| (1) | 21D | + | 31T | → | 42He | ( | 3.5 MeV | ) | + | n0 | ( | 14.1 MeV | ) | ||||||
| (2) | 21D | + | 21D | → | 31T | ( | 1.01 MeV | ) | + | p+ | ( | 3.02 MeV | ) | 50% | |||||
| (3) | 21D | + | 21D | → | 32He | ( | 0.82 MeV | ) | + | n0 | ( | 2.45 MeV | ) | 50% | |||||
| (4) | 21D | + | 32He | → | 42He | ( | 3.6 MeV | ) | + | p+ | ( | 14.7 MeV | ) | ||||||
The following values are used to check the meaning of the terms in the above equations:
electron mass, kg: 9.10938E-31
neutron mass, kg: 1.67493E-27
proton mass, kg: 1.67262E-27
deuterium mass, kg: 3.34449E-27
tritium mass, kg: 5.00827E-27
eV in Joule: 6.24219725343321E+18
unit of mass eV 8.37185E+25
speed of light, m/sec: 299792458
helium 3 mass, kg: 5.00823E-27
helium 4 mass, kg: 6.64648E-27
Mass Balance of Equation (1)
| (1) | 21D | + | 31T | → | 42He | ( | 3.5 MeV | ) | + | n0 | ( | 14.1 MeV | ) |
= 3.13563E-29 kg
Energy difference, J = 3.13563E-29* sq(299792458) = 2.25318E-12 Joules
Energy difference, MeV = 2.25318E-12*6.24E+18/E+06= 17.6 MeV
In the above equation, it was assumed that the fused nucleus was at zero velocity such that:
0 = (mass of He-4)* (velocity of He-4) + (mass of neutron)* (velocity of neutron)
The above momentum conservation equation yields:
(velocity of neutron)/(velocity of He-4) = - (mass of He-4)/(mass of neutron)
Squaring and multiplying by (mass of neutron)/ (mass of He-4) gives"
(Energy of neutron)/(Energy of He-4) = (mass of He-4)/(mass of neutron)
= 6.64648E-27/1.67493E-27 = 3.968213597(Energy of neutron) = 17.6(3.968213597/(1+3.968213597))
= 14.0574792 MeV
(Energy of He-4) = 17.6/(1+3.968213597)
= 3.542520799 MeV
As we see, we did not include the kinetic energies
of the reactants D and T. We only assumed that the binding energy
is the net energy output. Recall that this reaction has peek fusion cross section
at kinetic energy 100 KeV. The proper way to write equation (1) is therefore:| 21D | + | 31T | + E(D-T) | → | 42He | ( | 3.5 MeV | ) | + | n0 | ( | 14.1 MeV | ) | + E(D-T) |
Where E(D-T) is the kinetic energy of the original projectiles D and T. This raises this question:
How to determine E(D-T) in the case of beam-based fusion when either of the energies of D and T projectiles could be manipulated precisely?
For example, in beam-based fusion, one could either make both D and T have equal energies in the lab reference or make D possess greater speed, such that the net energy of D and T remains the same but the proportion of energies of D to T is different. I the projectiles D and T collide with net zero velocity after fusion, then:
How would we retrieve E(D-T) that was spent in energizing the projectiles?
Let us leave this question for now and turn to the next equation.
Mass Balance of Equation (2)
From the outset we must conserve the electronic charge by adding an electron on the right hand side.
| (2) | 21D | + | 21D | + | E | (D-D) | → | 31T | ( | 1.01 MeV | ) | + | p+ | ( | 3.02 MeV | ) |
|
Mass difference = 2* 3.34449E-27-5.00827E-27- 1.67262E-27 - 9.10938E-31 = 7.17906E-30 kg
Energy difference, J = 7.17906E-30* sq(299792458) = 6.45222E-13 Joules
Energy difference, MeV = 6.45222E-13*6.24E+18/E+06= 4.03 MeV
(Energy of proton) = 1.010647969 MeV
(Energy of T) = 3.0219763 Mev
(Energy of electron) = 0.000550 MeV
Again, to determine E(D-D), in beam-based nuclear fusion, the kinetic energy for the projectiles D and D is subjected to manipulating the individual energies of each projectile.
Let us now modify Equations (3) and (4) so as to represent the real reactants:
| (3) | 21D | + | 21D | → | 32He | ( | 0.82 MeV | ) | + | n0 | ( | 2.45 MeV | ) + E(D-D) | ||
| (4) | 21D | + | 32He | → | 42He | ( | 3.6 MeV | ) | + | p+ | ( | 14.7 MeV |
|
Query
What is the proper proportion of energies, of deuterium and tritium projectiles, in the laboratory frame of reference, that could peak the fusion cross-section, in beam nuclear fusion?
(Hint: beam-based nuclear fusion refers to using beams of ions of D and or T to induce nuclear fusion. This differs from bulk plasma-based fusion where the temperatures of the two projectiles are dependent on the temperature of the bulk plasma.)
Mohamed F. El-Hewie
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