NUCLEAR PHYSICS IN PHOTOGRAPHS

TRACKS OF CHARGED PARTICLES

IN PHOTOGRAPHIC EMULSIONS *)

C. F. Powell and G. P. S. Occhialini

ARTIFICIAL TRANSMUTATION OF ELEMENTS

The preceding photographs illustrated the elastic collision of fast particles with the nuclei of various elements, in which the incident particle was deflected from its original direction without penetrating into the target nucleus. We shall now consider collisions in which fast particles penetrate inside the nuclei of light elements and transform them into nuclei of other elements. Experiments of this type were first carried out in 1919 by Rutherford, who for this purpose used fast $\alpha$-particles from radium C′ and observed, by the scintillation method, the particles formed as a result of the disintegration.

For heavy elements, such as gold, Rutherford and his collaborators showed that, because of the large electrostatic repulsive forces of positive charges, even the fastest $\alpha$-particles from natural radioactive substances are unable to penetrate into the nuclei. For light elements, owing to the fact that their nuclei have a smaller charge, the magnitude of these forces (according to Coulomb’s law) is smaller, and therefore, under favorable collision conditions, a fast $\alpha$-particle can penetrate into the nucleus and split it. The probability that a given $\alpha$-particle, in passing through matter, will produce a nuclear disintegration is very small because of the small dimensions of nuclei. Therefore, in order to obtain a few successful collisions, it is necessary to bombard a layer of the substance serving as the target with many millions of $\alpha$-particles.

Our knowledge of the artificial transmutation of nuclei was considerably broadened by the work of Cockcroft and Walton (1932), who showed for the first time that artificially accelerated protons can also penetrate into light nuclei and split them. These experiments were

*) Continuation. See UFN, vol. XXXV, issue 2, p. 213 (1948).

NUCLEAR PHYSICS IN PHOTOGRAPHS

repeated using photographic plates to detect particles formed in disintegration. Photographs XVII–XIX (see at the end of the issue) are microphotographs of tracks obtained by this method.

The apparatus used in exposing the plates is shown schematically in Fig. 7. The primary particles—in this case deuterons, i.e., nuclei of heavy hydrogen—accelerated in the high-voltage Cockcroft apparatus, strike a thin film of one of the light elements. As a result, from those nuclei of the target that are disintegrated by the stream of primary deuterons, protons and other particles fly out. Particles formed in disintegration are emitted in all directions; some of them are stopped in the metallic support of the target or in the walls of the apparatus. However, a certain small fraction of them is emitted in such a direction that it reaches the mica or celluloid window. The latter is made sufficiently thin that the particles pass through it without noticeable slowing and reach the photographic plate, forming characteristic tracks in it.

Fig. 7. Diagram of an installation for obtaining, in photographic plates, tracks of particles arising when a target of light elements is bombarded by fast deuterons.

Photographs XVII–XIX are typical examples of the picture observed on plates exposed by the method described, if after development they are examined under a microscope. They were obtained for targets prepared respectively from lithium oxide, beryllium, and boron, the target material being deposited as a thin layer on the surface of a cooled metal block. In each photograph tracks from different types of particles are visible, since the penetration of a deuteron into a nucleus does not always cause one and the same type of disintegration. In addition, some elements, such as lithium and boron, contain several isotopes, and therefore it is necessary to study simultaneously the disintegration of nuclei of different types. This difficulty can be avoided if elements consisting of only one isotope are available.

As an example of this kind of complication one may cite photograph XVII, obtained in the bombardment of a lithium target; in it

visible, among others, are the tracks of \(\alpha\)-particles and protons arising in the following reactions:

\[ {}_{3}\mathrm{Li}^{6}+{}_{1}\mathrm{H}^{2}\to{}_{2}\mathrm{He}^{4}+{}_{2}\mathrm{He}^{4}, \tag{3} \]

\[ {}_{3}\mathrm{Li}^{7}+{}_{1}\mathrm{H}^{2}\to{}_{2}\mathrm{He}^{4}+{}_{2}\mathrm{He}^{5}, \tag{4} \]

\[ {}_{3}\mathrm{Li}^{6}+{}_{1}\mathrm{H}^{2}\to{}_{3}\mathrm{Li}^{7}+{}_{1}\mathrm{H}^{1}. \tag{5} \]

A very important feature of these reactions is that the energy of the particles emitted in the disintegrations is greater than the energy of the primary deuterons. Whence is this energy obtained?

At the present time the relative values of the masses of the nuclei of the light elements have been determined with great accuracy. If these values are substituted in equation (3), it turns out that the mass of the two initial nuclei will be greater than the mass of the two \(\alpha\)-particles. The actual values of these quantities are as follows:

\[ \left. \begin{aligned} {}_{3}\mathrm{Li}^{6}&=6{,}016917, \qquad &{}_{2}\mathrm{He}^{4}&=4{,}003860,\\ {}_{1}\mathrm{H}^{2}&=\frac{2{,}014725}{8{,}031642}\ \text{mass units}, \qquad &{}_{2}\mathrm{He}^{4}&=\frac{4{,}003860}{8{,}007720}\ \text{mass units}. \end{aligned} \right\} \tag{6} \]

Thus, in this transformation a certain amount of rest mass has disappeared, which must be replaced by an equivalent amount of energy. A careful determination of the amount of energy liberated in the disintegrations shows that its value \(E\) is determined with great accuracy by the equation \(E=c^{2}m\), where \(m\) is the magnitude of the vanished rest mass. Experiments of this kind are a common occurrence in modern physical laboratories. They provide the most convincing confirmation of the correctness of the theory of relativity. A more detailed description of the methods for interpreting observations of this type is given in Appendix E.

DISINTEGRATION OF THE NUCLEI OF EMULSION ATOMS

The experimental method described in the preceding paragraph was intended to detect and measure the energy of only one of the particles produced in the disintegration of a given nucleus. In those cases where there are only two products of disintegration, for example, a proton and a recoil nucleus, application of the law of conservation of momentum makes it possible to analyze the given disintegration from observations of a single track (see Appendix F). If, however, several particles are emitted in the disintegration of a nucleus, this is no longer possible. Consequently, in studying such disintegrations it is important to be able to register all the emitted particles. For this purpose the Wilson chamber can be used successfully. However, such experiments are very difficult, since many thousands of photographs of the tracks of fast particles have to be taken before one succeeds in obtaining a single photograph showing a disintegration. If, however, fast deuterons are directed onto a photographic emulsion, then in one

day of examining it under a microscope one can detect many hundreds of cases of such processes.

An ordinary emulsion consists of a limited number of elements, of which the most common are hydrogen, carbon, nitrogen, oxygen, bromine, silver, and iodine. Consequently, in ordinary plates one may expect the disintegration of nuclei only of these atoms. In special cases, however, one can introduce other elements into the emulsion by various methods.

We have already encountered one method for solving this problem in experiments on radioactivity—namely, when radioactive substances were introduced into the emulsion by treating the plate in a suitable solution. After the plate has been dried, the salt remains distributed in the gelatin. A similar result can be obtained if a thin layer of “charged” material is applied to the surface of a dried emulsion and then covered with a second layer of emulsion. The charged substance is thus enclosed between two layers of emulsion. It is possible that in the future, when emulsions are manufactured, it will prove possible to introduce into them fine insoluble particles of the desired element, and these particles will be such that they can be distinguished under the microscope from the developed silver grains by their size or color. Up to now, however, the most frequently used method consists in introducing a salt of the charged material into the emulsion during its manufacture. Sometimes such a procedure leads to undesirable results, since it affects the photographic properties of the plates obtained.

Because of the variety of elements present in the emulsion, it is not always possible to identify the nucleus that has undergone disintegration. Such a difficulty arose in the discussion of photograph XV, where it was impossible to determine whether the deuton had been elastically scattered by a nucleus of carbon, nitrogen, or oxygen. In the following examples, however, the experimental results make it possible to determine the character of the nuclear reaction that occurred, or else to give an explanation that is, in all probability, correct.

NEUTRONS

Up to now we have considered only such nuclear disintegrations as lead to the emission of charged particles. Observations confirm the point of view, already following from experiments on the radioactivity of heavy elements, according to which the nucleus should be regarded as consisting not of some homogeneous substance, but of separate particles; moreover, the division of the nucleus into its constituents may be caused as a result of a sufficiently strong disturbance of the nucleus.

Before the discovery of a new type of particle—neutrons—it was assumed that nuclei consist of protons and electrons. According to this view, the mass of the nucleus would be determined almost completely by its protons,

whereas the number of the latter would have had to be equal to the mass number of the nucleus. Since the charge number is smaller than the mass number, it was assumed that this difference is due to the negative charges of electrons contained in the nucleus. For example, it was assumed that the boron nucleus \({}_{5}\mathrm{B}^{10}\) contains 10 protons and \((10-5)=5\) electrons. Similarly, it was believed that \({}_{92}\mathrm{U}^{235}\) has 235 protons and 143 electrons. The presence of electrons made it possible to give a simple explanation of certain features of the \(\beta\)-decay of radioactive nuclei, i.e. the emission of electrons by nuclei.

However, over many years this theory encountered a number of serious difficulties. For example, it was possible to make a rough estimate of the size of the electron. The values obtained in this way turned out to be almost equal to the values of the sizes of heavy nuclei obtained from experiments on the elastic scattering of \(\alpha\)-particles. It was hard to imagine such a structure as the nucleus \({}_{92}\mathrm{U}^{235}\), which would contain 143 electrons, the diameter of each of which is comparable with the diameter of the nucleus itself, of which they are a part. Considerably more serious difficulties arose in the attempt to describe the properties of nuclei on the basis of wave mechanics.

This theory of the structure of nuclei was completely undermined, and its contradictions resolved, after Chadwick’s discovery in 1932 of a new particle, emitted in nuclear transformations, which was called the neutron*).

It is now known that the neutron has a mass almost equal to the mass of the proton, but, as its name indicates, it is devoid of charge. This particle is therefore denoted by the symbol \({}_{0}n^{1}\).

The assertion that a particle has no charge simply means that electric fields and the electric charges of neighboring particles do not act on it. The absence of charge leads to certain surprising properties of the neutron.

In considering the passage of an \(\alpha\)-particle through atoms, we used the analogy with the motion of a fast star through the solar system. If, in this case, the star moves with a sufficiently low velocity, then, owing to the action of gravitational forces, it may happen that one or more planets will be knocked out beyond the limits of the solar system. Similarly, electrons may be knocked out when a fast charged particle passes through an atom, owing to the action of electrostatic forces between the particles. The atoms are thereby ionized, and the energy of the fast charged particle flying through the substance gradually decreases. Fast neutrons, however, pass through the electronic shell of atoms quite freely, without imparting energy to them. As a result, they are capable of passing through

*) The now generally accepted idea that nuclei do not contain electrons, but consist of protons and neutrons, was first expressed by D. D. Ivanenko. (Ed. note)

thicknesses of matter thousands of times greater than those through which charged particles pass. A particle can be detected only insofar as it interacts with matter, and since neutrons do not ionize a gas, they cannot be detected by the usual electrical method or by their tracks in a Wilson chamber. This explains the somewhat belated nature of their discovery.

Another important consequence of the neutron’s lack of charge is that, unlike positively charged particles, it is not subjected to the action of considerable repulsive forces when approaching a nucleus. Therefore a neutron, even one possessing an insignificant velocity but moving in the appropriate direction, can penetrate into the nucleus and cause disintegration with the emission of charged particles. Another possibility is that the neutron collides elastically or inelastically with the nucleus, changing its direction and imparting to the nucleus a certain recoil. In the case of an inelastic collision, the struck nucleus passes into an excited state, and the energy required for this is obtained at the expense of the neutron’s store of kinetic energy. The latter therefore has a lower velocity after the collision, while the excited nucleus returns to the normal or a lower state, emitting a quantum of radiation.

Fast neutrons can be obtained in the most varied nuclear transformations. Thus, when a beam of α-particles falls on metallic beryllium, the following reaction takes place:

\[ {}_{4}\mathrm{Be}^{9} + {}_{2}\mathrm{He}^{4} \to {}_{6}\mathrm{C}^{12} + {}_{0}n^{1}. \tag{7} \]

As in the examples of nuclear transformations considered earlier, one can calculate the energy released in the reaction by knowing the masses of the corresponding nuclei and the mass of the neutron.

Two other important reactions leading to the emission of fast neutrons are represented by the equations:

\[ {}_{5}\mathrm{B}^{11} + {}_{1}\mathrm{H}^{2} \to {}_{6}\mathrm{C}^{12} + {}_{0}n^{1} \quad \text{and} \quad {}_{1}\mathrm{H}^{2} + {}_{1}\mathrm{H}^{2} \to {}_{2}\mathrm{He}^{3} + {}_{0}n^{1}, \tag{8} \]

where deuterons serve as the bombarding particles, and the targets consist respectively of boron and deuterium.

DETECTION OF NEUTRONS

Every charged particle passing through a photographic emulsion leaves a track in it, and therefore examination of the plate makes it possible to determine the number and ranges of the particles falling on a given area. In the case of neutrons, however, only a small fraction of the particles passing through the emulsion react with nuclei, producing disintegration products or recoil nuclei which in turn indicate the presence of neutrons. Under these circumstances, the following methods are used for detecting neutrons and for determining their energy.

Slow neutrons, having velocities of the same order of magnitude as the velocities of atoms at ordinary temperatures, are called “thermal neutrons” and are obtained when fast neutrons pass through matter. Fast neutrons, colliding with atomic nuclei, impart energy to them, thereby losing their own velocity. In the case of collisions with the nuclei of such light elements as hydrogen or carbon, the neutron loses a noticeable fraction of its energy already in a single collision. Over the course of only a few centimeters of its path, the neutron undergoes a sufficient number of collisions for its energy to fall to values characteristic of atoms at ordinary temperatures, i.e. of the order of \(0.03\ \mathrm{eV}\). Substances that serve to slow neutrons in this way are called “moderators” and play an important role in uranium piles.

Thermal neutrons have a high probability of penetrating the nuclei of certain isotopes of the light elements lithium and boron, namely \(^{6}_{3}\mathrm{Li}\) and \(^{10}_{5}\mathrm{B}\). These nuclei behave as though, for slow neutrons, they present a target of large area, and this area has a diameter considerably larger than the diameter corresponding to the ordinary “size” of nuclei. When slow neutrons penetrate these nuclei, the following reactions occur:

\[ {}^{6}_{3}\mathrm{Li}+{}^{1}_{0}n\to{}^{3}_{1}\mathrm{H}+{}^{4}_{2}\mathrm{He}, \tag{9} \]

\[ {}^{10}_{5}\mathrm{B}+{}^{1}_{0}n\to{}^{7}_{3}\mathrm{Li}+{}^{4}_{2}\mathrm{He}. \tag{10} \]

Since the neutron has a small velocity, its momentum will be small, and the two final nuclei of the reaction fly apart with equal and opposite momenta. The particles can be detected by methods of electrical counting from the ionization that they produce in a gas, or, if the reaction has occurred in a photographic emulsion or in the gas of a Wilson chamber, from the resulting tracks. For experiments in which the photographic method is used, lithium or boron is introduced into the emulsion during its preparation by adding suitable salts.

We see that in the case of fast neutrons their collisions with nuclei lead to the formation of recoil particles. If the latter are nuclei of large mass, then they receive only a small fraction of the energy of the primary particle, just as occurs in the collision of a fast proton with a nucleus. If, however, the interacting particles have equal masses, then on average the initial energy is divided equally between them. The mechanical problem of the collision of a neutron with a proton is almost identical with the problem of the collision of a proton with a proton. In both cases the particle masses are equal*), and after the collision they move in mutually perpendicular directions. Therefore, after development of the photoemulsion,

*) The masses of the neutron and proton are not quite equal; their values, referred to \( \mathrm{O}^{16}=16.0000 \), are respectively \(1.00894\) and \(1.00813\). For the question under discussion this difference plays no role.

through which the neutrons have passed, one can observe the tracks of recoil protons. In the case of a “head-on” collision of a neutron with a proton, the former is stopped, while the latter receives the entire kinetic energy of the incident neutron. Therefore, if one measures the ranges of recoil protons whose tracks are approximately parallel to the direction of motion of the incident neutrons, one can find the distribution of the neutrons according to their energies. It should be noted that, for the successful application of this method, knowledge is required of the direction of motion of the neutrons in the plate.

DISINTEGRATIONS CAUSED BY FAST NEUTRONS

The preceding method of determining the energy of a group of fast neutrons suffers from the drawback that, in addition to the ranges of the knocked-out protons, it is necessary to know the direction of motion of the incident particles, since otherwise it is impossible to determine the angles of motion of the particles after collision. This difficulty can be avoided if observations are made of disintegrations produced by fast neutrons as they pass through plates containing lithium or boron. We saw in photograph XXIII that the penetration of a slow neutron into the nucleus \({}_{5}\mathrm{B}^{10}\) leads to the reaction \({}_{5}\mathrm{B}^{10}+{}_{0}n^{1}\to{}_{2}\mathrm{He}^{4}+{}_{3}\mathrm{Li}^{7}\). A fast neutron, however, produces disintegrations with the emission of three particles, according to the equation:

\[ {}_{5}\mathrm{B}^{10}+{}_{0}n^{1}\to{}_{2}\mathrm{He}^{4}+{}_{2}\mathrm{He}^{4}+{}_{1}\mathrm{H}^{3}. \]

The total mass of all three resulting nuclei is less than the mass of the two initial particles, and therefore in this reaction energy of the order of \(0.35\) MeV is released. Consequently, if \(E\) is the energy of the two \(\alpha\)-particles and the triton emitted in the reaction under consideration, then the energy of the primary neutron was \(E+0.35\) MeV. Further, by the law of conservation of momentum, the vector sum of the momenta of all three particles must be equal to the momentum of the incident neutron. From this one can determine the direction of motion of the primary neutron. The magnitude of its momentum gives a second possibility for determining the velocity, and hence the energy, of the neutron, and the value obtained in this way must coincide with the value obtained by the first method.

Two examples of such a reaction are shown in photograph XXVIII; the smaller star was produced by a neutron with an energy of 13 MeV, arising as a result of the bombardment of boron targets by deuterons with an energy of 900 KeV. In this case the direction of the incident neutron was known. The values of the energy and momentum derived from the observations agree with those values which would be expected on the assumption that the disintegration was caused by a neutron with an energy of 13.4 MeV, emitted by such a source and entering the emulsion in a definite direction. The large star was obtained as a result of the action of cosmic rays at high altitude; measurement

energy and momentum of the three particles shows that this star was produced in a disintegration of the same type as in the case of the small star, with the difference, however, that the neutron had a higher energy, namely 35 MeV.

Disintegrations in lithium, caused by both slow and fast neutrons, lead to the formation of only two particles: \({}_{3}\mathrm{Li}^{6}+{}_{0}n^{1}\to{}_{2}\mathrm{He}^{4}+{}_{1}\mathrm{H}^{3}\). However, unlike the case of slow neutrons, the momentum of the fast neutron proves sufficient so that the two nuclear fragments do not fly apart exactly in opposite directions, and therefore it is possible to determine the point of origin of both tracks.

FISSION

The penetration of a slow neutron into the nucleus \({}_{3}\mathrm{Li}^{6}\) leads to its fission into two parts of almost equal mass, flying apart in opposite directions with great velocity. In a certain limited sense this process may be regarded as a prototype of the famous fission process on which modern possibilities for the liberation of nuclear energy are based. Photograph XXIX gives examples of tracks obtained when slow neutrons penetrate the nucleus of uranium \({}_{92}\mathrm{U}^{235}\); the latter was introduced into the emulsion by means of special treatment of it.

As in the case of lithium, the uranium nucleus, when a slow neutron penetrates it, divides into two almost equal parts. The kinetic energy of both fragments is large—of the order of 200 MeV—but owing to the large charge carried by each of them, they rapidly expend their energy and travel in the emulsion a distance of only a few microns.

An important feature of the process of uranium fission, which cannot be observed in the photographs, consists in the fact that, on the average, in the course of fission the disintegrating nucleus emits more than two neutrons of high energy.

In a sufficiently great thickness of uranium-containing material the neutrons will be slowed down by collisions with nuclei and may in turn lead to further fission of \(\mathrm{U}^{235}\) nuclei. This creates the possibility of a “chain reaction,” in which the number of neutrons rapidly increases and, along with it, the rate of liberation of energy as a result of disintegrations caused by neutrons also increases.

NUCLEONS

The discovery of the neutron led to a complete revolution in our views on the structure of nuclei. At the present time we believe that every nucleus consists of neutrons and protons, the total number of them being equal to the mass number \(A\), while the number of protons is equal to the charge num-

the number \(Z\) and the number of neutrons—\((A-Z)\). It is convenient to have one term for denoting both types of particles, i.e., protons and neutrons, since we no longer assume that they preserve their individuality in the nucleus. We believe that between neutrons and protons there occurs an exchange of charges, a certain rapid and continuous interchange

![Diagram of the composition of isotopes of the first five elements, with protons and neutrons shown schematically. Labels include: proton, deuteron, triton, helion or α-particle, lithium 6, lithium 7, beryllium 9, boron 10, boron 11; filled circles are protons and open circles are neutrons.]

Fig. 8. Composition of the isotopes of the first five elements of the periodic system. This diagram is purely schematic and shows only the number of protons and neutrons in the various nuclei. Nothing is yet known about the actual arrangement of nucleons in nuclei.

of roles, and therefore we speak of both particles as “nucleons” (i.e., nuclear particles). The composition of the nuclei of the first five elements of the periodic system, including the commonly occurring isotopes, is shown schematically in Fig. 8. At present we know very little about the true distribution of nucleons in the nucleus, and the figure given shows only the number of protons and neutrons in the various nuclei.

According to modern ideas, a nuclear transformation involving heavy particles is reduced merely to a redistribution of nucleons. Thus, in the reaction of slow neutrons with lithium, we depict the transformation by the following scheme (Fig. 9).

![Diagram illustrating redistribution of nucleons in the splitting of a \(^{6}_{3}\mathrm{Li}\) nucleus by a slow neutron: \(^{6}_{3}\mathrm{Li} + n \to \alpha\)-particle + triton.]

Fig. 9. Diagram illustrating the redistribution of nucleons in the splitting of the nucleus \({}^{6}_{3}\mathrm{Li}\) by a slow neutron.

The new view of the composition of nuclei has removed the difficulties associated with the supposed presence of electrons. When a nucleus decays and emits a negative electron, it is considered that the latter is born in this transformation, with the number of protons increasing, and the number of neutrons simultaneously decreasing by one. Thus, the electron is not regarded as a permanent constituent part of the structure emitting it. We saw earlier that analo-

...in a similar way we imagine the creation of a quantum of radiation during the transition of an atom from one energy level to a lower one.

An example of a $\beta$-decay process may be the nucleus ${}_{5}\mathrm{B}^{12}$, which, on decaying, gives ${}_{6}\mathrm{C}^{12}$, in accordance with the scheme in Fig. 10. A particularly interesting case is ${}_{3}\mathrm{Li}^{8}$, which, upon emitting an electron, decays into two $\alpha$-particles flying apart in opposite directions with equal velocities (see Fig. 10 and photographs XL and XLI).

Fig. 10. Two examples of changes in the composition of nuclei during beta decay.

The new theory of the structure of nuclei nevertheless left a whole series of problems unresolved. One of the most important is the question of the binding forces. It is necessary to elucidate the nature of the forces responsible for the stability of the system of nucleons composing any nucleus.

The masses of individual protons and neutrons are so small that the forces of gravitational attraction between them may be neglected. On the other hand, the only electrostatic forces acting between protons are the forces of repulsion of their positive charges, and therefore, if only these forces acted, nuclei would fly apart into their constituent parts. Under these circumstances it is necessary to assume that between nucleons there exist forces of a new type, which have received the name mesonic forces. Making a broad generalization, one may say that in the world of stars, in celestial mechanics, gravitational attractive forces predominate; in the world of atoms electrostatic forces predominate; and in the world of nuclei—mesonic forces. Our knowledge of mesonic forces can by no means be considered as complete as our knowledge of the other two types of forces, and much work remains ahead before it will be possible to say that the nucleus has at last been understood.

Mesonic Forces

From the general study of nuclei one may conclude that the mesonic forces between two nucleons do not obey the inverse-square law characteristic both of gravitational and of electrostatic forces. When the distance between two nucleons exceeds a certain very small value, the attractive forces must decrease much more rapidly than would be the case for the inverse-square law. This conclusion follows from the fact that the binding energy of a proton or neutron in a nucleus is of the order of 8 MeV and does not depend essentially on whether the nucleus is light or heavy. Thus, for example, in the case of the Earth’s body (and if one assumes that its density is the same everywhere), the greater the mass of the body, the greater is the force of attraction with which it acts on another body situated on its surface, and the greater must be the kinetic energy of the latter in order for it to be able to leave the field of gravitational attraction. In exactly the same way,

if the meson forces obeyed the inverse-square law, the binding energy of a neutron in a heavy nucleus would be greater than its binding energy in a light nucleus. The observed approximate constancy of the binding energy of heavy and light nuclei can be explained if one assumes that, in the binding of a particle to a nucleus, only the attraction of neighboring protons and neutrons plays a role, i.e., that the forces in the nucleus act over a very small distance. We encounter a similar phenomenon with the short-range forces in the study of molecules. Thus, in the hydrogen molecule, atoms are bound by the exchange of electrons between them. By analogy with this it is believed that meson forces are closely connected with the exchange of charges between the neutrons and protons of the nucleus.

NUCLEAR DISINTEGRATIONS ACCOMPANIED BY THE EMISSION OF MANY PARTICLES

The disintegrations with which we have become acquainted up to now were caused by particles having, on the nuclear scale, a relatively small energy—of the order of 10 MeV. The work that must be done in order to remove a nucleon from the nucleus—the so-called binding energy of a particle—is different for different nuclei, but on average is approximately equal to 8 MeV. Thus, having primary particles with energies less than 30 MeV, one should not expect that, as a result of the disintegration of a nucleus, more than three or four particles will be emitted, and some of them may be neutrons, which leave no tracks in the emulsion. At present, however, it is possible to obtain charged particles and photons having energies above 100 MeV and, consequently, to produce disintegrations with the emission of eight or ten particles.

Photographs XXXI–XXXIV show disintegrations caused by particles from the large synchro-cyclotron of the University of California, as well as by high-energy γ-rays from the betatron in Schenectady. The first of these installations gives deuterons with a maximum energy of 200 MeV, or α-particles with an energy of 400 MeV. The betatron gives a stream of fast electrons which, upon collision with matter, emit γ-rays whose maximum energy is equal to the energy of the electrons, i.e., 100 MeV.

When particles from the cyclotron strike a photographic plate, in the field of view of the microscope one can observe parallel tracks of a large number of particles. The energy of these particles is so great that the ionization per unit path is very small even for α-particles, and therefore the grain density of the tracks is also small.

For particles with the greatest attainable energy this effect is so sharply pronounced that their tracks can no longer be detected.

A complete analysis of disintegrations produced by particles with energies of this order is difficult, since the neutrons which must be emitted in the process remain unregistered. Therefore an investigation similar to that which was carried out for the disintegration of nitrogen caused by deuterons (see photograph XXII) is impossible.

Another difficulty is that, generally speaking, it is impossible to determine the type of nucleus that has undergone disintegration.

In photographing, mosaics had to be assembled, since most of the tracks are inclined to the plane of the emulsion. This is the only way to obtain a satisfactory image of a three-dimensional event on a flat surface.

DISINTEGRATIONS CAUSED BY COSMIC RADIATION

To study nuclear disintegrations caused by particles with energies exceeding those that can be obtained artificially, one must turn to natural sources. As is known, a weak rain of particles, arriving from the depths of space and called cosmic rays, continuously falls upon the upper layers of the atmosphere. According to modern views, at least some of these particles are protons with energies above \(10^8\) MeV. Interacting in the upper layers of the atmosphere with atomic nuclei, the fast primary particles create secondary rays; as a result, at lower altitudes there is a complex flux of protons, neutrons, photons, and other particles. These radiations can readily be detected even at sea level; however, their intensity and average energy increase rapidly with altitude. To detect these particles and the disintegrations of emulsion atoms that they produce, it is sufficient to expose photographic plates for a short period of time at a great height.

The events discussed below were detected in plates left on the Pic du Midi at an altitude of 2800 m; usually it is not even necessary to remove the plates from their packing boxes, since most rays easily pass through the packing material. Usually the plates are left in such a position that the surface of the emulsion lies in a vertical plane.

If a particle with an energy of the order of 1000 MeV collides with a nucleus, the latter may be completely broken up into its constituent nucleons, and some of them may be emitted in the form of complex groups, such as \(\alpha\)-particles or still heavier fragments. Examples of such disintegrations are shown in photographs XXXV–XLI. A primary particle of such energy would not leave a track in the emulsion even if it were charged, and therefore it is difficult to determine the type of particle responsible for such a disintegration. It is reasonable to suppose, however, that, irrespective of the nature of the particles causing the disintegrations, their actions will be very similar, since protons, neutrons, and \(\gamma\)-rays of such energy have momenta of the same order of magnitude. In one or two rare cases of disintegrations caused by particles of noticeably lower energy, determination of the type of primary radiation or of the nature of the heavy fragments formed in the disintegrations proved possible.

MESONS

It has been known for many years—chiefly thanks to the work of Anderson, Blackett, and Wilson with the Wilson chamber—that the penetrating, or “hard,” component of cosmic rays consists mainly of particles with a mass of the order of \(200\,m_e\). Because of their mass, intermediate between the masses of the electron and the proton, these particles were given the name mesons, or mesotrons. The deflection of fast mesons in magnetic fields showed that some of them are positively charged and approximately the same fraction negatively charged.

Theoretical considerations lead to the conclusion that the meson must have a very short lifetime and, upon disappearing, must emit a fast electron. The energy of this electron and, at the same time, of another particle which leaves no track in the Wilson chamber, but whose existence is necessary in order that the law of conservation of momentum be fulfilled in the decay, is taken at the expense of the disappearance of the meson’s rest mass. The short lifetime of mesons has been confirmed experimentally by electrical methods, and a value of the order of two microseconds was obtained. It is believed that mesons are closely connected with those forces whose existence was postulated in order to explain the binding of nucleons in the nucleus.

In addition to the numerous “stars” and single tracks from protons and \(\alpha\)-particles, in plates exposed to cosmic rays at great altitudes one can also notice a relatively small number of tracks similar to those shown in photographs XLII–XLVII. These tracks are distinguished by their low grain density and by the frequent changes in direction which the particles undergo in passing through the emulsion. It further turns out that if such a particle stops in the emulsion, then in the immediate vicinity of the end of its range a very sharp increase in grain density is observed.

All these features indicate that the mass of these particles is considerably less than the mass of the proton. Indeed, a particle of small mass undergoes greater deflections in collisions with nuclei than a heavy particle of the same velocity. Further, the initial velocity of such a particle will be greater than that of a proton with the same range, and therefore on the average it produces less ionization and, consequently, a smaller number of developed grains as it passes through the emulsion. A detailed study of the properties of a large number of tracks of this type made it possible to estimate the mass of this particle as \(230\,m_e\), which is approximately one eighth of the proton mass. It is therefore natural to suppose that at least some of the particles of small mass observed in photographic plates belong to the same type as the mesons discovered in Wilson chambers.

D. H. Perkins discovered that sometimes from the end of the track of a meson stopped in a photographic emulsion there emerges several

heavy charged particles. It is believed that in this process the meson penetrates into the nucleus and disappears there, imparting to the nucleus the energy of its rest mass, i.e. approximately 100 MeV, and causing a disintegration with the emission of several heavy particles. Since these mesons enter the nucleus while having only a small velocity, it is thought that they have a negative charge, since otherwise the nucleus would repel them, as it does any other positively charged particles. This point of view, however, has not yet been finally established. In any case, apparently, mesons are among the most effective of the known agents capable of causing the disintegration of the nuclei of heavy elements. The emulsions available at present do not make it possible to register fast electrons, and therefore from the photographs it is impossible to determine whether those mesons which, upon stopping, decay without producing nuclear disintegrations actually emit electrons.

The study of the behavior of mesons and of their connection with nuclear forces is one of the most urgent problems of nuclear physics. The present state of our knowledge does not permit any final conclusions to be drawn.

EMITTED MESONS

Mesons cannot be primary particles of cosmic rays, since their lifetime is short and they would decay during their flight through space. It is believed that they are produced when very fast protons and neutrons of cosmic rays pass through nuclei. Photograph XLVIII shows a low-energy meson emitted by a disintegrating nucleus. It is impossible to determine the nature of the particle that caused the disintegration, but such photographs illustrate the mode of formation of mesons which is perhaps the most widespread in the upper layers of the atmosphere. In the present case the meson stopped in the emulsion, and therefore it proved possible to determine the type of particle. It should be supposed that at great heights mesons will be emitted in such processes with considerably higher energies.

Up to now, on the plates investigated in the Bristol laboratory, eight such cases have been found. In some cases the meson, after stopping, reacts with a nucleus of the emulsion, forming secondary heavy particles.

SECONDARY MESONS

In several cases it was found (photographs XLIX, L) that a meson stopped in the emulsion emits another meson with considerable energy, of the order of 4 MeV. The number of observed events of this kind is too small for it to be possible to draw a definitive conclusion, but nevertheless it seems probable that the secondary meson

is always emitted with one and the same velocity. This follows from the fact that, as has been found up to now, the secondary mesons have almost identical ranges—610 μ, with small deviations from this value attributable to “straggling.” These observations are a strong argument in favor of the view that there exist different types of mesons; namely, primary ones, which we shall call π-mesons, and which, on decaying spontaneously, emit secondary, so-called μ-mesons.

In order to satisfy the law of conservation of momentum, it is necessary to assume that, in the decay of a π-meson, the momentum of the emitted μ-meson must be balanced by the momentum of some other particle emitted in the opposite direction. This second particle may be a photon or, possibly, a neutral meson, which, having no charge, passes through the emulsion without leaving a track. A count of the number of grains shows that the mass of the π-meson is considerably greater than the mass of the μ-meson. It may therefore be assumed that the kinetic energy of the secondary particles is obtained as a result of the disappearance of a certain fraction of the rest mass of the primary meson.

Probably, π- and μ-mesons possess charges of both signs. All μ-mesons observed up to now that were formed in the emulsion as a result of the decay of π-mesons have stopped in the emulsion without producing disintegrations. It is therefore possible that only positive π-mesons decay with the formation of a μ-meson, whereas negative π-mesons are captured by atoms and produce disintegrations with the emission of heavy particles.

APPENDIX A

Method of Processing Ilford Emulsions “Nuclear Research”

The development and fixing of these emulsions is associated with certain specific difficulties, owing to their considerable thickness and very high concentration of silver halide in comparison with ordinary photographic plates.

For emulsions 50 μ thick, the procedure described below gives satisfactory results. For thicker emulsions it is necessary substantially to increase the fixing time, especially if the plate has been overdeveloped. During development and fixing, the trays, whose dimensions should be considerably larger than the dimensions of the plate, should be rocked continuously. If many plates must be processed, it is desirable to use a mechanical device for rocking, which would ensure uniform washing of the emulsion with the solutions. These operations may be carried out by the light of a 40-watt gas-filled lamp covered with filter No. 2 “Wratten safe-light.”

Development. Development should be carried out for 33 minutes at a temperature of 18° C in the following solution:

Stock solution:

Component	Amount
Metol	5.5 g
Anhydrous sodium sulfite	180 g
Hydroquinone	22 g
Anhydrous sodium carbonate	120 g
Potassium bromide	10 g
Water	to 2500 cm³

When working, the stock solution should be diluted with water in a ratio of 1:4.

To obtain satisfactory plates, the development time is not an essential quantity, but it should lie within the range from 20 to 40 minutes.

The choice of development time depends on the desired degree of difference between the tracks of particles of different ionizing power; the shorter the development time, the weaker the tracks obtained, but the more noticeable the difference between them.

“Stop” bath: the plate is lowered for 10 minutes into a solution of anhydrous acetic acid in water (2% by volume).

Fixing. Fixing is carried out in the following solution until the turbidity from silver halide has disappeared completely and the plate has become transparent. In the case of emulsions 50 μ thick, this requires no more than 30 minutes.

Fixer:

Component	Amount
Saturated solution of hyposulfite	2 parts
Stock acid hardening solution	1 part
Water	8 parts

Composition of the stock acid hardening solution:

Component	Amount	Solution
Anhydrous sodium carbonate	125 g	solution A
Acetic acid	188 cm³	solution A
Both these substances are dissolved in water	650 cm³	solution A
Potassium alum	250 g	solution B
Dissolve in water at a temperature of 87° C	1600 cm³	solution B

Both solutions A and B are mixed at a temperature below 21° C and water is added to 2500 cm³.

Washing. The fixed plate is washed for one hour in a slow stream of running water; the surface of the emulsion is carefully wiped, for cleaning, with a piece of soft chamois, and it is left to dry in room air overnight.

APPENDIX B

Relation between range and energy of charged particles in Ilford emulsions of the “Nuclear Research” type

The relation between the mean range of a homogeneous group of protons or α-particles and their energy, for values of the latter in the interval from 0 to 15 MeV, was shown in Figs. 2 and 5.

For higher energies, only approximate values have been obtained, given in the following table (according to W. Camerini and K. M. G. Lattes):

Energy	Proton range (in microns)	α-particle range (in microns)
10	565	58
15	1135	117
20	1870	201
25	2750	315
30	3760	464
35	4925	553

The corresponding values for deuterons can be derived by a method based on the following considerations. When a charged particle passes through any medium, the energy loss per unit path depends only on its charge and velocity.

Let us consider a proton and a deuteron having one and the same velocity; the energy of the proton will then be only half the energy of the deuteron. Since both particles have the same charge \(+e\), the energy loss of the deuteron will be twice as great as the corresponding value for the proton, if both particles have undergone the same decrease in velocity. It follows from this that the distance traversed by the deuteron will be twice as great as the corresponding value for the proton for a given decrease in velocity, and this result does not depend on the particular values of the velocity. Consequently, the range of the deuteron is twice the range of the proton with the same velocity, i.e., twice the range of the proton having half the energy of the deuteron. Similarly, the range of a triton is three times the range of a proton having one third of its energy.

Thus, from the range–energy curve of protons in a given medium, one can obtain the corresponding curve for deuterons and tritons in the same medium.

APPENDIX C

Collision of a Proton with a Proton

Let us denote by \(m\) the mass of the incident proton and by \(v\) its velocity. Let, after the collision, its velocity be equal to \(v_1\), and let the direction of motion make an angle \(\Theta\) with the direction of the initial motion; similarly, let the velocity and the direction of motion for the struck proton be given by \(v_2\) and \(\Phi\).

Fig. 11. Momentum diagram.

Since in an elastic collision the kinetic energy is conserved, one may write

\[ \frac{1}{2}mv^2=\frac{1}{2}mv_1^2+\frac{1}{2}mv_2^2. \]

or

\[ v^2=v_1^2+v_2^2, \tag{a} \]

in addition, the momentum in the direction of the initial motion is conserved, i.e.

\[ mv=mv_1\cos\Theta+mv_2\cos\Phi \quad \text{or} \quad v=v_1\cos\Theta+v_2\cos\Phi. \tag{b} \]

Further, the momentum perpendicular to the initial direction of motion is also conserved, i.e.

\[ 0=mv_1\sin\Theta-mv_2\sin\Phi \quad \text{or} \quad v_1\sin\Theta=v_2\sin\Phi. \tag{c} \]

Squaring (b) and (c), we obtain:

\[ v^2=v_1^2\cos^2\Theta+v_2^2\cos^2\Phi+2v_1v_2\cos\Theta\cos\Phi, \tag{d} \]

\[ 0=v_1^2\sin^2\Theta+v_2^2\sin^2\Phi-2v_1v_2\sin\Theta\sin\Phi; \tag{e} \]

adding (d) and (e), we have:

\[ v^2=v_1^2(\sin^2\Theta+\cos^2\Theta)+v_2^2(\sin^2\Phi+\cos^2\Phi)+ \]

\[ +2v_1v_2(\cos\Theta\cos\Phi-\sin\Theta\sin\Phi) \]

or

\[ v^2=v_1^2+v_2^2+2v_1v_2\cos(\Theta+\Phi). \tag{f} \]

But, by (a),

\[ v^2=v_1^2+v_2^2. \]

Consequently,

\[ \cos(\Theta+\Phi)=0 \quad \text{and} \quad \Theta+\Phi=\frac{\pi}{2}. \]

The directions of motion after the collision form a right angle with one another.

The energies of the particles after the collision, \(\frac{1}{2}mv_1^2=E_1\) and \(\frac{1}{2}mv_2^2=E_2\), are simply related to the energy of the incident particle \(E_0\).

From equation (c):

\[ v_1 \sin \Theta = v_2 \cos \Theta,\quad \text{whence}\quad v_2 = v_1 \operatorname{tg}\Theta . \]

Since \(v^2 = v_1^2 + v_2^2\),

\[ E_0 = E_1 + E_1 \operatorname{tg}^2 \Theta = E_1 \sec^2 \Theta . \]

Consequently, \(E_1 = E_0 \cos^2 \Theta\). Similarly, \(E_2 = E_0 \sin^2 \Theta\).

In photograph XII the directions of motion of both protons immediately after the collision are \(\Theta = 33.6^\circ\) and \(\Phi = 56.8^\circ\), so that \((\Theta+\Phi)=90.4^\circ\). Further, the ranges of both protons after the collision are \(138.6\mu\) and \(41.8\mu\), and their energies, in accordance with the range–energy relation for protons, are \(4.35\) MeV and \(2.06\) MeV.

Since \(E_2 = E_1 \operatorname{tg}^2 \Theta\), one can calculate the value of \(\Theta\) corresponding to the observed values of \(E_1\) and \(E_2\). Thus, \(E_2/E_1 = 0.4736 = \operatorname{tg}^2 \Theta\), \(\operatorname{tg}\Theta = 0.6882\), and therefore \(\Theta = 34.5^\circ\), while \(33.6^\circ\) was observed directly.

Observations of this kind show that the direction of motion of a track in the emulsion can be determined with an accuracy of the order of \(\pm 0.5^\circ\).

It should be noted that the two protons are indistinguishable, and therefore it is impossible to determine which of them was initially at rest in the emulsion. By analogy with collisions of two billiard balls, one may speak of a “central,” or “head-on,” and of a “noncentral” collision. In an almost central collision the struck proton will be thrown forward (the angle \(\Phi\) is small) and it receives almost all the energy of the incident proton, which is deflected at an angle of almost \(90^\circ\) to the initial direction of motion, having only a small velocity. This case is indistinguishable from the case of a noncentral collision, in which the proton retains almost all the energy and undergoes only a small deflection.

APPENDIX D

Collision of a deuteron with a proton

Let a particle having velocity \(v\) and mass \(M_1\) collide with another particle of mass \(M_2\), which was initially at rest. Suppose further that, as a result of the collision, \(M_1\) is deflected through an angle \(\Theta\), and \(M_2\) through an angle \(\Phi\) (see Fig. 12).

Applying the principles of elementary dynamics, one can show that

\[ \operatorname{tg}^2 \Theta = \frac{\sin^2 \Phi}{M_1/M_2 - \cos^2 \Phi}. \]

If \(M_1\) is the mass of the deuton and \(M_2\) the mass of the proton, then instead of the true masses one may simply substitute the mass numbers of the nuclei, i.e., respectively 2 and 1; we then obtain

\[ \tg \theta=\frac{\sin \Phi}{2-\cos^2 \Phi}. \]

If one plots the curve of \(\tg \theta\) as a function of \(\Phi\), starting from \(\Phi=0\), it turns out that at \(\theta=30^\circ\) a maximum is reached, equal to \(1/\sqrt{3}\). Hence it follows that, in the collision of a deuton with a proton at rest, the deuton cannot be scattered through an angle greater than \(30^\circ\). Therefore the long track in photograph XIV must be assigned to an elastically scattered deuton. Measurements show that for this particular case \(\theta=23.5^\circ\) and \(\Phi=52.5^\circ\).

Fig. 12. Momentum diagram.

Substituting the latter value into the formula given above, we obtain \(\theta=23.2^\circ\). The residual range of the deuton after the collision is found to be \(104.8\,\mu\). Since the range of a deuton is twice as great as the range of a proton with energy half as large, we can determine from Fig. 5 the energy of a proton having a range \(52.4\,\mu\). The value obtained is \(2.43\) MeV, whence we conclude that the energy of the scattered deuton was \(4.86\) MeV. To a range of the struck proton of \(57.0\,\mu\) there corresponds an energy of \(2.50\) MeV. If the collision was elastic, then immediately before the collision the energy of the deuton was \(7.36\) MeV.

The momentum of a particle, \(Mv\), is equal to

\[ \sqrt{2ME}, \quad \text{where } E=\frac{1}{2}Mv^2 \]

is its kinetic energy. It is convenient to choose such units that for a nuclear particle with mass number \(N\) and energy \(E\) (in MeV), the momentum is equal to \(\sqrt{2NE}\). For the struck proton \(N=1\) and \(E=2.50\) MeV; consequently its momentum is

\[ \sqrt{2\cdot 1\cdot 2.50}=2.236 \text{ units}. \]

Similarly, the momentum of the scattered deuton is

\[ \sqrt{2\cdot 2\cdot 4.86}=4.409 \text{ units}. \]

The components of the momentum in the direction perpendicular to the initial direction of motion are, for the proton:

\[ 2.236\times 0.7934=1.775 \text{ units} \]

and for the deuton:

\[ 4.409\times 0.3987=1.760 \text{ units}, \]

Moreover, the discrepancy between these two values lies within the limits of experimental error.

The energy of the deuteron immediately before the collision, as was shown, was \(7.36\) MeV. The range of such a particle in the emulsion is \(229.2\mu\). The observed length of the deuteron trajectory in the emulsion up to the point of scattering is \(69.4\mu\), and consequently the initial range of the deuteron may be estimated as \(298.4\mu\). The value of the energy corresponding to this range, \(9.10\) MeV, may be compared with the mean energy of the deuterons falling on the plate, equal to \(8.90\) MeV.

The agreement between the experimental and theoretical values, which is observed in all cases, shows that the collision was correctly interpreted.

APPENDIX E

Collision of an \(\alpha\)-particle with a proton

If an \(\alpha\)-particle with mass number 4 collides with a proton having mass number 1 and initially at rest, then the scattering angles \(\Theta\) and \(\Phi\) (see Fig. 12) are related by the relation

\[ \operatorname{tg}\Theta=\frac{\sin^2\Phi}{4-\cos^2\Phi}. \]

One may analyze the event shown in photograph XVI by a method similar to that which was used in considering the collision of a deuteron with a proton; the resulting scattering angles and particle ranges are as follows:

\[ \Theta=8.0^\circ,\ \text{residual range of the } \alpha\text{-particle } 6.75\mu, \]

\[ \Phi=16.5^\circ,\ \text{range of the ejected proton } 101.0\mu. \]

Thus, the energy of the \(\alpha\)-particle at the point of scattering proves to be \(5.64\) MeV. The distance in the emulsion between the point of origin of the star and the point of scattering of the particle is \(7.4\mu\). Hence the initial energy of the \(\alpha\)-particle is found to be \(6.73\) MeV, which indicates that it arose as a result of the decay of the thorium A nucleus.

APPENDIX F

Interpretation of experiments on nuclear disintegration

Photographs XVII–XIX constitute definitive proof of the transformation of light elements under the action of fast deuterons. Following the first experiments of Cockcroft and Walton in 1932, further experiments of a similar type, but using electrical methods of particle detection, constituted one of the principal fields of research in nuclear physics. To illustrate the methods used in the analysis of such photographs, let us consider the data ob—

irradiated together with a plate on which fell particles arising in the disintegration of beryllium under the action of deuterons with an energy of 900 KeV (see photograph XVIII).

The lengths of all tracks appearing on the given area of the emulsion were first measured under a microscope with the aid of an ocular scale; the results obtained in this way by one observer over the course of 8 hours are presented in Fig. 13. This block diagram gives the distribution of the particles by ranges: along the abscissa is plotted the range in microns, and along the ordinate—the number of tracks of a given range. The abscissa axis is divided into intervals of one micron.

Fig. 13. Distribution by ranges of tracks obtained in the disintegration of beryllium by deuterons with an energy of 900 KeV

\[ \begin{aligned} a)&\quad \mathrm{Be}^{9}_{4}+\mathrm{H}^{2}_{1}\to \mathrm{Be}^{10}_{4}+\mathrm{H}^{1}_{1};\\ b)&\quad \mathrm{Be}^{9}_{4}+\mathrm{H}^{2}_{1}\to {}^{*}\mathrm{Be}^{10}_{4}+\mathrm{H}^{1}_{1}, \quad \text{the nucleus } \mathrm{Be}^{10} \text{ remains in an excited state;}\\ e)&\quad \mathrm{Be}^{9}_{4}+\mathrm{H}^{2}_{1}\to \mathrm{Be}^{8}_{4}+\mathrm{H}^{3}_{1};\\ d)&\quad \mathrm{Be}^{9}_{4}+\mathrm{H}^{2}_{1}\to \mathrm{Li}^{7}_{3}+\mathrm{He}^{4}_{2};\\ c)&\quad \mathrm{H}^{2}_{1}+\mathrm{H}^{2}_{1}\to \mathrm{H}^{3}_{1}+\mathrm{H}^{1}_{1}. \end{aligned} \]

Short tracks are formed by: deuterons elastically scattered by target nuclei, tritons from reaction (e), and lithium recoil nuclei from reaction (d).

Figure 13 shows that the tracks are distributed in separate groups. By studying the grain density of the tracks of any one of these groups, we can determine the nature of the corresponding particles. Thus, one of the tracks of group (a), shown in photograph XVIII, is easily identified with a proton. This group of particles could have appeared only in the disintegration of beryllium by deuterons. Since this element under natural conditions occurs as a single isotope with mass number 9, the type of the disintegrated nucleus is unambiguously determined,

and the transformation that led to the formation of protons may be represented by the following equation:

\[ {}_{4}\mathrm{Be}^{9}+{}_{1}\mathrm{H}^{2}\rightarrow X+{}_{1}\mathrm{H}^{1}, \]

where \(X\) is one of the resulting nuclei and is to be determined. To balance the total charge and the mass numbers on both sides of the equality, it is necessary to take the mass number of \(X\) as equal to 10 and its charge number as 4. Thus \(X\) is a beryllium isotope with mass number 10, and therefore the equation may be rewritten in the following form:

\[ {}_{4}\mathrm{Be}^{9}+{}_{1}\mathrm{H}^{2}\rightarrow{}_{4}\mathrm{Be}^{10}+{}_{1}\mathrm{H}^{1}. \]

The next question that must be considered is the determination of the amount of energy released in the transformation. First we determine the mean energy of the protons of group \((a)\), from the relation between range and energy. In this way one obtains \(4.76 \pm 0.02\) MeV, but this value does not give the whole energy released in the reaction, since the nucleus \({}_{4}\mathrm{Be}^{10}\) receives a certain kinetic recoil energy from the proton. Although the tracks of these recoil nuclei are not visible in the photograph, their energy can nevertheless be estimated by applying the principle of conservation of momentum.

In Fig. 14, \(OA\) represents the momentum of the incident deuteron and \(OB\) the momentum of the ejected proton. Since the initial and final momenta must be equal, it is necessary to assign to the nucleus \({}_{4}\mathrm{Be}^{10}\) such a momentum that, when added vectorially to \(OB\), gives a resultant equal to \(OA\), i.e. the momentum of the incident deuteron.

Fig. 14. Momentum diagram.

Thus the momentum of the recoil nucleus is represented in direction and magnitude by the segment \(OC\), which is parallel and equal in length to the segment \(AB\). Knowing the mass and momentum of the Be nucleus, one can determine its velocity and, consequently, its energy. In this way a value of \(0.66 \pm 0.01\) MeV is obtained, and therefore the total amount of energy released is \(4.52 \pm 0.03\) MeV. It may be noted that, as with a rifle when fired, the heavy particle receives only a small share of the energy. The question arises: what, in general, accounts for the observed release of energy, why is precisely this observed value of the energy obtained, and not some other?

We noted earlier that the true masses, unlike the charge and mass numbers, are not integers if the mass of the proton is taken as unity. Thanks to the work of Aston and

the masses of other nuclei have been measured with great accuracy. It is customary to express the mass of a nucleus of a given type in such units that the mass of the ordinary isotope of oxygen, having mass number 16, is also equal to 16. Then, in these units, the masses of the four nuclei participating in the transformation we are considering are

\[ {}_{4}\mathrm{Be}^{9}=9.014958,\qquad {}_{4}\mathrm{Be}^{10}=10.016622, \]

\[ {}_{1}\mathrm{H}^{2}=2.014725,\qquad {}_{1}\mathrm{H}^{1}=1.008131. \]

It is easy to see that the total mass of the initial nuclei is equal to 11.029683, while for the final nuclei it is equal to 11.024753. Thus a mass of \(0.00493\pm 0.00012\) units has disappeared*).

According to Einstein’s equation, such a disappearance of mass must be accompanied by the appearance of an equivalent amount of energy. Using the unit defined above as the unit of mass and \(1\ \mathrm{MeV}\) as the unit of energy, we find that the disappearance of \(0.001\) mass unit corresponds to the appearance of \(0.9312\ \mathrm{MeV}\) of energy. Consequently, the observed loss of mass of \(0.00493\pm 0.00012\) units must be accompanied by the appearance of an energy of \(4.59\pm 0.11\ \mathrm{MeV}\), which, within the limits of error, is equal to the value obtained experimentally by us.

An analogous consideration of the other groups of particles shown in Fig. 13 indicates that they can be attributed to the following reactions. Group \((c)\) consists of tritons produced in the reaction

\[ {}_{4}\mathrm{Be}^{9}+{}_{1}\mathrm{H}^{2}\rightarrow{}_{4}\mathrm{Be}^{8}+{}_{1}\mathrm{H}^{3}. \]

\({}_{4}\mathrm{Be}^{8}\) is not found in nature, and its mass is almost exactly equal to the mass of two \(\alpha\)-particles. It is quite probable that soon after its formation it decays in accordance with the equation \({}_{4}\mathrm{Be}^{8}\rightarrow{}_{2}\mathrm{He}^{4}+{}_{2}\mathrm{He}^{4}\), and since the mass difference between the two sides of the equation is so small, the \(\alpha\)-particles fly apart with a very small kinetic energy and therefore cannot be detected.

Group \((d)\) consists of \(\alpha\)-particles produced in the reaction

\[ {}_{4}\mathrm{Be}^{9}+{}_{1}\mathrm{H}^{2}\rightarrow{}_{3}\mathrm{Li}^{7}+{}_{2}\mathrm{He}^{4}. \]

The numerous short-range particles in photograph XVIII are predominantly deuterons elastically scattered by nuclei of the target atoms.

It remains to explain group \((e)\). Examination of the tracks shows that the particles of this group are protons, the appearance of which may be attributed to the reaction

\[ {}_{4}\mathrm{Be}^{9}+{}_{1}\mathrm{H}^{2}\rightarrow{}^{*}_{4}\mathrm{Be}^{10}+{}_{1}\mathrm{H}^{1}. \]

Here the symbol \({}^{*}_{4}\mathrm{Be}^{10}\) means that the beryllium isotope is formed in

*) The errors correspond to the errors in determining the masses of the nuclei participating in the reaction.

excited state. In this case one can calculate the energy released in the reaction by the same method as for the cases of group (a), and then, using Einstein’s relation, calculate also the mass of the excited nucleus. The value obtained in this way turns out to be greater than the mass of the nucleus \(_4\mathrm{Be}^{10}\) in the normal, or lower, state, and from the difference of the masses one can calculate the excitation energy of the nucleus. We have thus determined the energy of the excited state of \(_4\mathrm{Be}^{10}\) formed in the reaction under consideration. By similar methods the energy states of many stable and unstable nuclei have been determined, and the results obtained show that, like atoms, nuclei can exist only in certain discrete energy states.

As a rule, an excited nucleus quickly returns to the lower state, emitting the excess energy in the form of a quantum of radiation. The changes in energy, however, are large in comparison with those observed for the electronic systems of atoms, and therefore the energy of the photons is correspondingly higher. This radiation is capable of passing through considerable thicknesses of matter and has been given the name \(\gamma\)-radiation.

APPENDIX G

Determination of the Energy of Fast Neutrons

The masses of the neutron and the proton differ so little from one another that the mechanical problem of the collision between these two particles is no different from the analogous problem for the collision of a proton with a proton. Therefore the energy \(E_1\) of the recoil proton is related to the energy \(E_0\) of the incident neutron by the equation

\[ E_1 = E_0 \cos^2 \theta, \]

where \(\theta\) is the angle between the direction of motion of the recoil proton and the direction of the incident neutron. A proton knocked out at a small angle receives almost all the energy of the neutron \((\cos^2 5^\circ = 0.992)\). By measuring the ranges, and consequently the energies, of a large number of just such recoil protons, for which \(\theta < 5^\circ\), one can determine the energy distribution of the incident neutrons. This method is one of the most powerful for determining the energy of fast neutrons.

A scheme of one of the apparatuses for experiments of this type is shown in Fig. 15. A beam of fast deuterons from a high-voltage generator is passed through a round aperture in a metal plate. The particles that have passed through the aperture strike a target, which is placed on a cooled metal support. The neutrons formed in the transformation of the material of the target fly out in all directions, passing through the walls of the vacuum vessel. Some

Fig. 15. Diagram of the apparatus for investigating the energy distribution of neutrons arising from the disintegration of light elements by fast deuterons.

Labels in the diagram: deuteron beam; glass tube; photographic plate in a light-tight box; molybdenum plate; target; water-cooling tubes.

Fig. 16.

a decrease in the number of neutrons will occur because of collisions of a small fraction of the neutrons with the nuclei of the solid material of the apparatus, but the intensity of the emerging neutron beam changes little as a result of such processes.

If the photographic plates are arranged as shown in the figure, then part of the emitted neutrons will enter the emulsion parallel to its surface. It is possible to determine the direction of the neutron flux at any point of the emulsion if the exact geometrical orientation of the plate during exposure is specified. Then one can determine the direction of any proton track and its range, and hence the energy of the neutron responsible for it.

An example of data obtained in this way is given in Fig. 16, which shows the results of measurements of the energy distribution of neutrons produced when boron is bombarded by deuterons with an energy of 900 KeV. As can be seen, there exist several groups of neutrons with different energies. The groups of greatest energy arise in the reaction represented by the equation

\[ {}_{5}\mathrm{B}^{11}+{}_{1}\mathrm{H}^{2}\to{}_{6}\mathrm{C}^{12}+{}_{0}n^{1}, \]

where the nucleus is formed either in the lower state or in an excited state with an energy of 4.3 MeV. Some of the other groups are due to the reaction

\[ {}_{5}\mathrm{B}^{10}+{}_{1}\mathrm{H}^{2}\to{}_{6}\mathrm{C}^{11}+{}_{0}n^{1}, \]

where the nucleus ${}_{6}\mathrm{C}^{11}$ is formed in various energy states. The group with an energy of 2.2 MeV owes its origin to the reaction

\[ {}_{1}\mathrm{H}^{2}+{}_{1}\mathrm{H}^{2}\to{}_{2}\mathrm{He}^{3}+{}_{0}n^{1}. \]

In this reaction the deuterons undergoing splitting are obtained as a result of the stopping of the primary particles in the target.

NUCLEAR PHYSICS IN PHOTOGRAPHS

PHOTOGRAPH XVII

α-particles and protons produced in the disintegration of lithium by deuterons.

α-particles and protons produced in the disintegration of lithium by deuterons

A mosaic of 24 microphotographs showing tracks of α-particles and protons produced when a lithium target is bombarded by deuterons with an energy of 900 KeV. The tracks of α-particles have the form of almost solid black grains of silver. In this emulsion layer the protons are much thinner and therefore can be distinguished from the tracks of α-particles. The oblique lines on the plate are scratches or mechanical damage produced as a result of friction. They are easy to distinguish from particle tracks, since they are formed only on the surface of the emulsion.

PHOTOGRAPH XVIII

↓

Particles from the disintegration of beryllium by deuterons

A mosaic of microphotographs showing the tracks of particles produced when a thin beryllium target is bombarded by deuterons with an energy of 900 KeV. Most of the very short tracks are due to primary deuterons elastically scattered by the nuclei of the target atoms. Also visible are tracks of \(\alpha\)-particles, protons, and tritons; the reactions in which they arise are discussed in Appendix F.

NUCLEAR PHYSICS IN PHOTOGRAPHS

PHOTOGRAPH XIX

Particles from the disintegration of boron by deuterons

A mosaic of microphotographs of the surface of a plate exposed in a stream of particles produced as a result of the disintegration of boron by deuterons with an energy of 900 KeV. The number of tracks is too large for detailed analysis because of the long exposure, but the traces of $\alpha$-particles can be distinguished, some of which arise in the reaction

\[ {}_{5}\mathrm{B}^{10}+{}_{1}\mathrm{H}^{2}\to{}_{2}\mathrm{He}^{4}. \]

If one places in focus the more distant layers of the emulsion, one can detect the tracks of long-range protons, which alone are capable of penetrating deeply into the emulsion. Examples of the latter are shown in photographs VII, VIII, and IX.

PHOTOGRAPHS XX and XXI

Splitting of emulsion-atom nuclei

Photograph XX shows a splitting caused by a deuteron with an energy of 8.5 MeV and leading to the ejection of a fast proton. The recoil nuclei leave very short tracks. Application of the laws of conservation of momentum and energy indicates that this event probably corresponds to the reaction

\[ {}_{8}\mathrm{O}^{16}+{}_{1}\mathrm{H}^{2}\rightarrow{}_{8}\mathrm{O}^{17}+{}_{1}\mathrm{H}^{1}, \]

where the oxygen nucleus with mass number 17 is formed in the lower state. This event can also be ascribed to the reaction

\[ {}_{7}\mathrm{N}^{14}+{}_{1}\mathrm{H}^{2}\rightarrow{}_{7}\mathrm{N}^{15}+{}_{1}\mathrm{H}^{1}, \]

but then the nucleus \({}_{7}\mathrm{N}^{15}\) is formed in one of the excited states, with energy \(7\) MeV.

Photograph XXI shows the splitting, produced by a deuteron with an energy of about 8 MeV, of \({}_{6}\mathrm{C}^{12}\) into a \({}_{5}\mathrm{B}^{10}\) nucleus and an \(\alpha\)-particle, in accordance with the equation

\[ {}_{6}\mathrm{C}^{12}+{}_{1}\mathrm{H}^{2}\rightarrow{}_{5}\mathrm{B}^{10}+{}_{2}\mathrm{He}^{4}. \]

The recoil nucleus moves forward almost in the direction of the incident deuteron.

NUCLEAR PHYSICS IN PHOTOGRAPHS

PHOTOGRAPH XXII

Splitting of nitrogen into four α-particles under bombardment by deuterons

Three deuterons with an initial energy of 8.9 MeV enter the emulsion from the upper side of the photograph at a small angle of incidence to the emulsion. One of them enters the nucleus and leads to the ejection of four particles which, judging from the density of the grains of the tracks, are α-particles. From the ranges of the particles one can determine their energy and momentum. The vector sum of the momenta is equal to the momentum of the incident deuteron, and the total energy corresponds to that obtained in the transformation

\[ {}_{7}\mathrm{N}^{14} + {}_{1}\mathrm{H}^{2} \rightarrow {}_{2}\mathrm{He}^{4} + {}_{2}\mathrm{He}^{1} + {}_{2}\mathrm{He}^{1} + {}_{2}\mathrm{He}^{4}. \]

No other interpretation is consistent with the observations, and the identification is undoubtedly correct. This example illustrates the advantage of observations made on splittings occurring in the emulsion itself. If, for example, nitrogen is bombarded with deuterons in an apparatus similar to that shown in Fig. 7, then only one α-particle, appearing in any splitting, will be observed. Such observations do not make it possible to establish the details of the transformation.

K. F. POWELL and G. P. S. OCCHIALINI

PHOTOGRAPH XXIII

Tracks obtained in the fission of lithium and boron by slow neutrons

Tracks produced by the passage of slow neutrons through a photographic emulsion containing borate of lithium introduced into the emulsion during the preparation of the latter. The tracks are obtained as a result of two reactions (9) and (10) (see p. 390): the short-range ones from boron and the two long-range ones from lithium. Individual components of each track cannot be distinguished. The two nuclei formed in each transformation fly apart in opposite directions, since on a nuclear scale the initial momentum of the neutron is almost zero. In contrast to charged particles, which leave their own track in the emulsion, only one neutron in a thousand causes fission and thereby creates a visible track.

NUCLEAR PHYSICS IN PHOTOGRAPHS

PHOTOGRAPH XXIV

Splitting of Lithium by Slow Neutrons

The emulsion is “loaded” with a lithium salt. The interaction of slow neutrons with lithium, leading to the formation of two nuclei of almost equal mass, is in a certain sense a prototype of the famous process of “fission” of \(U^{235}\), which takes place under the action of slow neutrons. The neutrons are obtained from a radium–beryllium source. The \(\alpha\)-particles of radium and of its decay products, on colliding with a beryllium nucleus, cause the emission of neutrons in accordance with equation (7) (p. 389). Radium also emits \(\gamma\)-rays, which are responsible for the background of grains.

K. F. POWELL AND G. P. S. OCCHIALINI

PHOTOGRAPH XXV

Proton tracks produced in collisions of fast neutrons with hydrogen nuclei

Recoil protons produced by neutrons with an energy of 2.5 MeV in the reaction

\[ \mathrm{H}^{2}+\mathrm{H}^{2}\rightarrow \mathrm{He}^{3}+\mathrm{n}^{1}. \]

The neutrons passed through the emulsion parallel to its surface in the direction indicated by the arrow. Only protons ejected in the direction of motion of the neutrons have an energy equal to the energy of the incident neutrons. The photograph shows that protons ejected at appreciable angles have shorter ranges.

NUCLEAR PHYSICS IN PHOTOGRAPHS

PHOTOGRAPH XXVI

[Annotation in the photograph: “End.” Scale: \(100\,\mu\).]

Track of a recoil proton

A mosaic of microphotographs showing a recoil proton produced as a result of a collision with a fast neutron. The proton track begins in the lower left corner and ends in the upper right corner of the photograph. The neutrons were produced by bombarding boron with deuterons of energy 900 KeV; they are obtained as a result of the reaction

\[ {}_{5}\mathrm{B}^{11} + {}_{1}\mathrm{H}^{2} \rightarrow {}_{6}\mathrm{C}^{12} + {}_{0}\mathrm{n}^{1}. \]

It is precisely this nuclear reaction that is especially important, since it makes it possible to produce homogeneous groups of fast neutrons of known energy, which is significant for experiments in nuclear physics. Unlike fast charged particles, we cannot generate fast neutrons otherwise than in the form of secondary particles arising in nuclear transformations.

K. F. Powell and G. P. S. Occhialini

PHOTOGRAPH XXVII

Track of a proton and of radiothorium stars

The long track, running from top to bottom, was produced as a result of the collision of a neutron with an energy of 9 MeV with a proton of a hydrogen atom in the emulsion. The track passes through a larger number of radiothorium stars, and the photograph shows that the proton track is much thinner over most of its path than the tracks of α-particles. The possibility of distinguishing the tracks of particles of different ionizing power by their grain density is called the “analyzing power” of the emulsion.

PHOTOGRAPH XXVIII

Fission of boron by fast neutrons

Two examples of the fission of \(B^{10}\) into two \(\alpha\)-particles and triton. In both cases the triton moves downward from top to bottom. The fission with the lower energy was caused by a neutron with an energy of 13 MeV from a high-voltage installation, and the second by a cosmic-ray neutron with an energy estimated at 35 MeV.

\(100\mu\)

K. F. POWELL AND G. P. S. OCCHIALINI

PHOTOGRAPH XXIX

Fission tracks of \(U^{235}\)

Examples of tracks obtained in the fission of \(U^{235}\), caused by slow neutrons. Uranium was introduced into the plate by immersing it in a solution of uranyl nitrate. Generally speaking, it is impossible to determine by what point the fission tracks were produced, and consequently impossible to determine the ranges of the individual fragments. It may be noted that in a number of cases the fission fragments collided with silver bromide nuclei, forming forked tracks. In the case of (a) two tracks produced in such a collision form a right angle with one another. This event probably corresponds to a collision with a silver nucleus, which has mass number 109, since some of the fission fragments have mass numbers close to this value.

NUCLEAR PHYSICS IN PHOTOGRAPHS

PHOTOGRAPH XXX

Traces of the fission of a nucleus into three parts,
one of which is an α-particle

In this photograph an example is shown of a rare mode of fission, first observed in photographic plates by D. L. Livesey. The nucleus has split into three parts. The long fission track is thinner than the other two and is thought to have been produced by an α-particle, although this circumstance has not yet been definitively established. The second fission track, passing beneath the track of the long-range particle, is not connected with the first fission event.

K. F. POWELL AND G. P. S. OCCHIALINI

PHOTOGRAPH XXXI

Tracks of Particles Having High Energy

Tracks of $\alpha$-particles and deuterons from the 184-inch cyclotron; the particles enter the emulsion almost parallel to its surface. Both types of particles have the same velocities, and the difference in their specific ionization is clearly visible: the tracks of the $\alpha$-particles are the most conspicuous, while the tracks of the deuterons are very thin. The initial energy was about 200 MeV for the $\alpha$-particles and 100 MeV for the deuterons. Two or three $\alpha$-particles remained in the field of view.

NUCLEAR PHYSICS IN PHOTOGRAPHS

PHOTOGRAPH XXXI

Disintegrations caused by high-energy particles
from the synchro-cyclotron

Three examples of mosaics of microphotographs of disintegrations caused by fast deuterons with an energy of 160 MeV, obtained in the 184-inch cyclotron. The direction of motion of the deuterons is indicated by arrows; however, it is impossible to identify reliably the tracks of the particles that caused the disintegrations.

K. F. POWELL AND G. P. S. OCCHIALINI

PHOTOGRAPH XXXIII

Splittings caused by particles of high energy

Splittings caused by particles from a 184-inch cyclotron in the photograph on the right; the track of the particle that caused the splitting cannot be seen, but it probably was a deuteron with an energy of 160 MeV. The visible tracks of the other particles are probably $\alpha$-particles with energies of 320 MeV. The track of the primary particle in the photograph on the left is marked by the letter $d$ and may be made clearly visible if the page is tilted so that the grains of the track lie on the line of sight.

NUCLEAR PHYSICS IN PHOTOGRAPHS

PHOTOGRAPH XXXIV

Examples of photodisintegration of nuclei

Disintegration caused by high-energy $\gamma$-radiation from the 100 MeV betatron at Schenectady.

9 Uspekhi Fizicheskikh Nauk, Vol. XXXV, No. 3

PHOTOGRAPH XXXV

Splitting of an Oxygen Nucleus

Splitting of \(O^{16}\) into four \(\alpha\)-particles, caused by cosmic radiation. It is impossible to assert with certainty that the splitting was caused by a photon, but this is the most probable explanation. The total energy of all four \(\alpha\)-particles, determined from their ranges, is equal to \(26\ \mathrm{MeV}\). If it is assumed that the splitting of \(O^{16}\) is described by the equation

\[ {}_{8}O^{16}+h\nu \rightarrow 4_{2}He^{4}, \]

then the photon energy must have been equal to \(40\ \mathrm{MeV}\). Within the limits of experimental error, the momentum of such a quantum is equal to the vector sum of the momenta of the four \(\alpha\)-particles. The results, however, can also be reconciled with the supposition that the splitting was caused by a fast neutron, which was not captured and did not change the direction of its motion*). In the case under consideration, the neutron energy would have had to be equal to \(400\ \mathrm{MeV}\), and it is more probable that photodisintegration took place.

*) Photon momentum. In interpreting the event under consideration it was necessary to compare the energy released in the reaction with the vector sum of the momenta of the four \(\alpha\)-particles. The mass of \(O^{16}\) is equal to \(16.000\), and the mass of four \(\alpha\)-particles is \(16.015410\). Thus, in this transformation the rest mass increased by \(0.0154\) mass units, which is equivalent to \(14.4\ \mathrm{MeV}\). According to the measurements of the ranges of the \(\alpha\)-particles, the sum of their kinetic energies is equal to \(26\ \mathrm{MeV}\), and therefore the total energy given up by the incident particle is \(40.4\ \mathrm{MeV}\).

This energy can be compared with the momentum of the \(\alpha\)-particles responsible for the splitting. The momentum of a quantum with energy \(h\nu\) is equal to \(h\nu/c\), where \(c\) is the speed of light. Taking \(h\nu=40.4\ \mathrm{MeV}\), we find that, within the limits of measurement error, the photon momentum is equal to the vector sum of the momenta of the four \(\alpha\)-particles.

In the case of photons of low energy, as in the case of visible radiation, the momentum of individual photons is very small, and therefore it is difficult to measure the pressure which it produces when incident on some surface, if a light source of ordinary intensity is used.

PHOTOGRAPH XXXVI

Splitting of a heavy nucleus

The splitting of a nucleus, probably silver or bromine, by a cosmic-ray particle. Four $\alpha$-particles with a total energy of $64$ MeV are emitted, two of which are emitted in opposite directions with almost equal energies.

At the present time it is not possible to determine the nature of the primary particle that caused the splitting. The short track, emerging from the nucleus in the direction of the arrow, has a very low grain density. It has not yet been established by what particle this track was formed,

K. F. Powell and G. P. S. Occhialini

PHOTOGRAPH XXXVII

“Explosive splitting of a nucleus”

Splitting of a nucleus, probably of silver, by cosmic-ray particles. The energy of the particle that caused the splitting was apparently of the order of 100 MeV. It is possible to distinguish the tracks of seven protons, five $\alpha$-particles, and a number of heavier nuclear fragments. Most of the particles pass from the emulsion into the glass or leave the plate completely, as a result of which it was not possible to determine precisely their ranges and, consequently, their energies.

NUCLEAR PHYSICS IN PHOTOGRAPHS

PHOTOGRAPH XXXVIII

Tracks of particles of various types

Disintegration of a heavy nucleus by cosmic rays and the tracks of a meson and of a proton (left, above). Notable is the sharp difference in the scattering experienced by a meson and by heavy particles, as well as the more rapid change in the density of the grains of the meson track. The long track from the star is produced by an $\alpha$-particle.

K. F. POWELL AND G. P. S. OCCHIALINI

PHOTOGRAPH XXXIX

Star formed by cosmic rays

Disintegration of a heavy nucleus, in which a long-range α-particle (with an energy of 32 MeV) remains in the emulsion. In air, under normal conditions, an α-particle of this energy would have a range of approximately one meter.

PHOTOGRAPHS XL and XLI

“Hammer-like Tracks”

Two examples of splittings caused by cosmic rays, in which heavy nuclear particles are emitted. The latter, after stopping, decay into two $\alpha$-particles, which fly apart in opposite directions with equal velocity.

The photographs shown indicate that both heavy particles were nuclei of ${}^{8}\mathrm{Li}$. It is known that the latter emit simultaneously a $\beta$-particle and two $\alpha$-particles, in accordance with the equation

\[ {}^{8}_{3}\mathrm{Li} \to e_{-1} + {}^{4}_{2}\mathrm{He} + {}^{4}_{2}\mathrm{He}. \]

K. F. Powell and G. P. S. Occhialini

PHOTOGRAPH XLII

Tracks of mesons with and without disintegration particles

Tracks of two different mesons, moving from top to bottom. One of the mesons leads to the disintegration of a nucleus, in which two $\alpha$-particles and a proton are emitted. One may compare the tracks of these mesons with the tracks of protons shown in photographs VII–IX. Since the meson could penetrate into the nucleus while having a low velocity, it was apparently negatively charged, but this circumstance has not yet been definitively established.

NUCLEAR PHYSICS IN PHOTOGRAPHS

PHOTOGRAPH XLIII

Meson track

A mosaic of microphotographs of the track of a meson that entered the emulsion from above and stopped in the lower part of the photograph. Noteworthy are the frequent changes in direction, which on the whole create the impression of a certain overall curvature, and also the “thinness” of the track along most of the trajectory. The increase in ionization at the end of the range is clearly visible.

K. F. POWELL AND G. P. S. OCCHIALINI

PHOTOGRAPH XLIV

Disintegration caused by a meson

Disintegration caused by a meson, in which two fast particles, probably hydrogen nuclei, are emitted. The meson track is denoted by the letter m. The smaller photograph was obtained in a single exposure and illustrates the need to construct a mosaic in order to obtain a complete picture of the event under consideration.

NUCLEAR PHYSICS IN PHOTOGRAPHS

PHOTOGRAPH XLV

Splitting caused by a meson

Mosaic of microphotographs of a splitting caused by a meson, in which a proton and an $\alpha$-particle are emitted. Two shorter tracks are due to $\alpha$-particles or, possibly, to heavier nuclear fragments. In splittings of this type, in all probability, neutrons are emitted, which remain unobserved.

K. F. Powell and G. P. S. Occhialini

PHOTOGRAPH XLVI

Disintegration caused by a meson

Disintegration caused by a meson and resulting in the emission of four charged particles. Particles (a) and (b) are $\alpha$-particles, (c) is a proton. It is probable that (d) is also a proton with an energy of 15 MeV. We cannot say exactly what type of nucleus was disintegrated by the meson, but according to current views it was a nucleus of silver or bromine.

NUCLEAR PHYSICS IN PHOTOGRAPHS

PHOTOGRAPH XLVII

Splitting caused by a meson

Splitting caused by a meson, in which two fast particles are emitted and a short-range, strongly ionizing particle is produced. In this case, a very clearly noticeable increase in the ionization of the meson near the end of its track is also observed.

PHOTOGRAPH XLVIII

Ejection of a slow meson from a disintegrating nucleus

Scattering through small angles in track \(m\) and a rapid increase in the density of grains near its end make it possible to identify the meson with complete certainty.

PHOTOGRAPH XLIX

Primary and secondary mesons

The $\pi$ track shows that the particle moved from top to bottom and stopped in the lower part of the photograph. In doing so it produced a second particle, $\mu$, which moved from bottom to top. This second meson left the emulsion and therefore it is impossible to measure its range exactly, but the available data indicate that it lost most of its kinetic energy before leaving the emulsion. The small star from radioactive decay illustrates the small ionization in the meson tracks compared with the ionization from an $\alpha$-particle.

K. F. Powell and G. P. S. Occhialini

PHOTOGRAPH L

Primary and secondary mesons

Tracks of a primary \(\pi\)-meson and of a secondary, lighter \(\mu\)-particle. The track of the primary particle is too short for its mass to be determined accurately by counting the number of grains. The track of the secondary meson ends in the emulsion and, judging from the grain density, the initial energy of the particle was of the order of 4 MeV.