ALL-UNION NUCLEAR CONFERENCE

M. P. Bronstein, Leningrad

On September 24–30, 1933, the First All-Union Nuclear Conference, organized by the Leningrad Physico-Technical Institute, was held in Leningrad. At the conference a number of reports by Soviet and foreign scientists were heard, and lively discussions took place. Below are brief accounts of the most important reports, compiled from the notes that I kept as one of the secretaries of the conference. Readers of Uspekhi fizicheskikh nauk will be able to become acquainted with the contents of approximately three quarters of the communications presented at the conference. Some no less important communications—for example, the report by D. V. Skobeltsyn (Leningrad) on cosmic rays, the reports by S. E. Frisch (Leningrad) and Rasetti (Rome) on the mechanical and magnetic moments of nuclei, the reports by the Kharkov physicists (Sinyelnikov, Santsov, Leipunsky) on their work on methods for obtaining ionic and electronic streams of high voltages and on the disintegration of nuclei—cannot, for various technical reasons, be included in our survey. Despite this incompleteness, we hope that our survey will be of interest to those Soviet physicists who, for one reason or another, were unable to attend the conference.

The report of P. A. M. Dirac (Cambridge) was devoted to the theory of the positron. The discovery of the positron has again revived interest in one old theory of Dirac’s (1929), which predicted the existence of particles apparently possessing all the observed properties of positrons. It is known that in nonrelativistic theory the kinetic energy has the form \(W = \frac{1}{2}mv^{2}\) and, consequently, is always positive. But in Einstein’s special theory of relativity another relation holds, namely:

\[ W^{2}=m^{2}c^{4}+c^{2}p^{2} \]

where \(p\) is the momentum. Hence

\[ W=\pm\sqrt{m^{2}c^{4}+c^{2}p^{2}}. \]

Of the two possible signs of the square root, in classical theory the plus sign is always chosen. This is done so that a positive value should be maintained. But quantum theory cannot confine itself to simply ignoring such states of the particle in which the kinetic energy is negative. The wave equation has the property that the presence of any external perturbation will lead to transitions into states with negative kinetic energy. (In classical theory this was not the case, since \(W\) was either greater than \(mc^{2}\) or less than \(-mc^{2}\), and all classical dynamical variables are continuous and cannot jump across a gap of width \(2mc^{2}\).) Many have tried to eliminate, in quantum theory as well, the possibility of such transitions by modifying this theory. Thus, for example, Schrödinger did this, but his theory suffered from the defect that it did not satisfy the requirement of relativistic invariance, and one may think that any modification of quantum theory eliminating transitions into states with negative kinetic energy will suffer from the same fatal defect. Therefore there are two

possibility: either quantum theory is in general mistaken in predicting these transitions, or states with negative kinetic energy have physical meaning. Let us examine the first possibility. Every physical theory has its limits of applicability. In particular, quantum theory too ceases to be applicable when one has to operate with lengths of the order of \(e^2/mc^2\) (the classical electron radius), since the existence and stability of the electron are not explained by quantum theory. If this length \(e^2/mc^2\) is regarded as a wavelength, then, converting it into photon energy, we obtain about 60 million electron-volts. Therefore quantum theory cannot give correct results in the region of energies exceeding this number. But \(2mc^2\) has a magnitude of the order of only one million volts, and there is no reason for quantum theory to be incorrect in this region of energies. Therefore the first possibility seems unsatisfactory, and we must settle on the second. What physical meaning should be ascribed to states with negative energy? (The word “kinetic” will hereafter be omitted for brevity.) Every particle possessing negative energy has strange properties: the faster it moves, the higher is its energy, and therefore it must be very unstable and, spontaneously losing energy, acquire enormous velocities. There is no logical impossibility in this, but do such particles exist in nature? If, for example, such an electron moves through a given electromagnetic field, then it must be deflected in such a direction as if it had a positive charge (negative energy is equivalent to negative mass, and since only \(e/m\) enters into the equations of motion, one may leave the sign of the mass unchanged, changing only the sign of the charge). All this reminds us of the positron. But the positron has positive energy. This shows that the explanation must be somewhat modified. Therefore Dirac proposes the following hypothesis: let us suppose that in the real world almost all electronic states with negative energy are occupied by electrons, but that we never perceive such a distribution of electrons because of its homogeneity, and only levels unoccupied by electrons, being something exceptional, can be observed by us in the same way as we observe occupied states with positive energy. It is easy to see that such an unoccupied state with negative energy, or “hole,” in the region of negative states will behave like an ordinary particle with a positive charge. Indeed, the absence of negative energy may be interpreted as the presence of positive energy, since two minuses give a plus. Moreover, considering the motion of electrons sitting on those levels with negative energy near which a “hole” is located, it is easy to see that the “hole” will move in a given external field as a particle with positive charge. The question remains only of how a field produces a “hole,” but certain difficulties are connected with this question, and it will be discussed in detail below.

From Dirac’s hypothesis a number of consequences follow. If it is assumed that the positron is such a “hole,” then

1) the positron charge must be exactly equal to \(+e\),

2) its mass must be exactly equal to the mass of the electron, as Weyl and others have shown.

Is this confirmed by experiment? The ratio of the positron charge to its mass is known with an accuracy only up to 50%, but in any case the fact that it is known is not inconsistent with the above assertions.

3) An electron with positive kinetic energy cannot pass into an already occupied state with negative energy, but it can pass into an unoccupied state, i.e. into a “hole.” In this process “annihilation” occurs, i.e. the hole is filled and both particles (the electron with positive energy and the positron) cease to be observable. If, together with such a system, we choose a frame of reference in which the center of gravity of both particles is at rest, i.e. the resultant momentum is zero, then from the laws of conservation of energy and momentum it follows that, as a result of annihilation, at least two quanta of radiant energy must appear. The theory makes it possible to calculate the probability of annihilation. A lengthy calculation shows that the order of magnitude of this

of the probability that would be the same as if the electron and positron were two spheres with radius \(e^2/mc^2\), which have speed \(c\) and must annihilate in each collision with each other. The exact value of the annihilation probability depends on the speed of the positron. The theory leads to the result that the probable lifetime of a positron moving slowly through air (at a pressure of \(1\) atm and at normal temperature) is \(3\cdot 10^{-7}\) sec. If the positron is moving rapidly, then the lifetime is somewhat longer. Such a value of the lifetime is in agreement with experiment, since, on the one hand, it is sufficiently large for fast positrons to pass through a Wilson chamber without being destroyed on the way; on the other hand, it is so small that positrons cannot be as ordinary an object of laboratory study as negative electrons.

There is also the possibility of another annihilation process: in the presence of a third body that can take upon itself the excess momentum—for example, in the presence of an atomic nucleus—the laws of conservation of energy and momentum no longer require that two quanta be emitted, and therefore one quantum may also be emitted. The inverse process is as follows: one quantum in the field of an atomic nucleus is absorbed with the simultaneous creation of a positron and an ordinary electron with positive energy. It is easy to see that this process consists in ejecting, by a quantum, one of the electrons with negative energy and transferring it into a state with positive energy; i.e., in essence this is nothing other than the photoelectric effect. The quantitative theory of such a photoelectric effect was developed independently by Nayerls, Oppenheimer, and Fermi, but at present differs only in details. Oppenheimer proves that heavy nuclei will promote such a photoelectric effect more than light ones; moreover, on the basis of such a theory it is possible to give an interpretation of the anomalous scattering of gamma rays in good agreement with the measurements of Gray and Tarrant.

Let us now pass to the question of the field produced by a positron. In contrast to the question of the motion of a positron in a given field, the question of the field of the positron itself has not yet been finally elucidated. For we have an infinite multitude of states with negative energy. One may even assert that the number of electrons with negative energy in a unit volume must prove infinite. One must be able to work mathematically with this infinity. If we make the usual assumption: \(\operatorname{div} E = 4\pi \rho\), then for infinite \(\rho\) it loses all meaning. Therefore one has to make the assumption that some “normal” distribution of electrons over states does not produce any field, and that only the difference between the actual distribution and this normal one is the source of the field. It is natural to choose as the normal distribution that in which all states with negative kinetic energy are occupied and all states with positive kinetic energy are free—in this case it will turn out that the positron field coincides with the field of the charge \(+e\). However, the matter is so simple only in the absence of an external field: if such a field exists, then the states with a given kinetic energy prove to be nonstationary, and therefore the “normal” distribution must be defined in some other way. In addition, this distribution must be subtracted from the actual charge distribution; i.e., one must subtract one infinity from another, although such an operation is not defined in mathematics. Only in one case has it been possible to solve this problem satisfactorily, namely in the case of a static field (Dirac and Nayerls). For this Dirac applies his method, which in his book (Russian translation, p. 254) is called the “density method.” Each electron is characterized by a wave function \(\psi_r(q)\), where \(r\) is the number of the electron, and \(q\) its coordinates (including the spin coordinate as well). The density of electrons is represented in the form of a matrix

\[ \sum_r \psi_r(q')\psi_r(q'')=(q'|R|q''), \]

which thus determines the distribution of electrons. It must satisfy the condition

\[ R^2=R, \]

which gives that all eigenvalues of this matrix are equal either to zero or to one (the Pauli principle: in each state there is no more than one electron). The quantum equation of motion for this matrix has the form:

\[ i h \dot R = H R - R H, \]

where \(H\) is the Hamiltonian of one electron, including the external field and the field of the remaining electrons.

What distribution should be chosen as the “normal” one, i.e. the one that does not produce a field? Dirac and Paierls choose as such the distribution

\[ R_0=\frac{1}{2}\left(1-\frac{W}{|W|}\right) \]

where \(W\) is the kinetic-energy operator:

\[ \frac{W}{c}=\rho_1(\sigma,p)+\rho_3 mc. \]

(In order to construct \(R_0\) as a function of \(W\), one must choose such a “representation” in which the matrix \(W\) is diagonal; \(R_0\) will be diagonal in the same representation, and its eigenvalues corresponding to positive eigenvalues of \(W\) are, as is easy to see, zero, while the eigenvalues corresponding to negative \(W\) are equal to 1.) The reason why \(R_0\) is expressed in terms of \(W\), and not in terms of \(H\) (through the kinetic, and not the total energy), is that the Hamiltonian \(H\) contains the potential, to which one may, without changing its physical meaning, add any constant, whereas \(W\) is invariant with respect to all such changes of the potential.

The distribution \(R_0\) is stationary. Subtracting \(R_0\) from the distribution \(R\), satisfying the condition

\[ 0=H R-R H,\quad R^2=R, \]

we obtain a density that produces the field. If the external field is not very strong, then perturbation theory may be applied. The details of the calculations are of no interest, but something should nevertheless be noted. If one takes the diagonal element \(R-R_0\), with respect to all four coordinates, and sums over the spin values, one obtains the desired density. It is obtained in the form of a divergent integral, which may seem catastrophic, but in fact this divergence of the integral is by no means so malignant. For the theory, as we see, is not applicable to very large energies (greater than 60 million volts), and the integral diverges precisely because of very deep negative levels. If the integral is simply cut off near 60 million volts (i.e. near the momentum \(p=137\,mc\)), then it becomes finite, and meanwhile it is arranged so that the exact place where it is cut off is of no essential significance. As a result one obtains for the density the approximate expression

\[ \frac{e^2}{hc}\left\{\rho-\frac{2}{15\pi}\left(\frac{h}{mc}\right)^2\nabla^2\rho\right\} \]

where \(\rho\) is the density producing the external field. The expression just written is the density that is superposed on \(\rho\). Physically this may be expressed by saying that the charges producing the external field produce a certain deformation, or “polarization,” in the distribution of the electrons sitting on the negative levels, and the density of the “polarized” distribution is superposed on the initial density \(\rho\). The principal role is played by the first term, which simply neutralizes the \(\frac{1}{137}\) part of the charge producing the external field. Hence follows the astonishing result that the values of static charges measured by us are smaller by \(\frac{1}{137}\) than the actual values of these

charges. (The second term of the formula only leads to the fact that this compensating density of the polarized distribution does not exactly coincide spatially with the points where the charges that produce the external field are located, but is smeared around these points in small volumes with linear dimensions \(h/mc\).) It may be thought that, for rapidly moving charges, such a screening action of electrons with negative energy does not take place. This would not be strange, but physically it is quite natural, since at the point through which a charge passes very rapidly, the polarization simply does not have time to be established. If this is so, then there is a possibility of experimentally testing the conclusions of the theory by using those phenomena in which very fast electrons play a role (the scattering of very fast electrons, scattering of very hard gamma quanta by electrons). Since in the formulas describing these phenomena the electron charge occurs (in Klein–Nishina’s formula, for example, in the fourth power), then, substituting into these formulas the electron charge measured statically, we should obtain a discrepancy with experiment. For example, according to the Klein–Nishina formula one should obtain a value \(3\%\) smaller than in reality. As for Rutherford’s formula for the scattering of electrons, in it a discrepancy should also be obtained, owing, moreover, to the “polarization” density smeared at a distance of \(10^{-11}\ \text{cm}\) around the nucleus, which is equivalent to a departure from Coulomb’s law.

Dirac emphasizes that this work, done jointly with Peierls, was done so recently that not everything in it may have had time to become sufficiently clear. Therefore these results should still be considered preliminary. As for the theory as a whole, it is not fully completed, for it is necessary 1) to solve the problem of the “polarization” of a homogeneous background of electrons with negative energy in arbitrary electromagnetic fields, 2) to present the whole theory in relativistically invariant form.

The report of Frédéric Joliot (Paris) on neutrons contained a review of the experimental facts obtained in the laboratory of F. Joliot and his wife Mme Irène Curie-Joliot (daughter of the famous Mme Curie-Skłodowska, discoverer of radium). This laboratory possesses a very active polonium preparation (150 millicuries), giving more than a billion alpha particles per second. The intensity of neutron emission is determined by ionization produced by protons knocked out by the stream of neutrons from paraffin or from some gas containing hydrogen (methane, or still better butane). Gamma rays are emitted together with the neutrons, but in order for emission to occur, the energy of the alpha particles must exceed a certain threshold; for gamma rays this threshold is lower than for neutrons. Hence it follows that, at some intermediate alpha-particle energies, excitation of the nucleus is possible with subsequent emission of gamma rays, but without the ejection of neutrons (for nitrogen only gamma rays were observed). As the energy of the alpha particles increases, the intensities of the emitted neutrons and gamma rays increase, the two curves running parallel to one another. For beryllium-9 two groups of fast neutrons are observed (with energies \(4.5 \cdot 10^{6}\) and \(7.8 \cdot 10^{6}\) electron-volts).

Of very great interest in Joliot’s report was his presentation of his point of view on the mass of the neutron. It is known that J. Chadwick derived the value of the mass of the neutron from the energy balance of the reaction in which a neutron is knocked out of boron. The formula for this reaction, according to Chadwick, is:

\[ \mathrm{B}_{11} + \alpha \longrightarrow \mathrm{N}_{14} + \omega \]

(\(\alpha\) denotes an alpha particle, \(\omega\) a neutron; the numbers correspond to the atomic weights). Adding on the left the mass corresponding to the kinetic energy of the incident alpha particle, and on the right the analogous mass corresponding to the kinetic energy of the neutron (this energy, for this purpose, need only be known approximately), and knowing the masses of the nuclei \(\mathrm{B}_{11}\) and \(\mathrm{N}_{14}\) from Aston’s data, Chadwick found that the mass of the neutron lies between \(1.005\) and \(1.008\). Therefore he accepted that it is equal to \(1.0065\) (unit of mass = one quarter of the mass of the helium atom). But if one accepts

this, then, as Joliot asserts, gives rise to a number of difficulties: first, it follows that the neutron emitted in the reaction

\[ \mathrm{Li}_7+\alpha \longrightarrow \mathrm{B}_{10}+\omega \]

must have an energy of \(12\cdot 10^6\) volts. But experiment, as we have seen, does not give more than \(7.8\cdot 10^6\) volts. True, the American Crane obtains in experiments with Wilson’s chamber faster neutrons emitted in this reaction, but Joliot for a number of reasons considers his measurements erroneous. It is therefore necessary to suppose that in this reaction photons are also emitted, carrying away the excess energy. This would seem to remove this difficulty. But a second arises: according to Bainbridge’s measurements the mass of the \(\mathrm{Be}_9\) nucleus is equal to 9.011. This is greater than the sum of the masses of two alpha-particles and one neutron, if the neutron is assigned the Chadwick mass. Therefore beryllium ought to turn out to be radioactive, i.e. to emit an alpha-particle. Henderson’s recent experiments have shown that, contrary to the assertion made by Langer and Rait, beryllium is not radioactive. This is fatal for the Chadwick value of the mass. Joliot therefore makes the hypothesis that the neutrons are emitted not from the isotope \(\mathrm{B}_{11}\), but from the isotope \(\mathrm{B}_{10}\), and that the formula of the reaction is as follows:

\[ \mathrm{B}_{10}+\alpha \longrightarrow \mathrm{C}_{13}+\omega+\varepsilon_{+}. \]

where \(\varepsilon_{+}\) denotes a positive electron. From this one obtains for the mass of the neutron the number 1.011, and the difficulty in explaining the stability of beryllium-nine now disappears. As for the reaction of knocking a neutron out of beryllium, there too one has to suppose that some excess of energy is carried away by a photon or by a pair consisting of electrons of both signs. The value of the neutron mass proposed by Joliot leads to a difficulty in explaining the balance of the reaction

\[ \mathrm{Li}_7+\alpha \longrightarrow \mathrm{B}_{10}+\omega, \]

if one sets \(\mathrm{Li}_7=7.011\) (according to Bainbridge) and \(\mathrm{B}_{10}=10.008\) (according to Aston), but this difficulty, Joliot thinks, will be clarified if comparable results are obtained for \(\mathrm{Li}_7\) and \(\mathrm{B}_{10}\). For the time being, however, the knocking-out of neutrons from lithium is an argument in favor of Chadwick and against Joliot.

The question of the exact value of the neutron mass is very interesting, since it is connected with the question of which is more stable—the neutron or the hydrogen atom (see also the exposition of Perrin’s report).

The report of Francis Perrin (Paris) was devoted to the question of the structural elements of the atomic nucleus. There exist three different points of view: according to one of them, the oldest, the fundamental elements of which nuclei are built are protons and electrons. The neutron is regarded as a complex consisting of an electron and a proton. If this point of view is treated in the spirit of ordinary wave mechanics, it immediately leads to known difficulties: one obtains the wrong rotational moment and wrong statistics for nuclei of nitrogen (the so-called “nitrogen catastrophe”); it remains incomprehensible why the nucleus does not capture electrons from outside even when this is energetically favorable, etc. Another point of view, proposed by Jean Perrin (the speaker’s father), is that nuclei consist of neutrons and positive electrons (positrons). The proton itself is then interpreted as a composite particle consisting of a positron and a neutron. Beta decay is in this case interpreted as the simultaneous birth of a positron and an electron, of which the former is captured by a neutron and the other flies away. This point of view, when treated in the spirit of ordinary wave mechanics, also leads to difficulties. In particular, it would have to be assumed that positrons obey Bose statistics and do not possess spin. This would destroy the symmetry between positrons and electrons required by Dirac’s theory. It should be noted that all these difficulties are probably connected with the application of nonrelativistic wave mechanics, and this application is in any case illegitimate, since, confined within the volume of the nucleus (linear dimensions from \(10^{-13}\) to \(10^{-12}\) cm),

electrons or positrons must have velocities close to the speed of light. It is possible that in a future relativistic theory the contradiction between the two points of view will prove to be considerably smaller than now appears. Within the framework of the existing theory it is much more convenient to make use of a third point of view, namely: nuclei consist of protons and neutrons, which are elementary particles capable of transforming into one another. It is this point of view that Perrin chiefly uses.

Protons and neutrons form among themselves various combinations that retain within nuclei a more or less individual existence. Among such combinations is the alpha particle (“helion” in Perrin’s terminology). Since it is very stable, it is natural to suppose that all nuclei consist of alpha particles, neutrons, and no more than one proton (if there were another proton, then the two protons together with two neutrons would form an alpha particle). Further, Perrin draws attention to the fact that there do not exist nuclei in which the proton would be bound with alpha particles without any neutrons. From this Perrin concludes that the proton in the nucleus is always bound with a neutron—if not as part of the alpha particle itself, then at least in the form of a proton-neutron group. This group, which Perrin calls a “hydron,” is identical with the nucleus of the hydrogen isotope (“deiton,” in American terminology). It has not yet been possible to decide the question of the disintegration and ejection of the deiton, unless one counts one of the “forks” photographed by Blackett, which, according to Perrin, can be explained without contradiction to the law of conservation of energy only on the assumption that the nitrogen nucleus, colliding with an alpha particle, gives an oxygen nucleus and a deiton.

Perrin discussed the mass remaining for the neutron. According to Chadwick this mass is equal to 1.0065 (the unit of mass is one quarter of the mass of the helium atom); according to Joliot it is equal to 1.011. Against Chadwick, Perrin advances the consideration that the hydrogen atom is stable and does not turn into a neutron, although its mass is greater than 1.0065. It is true that against Joliot an analogous argument would be that no one has observed the transformation of a neutron into a hydrogen atom. But it may be that such a transformation does take place, Perrin says, for after all we do not observe the free neutron for any considerable length of time. Recently Lawrence, on the basis of the energy balance of the disintegrations he observed in collisions of deitons with various nuclei, attributed to the neutron even the mass 0.9995. However, Perrin does not consider this hypothesis serious. Lawrence proceeded from the assumption that the deiton, breaking up in this collision, gives a fast proton and a neutron, which also flies away. But according to Perrin it is possible that the neutron in this case does not fly away, but remains as part of the nucleus. In all cases investigated by Lawrence, such capture should in fact give a very stable combination, and by this, according to Perrin, the large energy of the departing proton is explained. (In the subsequent discussion A. I. Leipunskii and I. V. Kurchatov proposed, in order to test Perrin’s view, bombarding with deitons such nuclei as do not give stable combinations with a neutron, for example aluminum. If fast protons are obtained in this case as well, this speaks against Perrin’s interpretation.) Thus the mass determined by Lawrence pertains not to the free neutron but to the bound neutron. This mass (0.9995) is very close to the ordinary value of the difference between the masses of two isotopes of one and the same element, differing from one another by one neutron. (Let us note that the same interpretation of Lawrence’s experiments as Perrin gave was independently proposed by V. Elsasser.) Perrin considers the absence of radioactivity in the beryllium-nine nucleus to be the most decisive argument in favor of F. Joliot’s proposed value of the neutron mass. If this value is accepted, then beryllium-nine proves to be stable, although readily disintegrated. The slow group of neutrons emitted by beryllium, discovered by Pierre Auger, can be explained, as Elsasser indicated, by the assumption that the alpha particle is not captured in this process.

In conclusion Perrin gave various considerations as to how the protons and neutrons of the nucleus are distributed among the energy levels present there. It is possible that the assumption of the formation of the “maximum” number of particles,

is too crude, since the binding energy between alpha particles in the nucleus cannot be regarded as small in comparison with their internal binding energy. It is therefore more appropriate to speak of the nucleus as an aggregate of protons and neutrons, and not of alpha particles with a tail of neutrons and at most one proton.* In this way, in explaining the stability of the carbon nucleus (“carbon-12”), Perrin makes, from the author’s point of view, a very strange hypothesis: that the six protons and six neutrons of this nucleus sit on the second level, leaving the first, deeper level free (two protons and two neutrons can fit on this level). Perrin thinks that the absence or large rarity of beryllium-8 nuclei is explained by the impossibility of arranging the protons and neutrons of this nucleus so that there are no partially filled levels. From our point of view such an explanation is highly suspect.

Report by L. H. Gray (Cambridge) dealt with the anomalous scattering of gamma rays discovered by Chao and since then investigated by Tarrant and Gray. The question is of enormous interest in connection with the theoretical investigations discussed below. The principal results of Gray and Tarrant had been published at the time; in the present account of Gray’s report we shall note only what had not yet been published. First of all, the number of substances in which gamma-ray scattering was studied has now increased. Pb, Sn, Fe, Cu, K₂SO₄, C, H₂O were investigated. It turned out that the anomalous scattering varies very smoothly with atomic number. In particular, the fact that copper, having an odd atomic number, lies on the same curve as the elements of even atomic number may be regarded as a convincing argument against the hypothesis on the causes of scattering proposed by Gamow (artificial beta decay under the influence of gamma rays). Scattering by carbon, whose nucleus consists only of alpha particles, also speaks against Gamow’s hypothesis.

At present Gray no longer considers that the anomalously scattered rays are characteristic of any constituent element of the nucleus. The primary radiation acts directly on the secondary radiation, and for each quantum of the primary rays with energy greater than 2 million volts there are four or more soft quanta (of \( \frac{1}{2} \) million volts) in the scattered rays. Thus almost all the anomalously absorbed energy is converted into anomalously scattered energy, and the nuclear cross section for such a process is proportional to the square of the atomic number. The question of the existence and exact position of the “threshold,” which according to Gray, Tarrant, and Chao lay near 2 million volts, has not yet been resolved. (Colli, in a discussion of Gray’s report, pointed out that the Prague measurements did not confirm this result of Chao’s.) Gray now considers it possible that the “anomalous” absorption begins at 1 million volts but at first increases very slowly, as a result of which in the previous measurements the threshold for the nuclei turned out to be equal to 2 million volts.

The question of the causes of the discrepancy between the Cambridge measurements and Meitner’s measurements is still not clear. Lise Meitner, who has just arrived from Copenhagen, reports that Jacobsen obtained results different both from Meitner’s results and from those of Tarrant and Gray.

Recently Oppenheimer and Plesset have put forward the hypothesis that anomalous absorption is associated with the formation of “pairs” (electron + positron) near the nucleus, while anomalous scattering is associated with the annihilation of the positron following this. This would give a cross section proportional to the square of the nuclear charge and of approximately the same order of magnitude as in Tarrant and Gray. The threshold would be about 1 million volts (\(2mc^{2}\)). The formation of unequal energies of the positron and electron is possible, which could explain the hard component of Gray–Tarrant. Gray considers that the most decisive argument against the interpretation of Oppenheimer is the circumstance that, in the scattering of rays—

* Perrin gives a scheme of levels with their quantum numbers, calculated by Elsasser.

in which ThC″ (2.6 million volts) gives up not \(1/2.6\), but almost 100% of the energy of the quantum taking part in the absorption, in the form of anomalously scattered quanta. This contradiction disappears only if one assumes that each primary quantum produces not one, but two “pairs” of electrons and positrons with an energy of approximately 0.5 million volts each.

The question of anomalous scattering is one of the most interesting and most acute experimental questions of nuclear physics. In his information on the Copenhagen conference, p-p Weisskopf, among other things, mentioned that M. Delbrück derived from the theory of “holes” the qualitative result that the empty space near the atomic nucleus, being in a “polarized” (see Dirac’s report) state, can coherently scatter gamma rays. Therefore Meitner’s results also may perhaps be interpretable from the standpoint of Dirac’s theory. Meanwhile the experimenters themselves still do not agree among themselves as to exactly how anomalous scattering takes place.

Gamow’s report (Leningrad) contained an account of his recent work on nuclear levels. It is very plausible that the forces acting on particles inside the nucleus, comparatively small in the inner regions of the nucleus, rapidly increase near the surface of the nucleus, so that the force field has the form of a potential well with a flat bottom and steep walls. Approximating this model by a rectangular well with walls of infinite height, we obtain the possibility of calculating the levels of the particles moving in the nucleus. In real nuclei this scheme of levels will, of course, be strongly distorted, especially in its upper part. However, Gamow assumes that, generally speaking, the alternation of azimuthal quantum numbers characterizing the levels remains even under this distortion the same as before. The theory gives, for the case of a rectangular well, the following order: S, P, D, S, F, P, G, D . . . (counting in order of increasing energy). This scheme can be tested on RaC′. Rutherford’s measurements give approximate energies for the nine groups of long-range \(\alpha\)-particles of RaC′, which corresponds to the assumption of nuclear levels. Ellis’s measurements give the intensity of the nine gamma lines and, what is still more important, the coefficient of internal conversion, which makes it possible to distinguish dipole transitions from quadrupole transitions (Taylor and Mott). This gives the following scheme of levels:

\[ \begin{aligned} &\mathrm{S} — 0.000 \cdot 10^6\ \text{volts}\\ &\mathrm{P} — 0.612\\ &\mathrm{D} — 0.838\\ &\mathrm{S} — 1.426\\ &\mathrm{F} — 1.611\\ &\mathrm{P} — 1.743\\ &\mathrm{G} — (\text{not found})\\ &\mathrm{D} — 1.779 \end{aligned} \]

Not all levels correspond to long-range \(\alpha\)-particles, since at a large azimuthal quantum number the probability of emission of an \(\alpha\)-particle is very small (for P, 1.3 times; for D, 4 times; for F, 16 times; and for G, 105 times less than for the S-level). Of the 21 mathematically possible transitions, 11 are in fact observed and are in complete agreement with the selection principle. Of the remaining 10 transitions, two should have an intensity of zero because of the selection rule, four fall in an uninvestigated region of the spectrum, and four are probably not observed because of their very small intensity.

The presence of \(\alpha\)-particles with less energy than in the main group of \(\alpha\)-particles (fine structure) is connected, as is known, with the fact that after decay the nucleus remains in an excited state. The intensity decreases rapidly with increasing energy and with increasing azimuthal quantum number in the excited state. If in the ground state the \(\alpha\)-particle has azimuthal quantum number zero, then the intensities of the slow groups will accordingly be very small. But if in the ground state this number is not zero (and there are some indications that this is the case for actinium and probably also for C-products), then the nuclei of the initial element and of the product

decay have different angular momenta, and it may happen that for some slow groups the difference between the angular momentum of the initial nucleus in the excited state and that of the decay product is zero; this can lead to the fact that the coefficient of the $\alpha$-particle will be very large (perhaps even larger than in the main group). In such elements it is very easy to detect the fine structure of the $\alpha$-rays. This also explains the existence of intense fine-structure groups in the C-products, in radioactinium, and in actinon. In the other elements the fine structure lies at the limit of measurement.

It is noteworthy that the level differences in a nucleus excited in $\alpha$-decay are almost four times smaller than the level differences of the Ra C′ nucleus excited in $\beta$-decay, and this is an argument in favor of the view that in beta decay gamma radiation is connected with the excitation of protons.

F. Joliot made an interesting report on positrons. It is known that F. Joliot and I. Joliot observed the tracks of positive electrons even before Anderson’s discovery had been made. These tracks came from a neutron source $(\mathrm{Be}+\mathrm{Po})$, but were interpreted as tracks of Compton electrons directed toward this source, produced by secondary gamma rays which in turn arose as a result of the scattering of the primary gamma rays of beryllium. This rather improbable explanation had to be accepted, since the existence of positive electrons seemed too unlikely. At the present time it is already known that the radiation of $\mathrm{Be}+\mathrm{Po}$ is capable of producing positive electrons in various substances. The ratio of the number of positive electrons to the number of negative ones is 0.05 in the case of aluminum, 0.18 in the case of copper, in the case of lead, and 0.40 in the case of uranium. If a two-centimeter lead filter is placed between the irradiated substance and the source, the number of positrons decreases very sharply, whereas this lead filter stops only a few percent of the neutrons. This proves that the production of positrons is connected not with neutrons but with gamma rays (the excess of the number of negative electrons over positive ones is explained by the presence of Compton scattering). One might expect that the sum of the energies of both members of the pair (electron + positron) would be less by one million volts (by $2mc^2$) than the energy of the gamma quantum. Measurements in a Wilson chamber, however, show that the difference between the energy of the primary quantum and the quantity $2mc^2$ is only an upper limit for the sum of the energies of the members of the pair. The usual value of this sum is somewhat smaller, which, in Joliot’s opinion, indicates that part of the energy leaves in the form of a scattered quantum. In the discussion Dirac noted another possibility, which seems to him more probable, namely that the remaining energy is absorbed by the nucleus. Skobeltsyn pointed out that his Wilson photographs relating to the Compton effect in gases likewise indicate the presence of pairs, the energy of the positive electron systematically exceeding the energy of the negative electron of the same pair. Then Joliot dwelt on his recent investigations of the production of positrons when aluminum is bombarded by polonium alpha particles. From his experiments it follows that the disintegration of aluminum proceeds in some cases according to the formula

\[ \mathrm{Al}_{27} + \alpha \longrightarrow \mathrm{Si}_{30} + \pi \]

and in other cases according to the formula

\[ \mathrm{Al}_{27} + \alpha \longrightarrow \mathrm{Si}_{30} + \omega + \varepsilon + \]

where $\varepsilon+$ denotes a positron, and $\pi$ a proton. These experiments are very important, since one of their possible interpretations is that the proton is capable of splitting into a neutron + positron. Dirac points out that this splitting, perhaps by analogy with beta decay, occurs with violation of the law of conservation of energy. Dirac, however, says that the existence of an upper limit in the beta spectrum makes nonconservation of energy very unlikely. But conservation of energy can be reconciled with Ellis’s calorimetric measurements only if one introduces the hypothetical particles devised by Pauli—

the “neutrino” (a particle without charge and with a mass on the order of the electron mass), which flies out of the nucleus together with the beta electron and carries away some energy. (All other particles that might compensate for the smearing of the energies in the beta spectrum—neutrons and photons—would be absorbed by the calorimeter.)

Guido Beck (Prague) gave a report on his theory of the continuous beta spectrum. Beck draws attention to the following facts: 1) At distances from the nucleus that are far greater than the linear dimensions of the nucleus \((e^2/mc^2)\), but smaller than the Compton wavelength \((h/mc)\), Dirac’s theory, with its transitions to negative states, etc., must be applicable. Here pairs (positron + electron) must be formed, and from the uncertainty principle it follows that the place of birth of such a pair cannot be determined with an accuracy greater than \(h/mc\).

2) An experimental fact is the continuous character of the beta spectrum, and the “neutrino” represents the only means of saving the law of conservation of energy. However, the assumption of the “neutrino” does not diminish but increases the difficulties, since, possessing an insignificantly small mass, they will not be subject to wave mechanics inside the nucleus.

3) There exists a sharp boundary of the continuous beta spectrum.

Beck’s hypothesis is as follows: the nucleus loses a definite energy \(\Delta E\), passing from one energy state to another. If this energy exceeds \(2mc^2\), then at its expense, within a radius \(h/mc\) around the nucleus, a “pair” is formed. The kinetic energies \(W\) and \(W'\) of the negative and positive electrons together give \(\Delta E - 2mc^2\).

Therefore the energy of the negative electron assumes all possible values, but not exceeding \(\Delta E - 2mc^2\) (a sharp boundary!). The probabilities of particular values of \(W\) can be calculated on the basis of Dirac’s theory. The resulting distribution of energies among the negative electrons agrees well with the experimental curves of the beta spectra. As for the energy of the positron \(W'\), Beck postulates that it cannot be observed (like the “neutrino”): the positron is captured by the nucleus and its energy disappears without a trace. Beck’s theory provoked much controversy among the participants of the conference, but in general did not win recognition.