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199 lines
6.0 KiB
Plaintext
199 lines
6.0 KiB
Plaintext
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Host Element Fusion
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Unpublished Work
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Copyright (c) 1990
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Earl Laurence Lovings
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1. Proton (1 Hydrogen 1) Energy: 938.3 Mev = 1.007825 amu
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2. Neutron (1 Neutron 0) Energy: 939.6 Mev = 1.008665 amu
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3. Deuterium (2 Hydrogen 1) = 2.014102 amu
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4. [(1 Neutron 0) + (1 Hydrogen 1) - electron] =
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(939.6 Mev + 938.3 Mev - .511 Mev) = 1877.389 Mev =
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2.015447128 amu
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5. [(1 Neutron 0) - (1 Hydrogen 1) + electron] =
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(939.6 Mev - 938.3 Mev + .511 Mev) = 1.811 Mev =
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1.9441763 x 10 - 03 amu
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6. (105 Palladium 46) = 104.905064 amu
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7. (103 Rhodium 45) = 102.905511 amu
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The host element fusion experiment begins with a Palladium
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electrode submersed in Deuterium. A energy source is supplied,
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which enables the fusion process to begin. My theory on this
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subject is explained below:
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The Deuterium atoms are allowed inside the Palladium electrode due
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to the electric field on the electrode. Once the Deuterium atoms
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are inside, the Deuterium causes the Palladium to become unstable.
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This is done by this process:
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[ (105 Pd 46 - (1 Neutron 0 + 1 Hydrogen 1 - electron) +
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(1 Neutron 0 - 1 Hydrogen 1 + electron)] or,
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[ 104.905064 amu - 2.015447128 amu + 1.9441763 x 10-03 amu] =
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102.8916 amu.
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The closest element Palladium can try to become stable is
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(103 Rhodium 45).
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Take Palladium's new mass and subtract it with Rhodium's mass.
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(103 Rhodium 45) - 102.8916 amu, or
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102.905511 amu - 102.8916 amu = 1.394653 x 10-02 amu.
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To find out how many electrons that is equivalent to:
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(1.394653 x 10-02 amu x 931.5 Mev/amu)/(.511 Mev/electrons) =
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25.42309 electrons
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This is the amount of electrons required to be ionized to enable
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host element fusion with Deuterium.
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That is the first process of host element fusion.
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The second process begins when the ionized electrons from the
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palladium atom shields the deuterium atoms to allow host element
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fusion.
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The Equation:
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Q = [(2 Hydrogen 1) + (2 Hydrogen 1) + (25.423 e) -
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(1877.389 Mev) + (1.811 Mev) - (2 Hydrogen 1)] x 931.5 Mev or,
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Q = [(2.014102 amu + 2.014102 amu + .01394653 amu - 2.015447 amu
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+ 1.944176 x 10-03 amu - 2.014102 amu)] x 931.5 Mev
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Q = 13.55 Mev
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You must realize for this process to work for host element fusion,
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you have to have a host element before Deuterium will fuse.
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My equation also theoretically works for known Deuterium fusion
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processes.
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Known Equation:
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1. [(2 Hydrogen 1) + (2 Hydrogen 1)] -> (3 Helium 2) +
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(1 Neutron 0)] = or,
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[(2.014102 amu + 2.014102 amu - 3.016030 amu - 1.008665 amu) x
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(931.5 Mev/amu)] = Q = 3.27 Mev
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My Equation:
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Host Element = (3 Helium 2) + (1 Neutron 0)
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[(3 Helium 2) + (1 Neutron 0) - 1877.389 Mev + 1.811 Mev] =
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[3.016030 amu + 1.008665 amu - 2.015447 amu + 1.944176x10-03 amu]
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= 2.011192 amu
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The element whose mass is closest to the new unstable "element" is
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(2 Hydrogen 1)
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(2.014102 amu) - 2.011192 amu = 2.909899 x 10-03 amu excess mass
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convert to electrons
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(2.909899 x 10-03 amu x 931.5 Mev/amu) /(.511 Mev/electrons) =
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5.304445 electrons
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Now the fusion of Deuterium atoms
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[(2 Hydrogen 1) + (2 Hydrogen 1) + 5.30444e - 1877.389 Mev
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+ 1.811 Mev - (2 Hydrogen 1)] x 931.5 Mev =
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[(2.014102 amu + 2.014102 amu + 2.909899x10-03 amu - 2.015447 amu
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+ 1.944176x10-03 amu - 2.014102 amu)] x (931.5 Mev/amu)
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Q = 3.27 Mev
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Known Equation:
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2. [(2 Hydrogen 1) + (2 Hydrogen 1) -> (4 Helium 2)] = or,
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[(2.014102 amu + 2.014102 amu - 4.002603 amu)] x
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(931.5 Mev/amu) = Q = 23.85 Mev
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My Equation:
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Host Element = (4 Helium 2)
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[(4 Helium 2) - 1877.389 Mev + 1.811 Mev] =
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[(4.002603 amu - 2.015447 amu + 1.944176x10-03 amu)] = 1.9891 amu
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The element whose mass is closest to the new unstable "element" is
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(2 Hydrogen 1)
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(2.014102 amu) - 1.9891 amu = 2.500188 x 10-02 amu excess mass
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convert to electrons
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(2.500188 x 10-02 amu x 931.5 Mev/amu) /(.511 Mev/electrons) =
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45.57585 electrons
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Now the fusion of Deuterium atoms
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[(2 Hydrogen 1) + (2 Hydrogen 1) + 45.58 electrons - 1877.389 Mev
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+ 1.811 Mev - (2 Hydrogen 1)] x 931.5 Mev =
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[(2.014102 amu + 2.014102 amu + 2.500188x10-02 amu - 2.015447 amu
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+ 1.944176x10-03 amu - 2.014102 amu)] x (931.5 Mev/amu) =
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Q = 23.85 Mev
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Known Equation:
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3. [(2 Hydrogen 1) + (2 Hydrogen 1) -> (3 Hydrogen 1) +
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(1 Hydrogen 1) = or,
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[(2.014102 amu + 2.014102 amu - 3.016050 amu - 1.007825 amu)] x
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(931.5 Mev/amu) = Q = 4.03 Mev
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My Equation:
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Host Element = (3 Hydrogen 1) + (1 Hydrogen 1)
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[(3 Hydrogen 1) + ( 1 Hydrogen 1) - 1877.389 Mev + 1.811 Mev] =
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[3.016050 amu + 1.007825 amu - 2.015447 amu + 1.944176x10-03 amu]
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= 2.010372 amu
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The element whose mass is closest to the new unstable "element" is
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(2 Hydrogen 1)
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(2.014102 amu - 2.010372 amu) = 3.729582x10-03 amu excess mass
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convert to electrons
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(3.729582x10-03 amu x 931.5 Mev/amu)/(.511 Mev/electrons) =
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6.798641 electrons
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Now the fusion of Deuterium atoms
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[(2 Hydrogen 1) + (2 Hydrogen 1) + 6.799 electrons - 1877.389 Mev
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+ 1.811 Mev - (2 Hydrogen 1)] x 931.5 Mev =
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[(2.014102 amu + 2.014102 amu + 3.7296x10-03 amu - 2.015447 amu +
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1.944176x10-03 amu - 2.014102 amu)] x (931.5 Mev/amu) =
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Q = 4.03 Mev
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Known Equation:
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4. [(1 Hydrogen 1) + (1 Hydrogen) -> (2 Hydrogen 1) +(electron)=
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[(1.007825 amu + 1.007825 amu - 2 electrons - 2.014102 amu)]
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x (931.5 Mev/amu) = Q = .42 Mev
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My Equation:
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Host Element = (2 Hydrogen 1)
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[(2 Hydrogen 1) - 1877.389 Mev + 1.811 Mev] =
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[(2.014102 amu - 2.015447128 amu + 1.9441763x10-03 amu)] =
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5.990302x10-04 amu
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The element whose mass is closest to the new unstable "element"
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is (1 Hydrogen 1)
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(1.007825 amu - 5.990302x10-03 amu) = 1.007226 amu excess mass
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convert to neutrinos
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(1.007226 amu x 931.5 Mev/amu) / (.42 Mev/neutrinos) =
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2,233.884 neutrinos
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Now the fusion of 1 Hydrogen 1 atoms
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[(1 Hydrogen 1) + (1 Hydrogen 1) + 2,234 neutrinos - 2 electrons
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- 1877.389 Mev + 1.811 Mev - (1 Hydrogen)] x 931.5 Mev/amu =
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[(1.007825 amu + 1.007825 amu + 1.007226 amu - .001097 amu
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-2.015447128 amu + 1.9441763x10-03 amu - 1.007825 amu)]
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x 931.5 Mev/amu = Q = .42 Mev
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