Itzhak Shechtman

2002-03-20 11:09:27 UTC

Hi,

In a deuterino molecule (i.e. a molecule consisting of 2 deuterino atoms)

with reduced distance between the nuclei, the probability of fusion

reaction between them due to their closer proximity can be many orders of

magnitude larger than in the normal deuterium molecule.

Results are given below of a calculation using the expression given by

J.D.Jackson [Phys. Rev.,106(1957),330] for the fusion reaction rate, P

[sec**-1], between the nuclei of a deuterium molecule:

P = A/(4 pi a**3) * alpha/(2 pi) * Exp(-lambda)

where A is the reaction constant(in cm**3/sec), proportional to the

reaction cross section,sigma; a is the "ground-state" radius of the atom;

alpha is the harmonic oscillator parameter in units of a**-2(appearing in

the vibrational ground-state wave function as Exp[-1/2alpha(x-xsub0)**2],

where (x-xsub0) is the separation from equilibrium in units of a); lambda

is the Coulomb barrier penetration factor for the deuteron pair.

If we denote by B the ratio of the Bohr radius to the radius of the

electron in the dehydrino atom, the above expression raduces to:

P ~ 2.27*10**8 * B**5/2 * Exp[-109.34/B**1/2] .

On the basis of the deuteron-deuteron penetration factor and reaction rate

calculated by Jackson [loc.cit.,Tables 1 and 2]for a muon-deuterium

molecule, with the sum of nuclear radii taken to be 5*10**-13 cm, the

results are given for the range B = 1 - 207.

B = 1 P = 7.41*10**-40 sec**-1

B = 2 P = 3.40*10**-25 "

B = 3 P = 1.36*10**-18 "

B = 4 P = 1.31*10**-14 "

B = 5 P = 7.36*10**-12 "

B = 6 P = 8.23*10**-10 "

B = 7 P = 3.32*10**-8 "

B = 8 P = 6.68*10**-7 "

B = 9 P = 8.18*10**-6 "

B = 10 P = 6.91*10**-5 "

B = 20 P = 9.78*10**0 "

B = 50 P = 7.72*10**+5 "

B = 100 P = 4.05*10**+8 "

B = 207 P = 7.00*10**+10 "

Itzhak Shechtman

In a deuterino molecule (i.e. a molecule consisting of 2 deuterino atoms)

with reduced distance between the nuclei, the probability of fusion

reaction between them due to their closer proximity can be many orders of

magnitude larger than in the normal deuterium molecule.

Results are given below of a calculation using the expression given by

J.D.Jackson [Phys. Rev.,106(1957),330] for the fusion reaction rate, P

[sec**-1], between the nuclei of a deuterium molecule:

P = A/(4 pi a**3) * alpha/(2 pi) * Exp(-lambda)

where A is the reaction constant(in cm**3/sec), proportional to the

reaction cross section,sigma; a is the "ground-state" radius of the atom;

alpha is the harmonic oscillator parameter in units of a**-2(appearing in

the vibrational ground-state wave function as Exp[-1/2alpha(x-xsub0)**2],

where (x-xsub0) is the separation from equilibrium in units of a); lambda

is the Coulomb barrier penetration factor for the deuteron pair.

If we denote by B the ratio of the Bohr radius to the radius of the

electron in the dehydrino atom, the above expression raduces to:

P ~ 2.27*10**8 * B**5/2 * Exp[-109.34/B**1/2] .

On the basis of the deuteron-deuteron penetration factor and reaction rate

calculated by Jackson [loc.cit.,Tables 1 and 2]for a muon-deuterium

molecule, with the sum of nuclear radii taken to be 5*10**-13 cm, the

results are given for the range B = 1 - 207.

B = 1 P = 7.41*10**-40 sec**-1

B = 2 P = 3.40*10**-25 "

B = 3 P = 1.36*10**-18 "

B = 4 P = 1.31*10**-14 "

B = 5 P = 7.36*10**-12 "

B = 6 P = 8.23*10**-10 "

B = 7 P = 3.32*10**-8 "

B = 8 P = 6.68*10**-7 "

B = 9 P = 8.18*10**-6 "

B = 10 P = 6.91*10**-5 "

B = 20 P = 9.78*10**0 "

B = 50 P = 7.72*10**+5 "

B = 100 P = 4.05*10**+8 "

B = 207 P = 7.00*10**+10 "

Itzhak Shechtman