DISCUSSION ON THE STRUCTURE OF THE ATOMIC NUCLEUS

From the editors. On February 7, 1929, the Royal Society of London organized a discussion under the chairmanship of E. Rutherford. We print a translation of the stenographic report of this discussion, published in Proceedings of the Royal Society A. Vol. 123, p. 373.

E. Rutherford.

On March 19, 1914, the last discussion on the structure of the atom was held at the Royal Society—exactly 15 years ago. I had the honor of opening that discussion. Taking part in it were: Moseley, Soddy, Nicholson, Hicks, Stanley Allen, and J. J. Thomson. In my introductory remarks I set forth the theory of the nuclear atom and presented a number of arguments in favor of this theory. Moseley spoke about his investigations with X-rays. These investigations made it possible to determine the atomic numbers of the elements and showed how many empty places still remained between hydrogen (atomic number 1) and uranium (number 92). Soddy pointed to the existence of isotopes in radioactive families and drew attention to the remarkable observations of Sir J. Thomson and Dr. Aston, who, in studying neon with a mass spectrograph, obtained two parabolas. Soddy expressed the thought that, perhaps, the ordinary elements as well are mixtures of isotopes. You will agree, I think, that these remarks and hypotheses, expressed during the discussion 15 years ago, have not lost their relevance even at the present time. Thus, for example, Hicks and Stanley Allen pointed out the necessity of taking into account the magnetic fields of the nucleus. And between

At that time we still knew very little about this question, and even now our information on it is very scanty. What, then, was accomplished in the intervening period? Looking back, we see that the overcoming of the problem of the structure of the atomic nucleus proceeded along three new paths.

The first, and in some respects the most significant, was the proof of the existence of isotopes among the ordinary elements and the exact determination of the masses or weights of the atoms of individual isotopes, carried out chiefly by F. W. Aston. This led to the development of Moseley’s original ideas. Moseley’s experiments made it possible to determine the number of possible charges of the nucleus. Aston, in turn, showed that there exists a whole series of atoms possessing one and the same atomic charge, while their masses and the structure of their nuclei may be different.

The principal point clarified in Aston’s early work was that the masses of all elements, with the exception of hydrogen, are expressed approximately by whole numbers if the mass of oxygen is taken as equal to 16. But, as we now know, the most interesting thing is not the law of the integral character of the masses of the elements, but rather the exceptions to this law. I shall dwell on this point in more detail below, but for the present I shall say only the following in this connection: the existence of isotopes and the experiments on the artificial disintegration of light elements show quite convincingly that the particles making up the nucleus possess a mass approximately equal to 1. These particles are called protons. We believe that the proton is identical with the free nucleus of hydrogen. During the discussion in 1914 I pointed out that the hydrogen nucleus is almost certainly a positive electron—by analogy with the ordinary negative electron.

The next discovery was the proof of the artificial disintegration of elements when they are bombarded with \(\alpha\)-particles. In these experiments I was personally interested, as was Dr. Chadwick, who will report on the results obtained and on their relation to the structure of the nucleus. These experiments, so far as I know, were the first to show definitely that we

we can change the structure of the nucleus itself by means of external action. We know that in all those cases where we succeed in doing this, a proton possessing great speed is ejected. It is interesting to note that in the disintegration of radioactive elements there are always emitted either helium nuclei or electrons, whereas in the artificial disintegration of light elements, so far as we know, helium nuclei do not appear, but instead a proton is liberated. In general, experiments show that the “ultimate” constituent parts of the nucleus are protons and electrons, and that in the nuclei of heavier elements there are also secondary formations in the form of helium nuclei. I shall return below to this important question.

A third path toward mastery of the problem of the structure of the nucleus was the study of the wavelengths of the penetrating \(\gamma\)-radiation arising in the disintegration of a radioactive nucleus. \(\gamma\)-rays come from the nucleus itself; therefore the frequency of these rays, first determined by Dr. Ellis, gives us very important information about the oscillations of the particles composing the nucleus. I hope that Dr. Ellis will in his report also dwell on certain questions concerning the relation of the components of the nucleus to the outer electrons, for he has obtained a number of interesting results on this subject.

Chadwick and I in recent years have been occupied mainly with the question of the dimensions of the nucleus and of the force field in the vicinity of the nucleus. This question is extremely important, for we obviously cannot make any quantitative calculations until we know the nature and laws of the force field near the nucleus. One can judge this field from the scattering of \(\alpha\)-particles, and we have carried out a large number of experiments on various elements. Our methods are in principle extremely simple. An intense beam of \(\alpha\)-particles of definite velocity falls on a thin material plate, and by the scintillation method the number of \(\alpha\)-particles scattered within an angle of approximately \(135^\circ\) is counted. The number of \(\alpha\)-particles observed in this way is usually in the ratio \(1:10^5\) to the total number of \(\alpha\)-particles falling on the scattering foil. The velocity of the \(\alpha\)-particles falling on the foil can be varied by placing in front of their source

thin layers of mica. In this way the number of scattered $\alpha$-particles at various velocities was determined. If the usual laws of electrostatics held between the nucleus and the $\alpha$-particles, the number of scattered particles should have varied as $1/E^2$, where $E$ is the energy of the $\alpha$-particle. In studying all the elements, from copper with atomic number 29 to uranium with number 92, normal scattering was found, i.e. scattering which, within the limits of experimental accuracy, agrees with that which may be expected if, in that region of the atom into which the $\alpha$-particle penetrates, there exists an ordinary force field. Since it may be expected that, if the $\alpha$-particle penetrated inside the nucleus, the force field would change, it may be concluded that the radius of the copper nucleus must be less than the closest distance of approach of the $\alpha$-particle to the atom, equal in this case to approximately $10^{-12}$ cm. The corresponding distance for the uranium nucleus is approximately $3 \cdot 10^{-12}$ cm, for the fastest $\alpha$-particles. Since for none of the atoms considered was any deviation from the ordinary force field detected, we cannot determine the dimensions of the nucleus with any certainty. All that we can say is that the dimensions of the nucleus must be smaller than the closest distance between the $\alpha$-particle and the corresponding nucleus during their collision. If the $\alpha$-particle penetrated inside the nucleus, then the force field would probably change, and consequently so would the law of scattering.

Quite different results are obtained in scattering by lighter elements. It had long ago been shown that, when hydrogen is bombarded by $\alpha$-particles, the scattering obtained is abnormal. The same result was recently observed for helium. Bieler, and then Chadwick and I, investigated in detail the scattering of $\alpha$-particles by magnesium and aluminium (atomic numbers 12 and 13). I shall not dwell on the details of the experiments, but shall only describe the type of scattering curve observed by us.

Let us first suppose that the scattering is normal, i.e. that the number of $\alpha$-particles scattered through an angle within the limits of $135^\circ$ varies as $1/E^2$, where $E$ is the energy of the $\alpha$-particle. The ratio of the observed scattering to the theoretical for dif-

... velocities of the $\alpha$-particles is indicated in the drawing (Fig. 1) by the dotted line. Such a straight line was found for gold and for all the investigated intermediate elements up to copper. If, however, the scattering material is aluminum, then the inverse-square law is approximately obeyed for slow $\alpha$-particles. With increasing velocity of the $\alpha$-particles, the curve falls below the normal straight line, reaches a minimum, and then rises again. From the appearance of the curve one may conclude that, if it were possible to create still faster $\alpha$-particles, the curve would rise sharply above the normal line. In the drawing (Fig. 1) is plotted the curve of scattering of $\alpha$-particles by aluminum through a mean angle of $135^\circ$. A similar curve was found for magnesium. Probably still lighter elements would give the same type of curve. In some respects the scattering curves of $\alpha$-particles by hydrogen and helium are quite analogous to this curve.

Fig. 1.

It seems to me that the following explanation is the most natural and suitable one: to the ordinary electrical forces of repulsion between the nucleus and the $\alpha$-particle, at very small distances there are added forces of attraction as well, so that the resultant force is a combination of forces of repulsion and attraction. Bieler was the first to calculate the scattering under these conditions, assuming that the forces of attraction are inversely proportional to the fourth power of the distance.

Later Debye and Hardmeier investigated the scattering under the assumption that the forces of attraction are inversely proportional to the fifth power of the distance. This assumption has a definite physical meaning. When an $\alpha$-particle approaches the nucleus closely, forces appear that are capable of separating or polarizing the charged constituent parts of the nucleus; therefore an attractive force begins to act on the $\alpha$-particle.

force approximately inversely proportional to the fifth power of the distance from the center of the nucleus.

It seems to me that this point of view is quite well founded, for under the action of such great forces polarization of the nucleus must occur, and the resulting attraction may become very large when the $\alpha$-particle approaches the nucleus closely. Gardmeier showed that calculations made on this assumption are in agreement with observations on aluminum. It should be noted that these assumptions are somewhat artificial, since the $\alpha$-particle is taken to be a point charge, and the nucleus a sphere. Observations on the scattering of $\alpha$-particles by hydrogen and helium indicate that both the hydrogen nucleus and the helium nucleus are apparently surrounded by a force field of unknown origin, not subject to the usual laws. This region of anomalous forces is, in size, approximately equal to the aluminum nucleus, and, as far as can be judged from the results obtained, this region is not spherical, but rather resembles a flat ellipsoid. It is interesting to note that the “size” of the hydrogen nucleus, or proton, determined in this way, turns out to be even greater than that of the helium nucleus.

However we may interpret the experimental results, it is clear that nuclei cannot be regarded as points, and that they must be ascribed a definite volume or structure. There is a supposition, and perhaps it is correct, that this peculiar distribution of forces around $\alpha$-particles and protons is caused by magnetic forces which, according to modern views, may arise owing to an intrinsic magnetic moment attributed to the proton and perhaps also to the electron, and which has nothing in common with the true motions of the constituent parts of the nucleus.

I now come to a very important point in the development of this argument. Observations on the scattering of $\alpha$-particles by uranium show that within the limits of experimental error—which, unfortunately, were rather large—the scattering is quite normal when the shortest distance is approximately $3.5\cdot 10^{-12}$ cm; and this shows that the radius of the nucleus is even smaller than this value. At the same time, considering

the velocity with which the slowest α-particles spontaneously fly out of a uranium nucleus, we come to the conclusion that the dimensions of the nucleus reach \(6.5 \cdot 10^{-12}\) cm, i.e. a value approximately twice as large as follows from experiments on the scattering of α-particles. Thus we are faced here with a great difficulty: two, apparently not mutually contradictory, methods of estimating the magnitude of the nucleus give strongly divergent values. If we construct the nucleus on the basis of classical ideas and try to make this model quite complete, we shall come to the conclusion that, under any assumption concerning the system of forces, the force field around the uranium nucleus must consist of an attractive force at small distances and a repulsive force at large ones. The variation

Fig. 2.

of the potential of these forces is given in Fig. 2. The maximum of this curve corresponds to the distance at which the attractive and repulsive forces balance one another. Within this distance the attractive forces prevail, and the potential may take negative values. At large distances the potential of the electrostatic forces is inversely proportional to the distance. I think everyone will agree that the potentials in the vicinity of the nucleus must depend on distance in approximately this way. According to classical electrodynamics the potential must reach approximately 4 or 5 million volts at a distance of \(4\) or \(5 \cdot 10^{-12}\) cm, independently of our assumptions about the nature of the forces prevailing there. These considerations compel us to think,

that fast \(\alpha\)-particles can penetrate deeply into the uranium nucleus, whereas experiments on the scattering of \(\alpha\)-particles contradict such a conclusion.

In the past year attempts have appeared to overcome this difficulty with the aid of the ideas of wave mechanics. G. Gamow, whom we are very glad to welcome here, has undertaken this problem, as have also Gurney and Condon. Gamow assumes a change of the potential near the nucleus very similar to that shown in Fig. 2, but by means of calculations he shows that the maximum of the curve lies much closer to the nucleus, approximately at a distance of \(0.7\cdot 10^{-12}\) cm, instead of \(4\) or \(5\cdot 10^{-12}\) cm. Corresponding to this, the maximum potential also proves to be higher, equal to about 30 million volts, and to fall off very sharply near the nucleus. Thus the nucleus—assumed to be spherical—is surrounded by a very high force barrier. No \(\alpha\)-particle emitted from uranium can surmount this barrier; if some \(\alpha\)-particle were to leap over this barrier, it would fly out with an energy far exceeding the observed energy of \(\alpha\)-particles. But, according to wave mechanics, particles can perform feats which appear entirely impossible according to classical mechanics. According to wave mechanics, an \(\alpha\)-particle need not at all leap over the force barrier in order to fly out of the nucleus. This \(\alpha\)-particle, or, more precisely, the wave system which we identify with the \(\alpha\)-particle, seeps through the barrier and, finally, emerges with a kinetic energy equal to the total energy of the particle inside the barrier. I shall not dwell further on this new, interesting point of view, which Fowler and Gamow are further developing. We shall see that, according to this theory, the radius of the uranium nucleus turns out to be very small, equal to about \(7\cdot 10^{-13}\) cm, and in this small volume there must be accommodated 238 protons and 146 electrons.

This sounds implausible, but perhaps it is not impossible.

I shall now pass to the discussion of some of Dr. Aston’s results concerning the structure of the nucleus. You all know the principal result obtained by him, namely that

DISCUSSION ON THE STRUCTURE OF THE ATOMIC NUCLEUS

the unit of mass in the structure of the nucleus is the proton with a mass approximately equal to unity, whereas the proton (the hydrogen nucleus) in the free state has a mass of 1.0073, taking the mass of oxygen \(O = 16\). This difference in the mass of a free proton and of a proton inside the nucleus is explained by the interaction of the electromagnetic fields of the protons and electrons in the highly concentrated nucleus.

The newest theories teach us that there is a close connection between mass and energy, and that a decrease in mass is equivalent to a loss of energy. A free proton has a mass of 1.0073, while the mass of a proton inside the nucleus is very close to 1. This seemingly small loss in mass indicates

Fig. 3.

that in the transformation of a free proton into a constituent part of the nucleus, a large amount of energy was radiated, corresponding to approximately 7 million volts.

Now look at the curve in Fig. 3, drawn by Aston and showing, on a large scale, the deviations of the masses of isotopes from whole numbers. The curve, beginning with hydrogen, crosses the line of whole numbers and reaches a minimum for an atomic weight of about 120. Then it rises again and again crosses the line of whole numbers at about atomic weight 200. Suppose now that the nucleus was originally constructed from free hydrogen nuclei and electrons (I do not say that I agree with such an assumption). In transfor—

...of an atom with a mass equal to 120, containing 120 protons, there was lost an energy corresponding to 840 million volts, and this energy was probably radiated into space. From this point of view it is clear that, contrary to ordinary objections, the atom not only is not a storehouse of energy, but quite the opposite: in order to break down, for example, the nucleus of mercury into its constituent parts, into individual protons, we would have to perform a colossal amount of work, equal to a minimum of 1400 million volts.

All this is certainly correct if one assumes that all atoms are built of free hydrogen nuclei and electrons. You may, of course, ask the question: how is this point of view to be reconciled with the fact that, in the transformation of uranium into lead, an enormous quantity of energy is spontaneously liberated—more than 40 million volts?

If, however, one adheres to another point of view, the results are more satisfactory. Let us suppose that the basic unit in the structure of the nuclei of the heavier elements is not the proton, but the helium nucleus. Let us suppose that this nucleus is built of 4 protons and 2 electrons. In its formation an enormous quantity of energy has already been released, and the mass of this nucleus in the free state is equal to 4.0018. Note that the curve of the deviations of masses from whole numbers reaches a minimum for a mass equal to approximately 120, and again intersects the line of whole numbers for a mass equal to 200. If, for simplicity, we suppose that, beginning with mass 120, nuclei are built by the gradual addition of $\alpha$-particles, then it is easy to show that the mass of each $\alpha$-particle in this interval of atomic weights must be approximately equal to 4.006; we see that this mass exceeds the mass of an $\alpha$-particle in the free state. It is also beyond doubt that the increase in mass is not the same for all atoms; probably it is greater for masses near 200 than for masses near 120. If Aston’s curve is extrapolated to uranium, then it can be shown that the increase of mass existing there satisfactorily explains, in any case, the approximately large quantity of energy emitted by radioactive bodies.

DISCUSSION ON THE STRUCTURE OF THE ATOMIC NUCLEUS

Now we can already form for ourselves a picture of the gradual construction of atomic nuclei. Probably, in the light elements the nucleus consists of a combination of $\alpha$-particles, protons, and electrons, with the separate parts of the nucleus strongly attracting one another, partly as a result of perturbing forces, partly as a result of magnetic forces. As to the nature of these forces, for the time being we can only make one or another supposition. First of all, a very concentrated and firmly bound nucleus is formed, and this process is accompanied by the emission of energy. For an atomic weight approximately equal to 120, we have the smallest mass, which signifies the tightest binding. With a further increase in atomic numbers, the added particles prove to be bound less and less densely.

Thus one may suppose that the nucleus has a very dense structure near the center, with the density gradually decreasing with distance from the center. This whole system is surrounded by a force barrier, which ordinarily prevents the escape of $\alpha$-particles. It may be that this static point of view will not please my theoretical friends, who would like to grant the $\alpha$-particle complete freedom of movement inside the nucleus. Nevertheless this point of view is quite legitimate and is in full agreement with the ideas I have set forth. In other words, if we could take an instantaneous snapshot of the nucleus—with an exposure of about $10^{-23}$ seconds—we would see at the center, as it were, densely packed, firmly bound $\alpha$-particles, with the density decreasing as the distance from the center increased. Without doubt, all the $\alpha$-particles are in motion, and their waves are reflected from the force barriers, and sometimes also penetrate beyond the limits of the system. It seems to me that the point of view I have developed is quite well founded, and I hope that our theoretical friends will be able to describe the whole picture in greater detail. We must not only explain the structure of the nucleus from $\alpha$-particles—we must also find a place for the electrons; but to shut electrons into one cage with an $\alpha$-particle is not so easy. Nevertheless I am so confident in the ingenuity of our theoretical friends that I firmly believe that they

will overcome this difficulty in some way. I hope to hear their considerations on this matter today.

One more point. The point of view I have set forth explains, it seems to me, why atoms of heavy uranium cannot exist. With increasing mass the nucleus would acquire more and more energy and would become so radioactive that it would disappear. Apparently, the greater the store of energy in nuclei, the sooner they would disappear, and, probably, uranium and thorium are not by chance the only surviving representatives of heavy nuclei. Here it is not appropriate to touch upon the highly speculative question of how the nuclei of the elements were formed. Before undertaking to solve this question, we need to learn much more about the details of the structure of the nucleus itself.

F. W. Aston.

Besides the basic “whole-number rule,” there is still a whole series of results obtained with the mass spectrograph that bear upon the subject of the present discussion. Why are there no atoms with atomic masses equal to 2, 3, 5, 8, etc.? Why is there never fewer than two protons per one electron in the nucleus? In elements with even atomic numbers the number of isotopes tends to increase with the atomic number; nevertheless, the difference in the masses of the extreme isotopes is for some reason limited to about 10%. In elements with odd atomic numbers this limitation appears still more clearly. They never have more than two isotopes, and, beginning with atomic number 9, the masses of these isotopes always differ by 2 units, with the lighter isotope predominating. Here we have a fundamental difference between elements with even and odd atomic numbers. In both classes the number of electrons in the nucleus is for the most part even; indeed, the only exceptions to this rule are beryllium and nitrogen. As has long since been pointed out by Harkins, in nature elements with even atomic numbers strongly predominate. If one plots a curve expressing the abundance of the elements existing in nature as a function of their atomic weights,

we note clearly expressed maxima for atomic weights of the type \(8n\).

One of the very few direct methods for investigating the structure of nuclei is the measurement of their relative masses. Therefore every effort was made to increase the accuracy of these measurements. The limit that could be attained with the new mass spectrograph was, in the most favorable cases, equal to \(1/10000\). The percentage excess or deficiency in comparison with whole numbers was measured, taking the mass of oxygen as equal to 16. This quantity was named the “packing fraction” and is expressed in ten-thousandths.

In Figure 3 the packing fractions are plotted as functions of the atomic weights, so that the observational errors are the same at all points of the curve. The points denote the result of direct measurements. Beginning with number 20, they lie approximately on one curve, which has a minimum of about \(-10\), near iron and nickel. Light atoms with odd ordinal numbers lie on a curve rising for hydrogen to \(+778\); atoms with even numbers lie considerably lower. In this is manifested the fundamental difference between these two classes of atoms. From this one may conclude that light atoms with odd ordinal numbers are bound less strongly and therefore have a heavier external structure, which is not observed in the denser and more stable nuclei of helium, carbon, and oxygen.

If the masses of atoms are calculated from the packing fractions and then their arithmetical differences from whole numbers are plotted, we obtain the curve indicated in the upper part of the figure. Here the minimum lies approximately in the middle of the series of all known atomic weights. For this curve the errors are not the same at all its points, but, as is shown by the dotted curves, they increase greatly for atoms with large atomic mass.

Discussion. Chadwick.

When some elements are bombarded with \(\alpha\)-particles, hydrogen nuclei, or protons, are knocked out of them, which

can be detected by the scintillation they produce on a zinc sulfide screen. These protons appear as a result of the artificial disintegration of the nuclei of these elements. We suppose that the disintegration of the nucleus occurs when an $\alpha$-particle penetrates into the nucleus and is retained there, as a result of which a proton is emitted. The probability of disintegration is small; thus, for example, in a favorable case, when nitrogen is bombarded, 20 nuclei disintegrate for every $10^6$ $\alpha$-particles. Owing to the rarity of this effect, as well as to various experimental difficulties, the information obtained by us so far is still rather scanty. With the exception of carbon and oxygen, all the elements from boron to potassium inclusive disintegrate when bombarded with $\alpha$-particles and in so doing emit a proton possessing considerable energy. This means that the nuclei of all these elements contain protons. Carbon and oxygen, if they disintegrate at all, do not emit particles with energy exceeding the energy of the scattered $\alpha$-particles. It is possible that they disintegrate into helium nuclei, but there is as yet no proof of this.

Some protons liberated in artificial disintegration have very large energies; for example, the energy of the protons knocked out of aluminium by $\alpha$-particles from radium C exceeds the energy of the incident $\alpha$-particles by 40%. Thus in some cases energy is liberated in disintegration. There is a marked difference in the behavior of elements with even and odd atomic numbers. The protons emitted from odd elements have a much greater maximum energy than the protons from even elements. In a disintegration consisting only in the capture of an $\alpha$-particle and the emission of a proton, an element with an odd number is transformed into an element with an even number, and conversely. Considering the different behavior of even and odd elements, as well as their relative abundance in nature and their atomic masses, one may conclude that even elements are more stable than odd ones.

It is further of interest to compare the experiments on the artificial disintegration of elements with experiments on the scattering of $\alpha$-particles. The former show that particles with a range of 3–

3.5 cm are capable of disintegrating aluminium, since they can penetrate into the aluminium nucleus and be captured by it. On the other hand, experiments on scattering show that the enormous majority of these \(\alpha\)-particles are scattered according to the usual laws. These experiments say that even \(\alpha\)-particles with a range of 7 cm are not able to penetrate into the aluminium nucleus. An explanation of this apparent contradiction between the two series of experiments was recently given by Gamow, on the basis of wave mechanics.

If it is assumed that in disintegration an \(\alpha\)-particle is captured and that only the \(\alpha\)-particle, the nucleus, and the protons take part in the collision, then it should be expected that the proton will appear at a definite angle with a definite energy. Experiments with aluminium show that this is not the case and that protons fly out with very varied energies. For example, protons emitted at an angle of \(90^\circ\) to the direction of the striking \(\alpha\)-particles of radium C had energies lying in the interval from 0.3 to 1.1 of the energy of \(\alpha\)-particles with a range of 7 cm. Apparently, the most correct explanation of this fact can be given by assuming that not all nuclei of one and the same type have equal masses or equal total energies, so that both the disintegration of nuclei and the formation of new nuclei can occur with different energy. The fluctuations of mass proposed here are small; the greatest difference in the masses of aluminium nuclei reaches about 0.006 mass units.

K. Ellis.

In many radioactive disintegrations it has been observed that immediately after the emission of a particle from the nucleus there occurs the emission of characteristic radiation of high frequency. These \(\gamma\)-rays are for the most part connected with transformations in which \(\beta\)-rays are emitted. The \(\gamma\)-rays may be regarded as the characteristic spectrum of the nucleus, and their emission occurs as a consequence of the disturbance caused by the emission of an \(\alpha\)- or \(\beta\)-particle.

The \(\gamma\)-rays are homogeneous to an accuracy of \(1/1000\), and from this one may draw certain interesting conclusions about the nature

particles emitting them. Kuhn showed that, in all probability, electrons could not emit such homogeneous radiation of high frequency, and that the same may be said with sufficient confidence of protons. Therefore we have to ascribe the origin of $\gamma$-rays either to $\alpha$-particles or to the entire nucleus as a whole, for example when it is in the process of rotation. To clarify this question it would be very interesting if it were possible to derive a series of energy levels corresponding to $\gamma$-rays. There is no doubt that some system of levels exists, for in all spectra many relations of the type $\nu_1+\nu_2=\nu_3$ have been found. However, to establish some unified system of levels on the basis of frequency differences alone requires a more detailed knowledge of these frequencies than we possess at present. This difficulty can be overcome in two ways.

The first way is to determine the absolute intensities of $\gamma$-rays, i.e. to determine the probability that in each disintegration one quantum of $\gamma$-rays will be emitted. It is natural to assume that in each disintegration only one excitation of the system of levels occurs, so that, for example, $\gamma$-rays corresponding to a definite change of level cannot have an intensity exceeding unity. There are also more exact criteria which can provide a good test of proposed systems of frequency differences.

The second way of solving this problem apparently lies in the experiments of Aston and Ellis. They found indications of a connection between the nucleus and the system of electrons. Let us explain this in more detail. When the $\gamma$-rays of some radioactive substance fall upon a thin sheet of lead, the intensity of the photoelectrons is closely connected with the intensity of the $\gamma$-rays and with the absorption coefficients. Analogous photoelectric groups are also emitted by radioactive atoms. They may be interpreted as the result of partial internal absorption of $\gamma$-rays. If the nucleus and the system of electrons were as independent of one another as the nucleus and the sheet of lead, one could expect complete parallelism between the intensity of the photoelectric groups from

lead, and from a radioactive atom. In fact, just the opposite has been found; and although in this respect the experiments are still not sufficiently complete, it may be said with certainty that, with increasing frequency of the \(\gamma\)-rays, surprising oscillations occur in the ratio of internal absorption to external absorption, which indicates the existence of a connection between the nucleus and the electrons—this phenomenon in itself is already of interest for the study of the structure of the nucleus. Moreover, we hope that it will give us a new method for resolving the question of the system of levels, namely, that it will make it possible to distinguish different \(\gamma\)-rays from one another.

T. A. Gamov.

I should like to make a few remarks concerning the chairman’s concluding words, which treat the nucleus as an assemblage of \(\alpha\)-particles.

There are quite a number of indications that all the \(\alpha\)-particles constituting the nucleus (in the heavy elements there are up to 90 \(\alpha\)-particles in the nucleus) are in one and the same quantum state, with quantum number equal to unity. This does not contradict the Pauli principle, since \(\alpha\)-particles, possessing an even charge, obey the Bose–Einstein statistics. Such an assemblage of \(\alpha\)-particles, between which attractive forces act that vary rapidly with distance, may be treated as a small drop of water, in which the particles are held together by surface tension.

An exact solution of the problem with such a model is, of course, exceedingly difficult. However, the first, crude approximation, which I shall now present, points to a number of interesting properties of such a model.

Let us write down two equations:

1) an equation relating the energy of the \(\alpha\)-particles to the surface tension of the imaginary “drop of water,” and

2) the quantum conditions of ordinary quantum mechanics.

Then we obtain a relation between the “energy of the drop” and the number of \(\alpha\)-particles contained in the drop, i.e. the atomic weight of the nucleus. The general form of the curve illustrating this relation agrees well with Aston’s curve

for a defect of mass, where one obtains a negative energy of support, which changes spontaneously for large values of \(N\).

According to wave mechanics, this problem has to be solved by Hartree’s method with a self-consistent field.

For a nucleus consisting of \(\alpha\)-particles, one has to solve the equation (written for one particle):

\[ \frac{\partial^2 \psi}{\partial x^2} +\frac{\partial^2 \psi}{\partial y^2} +\frac{\partial^2 \psi}{\partial z^2} +\frac{8\pi^2 m}{h} \left[ E-(N-1)\int \psi \bar{\psi}\, U(r)\,dv \right]\psi=0; \]

here the second term in square brackets represents the action of the remaining \((N-1)\) \(\alpha\)-particles; \(U(r)\) is the energy of interaction of two \(\alpha\)-particles at a distance \(r\) from one another.

A solution of this equation, under strongly simplifying assumptions concerning the force field, was given by Dr. Hartree. He simply takes the second term in the brackets to be equal to \((N-1)K\psi\bar{\psi}\) (\(K\) being a constant). This assumption, often used in the theory of capillarity, amounts to neglecting the Coulomb repulsive forces between the \(\alpha\)-particles and to assuming that the “radius of action” of the attractive forces is small in comparison with the dimensions of the nucleus.

This equation has discrete characteristic numbers, giving the correct order of magnitude for the energy and radius of the nucleus (by a corresponding choice of \(K\)).

I hope that further calculations on the basis of this model and with a more precise knowledge of the forces of interaction between \(\alpha\)-particles, obtained at least from experiments on the scattering of \(\alpha\)-particles by helium, will give exact agreement with experiment, in any case for the light elements. Further difficulties arise for elements heavier than argon, where free nuclear electrons begin to play a role—and so far we know nothing about the behavior of nuclear electrons.

R. Fowler.

I should like to set forth for you in what way the new quantum theory may help us in the discussion of the structure and properties of the nucleus. This question was already indicated by the chairman in his introductory remarks. I should like to develop it somewhat. The first thing to keep in mind is that the new quan-

Quantum mechanics developed logically, on the basis of the properties of electrons in atoms. We must suppose that particles have many properties inherent in waves. Whether we call them particles or waves is a matter of taste. The choice of name most likely depends in each individual case on their state. Since particles resemble waves, we must expect, for example, that they will not always be reflected from barriers of a definite height. They can pass through a barrier, of course, only in certain cases. You may say that there is a finite probability for each of us to leave this room without opening the doors and, of course, without being thrown out through the window. (Then the nature of the reflection of particles from a force barrier was explained in detail, and a close analogy was drawn between such reflection and the optical phenomenon—the passage of light through a thin layer when the angle of incidence is greater than the limiting angle of total internal reflection.)

The fact that particles can pass through barriers of this kind is very important for explaining the phenomenon of the emission of $\alpha$-particles by heavy nuclei.

If we picture the nucleus, as we have already said here today, in the form of a certain small box, surrounded on all sides (in three dimensions) by a force barrier (Fig. 2), then it may be assumed that inside it there is an $\alpha$-particle, which must be represented as a wave whose energy is less than the potential energy in the upper part of the barrier. According to classical theory, the $\alpha$-particle would remain inside the barrier forever. But according to quantum theory there is a finite probability that the wave will pass through the thin wall and go off to infinity. This idea is the basis of the quantum theory of the emission of $\alpha$-particles. This idea was expressed independently by Gamow on the one hand and by Gurney and Condon on the other. All of them, and especially Gamow, developed it in considerable detail.

When the $\alpha$-particle passes through the barrier, it can, of course, no longer be identified with a standing wave. It will be correct to represent the $\alpha$-particle by a damped oscillation. We

we shall have, inside the barrier, a damped oscillation, i.e. a harmonic oscillation with the ordinary damping coefficient, and outside a very weak wave corresponding to the emission of an α-particle. In fact this problem can be solved very well, and the damping coefficient is obtained in the form of the imaginary part of the energy. This was done with great success by Gamow.

He found that, for these calculations, it is not of great importance what exact form you assume for the inner part of the barrier. The main, outer part of it is well known from experiments on the scattering of α-particles.

The probability for an α-particle to penetrate through the barrier depends strongly on the energy of the α-particle. The greater its energy, the thinner is the barrier through which it must pass, and the lower is its height. Therefore, evidently, there is a very close connection between the energy of the α-particle, which we judge from the energy of the emitted α-particle, and the probability for this α-particle to get outside, which we judge from the lifetime of the atom. This is precisely the Geiger–Nuttall law.

In conclusion I shall say that this is a very beautiful theory, and that we can be absolutely certain that in its general outlines it is correct. The great merit of this theory is that it gives the Geiger–Nuttall law quite independently of the details of the structure of the nucleus.

O. Richardson.

I should like to say a few words concerning the rotation of the nucleus.

It seems to me that a priori arguments show that in some cases the nucleus rotates. I imagine a system consisting of one proton and one electron, and allow them to combine. They form a hydrogen atom situated at the very lowest of the possible levels and, in doing so, emit radiation. From spectroscopic observations it is clear that in the normal atom the electron has an angular momentum equal to one half. Conseq—

Therefore, when a proton combines with an electron, the electron acquires an angular momentum equal to one half.

But if, in this process, nothing else changes, the law of conservation of angular momentum will be violated. Therefore I say that the nucleus of an unexcited hydrogen atom has an angular momentum equal to one half and opposite in sign to the moment of the electron. It may be objected that the emitted radiation can preserve the law of angular momenta intact, if it is sufficiently strongly elliptically polarized. However, this possibility is extremely improbable.

Spectroscopic data give us two kinds of evidence for the rotation of the nucleus. The first is the hyperfine structure. Back and Goudsmit have shown that every “line” in the spectrum of bismuth in reality represents a whole series of very fine lines. They studied the various transitions possible among these lines and found that there exists there a new kind of internal quantum number, having the value \(4^{1/2}\). The hyperfine splitting of the lines, and likewise the general structure of the spectrum, are also in agreement with this number—\(4^{1/2}\). This effect must be caused by the rotation of the nucleus, for the magnetic moment of the electron has already been used to explain the ordinary multiplet structure of spectra. Finally, in a strong magnetic field each Zeeman component splits into 10 equally spaced and equally intense lines, corresponding to the spatial quantization of the nucleus in this rotation.

From the hyperfine structure of the cesium spectrum Jackson succeeded in comparing the rotation of the nucleus with the rotation of the electron. It turned out that their magnitudes are of the same order.

The other spectroscopic evidence for the rotation of the nucleus is based on the alternation of the intensity of lines in bands of the spectra of diatomic molecules, such as, for example, \(H_2\), \(He_2\), \(N_2\), and \(O_2\). There is a theorem of Hund stating that the characteristic functions of molecules must be either all symmetric or all antisymmetric with respect to the nuclear coordinates. These functions contain as factors the characteristic functions of the motion of the electrons and

of rotation of the molecule. The characteristic functions of molecular rotation are symmetric with respect to the nuclear coordinates when the rotational quantum number \(p\) is even, and antisymmetric when \(p\) is odd. Thus, if \(p\) is odd, the electronic functions must be symmetric with respect to the nuclear coordinates in order that the entire characteristic function of the molecule be antisymmetric, and conversely.

When \(p\) is even, the conditions will be the reverse. It follows from this that in such bands the alternating lines characterize states which do not combine with one another. If the nucleus manifests itself in no way, part of the functions is absent, and in each band there is no alternation of lines. This, for example, occurs in the case of He\(_2\); from this one may conclude that the helium nucleus has no rotation. In the spectrum of H\(_2\), on the contrary, the lines have an alternating intensity corresponding to the mass ratio \(3:1\).

This is what was to be expected if one assumes that one series of lines comes from molecules in which the moment of the nucleus is equal to \(1/2+1/2=1\) quantum, and the other from molecules with the moment of the nucleus equal to \(1/2-1/2=0\).

With spatial quantization, in the first case three different combinations are obtained \((1,0-1)\), and in the second only one. The nitrogen molecule, which also contains an atom with an odd number of electrons, is similar in this respect to H\(_2\), whereas O\(_2\) rather resembles He\(_2\).

D. Hartree.

There is one inconsistency in this model of the nucleus, with a dense middle and less dense edges. It is difficult to explain why nuclei apparently become less radioactive with increasing atomic number, instead of, on the contrary, becoming more radioactive. Can this be explained without any additional assumptions?

E. Rutherford.

Unfortunately, we do not know from which part of the nucleus the \(\alpha\)-particles are taken. They may originate either from the inner-

or from the outer part of the nucleus. It is highly probable that, in such a tightly bound system as the nucleus, energy can pass very rapidly from one particle to another. The increase of radioactivity with decreasing atomic number is, of course, a very remarkable phenomenon, and as yet there is no explanation for it.

I should like to thank the speakers for their participation in the discussion and to express regret that time does not allow us to arrange an exchange of views on a number of different questions. Our knowledge concerning the nucleus is still in a very elementary state, although the advances achieved in this field are greater than I could have expected 15 years ago.