VbvNeuE-001
Boris V. Vasiliev About Quantum-Mechanical Nature of Nuclear Forces and Electromagnetic Nature of Neutrinos
Boris V. Vasiliev
About Quantum-Mechanical Nature of Nuclear Forces
and Electromagnetic Nature of Neutrinos
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38 CHAPTER 7. DEUTRON AND OTHER LIGHT NUCLEI<br />
Figure 7.1: Schematic representation of the structure of light nuclei. Dotted<br />
lines schematically indicate the possibility of a relativistic electron hopping between<br />
protons.<br />
To compare this binding energy with the measurement data, let us calculate<br />
the mass defect of the three particles forming the deuteron<br />
∆M 3 = 2M p + m e∗ − M d ≈ 3.9685 · 10 −27 g, (7.4)<br />
where M d is mass of deuteron.<br />
This mass defect corresponds to the binding energy<br />
E d = δM d · c 2 ≈ 3.567 · 10 −6 erg. (7.5)<br />
Using the relativistic electron mass in Eq.(7.4) does not seem obvious. However,<br />
this is confirmed by the fact that at the fusion reaction proton and neutron to<br />
form a deuteron<br />
p + n → D + γ (7.6)<br />
γ-quantum takes energy equal to 3.563 · 10 −6 erg [11]-[12].<br />
Thus the quantum mechanical estimation of the bonding energy of deuteron<br />
Eq.(7.3), as in the case of the hydrogen molecular ion, consistent with the experimentally<br />
measured value Eq.(7.5), but their match is not very accurate.<br />
7.0.2 Nucleus 3 2He<br />
As can be seen from the schematic structure of this nucleus (Fig.7.1), its binding<br />
energy is composed by three pairwise interacting protons. Therefore it can be<br />
assumed that it equals to the tripled energy of deuteron:<br />
The mass defect of this nucleus<br />
E He3 = 3 · E d ≈ 10.70 · 10 −6 erg. (7.7)<br />
∆M(He3) = 3M p + m e∗ − M He3 = 1.19369 · 10 −26 g. (7.8)<br />
Thus mass defect corresponds to the binding energy<br />
E ∆M(He3) = ∆M(He3) · c 2 ≈ 10.73 · 10 −6 erg. (7.9)<br />
Consent energies E He3 and E ∆M(He3) can be considered as very good.