no states with \(T=1\) follows from comparison of the binding energies of \(O^{16}\) and \(N^{16}\). In \(N^{16}\) there is an excited state with excitation energy 1.6 MeV. The corresponding level should also exist in \(O^{16}\), but so far it has not been identified.

The nucleus \(F^{16}\), which should be analogous to \(N^{16}\), is unstable because of its large Coulomb energy.

\(N^{17},\ O^{17},\ F^{17}\)

\(O^{17}\) and \(F^{17}\) are mirror nuclei with \(|T_\zeta|=\frac12\), while for \(N^{17}\), \(T_\zeta=-\frac32\). The ground states of \(O^{17}\) and \(F^{17}\), apart from the Coulomb energy and the neutron–proton mass difference, have the same energy, while the binding energy of \(N^{17}\) is \(8.8\) MeV less than the binding energy of \(O^{17}\). Therefore, in the ground states of \(O^{17}\) and \(F^{17}\), \(T=\frac12\). The first excited level of \(F^{17}\) is somewhat lowered in comparison with the corresponding level of \(O^{17}\). This, as in the case of \(C^{13}\) and \(N^{13}\),

Fig. 15.                                                     Fig. 16.

can be interpreted as the effect of the “swelling” of the nucleus upon excitation. The isotopic spin of the ground state of \(N^{17}\) is \(T=\frac32\). The corresponding level in \(O^{17}\) should be located at an energy of \(\sim 11.0\) MeV, and all lower levels of \(O^{17}\) and \(F^{17}\) must have \(T=\frac12\).

\(O^{18},\ F^{18}\)

For \(F^{18}\), \(T_\zeta=0\), while for \(O^{18}\), \(T_\zeta=-1\). Accordingly, in \(F^{18}\) states with \(T=0\) are possible, which cannot occur in \(O^{18}\). The isotopic spin of the ground state of \(O^{18}\) is equal to 1. This follows, for example, from the fact that, according to the shell theory of the nucleus, this nucleus has two neutrons outside the filled shells of \(O^{16}\), and a system of two identical nucleons has \(T=1\).

Comparing the binding energies of \(F^{18}\) and \(O^{18}\), it is easy to see that the ground state of \(F^{18}\) must have \(T=0\), while the first excited state with \(T=1\) must lie at an energy of \(\sim 1\) MeV. In \(F^{18}\) there is indeed a level at \(E=1.05\) MeV. This level appears in the reaction \(Ne^{20}(d\alpha)F^{18*}\). Since the isotopic spin of \(Ne^{20}\) is \(T=0\), it follows from this reaction that the state of \(F^{18}\) under consideration has \(T=0\). It is possible, however, that in this case, because of the large amount of admixtures, the selection rules for isotopic spin are violated, so that in fact the isotopic spin of the excited state of \(F^{18}\) with \(E=1.05\) MeV is \(T=1\). One may also suppose that the level of \(F^{18}\) with \(E=1.05\) MeV has \(T=0\), but that near it there is another level with \(T=1\).

Fig. 17.                  Fig. 18.

In order to choose between these two possibilities, it is necessary to measure the moment and the parity of the level of \(F^{18}\) with \(E=1.05\) MeV.

\(O^{19},\ F^{19},\ Ne^{19}\)

For \(F^{19}\) and \(Ne^{19}\), \(|T_c|=1/2\). From energy considerations it follows that the ground states of these nuclei are similar. The binding energy of \(O^{19}\), for which \(T_c=-3/2\), is 4.5 MeV less than the binding energy of \(F^{19}\). Hence it follows that in the ground states of \(F^{19}\) and \(Ne^{19}\), \(T=1/2\), while in \(O^{19}\), \(T=3/2\). The latter follows from the general tendency that the energetically most favorable states are those with minimal isotopic spin, and also from structural considerations analogous to those used in establish-

… of the isotopic spin of \(O^{18}\) and \(F^{18}\). From the difference in the binding energies of \(F^{19}\) and \(O^{19}\) we find that the first state with \(T=\frac{3}{2}\) should occur in \(F^{19}\) at an energy of \(\sim 7.3\) MeV. This level has not yet been observed at present.

In the region of nuclei with \(Z>10\), which we shall now consider, the law of conservation of isotopic spin apparently already loses its meaning, so that for the classification of states by isotopic spin one may, as was noted at the end of Chapter III, use only the static properties of nuclei (energy, considerations following from the shell model, etc.). Since in all the cases that we shall consider the method of identification will be the same, we shall illustrate it in detail with the example of \(F^{20}\), \(Ne^{20}\), and \(Na^{20}\), and for the other cases we shall give only the results of the identification.

\(F^{20},\ Ne^{20},\ Na^{20}\)

For \(Ne^{20}\), \(T_\xi=0\), so that states with \(T=0,1,2,\ldots\) are possible, while for \(F^{20}\) and \(Na^{20}\) the values of \(T_\xi\) are respectively \(-1\) and \(+1\), and states with \(T=0\) cannot occur in them.

Fig. 19.

In the figure: \(F^{20}(T_\xi=-1)\), \(Ne^{20}(T_\xi=0)\), \(Na^{20}(T_\xi=1)\); levels marked include \(3.0\), \(17\) levels, \(0.65\); \(17.81\), \(3.05\), \(4.85\), \(10.5\), \(11.7\), \(9.2\), \(7.87\), \(7.43\), \(7.33\), \(7.15\), \(6.75\), \(5.4\), \(4.3\), \(2.2\), \(1.65\), with spin-parity indications \(2^+\), \(2^+\), \(1^+\), \(4^+\), \(0^+\), \(10^+\).

The nucleus \(F^{20}\) is heavier than the nucleus \(Ne^{20}\) by \(7\) MeV. If their states were similar, then, because of the larger Coulomb energy, \(Ne^{20}\) should have been heavier than \(F^{20}\) by approximately \(4.5\) MeV. Consequently, \(Ne^{20}\),

has a different isotopic spin, which \(F^{20}\) cannot have. Hence we conclude that for \(Ne^{20}\) in the ground state \(T=0\). Conversely, comparing the masses of \(F^{20}\) and \(Na^{20}\), we see that their masses differ approximately by the Coulomb energy (after subtracting the mass difference of two neutrons and two protons). This means that the two nuclei are similar. It is easy to see that for them \(T=1\).

If, for example, \(T=2\), it would follow that there should exist a nucleus \(O^{20}\) \((T_\xi=-2,\ T=2)\) lighter than \(F^{20}\) (because of the smaller Coulomb energy). Then \(F^{20}\) would

Fig. 20.

Fig. 21.

decay according to the scheme \(F^{20}(\beta^+)O^{20}\), and the isotope \(O\) would be stable (it could transform into \(Ne^{20}\) only by double \(\beta\)-decay). Hence we also conclude that the value \(T=2\) (and still more \(T>2\)) is impossible for the ground state of \(F^{20}\). Having established the isotopic spin of the ground states of \(F^{20}\) and \(Na^{20}\), we immediately find that the first state with \(T=1\) is located in \(Ne^{20}\) at an energy \(\sim 10.5\) MeV.

\[ Ne^{21},\quad Na^{21} \]

These are mirror nuclei with \(|T_\xi|=\frac12\). The isotopic spin of the ground states is equal to \(\frac12\). The position of the first level with \(T=\frac32\) is unknown.

\[ Ne^{22},\quad Na^{22} \]

For \(Na^{22}\), \(T_\xi=0\), while for \(Ne^{22}\), \(T_\xi=1\). The ground state of \(Na^{22}\) has \(T=0\), and the first state with \(T=1\) (similar to the ground state of \(Ne^{22}\)) should be located at an energy \(\sim 0.6\) MeV. In this region there is a level of \(Na^{22}\) with energy \(0.59\) MeV. If this

level has \(T=1\), then its angular momentum must be equal to the angular momentum of the ground state of \(\mathrm{Ne}^{22}\), i.e. \(I=0+\). The angular momentum of this level has not yet been measured. The excited state of \(\mathrm{Ne}^{22}\) with excitation energy \(1.27\) MeV has, like the ground state, \(T=-1\).

\[ \mathrm{Ne}^{23},\ \mathrm{Na}^{23},\ \mathrm{Mg}^{23} \]

\(\mathrm{Na}^{23}\) and \(\mathrm{Mg}^{23}\) are mirror nuclei with \(|T_\xi|=1/2\), while for \(\mathrm{Ne}^{23}\), \(T_\xi=-3/2\). The ground and all excited states of Na and Mg with energies

Fig. 22.                    Fig. 23.

less than \(8\) MeV have \(T=1/2\). The first state with \(T=3/2\) (analogous to the ground state of \(\mathrm{Ne}^{23}\)) should be located at an energy \(\sim 8.0\) MeV.

\[ \mathrm{Na}^{24},\ \mathrm{Mg}^{24},\ \mathrm{Al}^{24} \]

For \(\mathrm{Mg}^{24}\), \(T_\xi=0\), and for \(\mathrm{Na}^{24}\) and Al, \(|T_\xi|=1\). The ground and first excited states of \(\mathrm{Mg}^{24}\) (up to \(\sim 10\) MeV) have \(T=0\). The first excited state with \(T=1\) (analogous to the ground states of Na and Al) in \(\mathrm{Mg}^{24}\) should be located at an energy \(\sim 10\) MeV. It is possible that this state is precisely the one to which the \(\beta\)-decay of \(\mathrm{Al}^{24}\) occurs. In favor of this is the small \(ft\) of this transition (see\({}^{4}\), p. 105).

\[ \mathrm{Na}^{25},\ \mathrm{Mg}^{25},\ \mathrm{Al}^{25} \]

For \(\mathrm{Mg}^{25}\) and \(\mathrm{Al}^{25}\), \(|T_\xi|=1/2\), while for \(\mathrm{Na}^{25}\), \(T_\xi=-3/2\). \(\mathrm{Mg}^{25}\) and \(\mathrm{Al}^{25}\) are mirror nuclei with a similar level system. The ground and first excited states of these nuclei have \(T=1/2\). The first

state with \(T=\frac{3}{2}\) (similar to the ground state of \(\mathrm{Na}^{25}\)) must have an energy \(\sim 8\) MeV.

\(\mathrm{Mg}^{26},\ \mathrm{Al}^{26}\)

For \(\mathrm{Al}^{26}\), \(T_\xi=0\), while for \(\mathrm{Mg}^{26}\), \(T_\xi=-1\). The ground state of \(\mathrm{Al}^{26}\) has \(T=0\), and the first state with \(T=1\) (similar to the ground state of \(\mathrm{Mg}^{26}\)) must lie at an energy \(\sim 0.6\) MeV.

Fig. 24.

Fig. 25.

In this region a level with energy \(0.46\) MeV has recently been discovered which, apparently, also has \(T=1\). This conclusion is confirmed by the results of a study of the \(\beta\)-decay \(\mathrm{Al}^{26}\to\mathrm{Mg}^{26}\) (\(\ln ft=3.3\)). The spin of this state must be equal to the spin of the ground state of \(\mathrm{Mg}^{26}\), i.e. \(I=0^+\).

\(\mathrm{Mg}^{27},\ \mathrm{Al}^{27},\ \mathrm{Si}^{27}\)

\(\mathrm{Al}^{27}\) and \(\mathrm{Si}^{27}\) have \(|T_\xi|=\frac{1}{2}\), while for \(\mathrm{Mg}^{27}\), \(T_\xi=-\frac{3}{2}\). The ground and first excited states of \(\mathrm{Al}^{27}\) and \(\mathrm{Si}^{27}\) are similar and have \(T=\frac{1}{2}\). The first state with \(T=\frac{3}{2}\) (similar to the ground state of \(\mathrm{Mg}^{27}\)) must lie at an energy \(\sim 7\) MeV, and the spin of this state must be \(\frac{1}{2}^+\). The similar character of the ground states of \(\mathrm{Si}^{27}\) and \(\mathrm{Al}^{27}\) is confirmed by the small value of \(ft\) for the transition \(\mathrm{Si}^{27}\to\mathrm{Al}^{27}\) (\(\ln ft=3.6\)).

\[ \mathrm{Mg}^{28},\ \mathrm{Al}^{28},\ \mathrm{Si}^{28},\ \mathrm{P}^{28} \]

For \(\mathrm{Si}^{28}\), \(T_z=0\); for \(\mathrm{Al}^{28}\) and \(\mathrm{P}^{28}\), \(|T_z|=1\); and for \(\mathrm{Mg}^{28}\), \(T_z=-2\). Correspondingly, in \(\mathrm{Mg}^{28}\) only states with \(T\geq 2\) can occur, and in \(\mathrm{Al}^{28}\) and \(\mathrm{P}^{28}\)—states with \(T\geq 1\). The ground state and the first excited states of \(\mathrm{Si}^{28}\) have \(T=0\). The first state with \(T=1\) (similar to the ground states of \(\mathrm{Al}^{28}\) and \(\mathrm{P}^{28}\)) should be at an energy of \(\sim 9.0\) MeV. The spin of the ground state of \(\mathrm{Al}^{28}\)

Fig. 26.              Fig. 27.

is equal to \((2^+\ \text{or}\ 3^+)\). Consequently, the corresponding state in \(\mathrm{Si}^{28}\), as well as the ground state of \(\mathrm{P}^{28}\), must likewise have spins \((2^+\ \text{or}\ 3^+)\). The similar character of the ground states of \(\mathrm{P}^{28}\) and \(\mathrm{Al}^{28}\) is confirmed by the results of studies of the \(\beta\)-decays \(\mathrm{P}^{28}\to\mathrm{Si}^{28*}\) and \(\mathrm{Al}^{28}\to\mathrm{Si}^{28*}\). Both of these transitions go to one and the same level of \(\mathrm{Si}^{28}\), with excitation energy 1.78 MeV. The first state with \(T=2\) and \(I=0^+\) (similar to the ground state of \(\mathrm{Mg}^{28}\)) should be located in \(\mathrm{Si}^{28}\) at an energy of \(\sim 15\) MeV, and in \(\mathrm{Al}^{28}\) and \(\mathrm{P}^{28}\) at an energy of \(\sim 6\) MeV.

\[ \mathrm{Al}^{29},\ \mathrm{Si}^{29},\ \mathrm{P}^{29} \]

For \(\mathrm{Si}^{29}\) and \(\mathrm{P}^{29}\), \(|T_z|=1/2\), while for \(\mathrm{Al}^{29}\), \(T_z=-3/2\). The ground and first excited states of \(\mathrm{Si}^{29}\) and \(\mathrm{P}^{29}\) are similar states with \(T=1/2\).

The first states with \(T=3/2\) (analogous to the ground state of \(\mathrm{Al}^{29}\)) must lie at an energy of \(\sim 8.5\) MeV, and their spins must be equal to the spin of the ground state of \(\mathrm{Al}^{29}\), i.e. \(I=5/2^+\).

\(\mathrm{Si}^{30},\ \mathrm{P}^{30}\)

For \(\mathrm{P}^{30}\), \(T_z=0\), and for \(\mathrm{Si}^{30}\), \(T_z=-1\). The ground state of \(\mathrm{P}^{30}\) has \(T=0\), while the ground state of \(\mathrm{Si}^{30}\) has \(T=1\). The first state with \(T=1\) in \(\mathrm{P}^{30}\), analogous to the ground state of \(\mathrm{Si}^{30}\), must lie at an energy of \(\sim 0.6\) MeV.

Fig. 28.

Fig. 29.

In this region a level with energy 0.8 MeV is known. If the isotopic spin of this level is equal to unity, then its moment must be equal to the moment of the ground state of \(\mathrm{Si}^{30}\), i.e. \(I=0^+\).

\(\mathrm{Si}^{31},\ \mathrm{P}^{31},\ \mathrm{S}^{31}\)

For \(\mathrm{P}^{31}\) and \(\mathrm{S}^{31}\), \(|T_z|=1/2\), and for \(\mathrm{Si}^{31}\), \(T_z=-3/2\). The ground and first excited states of \(\mathrm{P}^{31}\) and \(\mathrm{S}^{31}\) have \(T=1/2\). The analogous character of their levels is confirmed by the small value of \(ft\) for the \(\beta\)-transition \(\mathrm{Si}^{31}\to \mathrm{P}^{31}\) (\(\log ft=3.6\)). The first states with \(T=3/2\) (analogous to the ground state of \(\mathrm{Si}^{31}\)) must have an excitation energy of \(\sim 6\) MeV and spin equal to the spin of the ground state of \(\mathrm{Si}^{31}\) (\(I=1/2^+\) or \(3/2^+\)).

Si\(^{32}\), P\(^{32}\), S\(^{32}\), Cl\(^{32}\)

For S\(^{32}\) \(|T_\xi|=0\), for P\(^{32}\) and Cl\(^{32}\) \(|T_\xi|=1\), and for Si\(^{32}\) \(T_\xi=-2\). The ground and first excited states of S\(^{32}\) have \(T=0\). The first state with \(T=1\) (similar to the ground states of P\(^{32}\) and Cl\(^{32}\)) should lie at an energy \(\sim 8\) MeV and have spin

Fig. 30.

Fig. 31.

\(1+\)—the same as in P\(^{32}\). The first state with \(T=2\) (similar to the ground state of Si\(^{32}\)) should in S\(^{32}\) lie at an energy \(\sim 13\) MeV, and in P\(^{32}\) and Cl\(^{32}\) at an energy \(\sim 5\) MeV.

P\(^{33}\), S\(^{33}\), Cl\(^{33}\)

For S\(^{33}\) and Cl\(^{33}\) \(|T_\xi|=1/2\), and for P\(^{33}\) \(T_\xi=-3/2\). S\(^{33}\) and Cl\(^{33}\) are mirror nuclei with a similar system of levels. The ground and all lower states have \(T=1/2\). The first state with \(T=3/2\), corresponding to the ground state of P\(^{33}\), should have an excitation energy of \(\sim 5.0\) MeV.

P\(^{34}\), S\(^{34}\), Cl\(^{34}\)

For Cl\(^{34}\) \(T_\xi=0\), for S\(^{34}\) \(T_\xi=-1\), and for P\(^{34}\) \(T_\xi=-2\). The \(\beta\)-decay Cl\(^{34}\to\)S\(^{34}\) has a large probability \((\log ft=3.5)\). This means that the ground states of S\(^{34}\) and Cl\(^{34}\) are similar, and therefore both

have \(T=1\), while the first excited state (with energy \(0.14\) MeV) has \(T=0\) (\(I=3+\)). The ground and first excited states of \(\mathrm{S}^{34}\) have \(T=1\). The corresponding levels should also be present in \(\mathrm{Cl}^{34}\); in particular, it should have levels with \(T=1\) at energies approximately equal to \(2.2\) MeV (\(I=2+\)), \(3.4\) MeV, and \(4.9\) MeV. These levels have not yet been observed. The first state with \(T=2\) in \(\mathrm{Cl}^{34}\) and \(\mathrm{S}^{34}\), like the ground state of \(\mathrm{P}^{34}\), should have an excitation energy of approximately \(10.5\) MeV and have spin \(1+\).

Fig. 31.

Fig. 32.

The nucleus \(\mathrm{Cl}^{34}\) is an exception to the general empirical rule according to which the ground states of all nuclei with \(T_z=0\) have \(T=0\) [24]. However, this deviation from the general rule should not cause any particular surprise. Indeed, \(\mathrm{Cl}^{34}\) is an odd-odd nucleus and, as is seen from all the experimental material we have considered, the first states with \(T=1\) in odd-odd nuclei have an energy only slightly exceeding the energy of the ground state (for example, \(0.46\) in \(\mathrm{Al}^{26}\)). Therefore even a small perturbation associated with the Coulomb interaction may lead to a change in the order of the levels.

Let us note that, since with increasing mass number of the nucleus the difference between the first levels with \(T=0\) and \(T=1\) in odd-odd nuclei continually decreases, it is natural to expect that in odd-odd nuclei heavier than \(\mathrm{Cl}^{34}\) the ground state will have \(T=1\), and not \(T=0\).

S³⁵, Cl³⁵, A³⁵

For A³⁵ and Cl³⁵, \(|T_z|=1/2\), while for S³⁵ \(T_z=-3/2\). The ground states of A³⁵ and Cl³⁵ are similar and have \(T=1/2\). The first state with \(T=3/2\) (similar to the ground state of S³⁵) should lie at an energy of \(\sim 5\) MeV and have spin \(3/2\). In the literature one often encounters the assertion that the ground state of Cl³⁵ has \(T=3/2\). This is based on data on the magnetic moment of Cl³⁵. If this assertion were correct, it would contradict the hypothesis of charge independence of nuclear forces, since the ground state of S³⁵ would have to lie below the ground state of Cl³⁵ by the amount of the excess Coulomb energy in Cl³⁵, i.e. by approximately 5 MeV; in actual fact the masses of S³⁵ and Cl³⁵ almost coincide.

Fig. 34.

Fig. 35.

Since the experimental material considered by us supports the hypothesis of charge invariance, and there is no reason to doubt even its approximate validity, it follows from our arguments that the ground state of Cl³⁵ indeed has \(T=1/2\).

S³⁶, Cl³⁶, A³⁶

For A³⁶ \(T_z=0\), for Cl³⁶ \(T_z=-1\), and for S³⁶ \(T_z=-2\). The ground state of A³⁶ has \(T=0\); the first state with \(T=1\), similar to the ground state of Cl³⁶, has an excitation energy of \(\sim 6.0\) MeV,

and the first state with \(T=2\), similar to the ground state of \(S^{36}\), has an excitation energy \(\sim 11\) MeV. The first state with \(T=2\) in \(Cl^{36}\) is found at an energy \(\sim 5\) MeV.

\(S^{37},\ Cl^{37},\ A^{37},\ K^{37}\)

In \(A^{37}\) and \(K^{37}\), \(|T_\zeta|=\frac{1}{2}\); in \(Cl^{37}\), \(T_\zeta=-\frac{3}{2}\); and in \(S^{37}\), \(T_\zeta=-\frac{5}{2}\). The ground and first excited states of \(A^{37}\) and \(K^{37}\) are similar

Fig. 36.

Fig. 37.

and have \(T=\frac{1}{2}\). The similar character of these states is confirmed by the small value of \(ft\) \((\ln ft=3.4)\) for the \(\beta\)-transition \(K^{37}\to A^{37}\). The first state with \(T=\frac{3}{2}\) (an analogue of the ground state of \(Cl^{37}\)) should have an excitation energy \(\sim 5\) MeV and spin \(\frac{3}{2}^{+}\), while the first state with \(T=\frac{5}{2}\) (an analogue of the ground state of \(S^{37}\)) should have energy \(\sim 14\) MeV. The first state with \(T=\frac{5}{2}\) in \(Cl^{37}\) should have excitation energy \(\sim 9\) MeV.

\(Cl^{38},\ A^{38},\ K^{38}\)

In \(K^{38}\), \(T_\zeta=0\); in \(A^{38}\), \(T_\zeta=-1\); and in \(Cl^{38}\), \(T_\zeta=-2\). The ground state of \(K^{38}\), according to the latest experimental data, evidently has \(T=1\) and \(I=0^{+}\), while the first excited state \((E=0.38\ \text{MeV})\)

with \(I=2^+, 3^+\) has \(T=0\). If this identification is correct, then in \(K^{38}\) the same thing occurs as in \(Cl^{34}\), i.e., the energy of the first state with \(T=1\) becomes lower than the energy of the first state with \(T=0\). Thus, the ground states of \(A^{38}\) and \(K^{38}\) are similar. In \(K^{38}\) there should be, at energies \(\sim 2.2\) MeV and \(\sim 3.8\) MeV, states with \(T=1\) and \(I=2^+\) and 3, respectively, analogous to the corresponding states of \(A^{38}\). The first state with \(T=2\), similar to the ground state of \(Cl^{38}\), should in \(A^{38}\) and \(K^{38}\) have an excitation energy of approximately \(10.5\) MeV and have spin \(2^-\).

\(Cl^{39},\ A^{39},\ K^{39},\ Ca^{39}\)

In \(K^{39}\) and \(Ca^{39}\), \(|T_\zeta|=1/2\); in \(A^{39}\), \(T_\zeta=-3/2\), and in \(Cl^{39}\), \(T_\zeta=-5/2\). The ground states of \(K^{39}\) and \(Ca^{39}\) are similar and have \(T=1/2\). The first state with \(T=3/2\) (the analogue of the ground state of \(A^{39}\)) in these nuclei

Fig. 38.

should have an excitation energy of \(\sim 7\) MeV and spin \(7/2^-\). The first state with \(T=5/2\) (the analogue of the ground state of \(Cl^{39}\)) should be located at \(\sim 16\) MeV and have spin \(3/2^+\). The first state with \(T=5/2\) in \(A^{39}\) should have an excitation energy of \(\sim 9\) MeV and spin \(3/2^+\).

\(A^{40},\ K^{40},\ Ca^{40},\ Sc^{40}\)

In \(Ca^{40}\), \(T_\zeta=0\); in \(K^{40}\) and \(Sc^{40}\), \(|T_\zeta|=1\); and in \(A^{40}\), \(T_\zeta=-2\). The ground and first excited states of \(Ca^{40}\) have \(T=0\). The first state with \(T=1\) (the analogue of the ground states of \(K^{40}\) and \(Sc^{40}\)) should have an excitation energy of \(\sim 8\) MeV and spin \(4^-\), and the first state

...state with \(T=2\) (an analog of the ground state of \(A^{40}\)) should lie at an energy \(\sim 12.9\) MeV and have spin \(0+\). The first state with \(T=2\) should lie in \(K^{40}\) and \(Sc^{40}\) at energies \(\sim 4.9\) MeV and have spin \(0+\).

Fig. 39.

\[ A^{41},\ K^{41},\ Ca^{41},\ Sc^{41} \]

In \(Ca^{41}\) and \(Sc^{41}\), \(|T_\zeta|=1/2\); in \(K^{41}\), \(T_\zeta=-3/2\); and in \(A^{41}\), \(T_\zeta=-5/2\). The ground states of \(Ca^{41}\) and \(Sc^{41}\) have \(T=1/2\); the first state with \(T=3/2\) (an analog of the ground state of \(K^{41}\)) should have an excitation energy \(\sim 5.0\) MeV and spin \(3/2+\); the first state with \(T=5/2\) (an analog of the ground state of \(A^{41}\)) should lie at an energy \(\sim 14\) MeV. The first state with \(T=3/2\) in \(K^{41}\) should have an excitation energy \(\sim 9\) MeV.

\[ A^{42},\ K^{42},\ Ca^{42} \]

In \(Ca^{42}\), \(T_\zeta=-1\); in \(K^{42}\), \(T_\zeta=-2\); and in \(A^{42}\), \(T_\zeta=-3\). The isotope \(Sc^{42}\) has not been observed. The ground state of \(Ca^{42}\) has \(T=1\) and spin \(0+\). The first excited state with \(I=2+\) also has \(T=1\) and strongly resembles analogous states (\(T=1,\ I=2+\) and approximately the same energy) in the following odd-odd nuclei with \(T_\zeta=0\): \(Li^{6}\), \(Ne^{22}\), \(A^{38}\). The first state with \(T=2\) (an analog of the ground...

of the state of \(K^{42}\)) should have an energy \(\sim 9.6\) MeV and spin \(2^{-}\). The first state with \(T=3\) (the analogue of the ground state of \(A^{42}\)) should have an energy not less than \(15.6\) MeV, since it is known that the binding energy of \(A^{42}\) is less than the binding energy of \(K^{42}\).

Fig. 40.

Fig. 41.

The isotope \(Sc^{42}\) is unknown. This is the first case in which there is no nucleus with \(T_\zeta=0\). Hence one may conclude that the Coulomb energy in the nucleus has increased so much that the possibility arises of the direct decay
\[ Sc^{42}\to Ca^{41}+p. \]

\(K^{43},\ Ca^{43},\ Sc^{43}\)

For \(Sc^{43}\), \(T_\zeta=-\frac{1}{2}\); for \(Ca^{43}\), \(T_\zeta=-\frac{3}{2}\); and for \(K^{43}\), \(T_\zeta=-\frac{5}{2}\). The ground state of \(Sc^{43}\) has \(T=\frac{1}{2}\); the first state with \(T=\frac{3}{2}\) (the analogue of the ground state of \(Ca^{43}\)) should lie at an energy \(\sim 4\)–\(5\) MeV and have spin \(\frac{7}{2}\). The first state with \(T=\frac{5}{2}\) (the analogue of the ground state of \(K^{43}\)) should be in \(Ca^{43}\) at an energy \(\sim 5\)–\(6\) MeV, and in \(Sc^{43}\) at an energy \(\sim 10\) MeV.

\(K^{44},\ Ca^{44},\ Sc^{44}\)*)

For \(Sc^{44}\), \(T_\zeta=-1\); for \(Ca^{44}\), \(T_\zeta=-2\); for \(K^{44}\), \(T_\zeta=-3\). The ground state of \(Sc^{44}\) has \(T=1\); the first state with \(T=2\) (the analogue

\[ \text{*) Mass differences and spins for the subsequent nuclei are taken from decay schemes in Tables }^{25}. \]

of the ground state of \(Ca^{44}\)) has an excitation energy \(\sim 3\) MeV, and the first state with \(T=3\) (an analog of the ground state of \(K^{44}\)) has an energy \(>11\) MeV.

\(Ca^{45},\ Sc^{45},\ Ti^{45}\)

In \(Ti^{45}\), \(T_\zeta=-\frac{1}{2}\); in \(Sc^{45}\), \(T_\zeta=-\frac{3}{2}\); in \(Ca^{45}\), \(T_\zeta=-\frac{5}{2}\). The ground state of \(Ti^{45}\) has \(T=\frac{1}{2}\); the first state with \(T=\frac{3}{2}\) (an analog of the ground state of \(Sc^{45}\)) has an excitation energy \(\sim 5\) MeV, and the first state with \(T=\frac{5}{2}\) (an analog of the ground state of \(Ca^{45}\)) has an energy \(\sim 12\) MeV.

Fig. 42.

\(Ca^{46},\ Sc^{46},\ Ti^{46}\)

In \(Ti^{46}\), \(T_\zeta=-1\), while in \(Sc^{46}\), \(T_\zeta=-2\), and in \(Ca^{46}\), \(T_\zeta=-3\). The ground state of \(Ti^{46}\) has \(T=1\); the first state with \(T=2\) (an analog of the ground state of \(Sc^{46}\)) has an energy \(\sim 9\) MeV, and the first state with \(T=3\) (an analog of the ground state of \(Ca^{46}\)) has an energy less than \(16\) MeV.

\(Ca^{47},\ Sc^{47},\ Ti^{47},\ V^{47}\)

In \(V^{47}\), \(T_\zeta=-\frac{1}{2}\); in \(Ti^{47}\), \(T_\zeta=-\frac{3}{2}\); in \(Sc^{47}\), \(T_\zeta=-\frac{5}{2}\); and in \(Ca^{47}\), \(T_\zeta=-\frac{7}{2}\). The ground state of \(V^{47}\) has \(T=\frac{1}{2}\); the first state with \(T=\frac{3}{2}\) (an analog of the ground state of \(Ti^{47}\)) should have an excitation energy \(\sim 6\) MeV, the first state with \(T=\frac{5}{2}\) (an analog of the ground state of \(Sc^{47}\)) an energy \(\sim 14\) MeV, and the first state with \(T=\frac{7}{2}\) (an analog of the ground state of \(Ca^{47}\)) an energy \(\sim 23\) MeV.

\(Ca^{48},\ Sc^{48},\ Ti^{48},\ V^{48}\)

In \(V^{48}\), \(T_\zeta=-1\); in \(Ti^{48}\), \(T_\zeta=-2\); in \(Sc^{48}\), \(T_\zeta=-3\); and in \(Ca^{48}\), \(T_\zeta=-4\). The ground state of \(V^{48}\) has \(T=1\); the first state with \(T=2\) (an analog of the ground state of \(Ti^{48}\)) should have an energy \(\sim 3\) MeV, the first state with \(T=3\) (an analog of the ground state of \(Sc^{48}\)) should have an energy \(\sim 14\) MeV and, finally, the first state with \(T=4\) (an analog of the ground state of \(Ca^{48}\)) an energy \(\sim 21\) MeV.

\( \mathrm{Ca}^{49},\ \mathrm{Sc}^{49},\ \mathrm{Ti}^{49} \)

For \(\mathrm{Ti}^{49}\), \(T_z=-\frac{5}{2}\); for \(\mathrm{Sc}^{49}\), \(T_z=-\frac{7}{2}\); and for \(\mathrm{Ca}^{49}\), \(T_z=-\frac{9}{2}\). The first level with \(T=\frac{7}{2}\) lies in \(\mathrm{Ti}\) at an energy of \(\sim 9\) MeV; the first level with \(T=\frac{9}{2}\), at an energy of \(\sim 19\) MeV.

\( \mathrm{Ti}^{50},\ \mathrm{V}^{50},\ \mathrm{Cr}^{50} \)

For \(\mathrm{Cr}^{50}\), \(T_z=-1\); for \(\mathrm{V}^{50}\), \(T_z=-2\); and for \(\mathrm{Ti}^{50}\), \(T_z=-3\). The ground state of \(\mathrm{Cr}^{50}\) has \(T=1\); the first state with \(T=2\) (the analogue of the ground state of \(\mathrm{V}^{50}\)) should have an energy of \(\sim 8\text{–}9\) MeV, and the first state with \(T=3\) (the analogue of the ground state of \(\mathrm{Ti}^{50}\)) an energy of \(\sim 13\text{–}14\) MeV.

References

1. I. S. Shapiro, Uspekhi Fiz. Nauk 53, 7 (1954).
2. G. I. Zel'der, Uspekhi Fiz. Nauk 53, 455 (1954).
3. F. Ajzenberg and T. Lauritsen, Rev. Mod. Phys. 24, 321 (1952).
4. M. Endt and J. C. Kluyver, Rev. Mod. Phys. 26, 95 (1954).
5. C. K. Bockelman, C. P. Browne, W. W. Buechner, and A. Sperduto, Phys. Rev. 92, 664 (1953).
6. L. Radicati, Phys. Rev. 87, 521 (1952).
7. D. C. Peaslee and V. L. Telegdi, Phys. Rev. 92, 126 (1953).
8. L. Radicati, Proc. Phys. Soc. A66, 189 (1953).
9. L. Radicati, Proc. Phys. Soc. A67, 39 (1953).
10. D. Wilkinson, Phil. Mag. 44, 1019 (1953).
11. D. Wilkinson, Phil. Mag. 44, 1322 (1953).
12. Clegg, D. Wilkinson, Phil. Mag. 44, 1269 (1953).
13. D. H. Wilkinson, Phys. Rev. 90, 721 (1953).
14. B. S. Dzhelepov, Izv. AN, Ser. Fiz. 17, 391 (1953).
15. A. I. Baz' and Ya. A. Smorodinskii, JETP 27, 382 (1954).
16. R. Malm and D. R. Inglis, Phys. Rev. 95, 993 (1954).
17. T. W. Bonner and C. F. Coos, Phys. Rev. 96, 122 (1954).
18. J. Wilkins et al., Proc. Phys. Soc. A64, 1056 (1951).
19. D. H. Wilkinson and G. A. Jones, Phys. Rev. 91, 1575 (1953).
20. G. Jones and D. H. Wilkinson, Phys. Rev. 90, 722 (1953).
21. D. H. Wilkinson, Nature 172, 172 (1953).
22. C. Hsiao and V. Telegdi, Phys. Rev. 90, 494 (1953).
23. M. Gell-Mann, L. W. Jeledy, Phys. Rev. 172, 576 (1953).
24. D. C. Peaslee, Nuovo Cimento 10, 1349 (1953).
25. J. M. Hollander, I. Perlman, G. T. Seaborg, Rev. Mod. Phys. 25, 469 (1953).
26. B. S. Dzhelepov, Izv. AN SSSR, Ser. Fiz. (1954).

1. These expectations are based on the fact that precisely in this state the nuclei \(C^{12}\) are formed in the reaction \(O^{16}(\gamma\alpha)C^{12}\) (see the analogous reaction \(C^{12}(\gamma\alpha)Be^{8}\) in the section on \(Be^{8}\)). ↩↩