New Physics Based on Yoon's Universal Atomic Model

Congrats Newedana. You made Dr. Yoon richer by $180.
I received the book yesterday.

Before you go on to so many topics. First justify that when electrons
move. The electric field at the front move back to the back. Dr. Yoon
said the field concept is wrong and these field lines are flexible
enough that they can move around the electron. Try to find proof of
this first before going to the dynamics details. Experiments done
in accelerators seem to show results that the field are all around it
even at the front. Try this java.

http://webphysics.davidson.edu/Apple...etard_FEL.html

Let's focus on this first.

Anyway. Even if Dr. Yoon is wrong in his atomic model. He shines in his
gel crystal theory in biochemistry. I don't know how much lab work
Dr. Yoon did, but maybe some of the explanations can be found in
the convensional physics model. He just have to look deeper for a
QM explanation of the gel crystal system. Anyway. I'll try to find
a QM counterpart for it.

later,

p6



newedana wrote:
newedana wrote:
The energy emission taking place when orbital electron rings expand, c=
an be observed in the case when chemical explosives such as TNT (trinitroto=
luen) explodes. The outermost orbital electron rings of their component ato=
ms contributing to combine them, expand only a little bit in this case of e=
xplosion, due to dissociation of TNT to form various kinds of gas molecules=
, such as H2O, CO2, and NO2 etc.

It is well known that the explosion of only about 7 kgs of uranium 235=
produces such an enormous energy equivalent to that emitted by explosion o=
f TNT 20,000 metric tons. The mass ratio of these two explosive materials =
is about, 1 : 2.86x10^6. If the orbital electron rings in K shell of uraniu=
m atom with radial parameter, say, =CE=B3=3D1/100, expands to be the orbita=
l electron rings in K shell of newly created atoms that has radial paramete=
r, say, =CE=B3=3D1/99.28, then the ratio of energy capacity of these two or=
bital electron rings becomes the same as the mass ratio, 2.86x10^6 , as sho=
wn above when we estimate it with Eq.=E2=96=B3E=3DE'(1/r^4, previously post=
ed. The difference of radial parameter between these two electron rings is=
negligibly small, or =CE=94=CE=B3=3D1/99.28-1/100=3D1/13,789, but the rati=
o of their energy capacity is enormous, as shown above. However, this energ=
y emission comes only from the expansion of orbital electron rings in K she=
ll of uranium 235. Other orbital electron rings in L, M, N,. . . .shells of=
uranium 235 would also have to expand their orbital radii emitting huge en=
ergies also as in the case of electron rings of K shell. Thus the explosion=
of only 7 kg of uranium 235 gives rise to producing such a tremendous ener=
gy.

The fundamental mechanism of emitting energy from nuclear fusion of de=
uterons is exactly the same as that of nuclear fission of uranium 235. It i=
s also the expanding energy of electron rings. In the case of nuclear fissi=
on atomic electron rings expand, while in the case of nuclear fusion nucle=
ar electron rings associated in the structure of two deuterons expand, emit=
ting nuclear energy. When nuclear electron rings of two deterons combine to=
build a unified nuclear electron ring with pair electrons, they have to ex=
pand their orbital radii emitting energy, in order to bind four protons to =
build two neutrons and two protons in a helium nucleus. This is the nuclear=
fusion energy. It is the same as that when two hydrogen atoms combine to f=
orm a hydrogen molecule having molecular electron rings carrying pair elect=
rons, with their two single atomic electron rings, emitting energy.
A single nuclear electron ring that binds two protons in constructing =
a neuteron, can emit =CE=B3-rays at the nearest distance to its two nuclear=
protons when it breaks. Since this single nuclear electron ring can emit =
=CE=B3-rays with wavelength 0.005 =E2=84=AB, its radial parameter must be,=
=CE=B3=3D1/430, when we estimate it with the same equation, =E2=96=B3E=3DE=
'(1/r^4) I posted above. If these single nuclear electron rings expand thei=
r orbital radii and emit energy equivalent to that energy given by explosio=
n of 7 kgs of uranium 235, their radial parameter has to expand from =CE=B3=
=3D1/430 to =CE=B3=3D1/429.991. The distinction between them is, =E2=96=B3=
=CE=B3=3D1/429.991-1/430=3D1/20,544,014. It is an awfully small expansion c=
ompared to that in the case of nuclear fission. However the energy emission=
in the process of nuclear fusion is the same as that in the case of nuclea=
r fission. The ratio of atomic volume of deuterium and helium is, D : He =
=3D 14.1 : 31.8. The larger atomic volume of helium than deuterium is attri=
buted to the reason that the helium nucleus is stabilized in the lowest ene=
rgy level than deuteron. Nuclear fusion energy includes also the energy of =
orbital electron rings of deuterium when they expand their orbital radii to=
be those of helium. newedana