Nuclear Physics

by W Heisenberg

  Since energy is liberated when a deuteron is formed from a proton and a neutron, the mass of the deuteron must be less than the sum total of the masses of the proton and the neutron while these two particles still had a separate existence. Thus must apply to every nucleus composed of N neutrons and Z protons. This statement can be expressed in the form of an equation as follows:

  where | E | is the binding energy, taken as a positive magnitude, of the nucleus, which in the case of the deuteron is liberated in the form of a photon.

  It is convenient to use the above equation in a slightly modified form, referring it to the neutral atom, i.e. to the nucleus plus the extranuclear atomic structure, as a whole, instead of the nucleus alone. In this case, the hydrogen atom of mass number 1, with its one electron, must be substituted for the proton. As a result, the masses appearing on both sides of equation increase by the mass of the Z electrons possessed by the entire atom, on the one hand, and by the Z protons, on the other. The equation then will read:

  From this equation we can compute the binding energy of a nucleus if we know exactly the mass of the atom in question, the mass of the neutron and the mass of the hydrogen atom.

  The mass of an atom is thus always |E|/c2 less than the sum of the masses of its component parts. This difference in mass is called the mass defect of the nucleus. It equals |E|/c2, and therefore:

  Atomic masses are usually expressed in terms of the standard physical atomic mass unit (a.m.u.) which is approximately equal to the mass of a hydrogen atom, or of a neutron; it is exactly of the mass of the oxygen isotope 8O16. The mass defects are of the order of magnitude of 1/1,000 a.m.u., and therefore it is customary to express them in units of 1/1,000 a.m.u. (a.m.u.−3). The energy equivalent of 1 a.m.u.−3, i.e. 1 a.m.u.−3 × c2, happens to differ only slightly from the energy unit 1 Mev. To be exact:

  1 a.m.u.−3 is the equivalent of 093 Mev.

  Accordingly, the binding energies of nuclei are also of the order of magnitude of 1 Mev.

  Thus the physicist has two completely independent processes available for determining the binding energies: He can measure them directly, or can compute them from the mass defects. In the latter case, it is essential, of course, to be able to determine the masses of the atoms to a high degree of accuracy, in order to deduce the binding energy from them, for we are dealing with minute differences, measurable in 1/1,000ths of the masses only. Let it suffice here to say that this is accomplished with the aid of the mass spectrograph, an apparatus first developed by Aston. In the mass spectrograph, charged atoms are caused to pass through electric and magnetic fields, where they are deflected. As we have already seen, the extent of their deflections depends on the ratio of their charge e to their mass m, as well as on their velocity. The mass spectrograph is designed so that only particles having a certain velocity are singled out and become the objects of observation. The ratio e/m, can then be determined, and since the charge e of the particles is known, their mass m can be computed—or, to be more exact, their mass, m, can be compared with the mass of the oxygen isotope 8O16, which is exactly 16000 a.m.u. by definition. For the calculation of the binding energies, we need also the exact masses of the hydrogen atom and the neutron. The mass of the hydrogen atom is 100813 a.m.u., the mass of the neutron is 100895 a.m.u.

  Let us now consider, quantitatively, the changes in mass when a deuterium atom is formed. Before the formation of this atom, there was an individual hydrogen atom, 1H1, and an individual neutron, 0n1 ; after the combination of these two, there is a deuterium atom, 1D2, and a free photon, hv. This finding can be expressed by the following formula:

  1H1 + 0n1 → iDz + hv

  Now the mass of a deuterium atom is 20147 a.m.u., whereas the sum of the masses of the hydrogen atom and the neutron amounts to 100813 + 100895 = 20171 a.m.u. The mass of the deuterium atom is actually smaller than the sum of the masses of its component parts, and this difference is the mass defect, which amounts to 00024 a.m.u. (24 a.m.u.−3). But expressed in terms of its energy equivalent, this mass defect represents a binding energy of about 22 Mev, which is exactly the energy released in the form of a photon, as we have already pointed out. Thus we see that there are two totally independent methods for the calculation of the binding energy, and both methods yield the same answer, and therefore provide the best possible proof of the truth of the principle of the equivalence of mass and energy. This is an especially important fact, for here we are not dealing with electric fields, with reference to which this principle was well known even in pre-relativistic times, but with fields of a totally different nature.

  The findings just discussed show also why the atomic weights of the elements are not integral multiples of a basic unit, as Prout had conjectured. In the first place, the mass of the proton and the mass of the neutron differ slightly from each other. Furthermore, when they combine, a fraction of the sum total of their masses disappears; this fraction corresponds to their binding energy. So even if the mass of the proton were exactly equal to the mass of the neutron, the atomic weights of the elements would still not be integral multiples of a basic unit. This explains the slight deviations from integral numbers in the atomic weights of the lighter elements. But the very considerable deviations of the atomic weights of the heavier elements must be explained in another way. These deviations are more apparent than real and are due to the fact that the natural elements are, mostly, mixtures of various isotopes. Each isotope consists of atoms with a nucleus of a specific type, the mass number of which is almost exactly an integer. But the average mass number of this mixture of isotopes is not an integer.

  Thus we have established the desired criterion for the stability of a nucleus. A nucleus holds together because an amount of work would be required in order to break it up into its component parts. The work required is a measure of the binding energy, which, in turn, is equal to the mass defect of the nucleus, expressed in terms of energy. A nucleus would not be stable if it were possible to break it up without doing any work. We must, however, add that the other conservation laws must be taken into consideration too. According to the law of the conservation of electric charge, an atom cannot undergo any change that would alter the total charge present in the system. Hence, the change of a proton into a neutron, or of a neutron into a proton, cannot take place in the nucleus without compensation. Otherwise, many atoms which are actually considered stable, would have to be unstable. There is, for instance, a boron nucleus having the (not exact) mass number 12, and also a carbon nucleus having the mass number 12. The boron nucleus consists of 7 neutrons and 5 protons, whereas the carbon nucleus has 6 protons and 6 neutrons. Their respective symbols are 5B12 and 6C12. The mass of the boron nucleus, however, is slightly greater than that of the carbon nucleus. The difference is 0013 a.m.u., and consequently, the difference in binding energy is about 12 Mev. The carbon nucleus has a greater mass defect than the boron nucleus, so that its component parts are bound together more tightly than those of the boron nucleus. We can assume, therefore, that the boron nucleus is unstable and changes spontaneously into a carbon nucleus. This transmutation would liberate energy to the extent of 12 Mev. But this process can occur only when a neutron of the boron nucleus changes into a proton. However, by virtue of the law of conservation of electric charge, such a process is possible only if the newly formed elementary quantum of positive electricity is compensated for by the simultaneous formation of an elementary quantum of negative electricity and the latter removed from the nucleus. This could take place through the emission of an electron simultaneously with the change of the neutron into a proton. Actually, this boron nucleus is not a stable structure, but a radioactive one. It emits electrons—in other words, negative beta rays—and changes into a carbon nucleus.

  Nevertheless, this transformation would not be possible if the law of the conservation of angular momentum did not hold. As we have mentioned before, the angular momentum, or spin, of both a proton and a neutron is /2, to be considered positive or negative according
to the spatial orientation of the axis of rotation. Therefore, the angular momentum of a nucleus composed of an even number of particles is always an even multiple of /2. Both the carbon nucleus and the boron nucleus consist of 12 particles; consequently, the nuclear spin of each must be an integer multiple of . But pursuant to the law of the conservation of electric charge, an electron must be emitted during the transmutation process, and this electron, too, has an angular momentum (electronic spin) of /2. Therefore, the resulting carbon nucleus would retain an angular momentum (nuclear spin), the value of which is an odd multiple of /2. In this case, therefore, the angular momenta involved would not balance. But in this predicament we must remember that when we studied natural beta radiation, we found that similar difficulties of balancing the energies furnished us with a clue to the existence of the neutrino, ejected simultaneously with the emission of the electron. The neutrino, obviously, accounts for the conservation of the angular momentum. The electrons emitted by the boron nucleus display a continuous series of energy values; this fact leads us to the assumption that a neutrino is emitted simultaneously in this case, too. The facts that the boron nucleus is really unstable and that it eventually changes into a carbon nucleus through the emission of an electron and a neutrino, point to the conclusion that the neutrino also has an angular momentum or spin value of /2, opposite in direction, or mathematical sign, to that of the electron, so that it compensates for the electron.

  We have now gained a general view of the conclusions which can be drawn from the three conservation laws with respect to the stability of the nucleus. Our findings may be summed up briefly as follows: The nucleus of an atom will always change spontaneously into another nucleus if this process, firstly liberates energy, and secondly, is compatible with the laws of conservation of charge and angular momentum. It is true, of course, that this spontaneous transformation may occur after the lapse of a considerable period of time only, in other words, that its probability may be very small. But if either one of the conditions just mentioned is not given, the nucleus in question is a stable one.

  II. NUCLEAR STRUCTURE

  The conservation laws have enabled us to reach far-reaching conclusions concerning the stability of atoms, without resorting to any hypothesis about the conditions within the nucleus or the forces operative in it. Now let us attempt to formulate conclusions concerning the internal structure of the nucleus, on the ground of experiments, independently of specific assumptions about these internal forces. How are the protons and neutrons distributed within the nucleus? Can we compare a nucleus with, say, a drop of liquid, in which the molecules are packed with uniform density? Or should the nucleus be compared rather to a globular star cluster, in the centre of which the stars are extremely close to one another, but more widely separated with greater distances from the centre?

  In this connection, the mass defects render a very important service. From them we can calculate the binding energy, and we find that when reckoned for individual particles within the atom, the binding energy is approximately the same in all atoms. The lighter atoms—up to about aluminium—where the absolute magnitude is smaller, are an exception to this rule. In the other atoms, the binding energy of each nuclear particle is always between 6 and 9 Mev. It seems, therefore, that all nuclear particles are bound more or less equally firmly.

  We may draw a further conclusion from the size of the nucleus. We can determine, approximately, the diameter of a nucleus from the deflection of alpha particles in the substance in question, by ascertaining what fraction of them shows so strong a deflection that it cannot be accounted for by the effects of the external electric field of the nucleus, but can only be explained by direct collisions with the nucleus. The larger the nucleus, the more frequently will such collisions occur. These experiments have shown, for instance, that the diameter of a uranium nucleus consisting of 238 particles is about four times as large as that of the helium nucleus which consists of 4 particles. Their respective volumes are, therefore, as 1 to 43, in other words, the volume of the uranium nucleus is about sixty-four times the volume of the helium nucleus. However, the uranium nucleus has about sixty times as many component particles as the helium nucleus.

  These two facts—the approximate equality of the binding energies of the individual particles, and the approximate proportionality of the number of particles to the volume of the nucleus—warrant the conclusion that the protons and neutrons are distributed approximately uniformly throughout the nucleus, because if this were not the case, their binding energies would show considerable differences, both in individual parts of a nucleus and in different nuclear types. Moreover, this fact, in conjunction with the proportionality of the number of particles and the volume, indicates that this density distribution is the same in all nuclei, except, again, for the very lightest atoms. Therefore, we may speak of a homogeneous nuclear substance in all nuclei, which consists of a mixture of protons and neutrons packed with an approximately constant density. There is a slight difference here between different atomic species, with respect to the ratio of the number of neutrons to the number of protons.

  Therefore, we obtain a very accurate model of a nucleus by likening it to a drop of liquid. Just as water droplets of different sizes can form from water molecules, so drops of nuclear matter of different sizes—the different atomic nuclei—can form from protons and neutrons. This liquid-drop model has exactly the properties which can be observed in an atomic nucleus. For the molecules are packed with equal density of distribution throughout a drop of liquid, and all the molecules are bound together in it by the same quantity of energy. The knowledge of the existence of a universal, homogeneous nuclear substance facilitates greatly our understanding of nuclear structure.

  But this liquid drop has also other, still more subtle characteristics, and now we must investigate whether these have their analogies in the nucleus. Actually, in a liquid drop, not all molecules are bound together equally firmly. The molecules on the surface are linked with the others on one side only, and therefore, on the whole they are less tightly bound than the others. This explains the phenomenon of surface tension. Certain considerations concerning energy, which are absolutely analogous to those which we have already applied to atomic nuclei, explain the fact that this surface tension causes these drops to assume a spherical shape. For the surface energy of a drop is proportional to the area of the surface, and it consequently tends to make the surface area as small as possible. We must assume the presence of such surface tension in atomic nuclei, too. As a result of the lesser cohesion of the particles situated on the surface, this surface tension must produce a decrease of the total binding energy, and consequently, of the average energy per particle as well. Just as in a liquid drop, the surface tension is the cause of the spherical shape of atomic nuclei, too.

  There is, however, one essential difference between nuclear matter and a liquid. The liquid consists of molecules which are electrically neutral, whereas the nuclear matter contains not only neutrons but electrically charged protons, too. Therefore, our analogy must be one between nuclei and liquid drops containing electrically charged molecules with forces of repulsion operating between them. There is also an electric force of repulsion present in the nuclei of atoms.

  III. THE THREE TYPES OF NUCLEAR ENERGY

  We may therefore regard the energy content of a nucleus as the sum total of three components. The major part of this energy comes from the nuclear forces which make the cohesion of the nucleus possible. This energy is modified by the surface tension. Finally, a part of this total energy originates from the force of electric repulsion. Let us consider these three components separately and correlate them with the number of protons and neutrons on the basis of a study by v. Weizsäcker.

  We shall begin with the nuclear forces. These are the forces which bind the protons and neutrons together, and as we have seen, they are related to the fact that nuclei are capable of emitting both electrons and positrons. This phenomenon is evidently quite symmetrical a
s between protons and neutrons. A neutron can change into a proton, in which process an electron is emitted, and conversely, a proton can change into a neutron, in which case a positron is the by-product of the process. This fact warrants the conclusion that with respect to the nuclear forces, or to the nuclear field, there is no difference between protons and neutrons. Therefore, it must be possible to represent that part of the binding energy which is the product of the nuclear forces, by some symmetric function of the number of neutrons and protons. Now, if we first write this function in a general form, and then, near the point where the number of neutrons is equal to the number of protons, we stop with the second term, we obtain a simple equation for the binding energy per particle (in so far as this energy originates from the nuclear field), as follows:

  where Ev is that part of the total binding energy which originates in the nuclear field, and which is proportional to the volume. A and B are constants. Eυ/(N + Z), the binding energy per particle, is therefore equal to a constant—A when N = Z, i.e. when the number of protons is equal to the number of neutrons. But if N and Z are not equal to each other, small divergences occur and in that case the simplest symmetrical function of N and Z is the expression (N − Z)2. But since nuclear matter is homogeneous, the binding energy per particle can depend only on the N/Z ratio, which latter is obtained through division by (N + Z)2. The most general symmetrical function would be, obviously, more complicated. But if it is developed, according to ascending powers of (N − Z), in a Taylor series and broken off after the second term, we obtain exactly the equation given above. This approximation is sufficient, since in the cases of the nuclei observed, N and Z never differ much from each other.