Are other nuclei made up of basic hydrogen nuclei Type I and III,
Type I (\(p^+\), \(g^+\), \(T^+\))
Type III (\(T^+\), \(p^+\), \(g^+\))??
No, because we have Helium-3 \(^3He\), by adding \(p^{+}\) to the Type I hydrogen nucleus \(^3H\)
(\(p^+\), \(g^+\), \(T^+\))+\(p^{+}\)\(\rightarrow\)(\(p^+\), \(g^+\), \(T^+\), \(p^{+}\))
and Helium-2 \(^2He\), by adding (\(T^+\), \(p^{+}\)) to the Type III hydrogen nucleus \(^2H\)
(\(T^+\), \(p^+\), \(g^+\), \(T^+\), \(p^{+}\))
Remember that the hydrogen nucleus Type I nucleus is the heaviest, followed by Type III then Type II, from which we made the correspondence,
Type I (\(p^+\), \(g^+\), \(T^+\)) is the Hydrogen-3 nucleus
Type III (\(T^+\), \(p^+\), \(g^+\)) is the Hydrogen-2 nucleus
and
Type II (\(g^+\), \(T^+\), \(p^+\)) is the Hydrogen nucleus
and we may have, without adding \(p^{+}\) particles which would increase the atomic number, create other Hydrogen isotopes from the Type II and III nuclei types,
From the Type II (\(g^+\), \(T^+\), \(p^+\)) nucleus,
(\(g^+\), \(T^+\), \(p^+\), \(g^+\))
and
(\(g^+\), \(T^+\), \(p^+\), \(g^+\), \(T^{+}\))
From the Type III (\(T^+\), \(p^+\), \(g^+\)) nucleus
(\(T^+\), \(p^+\), \(g^+\), \(T^+\))
It is not possible to add to a Type I (\(p^+\), \(g^+\), \(T^+\)) nucleus because the next particle to be added is \(p^{+}\) which would increment the atomic number.
It is possible to reduce the existing nuclei types I and III by one particle without changing the atomic number,
From Type I (\(p^+\), \(g^+\), \(T^+\)),
(\(p^+\), \(g^+\))
and
(\(p^+\)), which is mass-less unless the single \(p^{+}\) spins and generates a \(g\) field.
From Type III (\(T^+\), \(p^+\), \(g^+\))
(\(T^+\), \(p^+\))
which is also mass-less unless the single \(p^{+}\) spins and generates a \(g\) field.
Do all these fit well with hydrogen isotopes already discovered? No! There are nine different nuclei here but only seven discovered isotopes. The assumption that the basic nuclei types are made up of all three positive particles may be wrong. It could be that, the nucleus
(\(T^+\), \(p^+\))
reduced from a Type III nucleus (\(T^+\), \(p^+\), \(g^+\)) is Hydrogen-2, \(^2H\),
(\(p^+\), \(g^+\))
reduced from Type I (\(p^+\), \(g^+\), \(T^+\)) nucleus is Hydrogen-3 \(^3H\), and
(\(p^+\))
reduced from (\(p^+\), \(g^+\)) is Hydrogen-1, \(^1H\).
In which case, there is still two extra isotopes.
Order, order, order in the court of hydrogen please.
Note: The difference in mass among the isotopes was previously attributed to the strength of the weak \(g\) field, which depended on the position of \(p^{+}\) particles (thus orbital radii) in the ordered set.
