Chronology Current Month Current Thread Current Date [Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

# Re: HOT FUSION

THIS IS A SLIGHTLY BETTER VERSION
I HOPE THAT THE UGLY =20 WILL NOT
ATTEAR ATE THE END OF YEACH LINE.

1) In electrostatic you might ask students to estimate
the coulomb barrier preventing positive D ions to fuse
at low temperatures. Here is my suggestion. First tell
students that in addition to repulsive forces the ions
attract each other by very strong nuclear forces. But
these forces do not obey the 1/r^2 law. They are
negligibly small when r is higher then the distance R
called range). For x<R the attractive nuclear forces are
much larger than repulsive electric forces. That would
be a good justification for defining the coulomb barrier,
CB, as the value of the electric potential at x=R.

The rest is trivial. Assume that R=3 F, for example,
and calculate CB. Note that F is the unit of length
(femtometer or 10^-15 m). My answers were 480 kV
and 360 kV, at R=3 F and 4 F, respectively. You may
or may not link this problem with the cold fusion
controversy.

2) If you do then consider addressing the screening
effect. Some scientists say that screening is possible
when D ions are embedded in metals, such as Pd or
Ti. Simply stated, and without trying to argue about
what causes screening (local clouds of electrons at
crystal’s boundaries?) one can simply declare:
screening consists of lowering of the coulomb barrier
by nearby electrons.

To illustrate screening do the following. Place one
deuteron at x=0 at treat the other deuteron, at x>0, as
a probe charge. That is what I did to calculate coulomb
x, for example, -2 F. The CB is now V=V1 + V2 (where
V1 is the positive part due to the deuteron and negative
V2 is the negative part due the electron). You will see
that CB approaches zero when electron is approaching
the origin, as it should be. My answer, for R=4 F, was
CB=120 kV for the electron placed at x=-2 F. And nothing
prevents you from introducing more than one screening
electron.

Suppose, for example, that a single electron at x=10 A
=100000 F is replaced by a negative particle of variable
charge. Assuming R=4 F, how does the magnitude of the
negative charge affect the coulomb barrier? It turns out
that the charge of only 250000*e is sufficient to eliminate
the coulomb barrier. Here my results;

charge in 10^5*e QB in kV
-----------------------------------------
2.1 0.0015
2.0 72
1.0 216
0.5 288
0.1 346
0.01 358
0.00 360 (an ideal two-body barrier)

I find them shocking. First because the distance of 10 A
is about ten times larger than the distance between atoms
in most metals. And second because the number of
electrons needed to eliminate the barrier (for that distance)
is a neglibible fraction of free electrons in each cubic micron.

4) Yes, I know that three or more particles would usually
not be at rest on the x axis. And I know that the nuclear
potential is not a rectangular well. My goal is to estimate
the orders of magnitude of CB, and to illustrate the idea
of screening.

5) By the way, we usually think that the so-called “free
electrons” in metals are uniformly distributed, like in
ionized gases. What evidence do we have for this?

6) No, I am not trying to poison yours with heretical
pseudo science.

On Sunday, Feb 15, 2004, Ludwik Kowalski wrote:

1) In electrostatic you might ask students=20
to estimate the coulomb barrier preventing=20
positive D ions to fuse at low temperatures.=20
Here is my suggestion. First tell students=20