Helium nucleus, formation

Figure 3.2 The formation of a helium nucleus from two protons and two neutrons results in a loss of 5 x 10 kilograms.

Helium, the second most abundant element in the universe after hydrogen, is rare on Earth because its atoms are so light that a large proportion of them reach high speeds and escape from the atmosphere. However, it is found as a component of natural gases trapped under rock formations (notably in Texas), where it has collected as a result of the emission of a particles by radioactive elements. An a particle is a helium nucleus (4He2+), and an atom of the element forms when the particle picks up two electrons from its surroundings. 

However, such processes occur relatively seldom, while E. Salpeter (Cornell University) showed that a two-step reaction should be more easily realisable (Kipper-hahn, 1980). A collision of two helium nuclei leads to the formation of a beryllium nucleus, which decomposes very rapidly to the starting materials unless it is hit by a further helium nucleus the newly-formed nucleus 12C is stabilized by radiation emission. The lifetime of the beryllium nucleus is only about 0.05 s (Hillebrand and Ober, 1982) thus, the density of the helium nuclei must be very high in order to give a high collision probability. 

One of the ways that nuclides with more than 83 protons change to reach the band of stability is to release two protons and two neutrons in the form of a helium nucleus, which in this context is called an alpha particle. Natural uranium, which is found in many rock formations on earth, has three isotopes that all experience alpha emission, the release of alpha particles. The isotope composition of natural uranium is 99.27% uranium-238, 0.72% uranium-235, and a trace of uranium-234. The nuclear equation for the alpha emission of uranium-238, the most abundant isotope, is... 

Stable miclides are those that have never been observed to decay spontaneously. Radionuclides, in contrast, undergo spontaneous disintegration, which ultimately leads to stable nuclides. The disintegration, or radioactive decay, occurs with the emission of electromagnetic radiation in the form of X-rays or gamma rays (y rays) with the formation of electrons, positrons, and the helium nucleus or by fission, in which a nucleus breaks up into smaller nuclei. 

While it is plausible that dislocations also act as internal surfaces which serve as centers for the initiation of decomposition, detailed studies of the role of dislocations require further examination. Recent advances in electron-microscope techniques may permit direct observation of the role of dislocations and impurities in the mechanisms of colloid or nucleus formation. Indeed, such studies on alkali halide crystals (at liquid-helium temperatures, to avoid radiation damage by the electron beam) confirmed the special role of dislocations in sensitizing the nucleation of colloids [52]. Such techniques should be possible for at least the (more stable) alkali azides. 

When a helium nucleus is formed, there is always some degree of loss of mass. If the loss of mass equals 3.1 x 10 kg during the formation of one mol of it, what is the binding energy ... 

TRANSMUTATION. The natural or artificial transformation of atoms of one element into atoms of a different element as the result of a nuclear reaction. The reaction may be one in which two nuclei interact, as in the formation of oxygen from nitrogen and helium nuclei (/3-particles), or one in which a nucleus reacts widi an elementary particle such as a neutron or proton. Thus, a sodium atom and a proton form a magnesium atom. Radioactive decay, e.g., of uranium, can be regarded as a type of transmutation. The first transmutation was performed bv the English physicist Rutherford in 1919. 

Only one combination of four nucleons is bound, 4He, with two protons and two neutrons. All other combinations of four nucleons are unbound. Moreover, 4He, or the a particle, is especially stable (very strongly bound), and the nucleons are paired to give a total spin 5=0. Interestingly, if we add a nucleon of either type to the a particle, we produce an unbound nucleus Thus, there are no stable nuclei with A = 5 as both 5He and 5Li break apart very rapidly after formation. This creates a gap in the stable masses and poses a problem for the building up of the elements in stars, which is discussed in Chapter 12. There are two bound nuclei with A = 6, 6He and 6Li, with the helium isotope decaying into the lithium isotope, the others are unbound. Continuing on, between mass 6 and 209, all mass numbers... 

Helium-6 is a radioactive isotope with fj/2 = 0.861 s. Calculate the mass defect (in g/mol) for the formation of a 6He nucleus, and calculate the binding energy in MeV/nucleon. Is a 6He nucleus more stable or less stable than a 4He nucleus (The mass of a 6He atom is 6.018 89 amu.)... 

Triple-alpha process The formation of a carbon-12 nucleus in the interior of a massive star through the essentially simultaneous fusion of three helium nuclei. 

Eventually, when the supply of hydrogen is exhausted, the core of the star again contracts with further heating until temperatures are reached where fusion of helium nuclei can occur. This leads to the formation of and O nuclei. In turn, when the supply of helium nuclei is depleted, further contraction and heating occur, until the fusion of heavier nuclei takes place. This process occurs repeatedly, forming heavier and heavier nuclei until iron nuclei are formed. Because the iron nucleus is the most stable of all, energy is required to fuse iron nuclei. This endothermic fusion process cannot furnish energy to sustain the star therefore it cools to a small, dense white dwarf. 

At this point, the synthesis of new elements stopped. The reason is that nuclei larger than those of hydrogen and helium can form only when like-charged particles (such as protons or helium nuclei) combine to form heavier nuclei. A hypothetical example is the formation of a beryllium nucleus by the combination of two helium nuclei ... 

One additional stage of element formation may also occur during helium burning, the reaction between an alpha particle and a carbon nucleus to produce an oxygen nucleus ... 

In the preceding sections we have discussed systems containing two nuclei, each with one stable orbital wave function (a Is function), and one, two, three, or four electrons. We have found that in each case an antisymmetric variation function of the determinantal type constructed from atomic orbitals and spin functions leads to repulsion rather than to attraction and the formation of a stable molecule. For the four-electron system only one such wave function can be constructed, so that two normal helium atoms, with completed K shells, interact with one another in this way. For the other systems, on the other hand, more than one function of this type can be set up (the two corresponding to the structures H- H+ and H+ H for the hydrogen molecule-ion, for example) and it is found on solution of the secular equation that the correct approximate wave functions are the sum and difference of these, and that in each case one of the corresponding energy curves leads to attraction of the atoms and the formation of a stable bond. We call the bonds involving two orbitals (one for each nucleus) and one, two, and three electrons the one-electron bond, the electron-pair bond, and the three-electron bond, respectively. 

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