- #1

- 16

- 0

IMO, it should be possible to initiate a nuclear fusion of H1-H2 and H2-H2 in crystals of Lithium Hydride induced by gamma photons emitted form decay of Na24.

Could anyone please verify if the following is correct?

By using nuclear photodisintegration effect, we need a gamma photon with energy high enough to overcame a threshold energy, which is specific to the reaction[1][4]:

H2 -> n0 + H1 - 2,2246 [MeV] *

Li6 -> He3 + H3 - 15,794 [MeV]

Li6 -> He4 + H2 - 1,4738 [MeV] *

Li6 -> He5 + H1 - 4,5883 [MeV]

Li7 -> He4 + H3 - 2,4676 [MeV] *

Li7 -> He5 + H2 - 9,6148 [MeV]

Li7 -> He6 + H1 - 9,9740 [MeV]

Gamma emission energy for radioactive decay of chosen radioisotopes is[2]:

Na24: 1.368626 [MeV] @ 99.99 [%]

Na24: 2.754007 [MeV] @ 99.85 [%] *

Y88: 1.836063 [MeV] @ 99.20 [%]

Therefore, for mentioned radioisotopes and photodisintegration channels of H2, Li6 and Li7 marked above with asterisks, we will gain some release of kinetic energy[1][4]:

Na24: (gamma @ 2,754 [MeV]) + H2 -> n0 + H1 + 0,5294 [MeV]

Na24: (gamma @ 2,754 [MeV]) + Li6 -> He4 + H2 + 1,2802 [MeV] *

Y88: (gamma @ 1,836 [MeV]) + Li6 -> He4 + H2 + 0,3622 [MeV]

Na24: (gamma @ 2,754 [MeV]) + Li7 -> He4 + H3 + 0,2864 [MeV]

According to the law of conservation of energy and momentum we should expect energy distribution like this[4]:

Na24: (gamma @ 2,754 [MeV]) + H2 -> ( n0 @ 0,2646 [MeV]) + (H1 @ 0,2648 [MeV])

Na24: (gamma @ 2,754 [MeV]) + Li6 -> (He4 @ 0,4285 [MeV]) + (H2 @ 0,8516 [MeV]) *

Y88: (gamma @ 1,836 [MeV]) + Li6 -> (He4 @ 0,1212 [MeV]) + (H2 @ 0,2410 [MeV])

Na24: (gamma @ 2,754 [MeV]) + Li7 -> (He4 @ 0,1231 [MeV]) + (H3 @ 0,1633 [MeV])

H1-H2 and H2-H2 fusion requires to exceed a threshold energy of 0,389 [MeV]. When achieved, the reaction will release extra energy[1][3][5]:

(H2 @ 0,389 [MeV]) + H1 -> He3 + 5,4935 [MeV] *

(H2 @ 0,389 [MeV]) + H2 -> He4 + 23,8465 [MeV] *

Therefore a two stage reaction should lead to nuclear fusion of H1-H2 and H2-H2 between stationary H1 or H2 in Lithium Hydride crystals with H2 @ 0,8516 [MeV] released in photodisintegration reaction of Li6:

1) Na24: (gamma @ 2,754 [MeV]) + Li6 -> He4 + (H2 @ 0,8516 [MeV])

2a) (H2 @ 0,8516 [MeV]) + H1 -> He3 + 5,96 [MeV]

2b) (H2 @ 0,8516 [MeV]) + H2 -> He4 + 24,31 [MeV]

The reactions if correct, have a very small cross sections and will occur very rarely. However I am just curious if my understanding of above is correct.

References:

[1] http://nrv.jinr.ru/nrv/webnrv/qcalc/

[2] https://www-nds.iaea.org/relnsd/vcharthtml/VChartHTML.html

[3] R. Bass, Nuclear Reactions with heavy ions, Springer-Verlag, NY, 1980

[4] Photodisintegration of Lithium Isotopes, Ward Andrew Wurtz, 2010

[5] Hydrogen Properties for Fusion Energy, P. Clark Souers, 1985

Many Thanks,

Toreno