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<DIV><FONT size=2 face=Arial>I have comme too realize that most of you seem to
prefer astrophysics subjects, but I want to ask your thoughts on a nuclear
subject which I think is tremendously important.</FONT></DIV>
<DIV><FONT size=2 face=Arial></FONT> </DIV>
<DIV><FONT face=Arial><FONT size=2>When nuclear physicists try to calculate the
details of what happens in any nuclear process, they never can get good
math. A central reason for tthat is that every nucleus which contains one
or more neutrons necessarily also has a huge amount of energy (the neutron
bindingn energy) of around 0.78 Mev which came from nowhere and therefore causes
all Energy Conservation calculations to be a problem. The impressive NIST
data base enabled me to find a surprising new approach. Consider a simple
process of a Tritium atom naturally beta-decaying into a Helium-<FONT
face="Times New Roman">3 atom. Using the NIST
data:</FONT></FONT></FONT></DIV>
<DIV><FONT size=2></FONT> </DIV>
<DIV>
<P>The beta decay of Tritium provides a simple and obvious example of where an
enormous blunder and many inappropriate complications exist in Nuclear Physics.
According to the very reliable government NIST database, a decaying
Hydrogen/Tritium atom, <B><SUP>3</SUP>H<SUB>1, </SUB></B>has a precise atomic
mass of <B><SUB></SUB></B></FONT><B><FONT size=4>3.0160492779 AMU
</FONT></B><FONT size=4>(Atomic Mass Units). A resulting </FONT>Helium-3 atom,
<B><SUP>3</SUP>He<SUB>2, </SUB></B>has a precise atomic mass of<B>
<SUB></SUB></B><B><FONT size=4>3.0160293201 AMU. </FONT></B><FONT size=4>The
NIST data also confirms that the Tritium atom does the following beta-decay,
</FONT><B><SUP>3</SUP>H<SUB>1</SUB> (Tritium) → <SUP>3</SUP>He<SUB>2</SUB>
(Helium-3) + radiation</B> with a half-life of 12.33 years (where the resulting
new electron simply begins to revolve around the new Helium nucleus to maintain
it as un-ionized.) It only takes a moment to examine the math of the precise
NIST scientific data regarding that Nuclear decay, <B><FONT size=4>3.0160492779
AMU → 3.0160293201 AMU </FONT></B><FONT size=4>to confirm that the<B>
</B>DIFFERENCE between the source and result atoms is <B>+0.0000199578
AMU</B></FONT><B> . That 0.0000199578 AMU is EXACTLY the NIST radiation which
also is described as the observed experimental radiation given off of 0.0185906
MeV.</B></P>
<P><B><FONT size=5>This simple math addition is EXACT! It scientifically
accounts for ALL the energy and mass involved in that entire atomic structure
and decay.</FONT></B></P>
<P>No one before had ever found even APPROXIMATE values to fit the math of the
processes. People have never even tried to do accurate math because they ASSUMED
that quite a few bundles of energy must also be in every atomic nucleus, such as
the Neutron Self-Binding Energy of 0.78235 MeV for EVERY neutron inside every
nucleus. According to traditional nuclear Physics, every Uranium atom
(allegedly) contains 146 Neutrons inside every nucleus which would then also
need to have spectacular amounts of Neutron Binding Energies inside every
nucleus, and they never could figure out WHERE all that energy could come from!
<STRONG>This recent approach provides EXACT mathematical solutions and
ELIMINATES many of their math complications!</STRONG></P>
<P>I have found that this amazing precision of energy conservattion is true for
thousands of nuclear processes. A web-page discusses many different types
of processes at <A
href="http://mb-soft.com/public4/nuclei7.html">http://mb-soft.com/public4/nuclei7.html</A></P>
<P><FONT size=2 face=Arial>There are some strange results from these
analyses. Neutrons (and neutron self-binding energies) may not even exist
in any neuclei.</FONT></P>
<P><FONT size=2 face=Arial>Carl Johnson</FONT></P>
<P><STRONG></STRONG></FONT></P></DIV></BODY></HTML>