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*From*: Ludwik Kowalski <kowalskil@MAIL.MONTCLAIR.EDU>*Date*: Sun, 15 Feb 2004 18:20:36 -0500

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

that in addition

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