From Current Literature

On the Existence of the “Bineutron”

The most convenient and effective method for studying the disintegrations of atomic nuclei by particles that are constituents of cosmic rays is the method of thick-layer photographic plates proposed by L. V. Mysovskii. Such plates make it possible to sum the action of cosmic rays over a long interval of time and yield a distinct picture of the disintegrations taking place. Since, however, the emulsion contains nuclei of various chemical elements (Ag, Br, H, C, N, O, as well as S and P), and, moreover, in most cases the track of the cosmic particle producing the disintegration is not visible, a final decoding of the disintegration reaction proves far from always possible.

Disintegration of a beryllium nucleus by a π-meson.

For an exact determination of the nature of the nuclei being disintegrated, A. P. Zhdanov proposed introducing into the emulsion of thick-layer photographic plates, during their preparation, those chemical elements whose interaction with cosmic rays must be studied. Under the influence of ultrasonic vibrations the substance introduced forms in the emulsion a suspension of uniformly distributed particles, considerably larger than the grains of the emulsion. Therefore, when examining the developed plate, one can without difficulty distinguish such particles from the grains of the photoemulsion. This makes it possible to count “stars” and “forks” emerging from such particles, the result of disintegrations of nuclei of the element introduced into the emulsion. Using this method, A. P. Zhdanov, P. I. Lukirskii, and Z. S. Sokolova studied dis-

splitting of beryllium nuclei by \(\pi^-\)-mesons.^2 The figure shows an example of such a disintegration. At the end of its range the \(\pi^-\)-meson is captured by one of the grains of the beryllium suspension.* In this event, only one strongly ionizing particle flies out of the nucleus, its range being 152 \(\mu\). The charge of the particle, determined from its specific ionization and range, is 3. Consequently, it is the nucleus of one of the lithium isotopes. If this nucleus belonged to \(\mathrm{Li}^8\), then at the end of the range there would necessarily appear a hammer-shaped track of two \(\alpha\)-particles flying in opposite directions. (\(_3\mathrm{Li}^8\), having emitted an electron, becomes \(_4\mathrm{Be}^8\), and the latter immediately disintegrates into two \(\alpha\)-particles.) Since there is no hammer-shaped track, it is natural to suppose that the nucleus formed belongs to \(\mathrm{Li}^7\). Then the disintegration reaction may be written in the form

\[ {}_4\mathrm{Be}^9 + \pi^- = {}_3\mathrm{Li}^7 + 2\,{}_0\mathrm{n}^1 + E_0 + E_1 + E_2, \]

where \(E_0\) is the energy of \(\mathrm{Li}^7\), and \(E_1\) and \(E_2\) are the energies of the neutrons. Substitution of the particle masses gives \(E_0 + E_1 + E_2 = 121.3\) MeV. The energy of the \(\mathrm{Li}^7\) nucleus, \(E_0\), determined from the range, is 28.2 MeV, and the momentum is 19.8 MeV/\(c\). The energy of both neutrons, \(E_1 + E_2\), is 93.1 MeV. Since the disintegration occurs under the action of a slow \(\pi^-\)-meson, which imparts no momentum to the \(\mathrm{Be}^9\) nucleus at rest, by the law of conservation of momentum the momentum of the \(\mathrm{Li}^7\) nucleus must be balanced by the sum of the momenta of both neutrons. This, however, proves possible only if both neutrons fly with the same velocity in one and the same direction. In that case their total momentum will have the greatest value, equal to 19.3 MeV/\(c\), i.e. capable of balancing the momentum of \(\mathrm{Li}^7\). If, however, the neutrons have different velocities or fly in different directions, their total momentum will be insufficient to compensate the momentum of \(\mathrm{Li}^7\). All this makes it possible to express the supposition that the two neutrons emitted in the disintegration of \(\mathrm{Be}^9\) by a \(\pi^-\)-meson form a single particle—the “bineutron.”

An experimental confirmation of the existence of the bineutron may also be taken to be the absence of a continuous \(\gamma\)-spectrum in the disintegration of deuterium by \(\pi^-\)-mesons.^3 It turned out that all \(\gamma\)-quanta arising in the reaction

\[ {}_1\mathrm{H}^2 + \pi^- = 2\,{}_0\mathrm{n}^1 + h\nu, \]

have an energy of \(\sim 130\) MeV. Such a result is possible only on the condition that both neutrons form a single particle—the “bineutron,” since, if two separately existing neutrons were formed, the \(\gamma\)-quantum could acquire any energy in the interval from 0 to 130 MeV.

V. Leshkovtsev

CITED LITERATURE

1. A. P. Zhdanov and K. I. Ermakova, DAN 70, 211 (1950).
2. A. P. Zhdanov, P. I. Lukirskii and Z. S. Sokolova, 80, 729 (1951).
3. L. Aamodt, I. Hadley and W. Panofsky, Phys. Rev. 80, 282 (1950).

* The black spots in the photograph are grains of the beryllium suspension introduced into the photoemulsion.