MECHANISM OF NUCLEAR FISSION. I

Ya. B. Zel’dovich and Yu. B. Khariton, Leningrad

In our preceding article1 the history of the discovery of a new type of radioactive process—fission of the uranium nucleus under the action of neutrons into two fragments of approximately equal mass—was set forth in detail. There too were presented the principal experimental facts established up to the present as a result of investigations by a number of laboratories and published in a series of papers that followed one another at a rapid pace during 1939 and the first half of 1940.

In the present article we shall dwell in greater detail on the question of the theoretical description of this entirely new type of radioactive process. The principal works in this field belong to Niels Bohr and his collaborator Wheeler[^2]. Simultaneously and independently, the theory was developed by Ya. I. Frenkel (Physico-Technical Institute, Leningrad)[^3]. In our exposition we shall follow Bohr’s more detailed paper, except for those parts of it which at present are subject to doubt.

All authors at present agree with the qualitative interpretation of the new type of phenomenon that was given by L. Meitner and Frisch[^4]. Developing the general scheme of nuclear reactions put forward by Bohr[^5], in which the nucleus is likened to a liquid drop, Meitner and Frisch note that for a sufficiently large nucleus the surface tension will no longer be able to withstand the action of the forces of electrostatic repulsion. With a constant ratio of electric charge to mass (as is approximately the case in the periodic system of the elements), the long-range forces of Coulomb repulsion, with a simultaneous increase in the dimensions of the mass and charge of the nucleus, grow more rapidly than the short-range forces of the “surface tension” of the nuclear liquid. Therefore, for a large nucleus one may expect a process analogous to the division of a large charged drop into smaller droplets.

Below we shall begin with the question of the energy balance of fission; then, in the second paragraph, we shall consider the question of the critical size of the nucleus and the present state of the question of the critical shape through which nuclei pass in fission; in the second part of the article, to be printed in the next issue, the questions of the kinetics of nuclear disintegration and the probability of other

competing with decay, i.e., questions determining the probability of fission of a nucleus that has captured a neutron; there we shall also deal with the state of the fission products and the theory of the processes of emission of radiation and neutrons following fission; and, finally, with the question of the fission of nuclei under the action of various other particles besides neutrons.

§ 1. ENERGY BALANCE OF THE NUCLEAR FISSION REACTION

We must first of all clarify the question of the change in the energy of a nucleus when it fissions into two parts approximately equal in their weight and charge. The chief difficulty in the case of the fission of heavy nuclei that interests us is the fact that the ratio of mass to charge in the periodic system is not quite constant. It slowly increases as charge and mass increase. Directly examining Mendeleev’s table, the atomic weights and ordinal numbers of the elements, we become convinced that when, for example, the uranium nucleus fissions into two equal parts with conservation of the total mass and total charge, we obtain two palladium nuclei with mass about 119–120 units of atomic weight, whereas the atomic weight of ordinary palladium is considerably smaller and is only 106.7. Thus, as a result of the fission of a heavy nucleus, we obtain two nuclei with an unusual ratio of charge and mass. On the one hand, this unusual ratio of charge and mass is the cause of the instability of such a fragment nucleus, leading to a series of further radioactive transformations of the fragments. It was precisely the investigation of these radioactive transformations that led to the discovery of fission processes. On the other hand, the unusual ratio of charge to mass in the fragment nuclei prevents us from directly using the presently rather extensive information accumulated on mass defects, i.e. on the internal energies and stability of ordinary isotopes of the elements.

We shall now have to find a method for estimating the energy of nuclei with a very unusual ratio of charge to mass.

Let us recall that the theory of relativity establishes a relation between the mass and energy of a body, which for our purposes may be written in the form \(E-E_0=(M-M_0)c^2\), where \(E\) and \(M\) are the energy and mass in one state, \(E_0\) and \(M_0\) the same in another state, and \(c\) the velocity of light. One unit of atomic weight is equal to an energy of

\[ \frac{(3\cdot 10^{10})^2}{6\cdot 10^{23}} = 1.5\cdot 10^{-3}\ \text{ergs} \]

per individual atom

\[ = \frac{(3\cdot 10^{10})^2}{96500\cdot 10^7} = 933\,000\,000\ \mathrm{eV} = 933\ \mathrm{MeV}, \]

and the mass of the electron is equal to an energy of 0.51 MeV. In what follows we shall speak of mass defects expressed in millions of electron-volts.

According to modern views, the nucleus consists of neutrons and protons. We shall express the nuclear charge \(Z\) in numbers of elementary charges; \(Z\) coincides with the ordinal number of the element in Mendeleev’s table. \(Z\) is equal to the number of protons contained in the nucleus and is an integer. The mass defects of elements relative to oxygen are less than 500 MeV; therefore, rounding off the atomic weights of isotopes of elements

to the integer \(A\), we directly obtain the number of heavy particles of which the nucleus consists, i.e., the sum of the number of protons and the number of neutrons.

Let us also note that when, for short, we speak of the mass defect of a nucleus, what is actually meant is the mass defect of the neutral atom, i.e., the nucleus \(+\ Z\) electrons. If we compare the mass defects of two nuclei with the same atomic weight (isobars), but with charges differing by one unit, then we obtain the energy of the process of transformation of one atom into another. Physically the transition from \(Z\) to \(Z+1\) is the emission of an electron, or \(\beta\)-activity; the reverse process is the capture by the nucleus of an electron from the shell (the \(K\)-electrons nearest the nucleus are captured). The existence in nature of different nuclei of a given atomic weight (isobars) is connected with the stability of the nucleus with respect to spontaneous processes of \(\beta\)-transformation and electron capture, which do not require external bombardment by any particles, do not require overcoming energy barriers, and therefore proceed comparatively easily. If the ordinary stable elements correspond, for a given atomic weight, to a minimum of the total energy of the nucleus, then for elements in which, at the given atomic weight, the nuclear charge deviates from this value, we may expect that the energy will be expressed by the following formula:

\[ M(Z,A)-M(Z_A,A)=\frac{1}{2}B_A(Z-Z_A)^2. \tag{1} \]

Fig. 1

In this formula \(Z\) is the charge of the nucleus under consideration. \(Z_A\) is the above-mentioned charge at which the energy of the nucleus is minimal for the given \(A\). Generally speaking, \(Z_A\) need not be an integer. Naturally occurring isobars group themselves around the value \(Z_A\). The values \(Z_A\) in Fig. 1 are given by the solid line.

\(^{1)}\) Near the minimum, in the expansion in powers of \((Z-Z_A)\), the first-order term is absent; we neglect terms of order higher than the second.

The quantity \(B_A\), characterizing the sharpness of the maximum, i.e., characterizing the change in energy upon deviation from the charge that brings the energy to a minimum, we cannot find from the existing experimental material on mass defects and energies of stable nuclei, since in stable nuclei the deviations of the quantity \(Z\) from \(Z_A\) are too small. However, we can approach the calculation of the quantity \(B_A\) theoretically by means of the following argumentation.

Let us try to determine what governs the usual relation between the charge and mass of a nucleus, i.e., the relation between the number of neutrons and the number of protons in the nucleus that corresponds to the minimum energy. In any theory of nuclear forces existing between neutrons and protons, one may expect that these forces reach saturation and give a minimum of energy when there is an equal number of neutrons and protons in the nucleus. And indeed, for light elements the ratio of mass to charge is quite close to 2, which corresponds precisely to \(Z - A = A\), to equality of the numbers of neutrons and protons. What determines the deviation of this ratio \(\dfrac{A}{Z}\) from 2, i.e., the deviation of the energy minimum from the conditions under which the number of neutrons and the number of protons are equal to each other? On the one hand, one must take into account the somewhat different intrinsic energy of protons and neutrons; on the other hand—and this is most essential—alongside the short-range “chemical” nuclear forces binding the nucleus, it is also necessary to take into account the electrostatic repulsive forces, which are the greater the larger the charge of the nucleus. It is precisely these forces that introduce an asymmetry between protons and neutrons and account for the fact that the minimum energy corresponds to a number of protons smaller than the number of neutrons. If the chemical energy of the nucleus is minimal when the two numbers are equal, then the presence of electrostatic energy will cause the minimum to shift toward nuclei in which the number of neutrons exceeds the number of protons. But the magnitude of the electrostatic energy of the nucleus is not difficult to take into account. Along with electrostatic forces, the difference between the intrinsic masses of the neutron and the proton is also fundamentally essential; the energy that could be released in the transformation of a neutron into a hydrogen atom outside the nucleus depends on this difference. This energy is known and amounts to only \(0.78\ \mathrm{MeV}\).

Let us write the expression for the mass of the nucleus in the following form:

\[ M(Z,A)=C_A+\frac{1}{2}B'_A\left(Z-\frac{1}{2}A\right)^2+\left(Z-\frac{1}{2}A\right)(M_p-M_n)+ \frac{3Z^2e^2}{5r_0A^{\frac{1}{3}}}, \tag{2} \]

where the quantity \(C_A\) does not depend on \(Z\), \(\left(Z-\dfrac{1}{2}A\right)\) is half the difference between the number of protons and the number of neutrons in the nucleus \(\left(\dfrac{Z-(A-Z)}{2}\right)\). In this formula the first two terms describe the chemical energy

of the nucleus; the form in which they are written corresponds to the fact that the chemical energy is minimal at \(Z = A - Z\); the third term is the difference between the proper masses of the neutron and the proton; the fourth term is the energy of electrostatic interaction. In this, we found the electrostatic energy of a sphere with constant volume charge density, making the usual assumption that the radius of the nucleus is equal to \(r_0 A^{\frac{1}{3}}\), where \(r_0\) is the radius per one particle. The value of \(r_0\) has long been known from the theory of \(\alpha\)-decay, in which the probability of decay depends on the magnitude of the energy barrier near the surface of the nucleus, i.e. depends essentially on the radius of the nucleus. The most probable value is

\[ r_0 = 1.48 \cdot 10^{-13}\ \text{cm}. \]

Differentiating formula (2) with respect to \(Z\) and setting it equal to zero, we find the value \(Z_A\) that brings the energy to a minimum, depending on \(B'_A\). Comparing it with the known data on the mean, most probable value of \(Z_A\) for stable elements, we can find the quantity \(B'_A\) in formula (2), characterizing the minimum of the chemical energy, and hence, by elementary calculations, the quantity \(B_A\) in formula (1).

Table 1 gives the values of \(Z_A\) and \(B_A\) of formula (1) as functions of the atomic weight. We shall not go here into further details of Bohr’s calculations, which take into account small oscillations of the energy depending on whether the number of neutrons and the number of protons in the nucleus are even or odd.

Table 1

With the aid of the estimate found for the energy of a nucleus as a function of its atomic weight and number of charges over a wide interval of variation of both quantities, we can now clarify the question of the energy released in the decay of a heavy nucleus.

Table 2 gives the results of such calculations for several typical nuclei.

In the third column of Table 2 is given the release of energy in the division of the initial nucleus (first column) into the products shown in the second column; in fission the sum of the charges and the sum of ...

atomic weights. However, the fission products obtained, owing to the unusual ratio of the number of protons to the number of neutrons, undergo further transformations; thus, the decay product of uranium, palladium, with atomic weight 120, must transform into the stable isotope of tin \(\mathrm{Sn}^{120}_{50}\), emitting four electrons (four \(\beta\)-particles). The additional energy released in this process is given in the last column of the table; for details on the \(\beta\)-activity of the fragments, see Part II.

Table 2

We see that up to an atomic weight equal to 100, nuclei are completely energetically stable with respect to fission1. Above this limit, the transformation of a nucleus into two fragments of equal mass, situated at a large distance from one another, becomes energetically favorable.

It is curious that fission of a nucleus into three equal parts becomes favorable beginning with an atomic weight of 110. In fission into three equal parts, uranium would release even somewhat more energy than in fission into two parts. The fission of uranium into ten equal parts would occur without the release and without the expenditure of energy.

We are especially interested in the energy relations in the case of uranium fission. As in Table 2, we consider the fission of the nucleus obtained when the nucleus of the principal isotope of uranium \(\mathrm{U}^{238}_{92}\) captures one neutron, yielding \(\mathrm{U}^{239}_{92}\).

In Fig. 2 is shown the amount of energy released in the fission of the uranium nucleus in various ways. Along the abscissa is plotted the number of protons in the fragment nucleus formed; along the ordinate, the number of neutrons in this nucleus. The atomic weight of the fragment, equal to the sum of both numbers, is constant along the straight lines cutting off an isosceles right triangle from the origin. The points in the figure denote stable nuclei. Finally, the ellipses shown in the figure are lines along which the energy release in uranium fission is constant and equal to the value written on the ellipse (in—

ened in millions of electron-volts). Having specified the charge and atomic weight of one of the fragments, we thereby completely determine, of course, the charge and atomic weight of the second fragment.

Thus, if one of the fragments is a normal ruthenium nucleus (charge 44, atomic weight 100—the lower asterisk), then the second nucleus must be cadmium (charge 48, atomic weight 139, number of neutrons 91—the upper asterisk). In the fission of a uranium nucleus into two such fragments, 150 MeV should be released.

It is not difficult to see that the points representing both fragments in Fig. 2 must always lie on a straight line passing through the centers of the ellipses, on both sides of the point corresponding to the most energetically favorable direction of splitting of the nucleus. Therefore, for the most energetically favorable direction of the fission process, the resulting nuclei both lie at a considerable distance from the Milky Way representing the stable isotopes. The transformation of the fragments into stable isotopes is associated with the emission of from three to six β-particles.

Fig. 2

The above-developed estimate of the energy of such unusual nuclei will later allow us to clarify completely the question of the subsequent radioactive transformations of the fragments.

The elementary calculations given above, combining the most general experimental data on the atomic weights of various nuclei and simple theoretical considerations concerning the nature of nuclear forces, show, in full agreement with experiment, that the fission of a heavy nucleus is the most powerful process in terms of the amount of energy released. The enormous energy of the fragments and their great ionizing power are widely used by experimentalists, who unmistakably distinguish fission from other radioactive processes.

§ 2. STABILITY OF A HEAVY NUCLEUS

The energetic stability of a nucleus with respect to fission into two fragments located at a large distance from one another still does not tell us directly about the possibility of such a process, because in reality fission will have to pass through a state in which both fragments are close to one another or have not even completely separated. Owing to the electrostatic repulsion of the fragments, the energy of such a state will be much greater than the energy of the state considered in the pre-

preceding paragraph, in which the fragments have moved far apart from one another and their interaction may be neglected. To clarify the question of the probability of nuclear decay and of the conditions under which the decay proceeds, we must approach more closely the very mechanism of the process and determine those intermediate phases through which the process of nuclear fission passes.

Before setting out the mathematical theory we shall present an interesting calculation due to I. I. Gurevich.

Representing the nucleus as a sphere composed of separate small spheres—protons and neutrons—we find the part of them lying on the surface; it turns out that, even with the closest packing, for a nucleus consisting of 238 particles, about 130 particles are on the surface—more than half the total number.

It is clear that, under such conditions, any calculations in which the energy is divided into volume and surface energy, or in which the change in energy upon a change in the shape of the nucleus is computed, cannot lay claim to accuracy. The calculations are illustrative in character; their results must, as far as possible, be checked by experiment. A sober estimate of the degree of approximation will make it possible to omit a number of calculations.

With the reservations made above, let us proceed to consider the energy of various forms of the nucleus.

Above, in formula (2), we considered the dependence of the energy on the charge for a nucleus of constant shape and size and constant total number of particles. The surface energy was then combined by us with other terms in the constant \(C_A\). Now, on the contrary, we shall consider changes in the shape of the nucleus at constant mass and charge and at a constant ratio of the number of neutrons to protons. The terms in the expression for the energy that depend on this ratio may be omitted.

We shall represent the total energy of a resting nucleus as the sum of electrostatic and surface energies

\[ E = W + Q. \tag{3} \]

Following Frenkel, Bohr, and Wheeler, we calculate the electrostatic energy for a body of a given shape whose volume is equal to

\[ \frac{4}{3}\pi r_0^3 A \]

[cf. § 1, formula (2)], with constant volume charge density, the total charge being equal to \(eZ\). The surface energy will be represented as the product of the surface area of the body by the (constant) value of the surface tension \(q\).

According to Finberg’s estimate\(^6\), the best value is

\[ 4\pi r_0^2 q = 14\ \mathrm{MeV}. \tag{4} \]

For interest, we note that from (4) it follows that \(q = 10^{20}\ \mathrm{dyne}/\mathrm{cm}^2\) (for water \(q = 80\), for liquid mercury \(q = 500\)).

Formula (3), with constant \(q\), as Ya. I. Frenkel rightly points out, is closer to the truth than the calculation of Frisch and Meitner, based on the influence of the charge on the surface tension.

We shall seek the dependence of the energy on the shape of the body. The electrostatic energy for a sphere is maximal; on the contrary, the surface ener-

reaches a minimum in the case of a sphere. The spherical shape always gives an extremum of the total energy. Let us investigate the energies of shapes close to a sphere.

It is obvious that the surface energy will only increase from any violation of the regular spherical shape of the nucleus, whereas the electrostatic energy of the nucleus, on the contrary, reaches a maximum for the spherical shape, in which the individual elementary charges are located closest to one another. Any perturbation of the spherical surface decreases the electrostatic energy and increases the surface energy. If the former is sufficiently large in comparison with the latter, the gain in electrostatic energy upon deformation of the nucleus may exceed the energy required to increase the surface, i.e. the work done against the forces of surface tension.

Frenkel considers ellipsoids of revolution. Denoting the semiaxes by \(a\) and \(b\), restricting ourselves to small deviations from the spherical shape, \(a-b \ll a\), and taking into account the constancy of the volume \(ab^2=\mathrm{const}\), we shall write his result in the following form:

\[ \left. \begin{aligned} W &= W_0\left[1-\frac{4}{45}\left(\frac{a-b}{a}\right)^2\right],\\ Q &= Q_0\left[1+\frac{8}{45}\left(\frac{a-b}{a}\right)^2\right], \end{aligned} \right\} \tag{5} \]

where \(W_0\) and \(Q_0\) refer to the sphere. Formulas (5) are equally valid for elongated \((a>b)\) and flattened \((a<b)\) ellipsoids of revolution. The total energy is

\[ E=E_0+\frac{4}{45}\left(\frac{a-b}{a}\right)^2(2Q_0-W_0). \tag{6} \]

Thus, the stability of the spherical shape of a charged drop depends on whether the electrostatic energy is greater or less than twice the surface energy. The sphere is stable when

\[ W_0<2Q_0. \tag{7} \]

More rigorously, the same result was obtained by Bohr and Wheeler. In general form they describe a small perturbation of the spherical surface by a sum of spherical functions and expand the total energy in a series in the coefficients of the functions. For \(W_0<2Q_0\), the minimum corresponds to equality to zero of all coefficients, i.e. to the unperturbed sphere. For \(W_0>2Q_0\), deformation of the sphere becomes energetically favorable.

The consideration of small perturbations has given us a stability criterion. What will occur under a strong perturbation, under considerable deformation of the sphere? Will the nucleus, for \(W_0>2Q_0\), find a stable shape different from the sphere? Are there limits of stability for \(W_0<2Q_0\)?

There are several theoretical works in the literature in which the authors, specifying the shape of the nucleus as an ellipsoid of revolution, seek the dependence of the energy of the nucleus, at given values of charge and mass, on the ratio of the axes of the ellipsoid, without restricting themselves to small deformations. However, when considering an ellipsoid of revolution we inevitably arriv-

we arrive at the result that even when the shape of the sphere becomes unstable, there exists some finite ratio of the axes of the ellipsoid corresponding to an energy minimum, i.e., as though stable; we arrive at the conclusion that a heavy nucleus can exist in a form substantially different from a sphere. It should be noted here at once that such a method of consideration is fundamentally incorrect, since it does not follow from anywhere, especially for finite large deformations of the nucleus, that it must remain an ellipsoid of revolution all the time. It is immediately obvious that the appearance of an energy minimum at finite deformation is connected with our artificial assumption that the nucleus must be an ellipsoid, because under this assumption we cannot obtain the transition from one nucleus to two separate fragments.

Thus, the method of considering an ellipsoid of revolution, though mathematically complicated, must for this reason be applied with great caution to the question of nuclear fission and has meaning only for small perturbations.

The existing information on surface tension leads, for uranium, to the relation \(W_0' = 1.71 Q_0\). Apparently the uranium nucleus (and, all the more, all other nuclei) is stable in the form of a sphere. A small deformation requires an expenditure of energy. Meanwhile, as we found in § 1, the formation from the uranium nucleus of two nuclei separated by a large distance is accompanied by an enormous release of energy. Taking the spherical state as one extreme point, and the separated fragments as the other end of the line, motion along which describes fission, we establish that for uranium and other nuclei the energy reaches a maximum somewhere in the middle of this line.

Bohr correctly points out that estimates of stability from values of the surface tension found by indirect methods are unreliable, and that the information of interest to us must be obtained from experimental data on fission by comparing them with theory. Let us continue the theoretical consideration of the question.

Let us find the energy of the fragments. The volume of a fragment is equal to \(\frac{1}{2} = 2^{-1}\), while the surface of a fragment is equal to \(2^{- \frac{2}{3}}\) of the surface of the initial nucleus, so that for equal fragments

\[ Q_1 = Q_2 = 2^{- \frac{2}{3}} Q_0, \tag{8a} \]

where \(Q_0\) still refers to the initial nucleus before fission.

The charge of a fragment is equal to half the initial charge; the radius has decreased by a factor of \(2^{\frac{1}{3}}\), so that

\[ W_1 = W_2 = \mathrm{const}\,\frac{Z^2}{R} = \mathrm{const}\,\frac{\left(\frac{Z_0}{2}\right)^2}{R_0 \cdot 2^{- \frac{1}{3}}} = 2^{- \frac{5}{3}} W_0 . \tag{8b} \]

Let us compare the energy of the system after fission, \(E'\), with the initial value \(E_0\):

\[ E' = W_1 + W_2 + Q_1 + Q_2 = 2^{-\frac{2}{3}} W_0 + 2^{\frac{1}{3}} Q_0, \tag{9} \]

\[ E' - E_0 = \left(2^{\frac{1}{3}} - 1\right) Q_0 - \left(1 - 2^{-\frac{2}{3}}\right) W_0. \tag{10} \]

Fission becomes energetically advantageous beginning with the value

\[ W_0 > \frac{2^{\frac{1}{3}} - 1}{1 - 2^{-\frac{2}{3}}} Q_0; \qquad W_0 > 0.7 Q_0. \tag{11} \]

The result obtained above (Table 2) for uranium and other nuclei, according to which the energetic possibility of fission is reached considerably earlier than the limit of stability of the spherical form, thus has a general character and follows from a comparison of (11) and (7).

We shall obtain substantial information by calculating the energy at the moment when the nucleus has already divided into two parts, but the fragments have not yet had time to fly apart and are in contact with each other. Obviously, such a state is a necessary link, whatever forms the fission process may pass through. Since the fragments are like-charged, the energy of the system at the moment of contact, \(E''\), is certainly greater than the energy \(E'\) in the state in which the fragments have flown apart.

Frenkel calculated \(E''\) for two spherical fragments in contact with each other. The intrinsic electrostatic and surface energies of the fragments retain their values (8a) and (8b). However, to them is added the electrostatic energy of interaction of the fragments

\[ W_{12} = \frac{\left(\frac{eZ}{2}\right)^2}{2R_0 \cdot 2^{-\frac{1}{3}}}, \tag{12} \]

where \(Z_0\) and \(R_0\) are the charge and radius of the original nucleus, and \(\frac{Z}{2}\) and \(R_0 \cdot 2^{-\frac{1}{3}}\) are the same for the fragments. Comparing with the expression \(W_0\), we find

\[ W_{12} = \frac{5}{24} \cdot 2^{\frac{1}{3}} W_0; \qquad E'' = E' + W_{12} = 2^{-\frac{2}{3}} W_0 + 2^{\frac{1}{3}} Q_0 + \frac{5}{24} 2^{\frac{1}{3}} W_0 = \]

\[ = \frac{17}{24} \cdot 2^{\frac{1}{3}} W_0 + 2^{\frac{1}{3}} Q_0; \tag{13} \]

\[ E'' - E = \left(2^{\frac{1}{3}} - 1\right) Q_0 - \left(1 - \frac{17}{24} \cdot 2^{\frac{1}{3}}\right) W_0. \tag{14} \]

For fission to be possible, from the condition

\[ E'' - E \le 0 \]

we find

\[ W_0 \geq \frac{\left(2^{\frac{5}{3}}-1\right)}{1-\frac{17}{24}\cdot 2^{\frac{1}{3}}}\,Q_0 = 2.4\,Q_0^{\,1)}, \tag{15} \]

The calculation leads to an unexpected result! It creates the impression that when \(\dfrac{W_0}{Q_0}\) is not much greater than 2, i.e., when

\[ 2 < \frac{W_0}{Q_0} < 2.4, \tag{16} \]

a heavy nucleus can no longer exist in the form of a sphere, but it also cannot decay, since it lacks the energy required to pass through the shape of two touching fragments. From these calculations Frenkel concluded that already for uranium condition (16) holds, and that the uranium nucleus therefore exists in a form substantially different from a sphere. This question was recently considered by Yu. A. Zysin and one of the authors of the present article1.

It turns out that Frenkel’s results are substantially connected with the arbitrary assumption that fission proceeds through the form of two touching spheres. It is evident that at the last moment of fission, when the fragments are only touching each other at one point, the energetically most favorable shape will be that of two elongated pears. An exact determination of this shape encounters very great mathematical difficulties, but the very extremal character of the problem makes such an exact calculation unnecessary. Considering ellipsoids of revolution elongated along the axis along which contact occurs, Zeldovich and Zysin showed that, with the most favorable choice of the ratio of the axes of the ellipsoid, the total energy of the system is less than the energy of the initial heavy nucleus not only for

\[ \frac{W_0}{Q_0}=2.4, \]

but also for

\[ \frac{W_0}{Q_0}=2 \]

and further, down to

\[ \frac{W_0}{Q_0}=1.64. \]

Consideration of shapes still more closely approximating the minimal ones, in particular pear-shaped asymmetric forms, can lead only to a further lowering of the obtained figure 1.64, which is not of great physical interest. Already from the result obtained one can draw the physically important conclusion for us that, in the case when near

\[ \frac{W_0}{Q_0}=2 \]

the spherical shape of the nucleus becomes unstable, decay of the nucleus proceeding through the shape of two touching elongated ellipsoids is entirely possible and is not forbidden energetically. Thus the arguments of Ya. I. Frenkel are refuted, and the assumption put forward by him concerning the possibility of the existence of nonspherical nuclei is deprived of its basis.

Above we have collected all the information on the energy relations in fission that could be obtained by simple calculations.

They are summarized in Figs. 3 and 4. In both figures the ordinate axis gives the energy of the system, and the abscissa axis gives the parameter \(\Phi\), describing the course of fission; this parameter is chosen so that it is equal to 0 for the spherical initial nucleus, takes the value 1 for the state of touching fragments, and 2 for fragments that have moved far apart. The solid lines show reliable results of calculations; the dashed lines connect them in the simplest possible way, i.e. they represent conjectures of a person with minimal imagination, adhering as far as possible to established facts.

Fig. 3 refers to the case

\[ 2<\frac{W_0}{Q}<2.4. \]

A small deviation from the spherical shape near \(\Phi=0\) leads to a decrease of the energy (line \(AB\)), in accordance with the instability of the sphere when

\[ \frac{W_0}{Q_0}>2. \]

Fig. 3

Fig. 4

At \(\Phi=1\) the point \(C_s\) for spherical fragments (Frenkel) lies above \(E_0\); at \(\Phi=2\), for the same fragments, the point \(D_s\) lies considerably below \(E_0\). The curve \(ABC_sD_s\) must inevitably have a minimum for \(0<\Phi<1\) at some point.

However, calculations for fragments having the shape of ellipsoids (points \(C_e\) and \(D_e\)) allow a monotonic curve \(ABC_eD_e\) to be drawn. Strictly speaking, one cannot assert that we have excluded the possibility of a minimum on the energy curve during fission, but in any case our calculations make the existence of such a minimum very improbable.

Fig. 4 is constructed for

\[ 2>\frac{W_0}{Q_0}>1.64. \]

The segment \(AB\), describing the effect of small deformations, shows an increase of energy. The положe-

the position of \(C_s\) and \(D_s\) is seen from the drawing; for us it is more important that the points \(C_e\) and \(D_e\) still lie below \(E_0\) ¹).

On the line \(ABC_eD_e\) there must necessarily exist a maximum of the energy at some point \(M\). The height of the maximum determines the excitation energy required for fission.

Whereas small deformations of the nucleus require an expenditure of energy, a sufficiently strong deformation will lead to a state in which the nucleus is unstable (this will occur beyond \(M\), for \(\Phi>\Phi_M\) in Fig. 4).

The existence of a maximum, the existence of a definite critical energy are extremely important for understanding the fission process. However, Fig. 4 is clearly insufficiently precise: we are trying to describe a change of shape by one parameter; without specification such a description is ambiguous. We shall improve matters by passing to the next figure.

Fig. 5a                Fig. 5b

In Fig. 5a the lines of constant energy are plotted as a function of two variables \(\alpha,\ \beta\), characterizing the shape of the nucleus. In reality one should imagine an analogous picture in a space of a very large number of dimensions, but even in Fig. 5a we can elucidate the main qualitative features of the process.

The origin of coordinates represents the spherical shape of the nucleus. Small deformations of it, i.e. small displacements from the origin in any direction, are associated with an increase of energy. However, at some finite deformation we arrive at a saddle point, and with further deformation the energy again falls.

The dotted line shows the path leading over the pass, through the saddle point. The height of the pass above the valley surrounding the origin represents that minimum energy which must be imparted to the nucleus in order that it may pass over the pass and disintegrate. In Fig. 5b the course of the energy is shown for the motion of a point along the dotted line of Fig. 5a, passing through the saddle.

In order that, in Fig. 4, the height of the energy maximum should correspond to the critical energy, it was necessary to choose the parameter \(\Phi\) so that

¹) By refining the calculation, we can probably lower the value of \(\dfrac{W_0}{Q_0}\), at which still \(E''<E_0\) and \(C_e\) lies below \(E_0\). The point \(D_e\) always lies below \(C_e\), since the fragments repel each other for any shape.

so that, when \(\Phi\) changes, the nucleus moves along the dotted line through the saddle of the energy surface.

The calculation of the shape of the nucleus at the turning point (saddle), needed for determining the critical energy, presents great mathematical difficulties, which no one has yet overcome. A rigorous formulation of the problem proceeds from the circumstance that in the sought state the energy is extremal: maximal with respect to one coordinate \((P)\) and minimal with respect to all the others.

Physically, an extremum of the energy means that mechanical equilibrium has been achieved.

In a liquid with a free surface and surface tension \(q\), the value of the pressure \(P\) at the surface of the liquid depends on the curvature of the surface:

\[ P=q\cdot g;\quad g=\frac{1}{r_1}+\frac{1}{r_2}, \tag{17} \]

where \(g\) is the mean curvature, the sum of the reciprocal radii of curvature in two perpendicular directions.

In a charged liquid, on each volume element, along with pressure forces, there act volume electrostatic forces; denoting the field strength by \(V\), the potential by \(\varphi\), and the charge density by \(d\), we find the equilibrium conditions:

\[ \left. \begin{aligned} \operatorname{grad} P&=-Vd=-d\,\operatorname{grad}\varphi,\\ P+\varphi\cdot d&=\text{const.} \end{aligned} \right\} \tag{18} \]

Finally, the equation of equilibrium gives, on the surface of the drop,

\[ q\cdot g+\varphi\cdot d=\text{const.} \tag{19} \]

In (19) \(q\) and \(d\) are constants of the nuclear liquid; \(g\) depends on the shape of the surface at the given place, and the potential \(\varphi\) at the given point of the surface can be calculated in the usual way for a prescribed spatial distribution of charge.

Equation (19) will then give a very complicated integro-differential equation of the surface bounding the nucleus.

Important results can be obtained by applying methods of similarity theory.

Below we shall understand by \(\Phi\) the entire set of parameters necessary for a complete description of the shape of the nucleus. The total energy of the system is

\[ E(\Phi)=W(\Phi)+Q(\Phi). \tag{20} \]

Just as with the variable \(\Phi\), the function sign in (20) has a symbolic character. \(E\), \(W\), \(Q\) are functionals of the shape of the surface.

We seek the “value” \(\Phi_e\) at which \(E(\Phi_e)\) is extremal (a saddle), and we are especially interested in the value of the critical energy \(E(\Phi_e)-E_0\), where \(E_0\) refers to the unperturbed nucleus (sphere). The functionals \(E\), \(W\), \(Q\) depend not only on the shape but also, naturally, on the charge, the magnitude of the surface tension, and the radius of the nucleus. For different nuclei the critical shape and the critical energy are different.

From dimensional considerations alone, even without writing out the expressions for \(W\) and \(Q\), it is easy to see that these quantities depend on the charge and the surface tension in the same way as \(W_0\) and \(Q_0\). Therefore they may be represented in the following form:

\[ W = W_0 \cdot a(\Phi);\quad Q = Q_0 \cdot b(\Phi), \tag{21} \]

where \(a\) and \(b\) are dimensionless functionals of the shape, common to all nuclei; thus, \(b\) is the ratio of the surface area of the body of shape \(\Phi\) to the surface area of a sphere of equal volume. Hence

\[ E = W_0 \cdot a(\Phi) + Q_0 \cdot b(\Phi) = Q_0 \left[\frac{W_0}{Q_0}a(\Phi)+b(\Phi)\right] = \]

\[ = Q_0 \cdot c\left(\frac{W_0}{Q_0}, \Phi\right). \tag{22} \]

In expression (22), \(c\) is a dimensionless functional depending, apart from the shape \(\Phi\), only on the dimensionless ratio \(\dfrac{W_0}{Q_0}\).

The ratio \(\dfrac{W_0}{Q_0}\) is the sole determining criterion of the problem, the only quantity that varies from one nucleus to another when we seek \(\Phi_e\). Finally we find

\[ \Phi_e = \Phi_e\left(\frac{W_0}{Q_0}\right);\quad E(\Phi_e)=Q_0 c'\left(\frac{W_0}{Q_0}\right), \]

\[ E(\Phi_e)-E_0 = Q_0 c'\left(\frac{W_0}{Q_0}\right)-W_0-Q_0 = \]

\[ = Q_0\left[c'\left(\frac{W_0}{Q_0}\right)-\frac{W_0}{Q_0}-1\right] = Q_0 f\left(\frac{W_0}{Q_0}\right), \tag{23} \]

where \(c'\) and \(f\) are no longer functionals, but ordinary functions of the variable \(\dfrac{W_0}{Q_0}\).

The character of the dependence of \(f\) on \(\dfrac{E_0}{Q_0}\) is shown in Fig. 6. In constructing \(f\) we proceed from consideration of the limiting cases.

For the value of the parameter \(\dfrac{W_0}{Q_0}=2\), the critical energy is equal to 0. For the value of the parameter equal to 0, i.e. in the case of fission of an uncharged drop, the fission must take place entirely at the expense of external forces overcoming the entire difference of surface energy between the initial drop and the two fragment drops; in this latter case the required energy, referred to \(Q_0\), is equal to \(2\cdot 2^{-2/3}-1=0.261\), whereas in the case of a charged nucleus a considerable part of the increase in surface energy is covered by the decrease in electrostatic energy1.

\[ f = f_1 = 0.261 - 0.108\frac{W_0}{Q_0}. \tag{24} \]

\(f_1\) characterizes the energy of touching spheres; \(f_1=0\) at \(\dfrac{W_0}{Q_0}=2.4\); see formula (15) and the dotted line in Fig. 6.

The middle part of the curve \(f\) in Fig. 6 between 0 and 2 was interpolated by Bohr with the aid of approximate calculation methods developed by him. However, this interpolation, especially in the most interesting region from 1.2 to 2, is apparently not very reliable, although we have not so far encountered anything better in the literature.

Bohr describes the deformation of the nucleus by spherical functions. The second spherical function \(P_2(\cos \theta)\) (where \(\theta\) is the latitude of the point under consideration on the surface of the sphere) describes the elongation of the sphere.

In Fig. 7, \(a\), the form is shown which corresponds to

\[ R(\theta)=R_0\left[1+a_2P_2\cdot(\cos\theta)\right];\quad a_2=0. \tag{25} \]

In Fig. 7, \(b\) and \(c\), the forms correspond to the perturbation taken in the form \(P_3(\cos\theta)\) and \(P_4(\cos\theta)\), respectively, with positive coefficients \(a_3\) and \(a_4\).

Fig. 6

Bohr seeks the critical form in the following way: having specified a definite value of \(a_2\), i.e. a definite elongation of the nucleus, Bohr finds the value \(a_4\) that minimizes the energy for the given \(a_2\). It turns out that \(a_4=a_4(a_2)<0\); comparing Fig. 7, \(a\) and Fig. 7, \(b\), we see that the conditions \(a_2>0\), \(a_4<0\) correspond to the form of Fig. 7, \(d\), in which a constriction is formed in the middle of the nucleus.

If \(a_4\) is expressed as a function of \(a_2\) from the condition of minimum energy, then, neglecting all other terms, one can find the energy as a function of \(a_2\) for a sequence of forms reasonably describing the deformation leading to division. Direct calculation gives the value of \(a_2\) that makes \(E[a_2,a_4(a_2)]\) a maximum, and the critical value of \(E\) itself.

All calculations contain the dimensionless parameter introduced above, \(\dfrac{W_0}{Q_0}\). It is obvious that the calculation is good only so long as the critical deformation is small, which takes place near \(\dfrac{W_0}{Q_0}=2\) (near the stability limit). Bohr’s result

\[ \frac{E_{\mathrm{cr}}-E_0}{Q_0} = f = \frac{49}{540}\left(2-\frac{W_0}{Q_0}\right)^3 - \frac{1421}{68850}\left(2-\frac{W_0}{Q_0}\right)^4 \tag{26} \]

is arranged in powers of the distance from the limiting value,

The curve in Fig. 6 was interpolated by Bohr, more or less arbitrarily, between the limiting laws (24) and (26).

In the practically interesting region Bohr actually uses expression (26). Formula (26) is very sensitive to \(\frac{W_0}{Q_0}\); therefore it is expedient to determine \(\frac{W_0}{Q_0}\) inversely from the experimental data on the critical fission energy. As we shall see, the fission energy of uranium with atomic weight 239, after capturing a neutron, is close to 6 MeV. The surface energy of heavy nuclei \(Q_0\) is of the order of 530 MeV. Comparing these values with formula (26), for uranium 239 Bohr finds \(\frac{W_0}{Q_0}=1.48\). This value is in quite reasonable agreement with our information on nuclear radii; nevertheless, too much significance should not be attached to it.

Fig. 7

There are indications that Bohr’s approximate calculation is inapplicable for \(\frac{W_0}{Q_0}<1.8\); at the same time, very recently the value itself of the critical fission energy of uranium 239 has been subjected to revision1.

However, even if we reconcile ourselves to the fact that the exact course of the curve \(f\left(\dfrac{W_0}{Q_0}\right)\) is unknown to us, the very notion of the existence of a smooth curve will enable us to arrange the nuclei of different elements in order of increasing difficulty of fission, and will enable us to embrace the totality of experimental facts.

Let us write out the expressions for \(W_0\) and \(Q_0\):

\[ W_0=\frac{3(Ze)^2}{5r_0 A^{\frac{1}{3}}}; \tag{27} \]

\[ Q_0=q\cdot 4\pi\left(r_0 A^{\frac{1}{3}}\right)^2. \tag{28} \]

Let us recall that \(r_0\) is the radius per one elementary particle, \(r_0 A^{\frac{1}{3}}\) is the radius of a nucleus of atomic weight \(A\), containing \(A\) elementary particles,

\[ \frac{W_0}{Q_0}=\frac{3e^2}{20\pi r_0^3 q}\cdot \frac{Z^2}{A}. \tag{29} \]

The factor \(\dfrac{3e^2}{20\pi r_0^3 q}\) is composed of constants common to all nuclei.

Thus, comparing the expression \(\dfrac{Z^2}{A}\) for different nuclei, we establish the following order in which they should be arranged according to increasing difficulty of fission\(^1\):

Table 3

Nothing is known about the fission of the last three substances, and they have been included in the table only for purposes of comparison.

fragments will be closer to one another. However, the main result,—namely, the critical ratio of electrostatic energy to surface energy, equal to 2, and the qualitative picture of Fig. 6,—remains thereby unshaken. This does not alter the considerations of Berestetsky and Migdal\(^9\).

\(^1\) The critical energy is equal to the product \(Q_0\cdot f\left(\dfrac{W'_0}{Q_0}\right)\). We assume that from one nucleus to another \(f\) changes much more sharply than \(Q_0\), and arrange the nuclei in decreasing values of the argument \(f\).

In Table 3, following Bohr, we have also attempted to estimate the critical fission energy of various nuclei.

Our estimate is based on an analysis of experimental data on fission under neutron bombardment; it should be regarded as more reliable than formula (26) itself, by means of which the extrapolation was carried out.

Before turning to a detailed consideration of the most important process—fission upon neutron capture—let us consider the question of the probability of spontaneous fission. As we have seen, there is every reason to think that for all existing atoms, in particular for the heaviest uranium nucleus, the value \(\frac{W_0}{Q_0}\) is appreciably less than two, less than the critical value; hence it follows that an energy barrier exists. The critical energy needed for fission (the difference between the energy of the saddle point and the energy of the unexcited nucleus) is supplied by neutron bombardment in the form of the binding energy (“condensation energy”) of the neutron and in the form of the kinetic energy of the neutron. At the same time, however, quantum mechanics establishes the possibility that the nucleus may pass through a state forbidden by classical theory—the so-called tunnel transition under the barrier. Simultaneously with the establishment of the correct point of view on the mechanism of fission, Frisch and L. Meitner noted that spontaneous fission of a nucleus by tunneling is very improbable, since the mass of the nucleus is very large.

The critical energy required for the decay of uranium is now established fairly well. However, calculation of the probability of barrier penetration requires knowledge not only of the critical energy (the height of the barrier), but also an estimate of the width of the barrier, i.e. the length of the tunnel. Here the width of the barrier enters into the expression for the probability of the process in the exponent. Its estimate is extremely difficult.

The calculation is a natural generalization of the theory of \(\alpha\)-decay. The expression for the penetrability of the barrier is given by the exponential

\[ \exp\left(-\sqrt{2A\left(E_{\mathrm{cr}}-E_0\right)}\cdot \frac{a}{\hbar}\right), \tag{30} \]

where \(A\) is the mass of the nucleus, and \(a\) is the effective width of the barrier.

As the pre-exponential factor, Bohr chooses the frequency of oscillation of the nucleus about the spherical shape.

The reciprocal probability of spontaneous fission (equal to the mean lifetime in the absence of other radioactive processes) is expressed by the formula

\[ t = 10^{-21}\exp\left(\sqrt{2A\left(E_{\mathrm{cr}}-E_0\right)}\cdot \frac{a}{\hbar}\right)\ \mathrm{sec}. \tag{31} \]

If the lifetime is expressed in years, \(A\) in atomic-weight units, the energies in millions of electron-volts, and the barrier width in units of \(r_0\) (see § 1), then the formula is transformed into the following form:

\[ t = 10^{-29+0.145\,a\sqrt{A\left(E_{\mathrm{cr}}-E_0\right)}}\ \text{years}. \tag{32} \]

For uranium, substituting \(E_{\kappa p}-E_0 \simeq 6\ \mathrm{MeV}\), \(A=238\), and taking

\[ a=\frac{3}{2}(238)^{\frac{1}{3}} \]

—the mean between the diameter and the radius of the nucleus—Bohr finds

\[ t=10^{-29+51}=10^{22}\ \text{years}=10^{30}\ \text{sec.} \tag{33} \]

For comparison, let us note that the fission time of a nucleus possessing a sufficient store of energy (not requiring tunnel penetration) is of the order of \(10^{-15}\) sec.

Bohr’s estimate leads to a lifetime enormous even in comparison with the lifetime for the \(\alpha\)-decay of uranium (\(4\cdot 10^9\) years). A spontaneous-fission time of \(10^{22}\) years would correspond to the formation of one pair of fragments per day in a mass of \(1\ \mathrm{kg}\) of uranium; it is unlikely that a process with such a probability can be observed, especially taking into account the existence of atmospheric (cosmic) neutrons and the difficulty of complete isolation from them.

The experiments of the Soviet physicists Petrzhak and Flerov, noted in our preceding article, \(^{10}\) show that in reality uranium fission proceeds with a significantly greater probability, with a half-life of the order of \(10^{16}\) years, which makes it accessible to observation with modern experimental technique.

The time \(10^{16}\) years is obtained if the observed number of fissions is referred to the principal isotope of uranium. Referring it to isotope 235 or 234, we obtain respectively \(10^{14}\) and \(10^{12}\) years.

From Bohr’s estimate it follows that the critical fission energy of the uranium isotopes is smaller: of the order of \(5.1\ \mathrm{MeV}\) for \(\mathrm{U}^{235}\) and \(4.8\ \mathrm{MeV}\) for \(\mathrm{U}^{234}\).

Leaving

\[ a=1.5\,A^{\frac{1}{3}}, \]

we obtain respectively \(10^{18}\) and \(10^{16}\) years for the isotopes.

Bohr’s estimate describes the experimental data of Petrzhak and Flerov if one assumes that the observed spontaneous fission is due to the light isotopes, and if for \(a\) (the barrier width) one chooses the value \(1.38\,A^{\frac{1}{3}}\) instead of

\[ 1.5\,A^{\frac{1}{3}}. \]

One should marvel at the tact with which, before the experiments of Petrzhak and Flerov, Bohr chose a reasonable value for the barrier width; the small discrepancies between the measured and Bohr-predicted probabilities of spontaneous fission are connected with the exceptional sensitivity of expressions (31), (32) to the factors standing in the exponent.

In the popular press one often encounters the assertion that the discovery of Petrzhak and Flerov explains why the periodic system of the elements does not extend further, but breaks off at element No. 92—uranium \(^{1}\). Formally this is not quite so: for the existence of the last element—uranium—\(\alpha\)-decay is much more important than spontaneous fission; it is very probable that for the nearest transuranium elements this ratio remains in force, and we do not observe them not because

\(^{1}\) The amount of uranium, of the order of \(10^{-6}\) of the whole mass of the terrestrial globe, is very large.

they divide too rapidly, and because of too rapid $\alpha$-decay. The discovery of spontaneous decay does not establish an exact boundary of the periodic system. But in a broader sense the connection between spontaneous fission and the boundaries of the periodic system is undoubted.

With a significant increase of $Z$ (by several units), the increase of $\dfrac{Z^2}{A}$ will lead to a sharp decrease of the critical energy, and then very soon, by virtue of the exponential dependence of expressions (31), (32), the probability of spontaneous fission will increase catastrophically.

If one bases oneself (for lack of a better estimate) on the estimate of the critical energy (26) and on the value following from it, $\dfrac{W_0}{Q_0}=1.48$ for uranium, then the absolute limit of stability $\left(\dfrac{W_0}{Q_0}=2\right)$ will be reached at $\dfrac{Z^2}{A}=48$, i.e. at $Z$ of the order of 125 (roughly we assume that $A$ increases proportionally to $Z$). Such a nucleus will live no longer than $10^{-20}$ sec. Let us now consider spontaneous fission by the mechanism of tunneling transition, discovered by Petrzhak and Flerov, and let us look for a nucleus with a lifetime of the order of 1000 years. From formulas (30), (31), (32) it follows that the corresponding fission probability will be reached at a critical fission energy of the order of $2 \dfrac{1}{2}—3$ MeV; for such a decrease of the critical energy it is necessary to have $\dfrac{Z^2}{A}$ of the order of 40—40.5, which corresponds to $Z$ of the order of 100—102. It is impossible to construct a stable nucleus with large $Z$ by increasing $A$, for the reasons indicated in § 1: in such a nucleus $\beta$-transformation processes will occur, $Z$ will increase at constant $A$, and the nucleus with large $Z$ and $\dfrac{Z^2}{A}$ will fission spontaneously.

From the probability of spontaneous fission observed experimentally one can draw important qualitative conclusions for what follows: according to a remark of I. I. Gurevich, the very effective width of the barrier, of the order of the nuclear radius, indicates that the critical deformation leading to fission can in no way be regarded as small, and that all calculations of the critical shape made under the assumption of small deformation can, at best, have only an illustrative character.

On the other hand, the negligible probability of the quantum-mechanical tunneling mechanism definitely indicates the classical (with very small corrections for quantum mechanics) course of energetic fission caused by neutron or other bombardment.

Let us return to the question of fission of nuclei under neutron bombardment. The modern views, developed by Bohr^5, are based on the fact that a heavy nucleus is a system consisting of many particles, with a large number of degrees of freedom. Any process occurring under one or another mode of excitation of the nucleus begins with the formation of an excited (heated) compound nucleus with a relatively large

with the lifetime. The lifetime of the excited nucleus, of the order of \(10^{-15}\) sec., is considerably greater than the time during which a neutron with an energy of several million electron-volts traverses a distance of the order of the nuclear radius, \(\frac{10^{-12}}{10^9}=10^{-21}\), and considerably greater than the period of oscillation of the nucleus, \(10^{-20}\) sec.

Such a relation makes it possible to speak, in neutron bombardment, of the formation of a compound nucleus in which the value of \(Z\) does not differ from the \(Z\) of the initial nucleus, while through the capture of the neutron \(A\) has increased by one unit.

Figure 8 gives the values of the critical fission energy (according to Bohr’s semi-empirical estimate) for various compound nuclei obtained upon capture of a neutron; the nuclei are arranged in a series according to increasing values of the parameter \(\frac{Z^2}{A}\). The reader will note that instead of \(^{238}_{92}\mathrm{U}\), \(^{235}_{92}\mathrm{U}\), \(^{231}_{91}\mathrm{Pa}\), the figure shows \(^{239}_{92}\mathrm{U}\), \(^{236}_{92}\mathrm{U}\), \(^{232}_{91}\mathrm{Pa}\).

Fig. 8

The numerical value of the energy read from this curve, and especially the magnitude of the slope of the curve, may differ appreciably from the true value. However, the order in which the elements are arranged by Bohr in the figure must undoubtedly be preserved. As is evident from the figure, the rare light isotopes of uranium should fission most easily. Protactinium occupies a position intermediate between the principal and the light isotopes of uranium. Finally, fission of thorium should require a substantially greater expenditure of energy and correspondingly is more difficult, occurring with smaller probability. At the same time, neutron capture not only changes the atomic weight of the nucleus but, at the same time—and this is most important—is also the principal source of energy for the fission process. Even in the case when fission is produced by slow, thermal neutrons, whose kinetic energy is negligible, the fission process differs very essentially from spontaneous fission. Indeed, on merging with the nucleus even the slowest thermal neutron releases a considerable amount of energy, of the order of 5–6 MeV—the condensation energy of the neutron. The compound nucleus that has captured the neutron proves to be energetically highly excited. Fission represents one of the possible reactions of such an excited system. Along with this, the reverse emission of a neutron is also possible. Capture followed by the reverse emission

neutron is nothing other than scattering of the neutron. If the nucleus captures a neutron of considerable kinetic energy, then, upon its re-emission, it is hardly probable that the neutron will carry away with it all the kinetic energy with which it arrived at the nucleus, leaving the latter in an unchanged state. It is much more probable that the evaporating neutron will carry away only part of the initial kinetic energy, leaving the nucleus excited. However, in the best case the excitation energy does not exceed the kinetic energy of the neutron, whereas, upon capture of the neutron, the condensation energy (binding energy) of the neutron is added to the kinetic energy. With neutrons whose energy does not exceed 4–5 MeV, inelastic scattering does not lead to fission.

Finally, the last possibility for the excited compound nucleus, still containing both the kinetic energy of the absorbed neutron and the heat of its condensation, is the radiation of energy in the form of a $\gamma$ quantum. As a result of the loss of energy, the total energy of the nucleus will be less than the heat of evaporation of a neutron; the nucleus can no longer evaporate a neutron back, nor can it fission. We thus obtain a relatively stable nucleus whose atomic weight has been increased by one unit.

The question of what exactly will occur under neutron bombardment of a given element is, above all, a question of competition, of the ratio of the probabilities of the three most important processes indicated above—fission, scattering of the neutron, i.e., capture followed by re-evaporation of the neutron, and, finally, capture of the neutron with radiation of energy in the form of a $\gamma$ quantum. In the following paragraph we shall consider in more detail the probability of each of the three processes listed above, applying for this purpose the method of the activated complex. The general uncertainty that still exists in the theory of fission does not allow quantitative conclusions to be drawn; however, application of the theory of the activated complex will allow us to establish, from general theoretical considerations, the character of the main dependences and, above all, the dependence of the prevailing direction of the process on the excitation energy.

The best proof of the power of theoretical analysis is Bohr’s explanation$^{11}$ of the confusing picture of the action on uranium of neutrons of different energies. This explanation (see § 4), which attributes different effects to different isotopes of uranium, has guided experimenters and has now received direct confirmation in experiments with isotopes separated by means of a mass spectrograph$^{12,13}$.

The content of the second part of the article, which will be printed in the next issue, will be an analysis of experimental data on fission and the behavior of fragments in the light of the theory of fission.

REFERENCES

1. Ya. B. Zel’dovich and Yu. B. Khariton, Uspekhi fizich. nauk, 23, 329, 1940.
2. N. Bohr and J. A. Wheeler, Phys. Rev., 56, 426, 1939.

1. Ya. I. Frenkel, JETP, 9, 641, 1939.
2. L. Meitner and O. R. Frisch, Nature, 143, 239, 1939.
3. N. Bohr, Advances in the Physical Sciences, 16, 425, 1936.
4. E. Feenberg (see note on fission), Phys. Rev., 55, 504, 1939.
5. Ya. B. Zel’dovich and Yu. A. Zysin, JETP, 10, 831, 1940.
6. J. K. Knipp and R. D. Present, Phys. Rev., 57, 751, 1940; ibid., p. 1188.
7. V. Berestetskii and A. Migdal, Doklady AN SSSR, 30, 701, 1941.
8. K. A. Petrzhak and P. G. Flerov, JETP, 10, 1013, 1940; Advances in the Physical Sciences, 25, 171, 1941.
9. N. Bohr, Phys. Rev., 55, 418, 1939.
10. A. O. Nier, E. T. Booth, J. R. Dunning, A. V. Grosse, Phys. Rev., 57, 546, 748, 1940.
11. K. H. Kingdon, H. C. Pollock, E. T. Booth, J. R. Dunning, Phys. Rev., 57, 749, 1940.

1. In a preliminary communication, Knipp and Present [8] indicate that from their calculations there follows the possibility of an asymmetric critical shape of the nucleus, which in this case describes the experimental fact that the fragments obtained in fission, as a rule, differ noticeably from one another in mass. Thus[^1a], fission with the ratio \(A_1:A_2=0.37:0.63\) is observed. More likely, for a symmetric shape we could expect that the masses and charges… ↩↩↩↩↩