Fusion in a pile and new radioactive pathways.          scroll down--->       

Possibilities in Nuclear physics resulting from the 1+1=2 Equation.

That is a result from the Particle-wave theory.             scroll down--->

The Neutron could be a Proton plus electron who is

to say that electrons don’t fall into the nucleus. Then Hydrogen would be a nucleus of two protons and one electron. When heated to the extreme the electron would go leaving two protons Helium (atomic fusion) then when it cooled down the nucleus capturing another electron reverting to hydrogen.

Uranium has a large number of these neutrons who might be protons with a electron bound. When the electrons go then this leaves too many protons---fission.

One could transform one element into another by adding electrons to the nucleus or taking them away at high temperature. Positrons would be a useful tool too.                                                                         scroll down---> 

 

Wave-Particle

Theory

Advantages:

1) Change one element into

Another (Gold, Platinum).

2) 10X bigger Atomic bomb or

nuclear energy with the 100% mass to energy conversion.

3) New radioactive pathways and

manufacturing isotopes.

4) A Good way of treating

Radioactive waste.

5) A understanding of

Radioactivity influencing the particles of the nucleus and decays even to the smallest particle.

6) A New understanding of the

nucleus and binding 1+1 that

can be generalized to all

atoms.

 

Alan Brett

alanbret@hotmail.com 

Wave-Particle Theory

Abstract:

A reaction in what all mass is converted (particle disintegrates) to energy. Caused by gamma radiation. Theory says one can force proton or electron decay using gamma radiation at the deBroglie wavelength of the particle.

Smaller particles, and this is important for particle physicists, will decay if hit by lower frequencies than gamma.

 

Done mathematically simply using Schrodingers wave equations FINDING a mathematical result that the above will only occur at a specific frequency of incoming photon—the deBroglie wavelength of the particle. One would have to account for Amplitude too and Potential and Kinetic energy.

Abstract: (continued)

Page two.

Of interest for high theory is one can redo the math treatment again for any energy E₁ + any energy E₂ producing any energy E_{3 .}

Eone + Etwo = Ethree (as time continues)

the math solves one variable and produces Eone=Etwo

producing a double Ethree.

1 + 1 = 2 (bound)

One thinks here of the neutron being so close in energy to the Proton and the binding that occurs in the nucleus. Plus radioactivity due to a asymmetric binding. And isotopes. And the smallest of particles and there binding. Also of interest to Nuclear Physicists is the binding of the hydrogen nucleus in fusion. And asymmetric binding in Uranium.

Section 1 Introduction

 

The before and after ENERGY conditions are placed in the Schrodinger equation to give the before and after WAVES.

See fig. 1

These waves are equal at the continuity point between before and after

See fig. 2

and this makes the waves equal at this point. This solves the variable in the initial conditions.

This solves the problem.

Page 3.

Mathematicle Treatment

Section 2 Energy Conservation

For fig. 1 page 1

E_before = E_after

Or from the figure a photon hits a particle (in a potential) and produces photons.

This is-

(1) hμ + mc² + V = xhμ

 

photon + particle + potential energy

= a unknown number of photons

E_after = xhμ this is x in units of (hμ)

This is any number of photons out because of additivity of Energy. But is that way to obtain a chain reaction.

 

hμ = energy of photon in

mc² = mass energy of particle

 

V = potential energy the particle is in

 

x =variable that determines E_after

 

 

Page 4

These energies are placed in the wave equations by Schrodinger, the time dependant ones, then with time I can describe the before and after conditions as waves that are continuous at the point between before and after.

Section 3 Initial Conditions

E_before = hμ + mc² + V

 

Ψ = ℮ ^-iEt/ħ Solution of the time dependant

Schrodinger equation.

 

Placing the energy equation in the above equation. We obtain.

 

(2) Ψ_initial = ℮^-i(mc²^+V)t/ħ + ℮^-ihμt/ħ

 

This is the addition of a particle wave and an electromagnetic wave ie. Absorption of the photon by the particle.

 

Folowing is the result of that absorption—

Section 4 Final Conditions

E_final = xhμ

So the wave out is--

Φ_final = x℮^-ihμt/ħ

= x (photon of energy hμ)

= a chain reaction—but it must be mentioned that their is the superposition of waves principle so long as they all add up to the above. So this may be anything out.

 

Page 5

Section 5 Continuity with time

 

 

(4) Ψ_initial= Φ_final (continuity condition)

At some point in time the wave before equals the wave after.

The Calculations

Section 6 Solve for hµ to obtain the frequency of the incident photon

(5) Ψ_initial= ℮^{-i(mc²+V)t/ħ} + ℮^-ihμt/ħ = x℮^-ihμt/Ћ= Φ_final

 

 

Next we solve this for x

 

Page 6

 

Solve for x---

(x-1) = ℮^{[-i(mc²+V)t/ħ + ihμt/ħ]}

 

Useing now equation 1 , Page 3

The energy conservation equation—

Where (x-1) =(mc² + V)/hμ

 

Equating these two at (x-1) and ln both sides to get rid of exp.

ln {(mc² + V)/hμ} = [mc² + V – hμ ](-it/ħ)

 

 

(6) hμ = mc² + V +( iħ/t) ( ln{ [mc² + V]/ hμ})  

 

Wave

Particle

Equation

 

The simplest solution is—

(7) hμ =mc² +V

 

This gives the initial frequency of photon needed to decay the particle.

 

 

 

Page 7

Section 7 Conclusions

Here is the frequency of the photon needed to decay a particle.

hμ = mc² + V

Where m is the rest mass (relativistic) and V is the potential energy of the particle.

This result ties in with de Broglie particle waves,

De Broglie gives

 

λ_deBroglie=h/p = h/mc (P=E/c)

(8) so hμ_deBroglie= mc²

 

like the top of the page

from our calculations, is there in established theory.

We must remember the deBroglie wave of a charged particle is also an electromagnetic wave.

For conservation of spin and charge small particles might be produced.

 

Page 8

Section 8 The neutron

Like mentioned earlier one can generalize the initial equations to—

(9) E₁ + E₂ = E₃

Our treatment

That isE₁ + E₂ = xE₁

Solution x = 2

And E₁=E₂

 

(10) E₁ + E₁ =2E₁

This is then a law that all particles at the quantum level will obey. So two protons OF THE SAME ENERGY combine. Two electrons OF THE SAME ENERGY bond in one orbit. The neutron is more or less THE SAME ENERGY as the Proton. Here is a peek at the structure of the nucleous with this binding occurring and radioactivity relieving the energy differences. And is interesting for isotopes being out of balance of 1+1=2.

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