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On the Processes of Formation of H$^3$ and He$^3$ in Nature
In recent years the existence in nature of the radioactive isotope of carbon C$^{14}$ has been demonstrated; it is formed as a result of reactions of cosmic-ray neutrons with the nuclei of atmospheric nitrogen.
The hypothesis that the light isotope of helium He$^3$, present as a constituent of atmospheric helium, was formed as a result of the decay of tritium H$^3$, which arises in the atmosphere under the action of cosmic radiation, has been confirmed by studies of the isotopic composition of the water of lakes and the ocean, and of atmospheric air$^1$.
In the first experiments, samples of highly concentrated heavy water obtained from surface waters were investigated. The content of the superheavy radioactive isotope of hydrogen—tritium—in them was expected to exceed its concentration in ordinary water by approximately a million times. It was found that these samples were indeed radioactive.
Their activity corresponded to an initial concentration of natural tritium of the order of 1 atom per $10^{18}$ atoms of ordinary hydrogen H$^1$.
This concentration was observed in the water of Norwegian lakes. The concentration of tritium in ocean water is lower by about two orders of magnitude. The number of tritium atoms in air, referred to the number of atoms of atmospheric hydrogen$^2$, is higher than or equal to $1:10^{14}$.
In view of the fact that the half-life of tritium, which transforms into He$^3$ with the emission of an electron, is only about 12 years, an indicated concentration can be maintained only by a sufficiently intense process of continuous formation of H$^3$.
Even before the determination of the concentration of tritium in water and air, it had been suggested that nuclear reactions of fast neutrons with nitrogen,
$$ \mathrm{n} + \mathrm{N}^{14} = \mathrm{C}^{12} + \mathrm{H}^{3} \tag{1} $$
may be one of the principal causes of the appearance of nuclei of natural tritium$^3$. The reaction was observed under laboratory conditions in 1941$^4$. However, the conditions of the experiment did not make it possible to estimate the reaction cross section with the required accuracy.
Below a method will be described that made it possible to estimate the cross section of the reaction \((n,\mathrm{H}^3)\) for fast neutrons arising in the fission of uranium nuclei, with an accuracy of up to 25 percent. Analyzing modern data on the number of fast neutrons in cosmic radiation and the number of \(\mathrm{H}^3\) nuclei generated in “stars” caused by cosmic particles of high energy, the author estimated the concentration of natural tritium. The result, to within an order of magnitude, agrees with direct experimental data\(^{1,2}\) on the amount of tritium in water and in the atmosphere.
1. DETERMINATION OF THE CROSS SECTION OF THE REACTION \(\mathrm{N}^{14}(n,\mathrm{H}^3)\mathrm{C}^{12}\) UNDER LABORATORY CONDITIONS
The experiments consisted in irradiating nitrogen with a known number of fast neutrons and subsequently determining the number of tritium nuclei accumulated as a result of the reaction \(\mathrm{N}^{14}(n,\mathrm{H}^3)\mathrm{C}^{12}\).
A steel vessel was filled with a mixture of nitrogen and hydrogen, or with nitrogen, hydrogen, and ammonia. To eliminate slow neutrons the vessel was covered with a layer of cadmium.
The source of fast neutrons was a plate of \(\mathrm{U}^{235}\) placed in a Brookhaven reactor. The gas irradiation was carried out continuously for 4–5 days. Then the gas was released from the vessel into a Wilson chamber or into a system for filling gas-discharge counters.
When working with the Wilson chamber, hydrogen and alcohol were added to the gas. Each photographic image contained on average 5 clearly distinguishable tracks of electrons from the \(\beta\)-decay of tritium; to obtain good statistics, one hundred photographs were sufficient. The background of slow-electron tracks, which was difficult to distinguish from tritium tracks, did not exceed one per photograph.
The Wilson chamber was calibrated by introducing known small doses of tritium. The calibration was carried out by admitting small quantities of deuterium whose tritium concentration was known and was \((3.6 \pm 0.1)\cdot10^{-12}\).
In several series of experiments the irradiated gas, directly or after observations in the Wilson chamber, was used to fill counters. In this case hydrogen was separated from the other components of the gas mixture by means of a palladium tube. A certain amount of butane was added to the hydrogen. The counters were placed in a special low-background apparatus consisting of a protective shield and counters connected in anticoincidence.
The results obtained with the counters were systematically lower, by about 10%, than the data obtained with the Wilson chamber. It was possible to establish that the discrepancy was explained by the formation of ammonia molecules that did not pass through the palladium leak into the counters.
The neutron flux was determined by three independent methods, which gave well-agreeing results.
a) The flux of thermal neutrons near the \(\mathrm{U}^{235}\) plate serving as the source of fast neutrons was determined from the activation of sodium, compared with the activity obtained in a reactor with a known flux. From the flux of thermal neutrons the flux of fast neutrons from fission was determined by direct calculation.
b) The flux of fast neutrons was determined directly from the activation of sulfur (reaction \(\mathrm{S}^{32}(n,p)\mathrm{P}^{32}\)), whose threshold is equal to \(1\ \mathrm{MeV}\).
The cross section of this reaction for fission neutrons is equal to \(50\cdot10^{-27}\ \mathrm{cm}^2\). The long half-life period (14 days) made it possible to obtain the average value of the flux during the irradiation time independently of fluctuations in the reactor operating regime.
c) From the activation of magnesium in the reaction \(\mathrm{Mg}^{24}(n,p)\mathrm{Na}^{24}\), whose threshold, \(4.7\ \mathrm{MeV}\), is very close to the threshold for tritium formation in nitrogen.
The relation between the cross section, the neutron flux, and the amount of tritium formed may be written as
\[ \frac{N_{\mathrm T}}{N_{\mathrm H}} = \left(\frac{N_{\mathrm N}}{N_{\mathrm H}}\right)tF \int_{4.4\,\mathrm{Mev}}^{\infty}\sigma(E)n(E)\,dE, \tag{2} \]
where \(\frac{N_{\mathrm T}}{N_{\mathrm H}}\) is the ratio of the number of tritium and hydrogen nuclei; \(\frac{N_{\mathrm N}}{N_{\mathrm H}}\) is the ratio of the number of nitrogen and hydrogen nuclei, equal in the present case to \(0.55\); \(t\) is the irradiation time in seconds; \(F\) is the flux of fast neutrons, averaged over the cross section of the vessel; \(\sigma(E)\) is the cross section as a function of energy, and \(n(E)\,dE\) is the energy distribution of neutrons in the fission of \(\mathrm{U}^{235}\).
According to data\(^{6,7}\), in the energy region up to \(17\ \mathrm{Mev}\), \(n(E)\,dE\) has the form
\[ n(E)\,dE = \left(\frac{2}{\pi e}\right)^{1/2} e^{-E}\operatorname{sh}\left[(2E)^{1/2}\right]\,dE, \tag{3} \]
where \(E\) is expressed in \(\mathrm{Mev}\). According to equation (3), \(8.5\%\) of the neutrons arising in fission have an energy exceeding \(4.4\ \mathrm{Mev}\). If \(\sigma\) is taken outside the integral and this quantity is denoted by \(\bar{\sigma}\) (the mean cross section for neutrons with energies above \(4.4\ \mathrm{Mev}\)), then expression (3) takes the form
\[ \frac{N_{\mathrm T}}{N_{\mathrm H}} = \frac{N_{\mathrm N}}{N_{\mathrm H}}\,tF\bar{\sigma}\times(0.085). \tag{4} \]
The mean value of the cross section \(\bar{\sigma}\), obtained by the author\(^{5}\), is
\[ (11\pm 2)\cdot 10^{-27}\ \mathrm{cm}^{2}. \]
If the Coulomb coefficient for the escape of tritium from \(\mathrm{C}^{12}\) is taken into account, the expression for the cross section has the form:
\[ \bar{\sigma}'=\sigma_0\exp\left(-\frac{Z_1Z_2 e^2}{\hbar v_{\mathrm T}}\right) = \sigma_0\exp\left[-\frac{1.51}{(E-4.4)^{1/2}}\right], \tag{5} \]
where \(E\) is the energy of the bombarding neutrons in \(\mathrm{Mev}\). Finally,
\[ \bar{\sigma}'=(46\pm 9)\exp\left[-\frac{1.51}{(E-4.4)^{1/2}}\right]\cdot 10^{-27}\ \mathrm{cm}^{2}. \tag{6} \]
2. FORMATION OF TRITIUM IN THE ATMOSPHERE
The existence of neutrons in the atmosphere is due to their production in the “stars” of cosmic radiation. The rate of neutron formation may\(^{5}\) be considered equal to the sum, referred to unit time, of the numbers of \(\mathrm{C}^{14}\) nuclei formed, of neutrons escaping beyond the atmosphere, and of neutron losses due to the \((n,\alpha)\) reaction and other types of reactions in air.
The lower part will consider slow neutrons having energies up to \(20\ \mathrm{Mev}\). The process of formation of “stars” will not be discussed.
The mean rate of formation of \(\mathrm{C}^{14}\) is\(^{8}\) \(2.33\ \mathrm{cm}^{-2}\cdot\mathrm{sec}^{-1}\). The number of neutrons escaping into the space surrounding the Earth can be calculated from Yuan’s data\(^{9}\) on the number of slow neutrons in air as a function of altitude. At a pressure of about \(10\ \mathrm{cm}\) Hg there is a broad maximum, beyond which, in the pressure region above \(20\ \mathrm{cm}\), there follows an exponential
tial decrease. The exponential follows the curve measuring the number of “stars,” which decreases exponentially from the upper boundary of the atmosphere.
The difference between the area under the exponential up to the boundary of the atmosphere and the curve obtained by Yuan for the number of slow neutrons gives the number of neutrons escaping into space.
If the energy of the neutrons arising in “stars” did not exceed \(0.4\) ev, then Yuan’s curve would run parallel to the star-production curve. The difference between the areas under the curves is equal to the area under the curve for slow neutrons. From this one may conclude that the number of escaping neutrons is approximately equal to the number of neutrons with energy below \(0.4\) ev that cause the formation of \(C^{14}\). According to \({}^{10}\), half of the \(C^{14}\) is produced by neutrons with energy below \(0.4\) ev; the rate at which neutrons escape beyond the atmosphere may therefore be taken as equal to
\[ 1.1\ \mathrm{cm}^{-2}\cdot \mathrm{sec}^{-1}. \]
The number of neutrons participating in reactions of a type other than \((n,p)\) is estimated to be less than
\[ 1.0\ \mathrm{cm}^{-2}\cdot \mathrm{sec}^{-1}. \]
Thus, the averaged rate of neutron formation above the earth’s surface \(P_n\) is \({}^{7}\)
\[ P_n = 2.2 + 1.1 + 1.0 = 4.3\ \text{neutrons}\ \mathrm{cm}^{-2}\cdot \mathrm{sec}^{-1}. \tag{7} \]
The number of collisions necessary to reduce the neutron energy from \(E\) to \(4.4\) Mev is
\[ \nu = \frac{\ln(E - 4.4)}{\rho}, \tag{8} \]
where \(\rho\) is the logarithm of the mean ratio of the energies before and after the collision; for air \(\rho = 0.130\).
The probability of formation of \(H^3\) as a result of a collision is equal to
\[ 0.80\frac{\sigma}{\sigma_s}, \]
where \(\sigma\) is the cross section of the reaction
\[ n + N^{14} = C^{12} + H^3, \]
\(\sigma_s\) is the scattering cross section \((1.5\cdot 10^{-24}\ \mathrm{cm}^2)\), and \(0.80\) is the concentration of nitrogen in air.
The fraction of neutrons participating in \(\nu\) collisions without formation of \(H^3\) is
\[ f=\left(1-0.80\frac{\sigma}{\sigma_s}\right)^\nu \tag{9} \]
for \(E = 10\) Mev, \(\nu = 6.3\).
If, in accordance with the experimental data, one takes
\[ \sigma = 11\cdot 10^{-27}\ \mathrm{cm}^2, \]
then \(f = 0.963\), which means that \(3.7\%\) of all neutrons form tritium. Allowance for corrections (see equations (5) and (6)) increases the fraction of neutrons forming tritium to \(4.7\%\).
If, instead of the assumption that the initial neutron energy is \(10\) Mev, one uses the energy distribution according to Bagge \({}^{11}\), the amount of tritium formed will prove to be smaller by \(40\%\).
The author \({}^{5}\) believes that from 3 to \(5\%\) of the total number of neutrons formed above the earth’s surface \((P_n)\) participate in the reaction
\[ N^{14} + n = C^{12} + H^3. \]
The rate of formation of \(H^3\) estimated above should be compared with the rate of tritium formation in other processes.
The most effective process of tritium formation is the direct knockout of \(H^3\) nuclei in “stars.”
The number of neutrons in “stars” is approximately equal to the number of particles with unit electric charge.
The ratio of the mean number of protons to the number of deuterons and tritons was determined by observing the tracks of particles in “stars” in photo-
emulsions$^{12}$. In addition, this ratio was measured for particles formed in the bombardment of Be and C by protons with an energy of 330 MeV$^{13}$.
The largest relative fraction of tritons is observed among particles in “stars” caused by cosmic radiation having an energy of 15–50 MeV.
The ratio protons : deuterons : tritons in this case is 3 : 2 : 1, i.e., 16% of the singly charged particles are tritons.
According to the results$^{13}$, in the bombardment of carbon the ratio
\[ \frac{\mathrm{H}^3}{\mathrm{H}^1+\mathrm{H}^2+\mathrm{H}^3} \]
is equal to 12%.
If tritons make up a fraction lying between 7 and 16% of the total number of singly charged particles, the total rate of formation of $\mathrm{H}^3$ may be considered to lie between $0.30$ and $0.70\ \mathrm{cm}^{-2}\cdot\mathrm{sec}^{-1}$.
The reaction of slow neutrons with $\mathrm{Li}^6$ in the ocean gives less than $10^{-6}$ tritons $\mathrm{cm}^{-2}\cdot\mathrm{sec}^{-1}$; the capture of neutrons by deuterium gives less than $5\cdot10^{-9}$ tritons $\mathrm{cm}^{-2}\cdot\mathrm{sec}^{-1}$. Reactions involving neutrons arising in the spontaneous fission of heavy nuclei can give less than $10^{-11}$ tritons $\mathrm{cm}^{-2}\cdot\mathrm{sec}^{-1}$.
The comparatively small number of primary particles in cosmic radiation (about $0.10\ \mathrm{cm}^{-2}\cdot\mathrm{sec}^{-1}$) and their high energy make it possible to conclude that the number of tritons arriving as part of the primary component is relatively small.
3. CONCLUSION
The principal processes considered above—direct formation of $\mathrm{H}^3$ nuclei in “stars” and the reaction $\mathrm{N}^{14}+n=\mathrm{C}^{12}+\mathrm{H}^3$—lead to the formation of a number of tritons lying between 0.4 and 0.9 per $\mathrm{cm}^2$ in 1 sec. The formation of tritons occurs mainly in the layer of the atmosphere at pressures between 0.5 and 30 cm Hg.
The tritium formed is oxidized, descends with atmospheric precipitation to the surface of the Earth, and ultimately enters the waters of the oceans. The average annual amount of precipitation over the ocean is 110 cm; over land it is 66 cm. The area of the oceans occupies 71% of the Earth’s surface; 25% of the precipitation falling on land enters the oceans. Thus, from each sq. cm of the Earth’s surface in 1 year there enters the ocean
\[ [110(0.71)+66(0.25)(0.29)]=84\ \text{cm} \]
of precipitation.
Rain water should have an average ratio of the amounts of $\mathrm{H}^3$ and $\mathrm{H}^1$ lying between 2 and $5\times10^{-18}$, which agrees well with data on the composition of the surface layer of lake water$^{1}$. Tritium entering ocean water forms $\mathrm{He}^3$ with time. Allowance for this circumstance gives a value of the ratio of the concentrations of $\mathrm{H}^3$ and $\mathrm{H}^1$ in ocean water lying between 1 and $2\cdot10^{-20}$.
The total continuously replenished and expended “reserve” of natural tritium may be estimated at 50–110 million curies.
The light isotope of helium, $\mathrm{He}^3$, accumulates in the atmosphere through the decay of tritium and direct formation in “stars.”
The number of $\mathrm{He}^3$ nuclei formed in “stars” is approximately equal to the number of $\mathrm{H}^3$ nuclei. The total number of $\mathrm{He}^3$ nuclei formed should lie between $0.7$ and $1.5\ \mathrm{cm}^{-2}\cdot\mathrm{sec}^{-1}$.
Undoubtedly, $\mathrm{He}^3$ escapes beyond the limits of the atmosphere. Taking into account that there are $10^{40}$ moles of $\mathrm{He}^3$ in the atmosphere, the average time until an atom of $\mathrm{He}^3$ escapes from the atmosphere is from 3 to 6 million years.
V. V.
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