Vbm Fusion Reactor H – H Cycle
Abstract:
Injection of bunches of charged particles,If the bunches of charged particles of same species (protons) are injected to a point „F‟, where two magnetic fields
(perpendicular to each other) are applied ,the charged particles (the protons) of the first bunch will undergo to a confined circular path and will pass through this point “F“ (point of injection ) by time and again and thus the confined protons will be available for the protons of the later injected bunch (reaching at point “F“) to be fused with at point “F“.
Occurrence of fusion at point “F“
As the proton of later injected nth bunch reaches at point “F”, it fuses with the proton of the first injected bunch (that has already confined) passing through the point “F“.Confinement of the produced useful charged nucleus:At point “F“, the two protons fuse and form a compound nucleus. The compound nucleus splits and the deuteron (and the positron) is produced. The produced deuteron, due to applied magnetic fields, undergo to a circular orbit. The produced deuteron starts its circular motion from point “F“ (the point of production of deuteron) and pass through this common tangential magnetic field point “F” (or the point of production of nucleus) by time and again during its circular motion .Thus the produced druteron is confined and so the produced deuteron will be available at point “F“ (the point of injection) for the proton of later injected bunch (that is reaching at point “F“) to be fused with.Exhausting the produced non – useful charged nuclei:
The produced positron annihilates with free electron and produce two gamma ray photons that in turn heat the tokamak.
As the proton of later injected bunch reaches at point “F“, it fuses with the confined deuteron (passing through
the point “F“) and form the helium -3 nucleus.
The produced helium -3 nucleus starts its circular motion from point “F“ (0,0,0)or the point p1(x1 , y1, z1) and reaches at point p2 (x2 , y2 , z2) located on the circumference of the circle to be followed by the helium -3 nucleus. As the helium -3 nucleus reaches at point p2 (x2, y2 , z2), it enters into the mouth of the horse pipe that is located at the point p2 (x2 y2 z2) and thus the helium -3 nucleus is extracted out of the tokamak with the help of vacuum pump attached to the another end of horse pipe. Thus we can establish a steady state controlled nuclear fusion reactor based on H-H cycle.
PRINCIPLE: HOW FUSION OCCURS
Verdict: Various charged particles fuse to form a homogeneous compound nucleus. The homogeneous compound nucleus is unstable. So, the central group of quarks [that which with gluons and other groups of quarks compose the homogeneous compound nucleus] with its surrounding gluons to become a stable and the just lower nucleus [a nucleus having lesser number of groups of quarks and lesser mass (or gluons) than the homogeneous nucleus] than the the other nearby located groups of quarks with their surrounding gluons and rearrange to form the ‘A’ lobe of the heterogeneous compound nucleus. While the remaining groups of quarks [the groups of quarks that are not involved in the formation of the lobe ‘A’] to become a stable nucleus includes their surrounding gluons (or mass) [out of the available mass (or gluons) that is not involved in the formation of the lobe ‘A’] and rearrange to form the ‘B’ lobe of the heterogeneous compound nucleus. The remaining gluons [the gluons (or the mass) that are not involved in the International Journal of Innovative Research and Advanced Studies (IJIRAS) ISSN: 2394-4404
formation of any lobe] keeps both the lobes joined them together. Thus, due to formation of two lobes within into the homogeneous compound nucleus, the homogeneous compound nucleus transforms into the heterogeneous compound nucleus.
The heterogeneous compound nucleus, due to its instability, splits according to the lines parallel to the direction of the velocity of the compound nucleus ( ) into three lobes. Where the each separated lobe represent a separated particle. So, the two particles that represent the lobes ‘A’ and ‘B’ are stable while the third particle that represent the remaining gluons (or the reduced mass) is unstable. According to law of inertia each particle that has separated from the compound nucleus has an inherited velocity ( ) equal to the velocity of the compound nucleus ( ).
Principle: How to confine the desired charged particles Verdict: Various charged particles with different momentum by charge ratio when injected to a point ‘F’ where two uniform magnetic fields perpendicular to each other are applied the charged particles follow the confined circular paths of different radii passing though the common tangential magnetic field point ‘F’ (or the point of injection) by time and again.Where,
r α
Where, the radius of the circular orbit followed by the confined charged particle is directly proportional to the momentum by charge ratio.
Or
r = K
Where,
EK = Kinetic energy of the confined particle.
Fr = The resultant force (net force) acting on the charged particle due to the magnetic fields.
By how we can apply the principle:
A. INJECTION OF BUNCHES OF CHARGED PARTICLE
if the bunches of charged particles of same species (Protons) are injected to a point ‘F’ where the two magnetic fields are applied, the charged particles (the Protons) of the first bunch will undergo to a confined circular path and will pass through this point ‘F’ [point of injection] by time and again and thus the confined protons will be available for the protons of later injected bunch (reaching at point ‘F‘) to be fused with at point ‘F’.
OCCURRENCE OF FUSION AT POINT ‘F’
As the proton of later injected nth bunch reaches at point ‘F’, it fuses with the proton of the first injected bunch (that has already confined) passing through the point ‘F’.
C. CONFINEMENT OF THE PRODUCED USEFUL CHARGED NUCLEUS
At point ‘F’, the two protons fuse and form a compound nucleus. The compound nucleus splits and the deuteron (and the positron) is produced. The produced deuteron, due to applied magnetic fields, undergo to a circular orbit. The produced deuteron starts its circular motion from point ‘F’ [the point of production of deuteron] and pass through this common tangential magnetic field point ‘F’ [or the point of production of nucleus ] by time and again during its circular motion . Thus the produced deuteron is confined and so the produced deuteron will be available at point ‘F’ (point of injection) for the proton of later injected bunch (that is reaching at point ‘F’) to be fused with.
EXHAUSTING THE PRODUCED NON – USEFUL CHARGED NUCLEI
The produced positron annihilates with free electrons and produce two gamma ray photons that in turn heat the tokamak.
The produced helium -3 nucleus starts its circular motion from point ‘F’ (0,0,0) or the point p1 (x1 , y1 , z1) and reaches at point p2 (x2 , y2 , z2) located on the circumference of the circle to be followed by the helium-3 nucleus . As the helium-3 nucleus reaches at point p2 (x2 , y2 , z2) , it enters into the mouth of the horse pipe that is located at the point p2 (x2 , y2 , z2) and thus the helium-3 nucleus is extracted out of the tokamak with the help of vacuum pumps .Thus we can establish a steady state controlled nuclear fusion reactor based on H-H cycle.
I. ION SOURCE
Source: Ion source is a duoplasmatron that produce the 3 x 1019 Protons per second. The produced bunches of protons enters into a wideroe – Type RF linac.
Figure 1
PARTICLE ACCELERATOR
MINIMUM KINETIC ENERGY (Em) REQUIRED FOR FUSION
Tunneling – tunneling is a consequence of the Heisenberg uncertainty principle which states that we know the velocity of the particle the less we know about its position in the space and vice versa
The uncertainty in the position is such that hen a proton collides with another proton , it may find itself on the other side of the coulomb barrier and in the attractive potential well of the strong force .
Work done to overcome the coulomb barriers U = kz1z2q2 / r0 So, the kinetic energy of the particle should be equal to Em = ½ mv2 = kz1z2q2 / r0
Rewriting the kinetic energy of the particle in terms of momentum ½ mv2 = p2 / 2m = (h/λ) 2 / 2m
If we require that the nuclei must be closer than the de-broglie wavelength for tunneling to take over nuclei to fuse. (ro = λ)
kz1z2q2 / r0 = kz1z2q2 / λ
where,
½ mv2 = (h/λ) 2 / 2m = kz1z2q2 / λ
So, h2/λ22m = kz1z2q2 / λ
Or lambda (λ) = ½ h2/ kz1z2q2m
If we use this wavelength as the distance of closest approach, the kinetic energy required for fusion is –
Em = ½ mv2 = kz1z2q2 / r0 = kz1z2q2 / λ = kz1z2q2 x 2kz1z2q2m / h2
Em = 2k2 z12 z22 q4m / h2
eq(1)
Where m is the mass of the penetrating (injected) nucleus.
MINIMUM KINETIC ENERGY REQUIRED FOR PROTON – PROTON FUSION
Em = 2 K2 Z 1 2 Z 2 2q4m from eq.(1)
2
h
Z1 = Z2 = 1
= mp = 1.6726 x 10-27 kg
EP-P = 2 X (9X109)2 X 12 X 12 X (1.6 X 10-19)4 X 1.6726 X 10-27
J 10-34)2
(6.62 X
EP-P = 1775.77132 X 1018 X 10-76 X 10-27 J
43.8244 X 10-68
EP-P = 40.52015133 X 10-17 J(1ev = 1.6 x 10-19 J)
EP-P = 25.3250 X 102 ev
EP-P = 2.5 kev eq.(2)
EP-P = 0.0025 Mev (1ev = 1.6 x 10-19 J)
C. MINIMUM KINETIC ENERGY REQUIRED FOR PROTON –DEUTERON FUSIONvo = vmax = 40 KV
Ttr = sinψ0 = 0.64
K4 = 4x 1.6 x 10-19 x 40x 0.64 KJ
K4 = 102.4 Kev = 0.1024 Mev eq.(4)
Length of the wideroe – type RF linac
A. LENGTH OF THE FIRST DRIFT TUBE
L1 = n½ x /
frf
Where, frf = 7 x 106 HZ , n = 1
L1 = x / -27 m
L1 = x /
L1= X
L1 = 1.5807 X 10-1
L1 = m
L1 = 0.1580 m eq.(5)
L2 = x L1
L2 = 1.4142 x 0.1580 m
L2 = 0.2234 m eq.(6)
L3 = x L1
= 1.732 x 0.1580
= 0.2736 m eq.(7)
L4 = x L1
= 2 x 0.1580 M
= 0.316 m eq.(8)
L = L1 + L2 + L3 + L4
L = 0.158+0.2234+0.2736+0.316 m From eq.(5,6,7,8)
L = 0.971 m
CONCLUSION: A wideroe – type RF linac accelerates the each proton with 102.4 kev enrgy.
NOTE: A wideroe - type RF linac is attached to the ion source [see fig. (1)]
IV. THE TOKAMAK
The takamak has two parts – one is the main tokamak and the another is the extended tokamak.
The points A, B , C , D , P , Q , R , and S represents the corners of the walls of the main tokamak while all the other remaining points represents the corners of the walls of the extended tokmak.
The tokmak is made up of steel.
The graphite of the boron is used as the inner liner of the tokamak to absorb the thermal neutrons. see fig . (2)
The main tokamak with its extension
EP-D = EP-P X Z22
Z2 = 1
EP-D = 2.5 X 12 kev
EP-D = 2.5 kev eq.(3)
EP-D = 0.0025 mev
D. FOR A WIDEROE – TYPE RF LINAC
Kn = n q vo Ttr
If parameters are Extended walls of the tokamak The Main tokamak
n = 4, q = 1.6 x 10-19 c Figure 2
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International Journal of Innovative Research and Advanced Studies (IJIRAS) ISSN: 2394-4404
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The points A, B, C, D, P, Q, R and S make the main Figure 3
tokamak while all the other points make the arm of the Here MF = 0.80 m and LF = 0.80 m
main tokamak (or the extension of the tokamak). AM = 0.80 m and MD = 0.80 m
Where, PL = 0.80 m and LS = 0.80 m
AB = BC =CD = DA = AP =PQ = QB = QR= RS = SP AP = AB = AD = =1.6 m
= SD =RC = 1.6 m Magnetic field coils
And also VBM fusion reactor has two pairs of semicircular
The length of each extended wall = 1.6 m magnetic field coils. out of them, one pair of semicircular
The breadth of the extended wall = 1.6 m magnetic field coils is vertically erected while another pair of
Total surface area of the tokamak semicircular magnetic field coils is horizontally lying.
1 Vertically erected magnetic field coils:
A. SURFACE AREA OF THE WALLS OF THE MAIN In a VBM fusion reactor, there are two vertically erected
TOKAMAK semicircular magnetic field coils that act as a helmholtz coil.
The distance between the two vertically erected
SURFACE AREA OF THE WALLS semicircular coils is equal to the radius of any one of the
m2 semicircular magnetic field coil.
ABCD = length x breadth = 1.6m x 1.6m = 2.56 i.e. d = r = 2.5 m
PQRS = 1.6m x 1.6m = 2.56 m2 The vertically erected semicircular magnetic field coils
APQB = 1.6m x 1.6m = 2.56 m2 acting as a helmholtz coil produce a uniform magnetic field ()
DSRC = 1.6m x 1.6m = 2.56 m2 parallel to y – axis. [see fig . (4)]
BQRC = 1.6m x 1.6m = 2.56 m2
So, the total surface area of the main tokamak = 12.80 m2 B. HORIZONTALLY LYING MAGNETIC FIELD COILS
eq.(9)
The points APSD do not represent a wall. It is a blank In a VBM fusion reactor, there are two horizontally lying
place that allows the injected protons to enter into the main
semicircular magnetic field cols that acts as a helmholtz coil.
tokamak. (or the region where the magnetic fields are applied.
The distance between the two horizontally lying
SURFACE AREA OF THE EXTENDED WALLS OF semicircular magnetic field coils is equal to the radius of any
one of the semicircular magnetic field coil.
THE TOKAMAK
i.e. d = r = 2.2 m
The surface area of the each extended wall The horizontally lying semicircular magnetic field coils
acting as a helmholtz coil produce a uniform magnetic field (
= 1.6m x 1.6m =2.56 m2
) parallel to z –axis . [see fig . (5)]
Total no of extended walls = 5
Magnetic field due to a semicircular coil at point x is –
Total surface area of the extended walls
B1 = µ o /4π x πR2ni / (R2 + x2) 3/2
= surface area of the extended wall x total no of extended Magnetic fields due to a semicircular coil at the x, If x =
walls
R/2
= 2.56 m2 x 5 B1 = µ o /4π x 2 ni / (R 2 2 / 4) 3/2 [ x = R / 2 ]
2 eq.(10) πR + R
= 12.80 m = 8 / 5 x µ o ni / 4R eq (11)
Total surface area of the tokamak = surface area of the
So, the magnetic field in the mid plane of the two
main tokamak + surface area of the extended walls
semicircular coils acting as a helmholtz coil is
= 12.80 m2 +12.80 m2
BT = B1 + B2
from eq. (9, and 10) respectively
= 2 B [B1 = B2 = B]
= 25.6 m2 from eq. (11)
The location of the point of injection ‘F‘ [or the center of = 16 / 5 x µ o ni / 4R
= 1.43 x µ o ni / 4R
fusion ‘ F’]
= 1.43 Bcenter [Bcenter = µ o ni / 4R] eq.(12)
point of injection is the point ‘F‘ through which the
Magnetic field (Bz) in the mid plane of the two
injected charged particles enters into the main tokamak which
horizontally lying semicircular coils acting as a helmholtz
in turn is covered up by the two magnetic fields perpendicular
coil is
to each other.
Bz = 1.43 B from eq.(12)
Centre
B = µ 0 ni
Centre 4 R
Where, µ 0 = 4 π x 10-7
n = 490 turns
i = 10 KA = 104 Amperes
R = 2.2 m
Bcentre = 4 x 3.14 x 10-7 x490 x 104 Tesla
4 x 2.2
Bcentre = 0.69936 Tesla
BZ = 1.43 Bcentre
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BZ = 1.43 X 0.69936 Tesla
= 1 Tesla eq.(13)
see fig. (4)
The horizontally lying semicircular magnetic field coils acting as a helmholtz coil.
Horizontally lying semicircular coil
Figure 4
The magnetic field coils are exterior to the main tokamak so, the area covered up by points 1, 2, 3 ,4 , 5, 6, 7 and 8 is greater than the area covered up by the points A, B, C,
D, P, Q, R and S of the main tokamak.
Magnetic field (BY) in the mid plane of the two vertically erected semicircular coils acting as a helmholtz coil is –
By = 1.43 Bcentre from eq.(12)
Bcentre = µ0 nI
4 R
Where,
n = 557 turns
i = 10 KA = 104 Amperes
R = 2.5 m
Bcentre = 4 x 3.14 x 10-7 x 557 x 104 Tesla
4 x 2.5
Bcentre = 699.592 x 10-3 tesla
= 0.699592 Tesla
By = 1.43 Bcentre
= 1.43 x 0.699592 Tesla
By = 1 Tesla eq.(14)
see fig.(5)
The vertically erected semicircular magnetic field coils that act as a helmhotz coil
The vertically erected magnetic field coils are exterior to the horizontally lying semicircular magnetic field coils which are in turn exterior to the main tokamak.
The directions of magnetic fields
The direction of flow of current in the horizontally lying semicircular coils is clockwise so that the direction of the
produced magnetic field is according to negative z – axis (i. e. downward)
As Bz = 1 Tesla from eq.(13)
So = -1 Tesla eq.(15)
see fig (6)
The direction of flow of current in the vertically erected magnetic coils is anticlockwise so that the direction of the produced magnetic field is according to positive y –
axis.
As By = 1 Tesla from eq.(14)
So = 1 Tesla eq.(16)
see fig.(6)
The direction of flow of current in the magnetic field coils.
Figure 6
In the horizontally lying coils the current flows in clockwise direction while in the vertically erected coils the current (i) flows in anticlockwise direction.
Magnetic fields within into the main tokamak:
We have denoted the presence of magnetic fields by the cross [x] sign.
vertically erected semicircular coil
Figure 5 Figure 7
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The uniform magnetic fields [and] are applied within into the main tokamak only.
Within into the ‘extension of the tokamak ‘there the no any magnetic field is applied. so, that until the injected proton reaches at point ‘ F ‘ [located within into the ‘ APSD ‘ area of the main tokamak] the injected proton is
not influenced by the magnetic lines of force.
The direction of the applied magnetic fields [and] in the main tokamak.
Nature of center of fusion (F)
As the magnetic field is tangential in nature so the point ‘F‘ [the center of fusion] that is located within into the magnetic fields is a tangential point of a number of circular orbits (followed or to be followed by the charged particles) of different radii.
see fig (10)
The circular orbits followed by the confined particles –
The confined proton and the confined deuteron passes through the tangential magnetic field point ‘F‘.
Figure 8
Where
= 1 Tesla eq.(16)
= -1 Tesla from eq.(15)
Where, and are perpendicular to each other
V. CENTER OF FUSION
Center of fusion is actually a point where two charged particles fuse.
For the VBM fusion reactor – The center of fusion is a point from where a charged particle (either it is injected or produced) undergoes to a confined circular path and passes from this point by time and again and thus available for another injected particle (reaching at this point ‘F‘) for fusion.
The location of center of fusion [or the point of injection]: The center of fusion is a point located within into the APSD area of the main tokamak where the two magnetic fields perpendicular to each other are applied and where the charged particle is injected to.
Within into the tokamak, the center of fusion is the first point from where the injected charged particle experiences magnetic lines of force and starts its circular motion. Thus the center of fusion and the point of injection of charged particle are the same – the point ‘F‘
see fig.(3)
Number of centers of fusion: The point ‘F‘
[From where a proton starts it’s circular motion] is acting
as a center of fusion.
so, the total number of centers of fusion is equal to the number of protons that contains the bunch.
Figure 9
The outer circular path denotes the circular orbit followed by the confined deuteron
The inner circular path represents the circular orbit followed by the confined proton.
Both the circular orbits lies in the plane made up of positive x – axis, negative y-axis and the negative z
– axis.
Center of fusion (F) is a platform where the fusion is a certainty:
Form the point ‘F‘ [The center of fusion] the proton (s)
of earlier bunch will undergo to confined circular path and will pass through this point ‘F‘ by time and again until it fuses with the proton of later injected bunch.
Similarly, the center of fusion [the point ‘F‘] also governs the produced charged particle (the deuteron) to undergo to a confined circular orbit and pass through this point ‘F‘ by time and again and thus tends the produced deuteron to be fused with the proton of later injected bunch.
Thus, the center of fusion [the point ‘F‘] avails us a platform where the fusion is a certainty.
Or, within into the tokamak, the center of fusion [the point ‘F‘] is the only and only point where the fusion reactions occur.
see fig.(9,10)
Center of fusion in the view of magnetic fields
By the view of magnetic fields the center of fusion is a
point where the two uniform magnetic fields [and] are perpendicular. But within into the region covered – up by the main tokamak, at each and every point the ratio of two perpendicular magnetic fields [and] is constant. so, the each
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and every point within into the region covered – up by the A. INJECTION OF THE PROTON
main tokamak can act as a center of fusion .
VI. CENTER OF PLASMA
The center of plasma [Cpm] is the center of the circular orbit followed by the charged particle.
Thus the center of plasma [Cpm] differs particle by particle.
But the center of fusion [the point ‘F‘] is a common point that is located on the circumferences of all the circular orbits of different radii followed by the charged particles.
see fig (10)
Center of fusion and the center of plasma
The center of fusion [the point ‘F‘] is a common tangential magnetic field point of a number of circular orbits followed by the confined particles.
While the center of plasma is the center of the circular orbit followed by the particle.
Figure 11
The injected proton reaches at point ‘F‘
The velocity of the proton make angle 30o with x – axis, 60o angle with y-axis and the 90o angle with z – axis.
B. VELOCITY OF THE PROTON
Each proton is injected into the tokamak with the kinetic energy equal to 0.1024 Mev. so, the velocity of the proton
V = [ 2E Mp]1/2
E = 0.1024x1.6x10-13 J from eq.(4)
mp = 1.6726x10-27 Kg
= 2x0.1024x1.6x10-13 J 1/2
1.6726x10-27 kg
Figure 10
Where,
CH = Center of the circular orbit followed by the proton. Cd = Center of the circular orbit followed by the
deuteron.
Che-3 = Center of the circular orbit followed by the helium
– 3 nucleus.
F= A tangential point of a number of circular orbits followed by the confined particles or to be followed by the undesired particles (he – 3 ash).
VBM plasma: RF linac injects the bunches of the protons into the tokamak at point F. such that each proton makes angle 300 with the x –axis, 600 angle with the y-axis and the 900 angle with the z-axis. RF linac injects each proton with 102.4 kev energy.
Confinement of protons of 1st bunch:
As the proton (s) of first bunch reaches at point F into the tokamak, it experiences a centripetal force due to magnetic fields and hence it follows a confined circular orbit passing through the point of injection (F) by time and again.
see fig (11,12)
0.32768x1014 ½ m/s
1.6726
[0.19591055841 x 1014] ½ m/s
= 0.4426 x 107 m/s eq.(17)
C. COMPONENTS OF THE VELOCITY OF THE PROTON WITH WHICH IT ENTERS INTO THE TOKAMAK AND REACHES AT POINT
As each proton is injected into the tokamak making angle 30o with x – axis , 60o angle with y-axis and 90o angle with z
– axis.
So,
1 = V cos α
V = 0.4426 x 107 m/s from eq.(17)
Cos α = cos 30o = 0.866
= 0.4426 X107 X 0.866 m/s
= 0.3832 x 107 m/s eq.(18)
= V cosβ
cos β = cos 60o = 0.5
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= 0.4426 X107 X 1/2 m/s
= 0.2213 x 107 m/s eq.(19)
3 = V cos y
cos y = cos 90o = 0
= V x 0 m/s
=0 m/s eq.(20)
MOMENTUM OF THE PROTON WITH WHICH THE PROTON IS INJECTED INTO THE TOKAMAK AND REACHES AT POINT F
= -0.61312 x 10-12 N eq.(26)
3. Fx = q Vy Bz sin θ
= 0.2213 x 107 from eq.(19)
= -1 Tesla
sin θ = sin 90o = 1
Fx = 1.6 x 10-19 x 0.2213 x 107 x 1 x 1 N
=0.35408 x 10-12 N
Form the right hand palm rule, the direction of the force is according to positive x axis, so,
= 0.35408 x 10-12 N eq.(27)
see fig.(12)
P = mv
v = 0.4426 x 107 m/s from eq.(17)
m = 1.6726 x 10-27 kg
P = 1.6726 x 10-27 x 0.4426 x 107 kg m/s
P = 0.7402 x 10-20 kg m/s eq.(21)
E. COMPONENT OF THE MOMENTUM OF THE INJECTED PROTON WITH WHICH THE PROTON REACHES AT POINT F
As each proton is injected into the tokamak making angle
30o
With x-axis, 60o angle with y-axis and 90o angle with z-axis.
So,
1. = P cos α
p= 0.7402 x 10-20 kg m/s from eq.(21)
Cos α = cos 30o = /2 = 0.866
= 0.7402 X10-20 X 0.866 kgm/s
= 0.6410 x 10-20 kgm/s eq.(22)
2. = P cos β
cos β = cos 60o = 0.5
= 0.7402 X10-20 X 1/2 kgm/s
= 0.3701 x 10-20 kgm/s eq (23)
= P cos y
cos y = cos 90o = 0
= P x 0 kg m/s
=0 kgm/s eq.(24)
FORCES ACTING ON THE PARTICLE – PROTON
(WHEN THE PROTON IS AT POINT ‘F ‘)
1. Fy = q Vx Bz sin θ
= 0.3832 x 107 m/s from eq.(18)
= -1 Tesla from eq.(15)
sin θ = sin 90o = 1
q = 1.6 x 10-19 c
Fy = 1.6 x 10-19 x 0.3832 x 107 x 1 x 1
= 0.61312 x 10-12 N
Form the right hand palm rule, the direction of the force
is according to negative y-axis, so,
= -0.61312 x 10-12 N eq.(25)
2. Fz = q Vx By sin θ
= 1 Tesla from eq.(16)
sin θ = sin 90o = 1
Fz = 1.6 x 10-19 x 0.3832 x 107 x 1 x 1 N
=0.61312 x 10-12 N
Form the right hand palm rule, the direction of the force is according to negative Z- axis, so,
Forces acting on the particle – proton (when the proton is at point ‘F‘)
Figure 12
where the point ‘ F ‘ is the point where all the three axes
intersect each other
4. Resultant force (Fr) acting on the proton:
FR2 = Fx2 + Fy2 + Fz2
Fz = Fy = F = 0.61312 x 10-12 from eq.(25,26)
Fx = 0.35408 x 10-12 N from eq.(27)
FR 2 = 2F2 + Fx2
FR 2 = 2(0.61312 x 10-12 ) + (0.35408 x 10-12 ) 2 N2
FR 2= 2 x0.3759161344 x10-24+0.1253726464 x10-24 N2
= 0.7518322688 + 0.125372646 N2
FR2 = 0.7518322688 x 10-24+ 0.1253726464 x10-24 N2
FR2 = 0.8772049152 x 10-24 N2
-12 Neq.(28)FR=0.9365x10
see fig.(12)
RADIUS OF THE CIRCULAR ORBIT FOLLOWED BY THE PROTON
r = E = ½ mv2 = 0.1024x 1.6x10-13 J from eq.(4)
mv2 = 0.32768 x 10-13 J
Fr = 0.9365 x 10-12 N from eq.(28)
r = 0.32768 x10-13 J
0.9365 x 10-12 N
r = 0.3498 x 10-1 m
= 3.498 cm = 3.498 x10-2m eq.(29) see fig.(13)
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H. TIME PERIOD (T) OF THE PROTON
T=2π/V
r = 3.498 x 10-2 m from eq.(29)
V = 0.4426 x 107 m/ s from eq.(17)
= 2 x 3.14 x 3.498 x 10-2 m
0.4426 x 107 m/s
= 21.96744 x 10-9 s = 49.6327 x 10-9 s
0.4426
= 4.96 x 10-8 s eq.(30)
CONCLUSION: the confined proton passes the point ‘F’
by the after each 4.96 x 10-8 second.
CONFINEMENT OF THE PROTON: THE CIRCULAR ORBIT FOLLOWED BY THE CONFINED PROTON
The circular orbit followed by the confined proton Figure 13
By seeing the directions of forces[ , ] acting on the proton [ when the proton is at point ‘ F ‘], we reach at the conclusion that the circular orbit followed by the confined proton lies in the plane made up of positive x – axis , negative y – axis and the negative z – axis.
Cp = center of the circular path followed by the proton
‘F‘ is the center of fusion or the point where the proton is injected to. Here ‘F‘ is the point where all the three axes intersect each other.
The line segment is the radius of the circular orbit
followed by the confined proton and is equal to 3.498 x 10-2 m.
= the resultant force acting on the particle when the particle is at point F.
It is the resultant force ( ) By virtue of which the particle starts its circular motion from point F and undergo to a confined circular path
Angles that make the resultant force () [acting on the proton when the proton is at point ‘F‘] with positive x, y and z-axis.
with x –axis
Cos α = FR cos α / FR = / FR
= 0.35408 x 10-12 N from eq.(27)
FR = 0.9365 x 10-12 N from eq.(28)
= 0.35408 x 10-12 N
0.9365 x 10-12 N
Cos α = 0.3780 eq.(31)
α = 67.8 degree [cos ( 67.8 ) = 0.3778 ]
2. With y – axis
Cosβ = FR cos β / FR = / FR
= -0.61312 x 10-12 N from eq.(25)
- 0.61312 x 10-12 N
0.9365 x 10-12 N
Cos β = -0.6546 eq.(32)
β=130.8 degree [cos (130.8)= -0.6534 ]
with z – axis
Cos y = FR cos y / FR = / FR
= -0.61312 x 10-12 N from eq.(26)
= -0.61312 x 10-12 N
0.9365 x 10-12 N
Cos y = -0.6546 eq.(33)
y =130.8 degree[cos(130.8) = -0.6534 ] see fig (14)
Figure 14
Angle that make the resultant force () with respect to positive x, y and z axis, when particle – proton is at point F.
Where
˜ 67.8 Degree
˜ 130.8 Degree y ˜ 130.8 Degree
The direction cosines of the line P1 P2
The line P1 P2 is the diameter of the circle followed (or to be followed) by the particle.
The points P1 (x1 y1 z1) and P2 (x2 y2 z2) make the line P1 P2.
The particle starts its circular motion from the point ‘F’ (-the center of fusion where the particle is either injected or produced).
So, we have denoted the Cartesian coordinates for the
Point ‘F’ as (0, 0, 0).
Here the point F (0, 0, 0) and the point P1 (x1 y1 z1) are the same.
So, the direction cosines of the line P1 P2 are:
l = cos α = x2 – x1 / d where,
d = 2 x radius of the circle
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cos α = cos component of the angle that make the resultant force ( ) [acting on the particle when the particle is at point F] with the positive x - axis.
m = cos β = y2 –y1 / d Where,
cos β = cos component of the angle that make the
resultant force ( ) [ acting on the particle when the particle is at point F ] with the positive y - axis .
n = cos y = z2 –z1 / d Where,
cos y = cos component of the angle that make the
resultant force ( ) [acting on the particle when the particle is at point F ] with the positive z - axis.
THE CARTESIAN COORDINATES OF THE POINTS P1 ( X1 , Y1 , Z1 ) AND P2 ( X2 , Y2 , Z2 ) LOCATED ON THE CIRCUMFERENCE OF THE CIRCLE OBTAINED BY THE PROTON
1. cos α = x2 - x1
d
d = 2 x r
r = 3.498x10-2 m from eq.(29)
= 2x3.498x10-2 m
= 6.996 x 10-2 m
Cos α = 0.37 from eq.(31)
x2 - x1 = d x cos α
x2 - x1 = 6.996x10-2x0.37 m
x2 - x1 = 2.5885 x 10-2 m
x2 = 2.5885 x 10-2 m[x1 = 0 ] eq(34)
cos β = y2 - y1 d
cos β = -0.65 from eq.(32)
y2 - y1 = d x cos β
y2 - y1 = 6.996x10-2x (-0.65) m
y2 - y1 = -4.5474 x 10-2 m
y2 = -4.5474 x 10-2 m [y1 , = 0 ] eq(35)
cos y = z2 - z1 d
cos y = -0.65 from eq.(33)
z2 - z1 = d x cos y
z2 - z1 = 6.996x10-2x (-0.65) m
z2 - z1 = -4.5474 x 10-2 m
z2 = -4.5474 x 10-2 m[z1 , = 0] eq(36)
The point ‘F‘ and the point ‘P1‘ are same. from where each particle starts its circular motion. so, the coordinates of point p1 (x1 , y1 , z1) are equal to P1 (0 , 0, 0).
see fig.(15)
The cartesian coordinates of the point p1 (x1, y1, z1) located on the circumference of the circle obtained by the proton.:
The line segment ___ is the diameter of the circle followed by the confined proton P1P2
Figure 15
CONCLUSION: the proton is confined. The proton starts its circular motion from point p1 [or the point ‘ F ‘ (0 , 0 , 0)] and reaches at point p2 (x2 , y2 , z2) and then again reaches at point p1 (x1 , y1, z1) to complete the circle of radius 0.03498 m. the proton keeps on following the confined circular orbit uninterruptedly and pass through this point ‘F’ [or the point p1 (0, 0,0) by time and again until it fuses with the proton of the later injected bunch reaching at point ‘F‘.
The confined proton fuses with the injected proton at only and only point ‘F‘ and form the compound nucleus at point ‘F‘.
VII. FUSION REACTIONS
1. Proton - proton fusion
11H + 11 H 12 D+ Ѵe + e+
[injected ] [ confined ] [ confined ]
2. Annihilation of positron and an electron
e+ + e- у + у rays
3. Proton – deuteron fusion
11H + 2 1 D 23 He + y rays
[injected ] [ confined ] [ not confined ]
4. Proton – deuteron – proton fusion
11H + 21 D + 11H[ 43 Li ] 32 He + Δm + 11H
[injected] [confined] [confined] [not confined] [not confined]
HOW FUSION OCCURS
FORMATION OF COMPOUND NUCLEUS
As the proton of Nth bunch reaches at point ‘F‘, it fuses with the confined proton [the proton of first bunch that has already confined and passing through the point ‘F‘] to form a compound nucleus.
THE SPLITTING OF COMPOUND NUCLEUS
The compound nucleus splits into three particles. out of three particles, two are stable nuclei while the third one (the
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reduced mass) is unstable. By the law of inertia, each particle that has separated from the compound nucleus has an inherited velocity( ) equal to the velocity of the compound nucleus( )
PROPULSION OF THE PARTICLES
The reduced mass converts into energy and the total energy (ET) is carried away by the neutrino. The neutrino do not interact with any particle. So, neither the deuteron nor the positron is propelled.
FOR PROTON – PROTON FUSION REACTION
11H + 11 H 12 D + Ѵe + e+
Formation of compound nucleus [12 D]:
INTERACTION OF NUCLEI
The injected proton as reaches at point F, it interacts [experiences a repulsive force due to the confined proton passing through the point F] with the confined proton at point F. The injected proton overcomes the electrostatic repulsive force and – a like two solid spheres join - the injected proton dissimilarly joins with the confined proton.
See fig.(16)
NOTE: [12 D] is a compound nucleus formed due to proton – proton fusion.
Interaction Of Nuclei
Interaction of nuclei-1 Interaction of nuclei-2
Figure 16
b. FORMATION OF THE HOMOGENEOUS
COMPOUND NUCLEUS
The constituents (quarks and gluons) of the dissimilarly joined nuclei (Protons) behave like a liquid and form a homogeneous compound nucleus. Having similarly
distributed groups of quarks with similarly distributed surrounding gluons.
Thus within the homogeneous compound nucleus – each group of quarks is surrounded by the gluons in equal proportion. So, within the homogeneous compound nucleus there are 2 groups of quarks surrounded by the gluons.
See fig.(17)
Homogeneous Compound Nucleus
Figure 17
where, = velocity of the compound nucleus
α = 300 from eq.(53)
β = 600 from eq.(54)
y = 900 from eq.(55)
FORMATION OF LOBES WITHIN INTO THE HOMOGENEOUS COMPOUND NUCLEUS OR
THE TRANSFORMATION OF THE HOMOGENOUS COMPOUND NUCLEUS INTO THE HETEROGENEOUS COMPOUND NUCLEUS
Emitting a positive charge, an up quark (u) of a group of quarks (uud) converts into a down quark (d). Thus the group of quarks (uud) that compose the proton converts into a group of quarks (udd) that compose the neutron.
The converted group of quarks (udd) with its surrounding gluons to become a stable and the next higher nucleus (deuteron) than the reactant one (the Proton) includes the another group of quarks (uud) with its surrounding gluons and rearrange to form the ‘A‘ lobe of the heterogeneous compound nucleus.
While, on the other hand, the emitted positive charge to become a stable nucleus (positron) includes its surrounding gluons or mass [out of the available mass (or gluons) that is not included in the formation of the lobe ‘A‘] and rearrange to form the ‘B‘ lobe of the heterogeneous compound nucleus .
Thus, due to formation of two dissimilar lobes within into the homogeneous compound nucleus, the homogeneous compound nucleus transforms into the heterogeneous compound nucleus.
See fig.(18,19,20)
The transformation of the homogenous compound nucleus into the heterogeneous compound nucleus:
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Formation Of Lobes -1 FORMATION OF LOBES
Figure 18
Heterogeneous compound nucleus – An up quark (u) emits a positive charge and converts into a down quark (d).
The transformation of the homogenous compound nucleus into the heterogeneous compound nucleus:
Formation Of Lobes -2
Figure 19
The axis along which the groups of quarks are arranged to The emitted positive
Charge reaches on the other end of the nucleus.
The transformation of the homogenous compound nucleus into the heterogeneous compound nucleus:
Formation Of Lobes - 3
Within into the homogeneous compound nucleus the greater nucleus is the deuteron and the smaller one is the
positron while the remaining space represents the remaining gluons
The greater nucleus is the lobe ‘A ‘while the smaller nucleus is the lobe ‘B’.
FINAL STAGE OF THE HETEROGENEOUS COMPOUND NUCLEUS
The process of formation of lobes creates void (s) between the lobes. So, the remaining gluons [the gluons (or the mass) that are not involved in the formation of any lobe] rearrange to fill the void (s) between the lobes. Thus the remaining gluons form a node between the dissimilar lobes of the heterogeneous compound nucleus.
Thus, the reduced mass (or the remaining gluons) keeps both the dissimilar lobes of the heterogeneous compound nucleus joined them together.
So, finally, the heterogeneous compound nucleus becomes like a dumb – bell.
See fig.(21,22)
Final stage of the heterogeneous compound nucleus:
Final Stage - 1
Figure 21
Final stage of the heterogeneous compound nucleus:
Final Stage - 2
Figure 22
Figure 20
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Calculations for the compound nucleus [21 D]:
JUST BEFORE FUSION, THE LOSS IN THE KINETIC ENERGY OF THE INJECTED PROTON
As the injected proton reaches at point ‘F’, the injected proton fuses with the confined proton (passing through the point ‘F’) to form the compound nucleus, at point ‘F’.
Just before fusion, to overcome the electrostatic repulsive force, the injected proton loses its energy equal to 2.5 Kev. [from eq.(2)]
So, the kinetic energy of the proton just before fusion is -
COMPONENTS OF THE MOMENTUM OF THE INJECTED PROTON JUST BEFORE FUSION ARE
= Pb cos α = mvb cos α m = 1.6726 x 10-27 kg
Vb cos α = 0.3785 x 107 m/s from eq.(38)
= 1.6726 x 10-27 x 0.3785 x 107 kgm/s
= 0.6330 x 10-20 kgm/s eq.(41)
2 = Pb cos β = mvb cos β
Vb cos β = 0.2185 x 107 m/s from eq.(39)
= 1.6726 x 10-27 x 0.2185 x 107 kgm/s
= 0.3654 x 10-20 kgm/s eq.(42)
Eb = [EP - Eloss]
EP = 102.4 kev from eq.(4)
Eloss = EP-P = 2.5 kev from eq.(2)
Eb = [102.4 -2.5] kev
99.9 kev
0.0999 Mev
VELOCITY OF THE INJECTED PROTON JUST BEFORE FUSION
½
Vb = 2E b
Mp
Eb = 0.0999 Mev
= 0.0999x1.6x10-13 J
mp =1.6726x10-27 kg
= 2x0.0999x1.6x10-13 J ½
1.6726x10-27 kg
= 0.31968x10-13 ½ m/s
1.6726x10-27
= [0.19112758579x1014] ½ m/s
= 0.4371 x 107 m/s eq.(37)
COMPONENTS OF THE VELOCITY OF THE INJECTED PROTON JUST BEFORE FUSION ARE
1 = Vb cos α
Vb = 0.4371 x 107 m/s from eq.(37)
Cosα = cos 30o =/2 = 0.866
= 0.4371 x 107 x 0.866 m/s
=0.3785 x 107 m/s eq.(38)
2 = Vb cos β
Cos β = cos 60o = 0.5
= 0.4371 x 107 x 0.5 m/s
= 0.2185 x 107 m/s eq.(39)
3 = Vb cos y
Cos y = cos 90o = 0
= Vb x 0 m/s
= 0 m/s eq.(40)
3 = Pb cos y = mvb cos y
Vb cos y = 0 m/s from eq.(40)
= 1.6726 x 10-27 x 0 kgm/s
= 0 kgm/s eq.(43)
COMPONENTS OF THE MOMENTUM OF THE COMPOUND NUCLEUS (PCN)
The injected proton penetrates the confined proton. So, Just before fusion, there is a loss in kinetic energy of the injected proton but the kinetic energy of the confined proton remains same. So, the momentum of the confined proton remains same as with which it was injected to.
X – component of the momentum of the compound nucleus ( ) =
X – component of X – component of the
the momentum momentum of the
of the confined proton + injected
at point F Proton just before fusion
at point F
= [0.6410 x 10-20] + [0.6330 x 10-20] kgm/s from eq.(22 and 41) respect.
= 1.274 x 10-20 kgm/s
• = PCN cos α = 1.274 x 10-20 kgm/s eq.(44)
y– component of the momentum of the compound nucleus ( ) =
y – component of y – component of the
the momentum momentum of the
of the confined + injected
proton Proton just before
at point’ F ‘ fusion
at point ‘F’
• = [0.3701 x 10-20] + [0.3654 x 10-20] kgm/s from
eq.(23 and 42)
= 0.7355 x 10-20 kgm/s
= PCN cos β = 0.7355 x 10-20 kgm/s eq.(45)
z – component of the momentum of the compound nucleus ( )=
z – component of z – component of the
the momentum momentum of the
of the confined injected
proton Proton just before
at point F fusion
at point F
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= [ 0 ] + [ 0 ] kgm/s from eq.(24 and 43) respectively
0 kgm/s
• = PCN cos y = 0 kgm/s eq. (46)
MASS OF THE COMPOUND NUCLEUS (M)
M = 2mp
= 2 x 1.6726 x 10-27 kg
= 3.3452 x 10-27 kg eq.(47)
COMPONENTS OF THE VELOCITY OF THE COMPOUND NUCLEUS
1 = VCN cos α = /M = P CN cos α
M
Px = PCN cos α = 1.274 x 10-20 kgm/s from eq. (44)
M = 3.3452 x 10-27 kg from eq.(47)
= VCN cos α = 1.274 x 10-20 kgm/s 3.3452 x 10-27 kg
= VCN cos α = 0.3808 x 107 m/s eq.(48)
2 = VCN cos β = = /M = P CN cos B
M
Py = PCN cos β = 0.7355 x 10-20 kgm/s from eq.(45)
= VCN cos β = 0.7355 x 10-20 kgm/s
3.3452 x 10-27 kg
= VCN cos β = 0.2198 x 107 m/s eq.(49)
3 = VCN cos y = = /M = P CN cos y
M
Pz = PCN cos y = 0 kgm/s from eq.(46)
= VCN cos y = 0 kgm/s M kg
= VCN cos y = 0 m/s eq.(50)
h. VELOCITY OF THE COMPOUND NUCLEUS
(VCN)
VCN2 = Vx2 + Vy2 + Vz2
Vx = 0.3808 x 107 m/s from eq.(48)
Vy = 0.2198 x 107 m/s from eq.(49)
Vz = 0 m/s from eq.(50)
VCN2 = (0.3808 x 107)2 + (0.2198x 107)2 + (0)2 m2/s2
(0.14500864 x 1014) + (0.04831204 x 1014) +0 m2/s2
VCN2 = (0.19332068 x 1014) m2/s2 eq.(51)
VCN = 0.4396 x 107 m/s eq.(52)
ANGLES THAT MAKE THE VELOCITY OF THE COMPOUND NUCLEUS ( VCN ) WITH POSITIVE
X , Y , AND Z AXES AT POINT ‘F‘
1 with x – axis
cos α = /VCN = VCN cos α/ VCN
VCN cos α = 0.3808 x 107 m/s from eq.(48)
VCN = 0.4396 x 107 m/s from eq.(52)
• cos α = 0.3808 x 107 m/s = 0.8662 eq.(53)
0.4396 x 107 m/s
• α = 30o [cos 30o = 0.8660]
2 with y– axis
cos β = /VCN = VCN cos β / VCN
VCN cos β = 0.2198 x 107 m/s from eq.(49)
cos β = 0.2198x 107 m/s = 0.5
7
0.4396 x 10 m/s
• β = 60o [ cos 60o = 0.5 ] eq.(54)
3 with z– axis
cos y= /VCN = VCN cos y / VCN
VCN cos y = 0 m/s from eq.(50)
cos y = 0 m/s = 0
0.4396 x 107 m/s
• y = 90o eq.(55)
VIII. THE SPLITTING OF THE HETEROGENEOUS COMPOUND NUCLEUS
The heterogeneous compound nucleus, due to its instability, splits according to the lines parallel to the direction of the velocity of the compound nucleus ( ) into the three particles – the deuteron, the positron and
the reduced mass ( m).
Out of them, the two particles (deuteron and positron) are stable while the third one (reduced mass) is unstable.
According to the law of inertia, each particle that has separated from the compound nucleus, has an inherited velocity ( ) equal to the velocity of the compound nucleus ().
So, for conservation of momentum
M = (md + Δm + me+) eq.(56)
Where,
= mass of the compound nucleus
velocity of the compound nucleus md = mass of the deuteron
Δm = reduced mass
me+ = mass of the positron See fig.(23)
The splitting of the heterogeneous compound nucleus
splitting -1 splitting -2
The compound nucleus (The particles produced
(before splitting) after splitting)
Figure 23
The heterogeneous compound nucleus splits into three particles – the deuteron, the reduced mass (Δm) and the positron.
= inherited velocity of the particle
= velocity of the compound nucleus
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Positron is a stable particle in the presence of the magnetic field.
Inherited velocity ( ) of the particles
Each particle that has separated from the compound nucleus has an inherited velocity ( ) equal to the velocity of the compound nucleus.
THE INHERITED VELOCITY OF DEUTERON
• = = 0.4396 x 107 m/s form eq.(52)
Components of the inherited velocity of the deuteron
1. = Vinh cos α = VCN cos α = 0.3808 x 107 m/s from
eq.(48) eq.(57)
= Vinh cos β = VCN cos β = 0.2198 x 107 m/s
from eq.(49) eq.(58)
= Vinh cos y = VCN cos y = 0 m/s from eq.(50) eq.(59)
b. THE INHERITED VELOCITY FOR THE POSITRON
= = 0.4396 x 107 m/s
Components of the inherited velocity of the positron
1. = Vinh cos α = VCN cos α = 0.3808 x 107 m/s
from eq.(48) eq.(60)
2. = Vinh cos β = VCN cos β = 0.2198 x 107 m/s
b. INHERITED KINETIC ENERGY OF THE
REDUCED MASS
Einh = Δm V2 inh= Δm V2 CN
2 2
V2 CN = 0.19332068 x 1014 m2/s2 from eq.(51)
m = 0.000747225 x 10-27 kg from eq.(63)
Einh = ½ x 0.000747225 x 10-27 x 0.19332068 x 1014 J
Einh = 0.00007222702 x 10-13 J
Einh = 0.000045 Mev [1Mev = 1.6 x 10-13 J] eq.(64)
RELEASED ENERGY (ER)
ER = Δmc 2
Δm = 0.00045 amu from eq.(63)
ER = 0.00045 x 931 Mev
ER = 0.41895 Mev eq. (65)
TOTAL ENERGY ( E T )
E T = Einh + ER from eq. (64 and 65 )
E T = ( 0.000045 ) + ( 0.41895 ) Mev from eq. (58)
E T = 0.418995 Mev eq.(66)
CONCLUSION: The total energy (ET) is carried away by the neutrino. The produced neutrino does not interact with any produced nucleus. So, neither the deuteron nor the positron is propelled by the neutrino.
See fig. (24)
from eq.(49) eq.(61)
3. = Vinh cos y = VCN cos y = 0 m/s from eq.(50)
eq.(62)
THE INHERITED VELOCITY OF THE REDUCED MASS
= = 0.4396 x 107 m/s
IX. PROPULSION OF THE PARTICLES
The total energy (ET) is carried away by the neutrino. The produced neutrino does not interact with any produced nucleus. So, neither the deuteron nor the positron is propelled by the neutrino.
REDUCED MASS
= [2mp] – [md +me+] mp = 1.00727 amu
md = 2.01355 amu me+ = 0.00054 amu
m = [ 2x1.00727 ] – [ 2.01355 + 0.00054 ] amu
Δm = [ 2.01454 ] – [ 2.01409 ] amu
Δm = 0.00045 amu
m = 0.00045 x 1.6605 x 10-27 kg [1amu = 1.6605 x 10-27
kg]
Δm = 0.000747225 x 10-27 kg
i.e. = 0.00045 amu = 0.000747225 x 10-27 kg eq.(63)
Propulsion Of The Particles
Figure 24
As the neutrino does not interact with any particle, so we have shown the direction of the velocity of the produced neutrino in such a way that it seems that neutrino does not interact with any particle.
VѴ = Final velocity of the neutrino.
The total energy ET is carried away by the neutrino and the produced neutrino does not interact with any particle and hence neither the deuteron nor the positron is propelled.
It is the inherited velocity ( ) of the reduced mass by virtue of which the produced neutrino can exceed the speed of light.
e. INCREASED ENERGY (EINC) OF THE PARTICLES
As the reduced mass converts into energy. The total energy (ET) is carried away by the particle – neutrino.
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The produced neutrino does not interact with any of the stable nuclei (deuteron and the positron)
Hence the increased energy (Einc) of the each final nucleus (the deuteron and the positron) is zero. or
Einc = 0 J eq.(67)
Increased velocity of the particles
As the increased energy (Einc) of the each particle (the deuteron and the positron) is equal to zero. that is
Einc = 0 J
So, the increased velocity (Vinc) of the each particle (the deuteron and the position) is also equal to zero. or
Vinc = 0 m/s eq.(68)
COMPONENTS OF THE INCREASED VELOCITY (VINC ) OF THE PARTICLE – THE DEUTERON
As the Vinc = 0 m/s
So,
1 = Vinc cos α = 0 m/s eq.(69)
2 = Vinc cos β = 0 m/s eq.(70)
3 = Vinc cos y= 0 m/s eq.(71)
COMPONENTS OF THE INCREASED VELOCITY (VINC ) OF THE PARTICLE – THE POSITRON
As the Vinc = 0 m/s
So,
1 = Vinc cos α = 0 m/s eq.(72)
2 = Vinc cos β = 0 m/s eq.(73)
3 = Vinc cos y = 0 m/s eq.(74)
COMPONENTS OF THE FINAL VELOCITY OF THE PARTICLES
For The Deuteron
According Inherited Increased Final velocity
To - Velocity ( ) Velocity ( ) = +
X – axis = 0.3808x107m/s = 0 m/s =0.3808x107m/s
eq.(75)
y – axis = 0.2198x107m/s = 0 m/s =0.2198x107m/s
eq.(76)
z – axis = 0 m/s = 0 m/s =0 m/s eq.(77)
md = 3.3434 x 10-27 kg
= 0.3808 x107 m/s from eq.(75)
= 3.3434 x 10-27 x 0.3808 x 107 kgm/s
= 1.2731 x10-20 kgm/s eq.(81)
2 = md
= 0.2198 x107 m/s from eq.(76)
= 3.3434 x 10-27 x 0.2198 x 107 kgm/s
= 0.7348 x10-20 kgm/s eq.(82)
3 = md
= 0 m/s from eq.(77)
= md x 0 kgm/s
= 0 kgm/s eq.(83)
FINAL VELOCITY (VF ) OF THE DEUTERON
Vf2 = Vx2 + VY2 +VZ2
Vx= 0.3808X107 m/s from eq.(75)
Vy= 0.2198X107 m/s from eq.(76)
Vz= 0 m/s from eq.(77)
Vf2 = (0.3808X107)2 + (0.2198X107)+(0)2 m2/S2
Vf2 = (0.14500864X1014)+(0.04831204X1014)+0 m2/S2
Vf2 =0.19332068X1014 m2/S2 eq.(84)
Vf =0.4396x107 m/s eq.(85)
Final kinetic energy of the deuteron
E= ½ md Vf2
Vf 2=0.19332068x1014 m2/s2 from eq.(84)
= ½ X3.3434X10-27 X 0.19332068X1014 J E = 0.32317418075X10-13 J
E = 0.3231X10-13 J eq.(86)
E = 0.2019 Mev
• E = ½ md Vf 2 = 0.3231 X 10-13 J from eq.(86)
= md Vf 2 =2 X 0.3231X10-13 J
= md Vf 2 =0.6462X10-13 J eq.(87)
k. FORCES ACTING ON THE PARTICLE –
DEUTERON [ WHEN THE DEUTERON IS AT POINT ‘ F ‘]
from eq.(57,58,59) from eq.(69,70,71)
respectively respectively
Table 1
For The Positron
According Inherited Increased Final velocity
To - Velocity ( ) Velocity ( ) = +
X – axis = = 0 m/s =0.3808x107m/s
0.3808x107m/s eq.(78)
y – axis = = 0 m/s =0.2198x107m/s
0.2198x107m/s eq.(79)
z – axis = 0 m/s = 0 m/s =0 m/s eq.(80)
from eq.(60,61,62) from eq.(72,73,74)
respectively respectively
Table 2
COMPONENTS OF THE FINAL MOMENTUM OF THE DEUTERON
1 = md
1 Fy = q Vx BZ sin θ
q = 1.6 x 10-19
= 0.3808 x 107 m/s from eq.(75)
= - 1 Tesla from eq.(15)
sin θ = sin 90o = 1
• FY = 1.6 x 10-19 x 0.3808 x 107 x 1 x 1 N
= 0.60928 x 10-12 N
Form the right hand palm rule, the direction of force is
according to negative y axis.
So, = - 0.60928 x 10-12 N eq.(88)
2 Fz = q Vx By sin θ
= 1 Tesla from eq.(16)
sin θ = sin 90o = 1
Fz = 1.6 x 10-19 x 0.3808 x 107 x 1 x 1 N
0.60928 x 10-12 N
Form the right hand palm rule, the direction of force is
according to negative z axis.
So, = - 0.60928 x 10-12 N eq.(89)
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3 Fx = q Vy BZ sin θ
= 0.2198 x 107 m/s from eq.(76)
-1 Tesla
sin θ = sin 90o = 1
Fx = 1.6 x 10-19 x 0.2198 x 107 x 1 x 1 N
0.35168 x 10-12 N
Form the right hand palm rule, the direction of force is
according to positive x axis.
So, = 0.35168 x 10-12 N eq.(90)
See fig. (25)
Forces acting on the particle – deuteron [when the deuteron is at point ‘F‘]
Figure 25
4. Resultant force ( FR ) acting on the deuteron
FR2 = Fx2 + FY2 +FZ2
Fy = Fz = F = 0.60928 x 10-12 N from eq.(88 and 89)
Fx = 0.35168 x 10-12 N from eq.(90)
FR2 = 2F2 + Fx2
FR2 = 2 x (0.60928 x 10-12)2 + (0.35168 x 10-12)2 N2
= 2 x (0.3712221184 x 10-24) +(0.1236788224 x 10-24) N2
FR2 = (0.7424442368 x 10-24 + 0.1236788224 x 10-24) N2
FR2 = 0.8661230592 x 10-24 N2
FR = 0.9306 x 10-12 N eq.(91)
See fig.(25)
RADIUS OF THE CIRCULAR ORBIT FOLLOWED BY THE DEUTERON
R = mv2/ FR
mv2 = 0.6462 x 10-13 J from eq.(87)
FR = 0.9306 x 10-12 N from eq.(91)
• R = 0.6462 x 10-13 J
0.9306 x 10-12 N
R = 0.6943 x 10-1 m
R = 6.943 cm = 0.6943 x 10-2 m eq.(92)
See fig.(26)
m. TIME PERIOD (T) OF THE CONFINED
DEUTERON
T = 2πr / V
r = 6.943 x 10-2 m from eq.(92)
V = 0.4396 x 107 m/ s from eq.(85)
T = 2 x 3.14 x 6.943 x 10-2 m
0.4396 x 107 m/s
T = 43.60204 x 10-9 s = 99.1857 x 10-9 s
0.4396
T = 9.9185 x 10-8 s eq.(93)
CONCLUSION: the confined deuteron passes the point ‘F’ by the after each 9.9185 x 10-8 second.
The circular orbit followed
Figure 26
by the confined deuteron
Where,
Cd = center of the circular orbit followed by the confined deuteron
= Resultant force
F = the center of fusion or the point where deuteron is produced.
By seeing the directions of forces [ , ] acting on the deuteron [ when the deuteron is at point ‘ F‘], we reach at the conclusion that the circular orbit followed by the confined deuteron lies in the plane made up of positive x
– axis, negative y – axis and the negative z – axis.
The deuteron is confined.
The line segment is the radius of the circular orbit followed by the confined deuteron and is equal to 6.943 x 10-2 m.
ANGLES THAT MAKE THE RESULTANT FORCE (FR) [ACTING ON THE DEUTERON WHEN THE DEUTERON IS AT POINT ‘ F ‘] WITH POSITIVE X , Y
AND Z-AXES 1 with x- axis
Cos α = FR cos α / Fr = / Fr
Confinement of deuteron: The circular orbit followed by the confined deuteron
= 0.35168 x 10-12 N from eq.(90)
Fr = 0.9306 x 10-12 N from eq.(91)
Cos α = 0.35168 x 10-12 N
0.9306 x 10-12 N
Cos α = 0.3779 eq.(94)
• α = 67.8 degree [cos ( 67.8 ) = 0.3778]
2 with y- axis
Cos β = FR cos β / Fr = / Fr
= -0.60928 x 10-12 N from eq.(88)
Fr = 0.9306 x 10-12 N from eq.(91)
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Cos β = -0.60928 x 10-12 N
0.9306 x 10-12 N
Cos β = - 0.6547 eq.(95)
• β = 130.8 degree [cos (130.8) = - 0.6534]
3 with y- axis
Cos y = FR cos y / Fr = / Fr
= -0.60928 x 10-12 N from eq.(89)
Cos y = -0.60928 x 10-12 N
0.9306 x 10-12 N
• Cos y = -0.6547 eq.(96)
• y = 130.8 degree
See fig.(27)
Figure 27
Angles that make the resultant force ( ) [acting on the particle, when the deuteron is at point ‘F’] with respect to
positive x, y, and z- axes.
Where,
α = 67.8 degree from eq.(94)
β = 130.8 degree from eq.(95)
Y = 130.8 degree from eq.(96)
‘F‘ is the point where all the three axes meet or intersect each other.
THE CARTESIAN COORDINATES OF THE POINTS P1 (X1 , Y1 , Z1) AND P2 ( X2 , Y2 , Z2)
LOCATED ON THE CIRCUMFERENCE OF THE CIRCLE OBTAINED BY THE DEUTERON
cos α = x2 - x1 d
d = 2 x r
r = 6.943x10-2 m from eq.(92)
= 2x6.943x10-2 m d = 13.886 x 10-2 m
Cos α = 0.37 from eq.(94)
x2 - x1 = d x cos α
x2 - x1 = 13.886 x10-2x0.37 m
x2 - x1 = 5.1378 x 10-2 m
x2 = 5.1378 x 10-2 m [x1 = 0] eq.(97)
cos β = y2 - y1 d
cos β = -0.65 from eq.(95)
y2 - y1 = d x cos β
y2 - y1 = 13.886 x10-2x (-0.65) m
y2 - y1 = -9.0259 x 10-2 m
y2 = -9.0259x 10-2 m [ y1 , = 0] eq.(98)
1 cos y = z 2 - z 1
d
cos y = -0.65 from eq.(96)
z2 - z1 = d x cos y
z2 - z1 = 13.886 x10-2x (-0.65) m
z2 - z1 = -9.0259 x 10-2 m
z2 = -9.0259 x 10-2 m [z1 , = 0] eq.(99)
The cartesian coordinates of the points P1 (x , y , z) and P2 (x2 , y2 , z2) located on the circumference of the circle obtained by the deuteron
The line P1P2 is the diameter of the circle.
Figure 28
Conclusion: the deuteron is confined. The deuteron starts its circular motion from. point p1 [or the point ‘F‘ (0 , 0, 0)] and reaches at point p2 (x2 , y2 , z2) and then again reaches at point p1 (x1, y1, z1) to complete the circle of radius 0.06943 m. the deuteron keeps on following the
confined circular orbit uninterruptedly and pass through this point ‘F’ [or the point p1 (0, 0, 0) by time and again until it fuses with the proton of the later injected bunch reaching at point ‘F‘.
The confined deuteron fuses with the injected proton at only and only point ‘ F ‘ and form the compound nucleus
at point ‘ F ‘.
Final kinetic energy of the positron
E = ½ me+ Vf2 = ½ me+ V2 inh = ½ me+ V2CN
me+ = 9.1 x 10-31 kg
V2CN = 0.19332068 x 1014 from eq.(84)
= ½ x 9.1 x 10-31 x (0.19332068 x 1014) J
E = 0.879609094 x 10-17 J
0.5497 x 10-4 Mev
me+ Vf2 = me+ V2inh = me+ V2CN
V2 CN = (0.19332068 x 1014) m2/s2 from eq.(84)
me+ Vf2 = 9.1 x 10-31 x 0.19332068 x 1014 J
= 1.7592 x 10-17 J eq.(100)
p. FORCES ACTING ON THE PARTICLE –
POSITRON
1 Fy = q Vx BZ sin θ
q = 1.6 x 10-19
= 0.3808 x 107 m/s from eq.(78)
= - 1 Tesla from eq.(15)
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sin θ = sin 90o = 1
• FY = 1.6 x 10-19 x 0.3808 x 107 x 1 x 1 N
= 0.60928 x 10-12 N
Form the right hand palm rule, the direction of force is
according to negative y axis.
So, = - 0.60928 x 10-12 N eq.(101)
2 Fz = q Vx By sin θ
= 1 Tesla from eq.(16)
sin θ = sin 90o = 1
FY = 1.6 x 10-19 x 0.3808 x 107 x 1 x 1 N
0.60928 x 10-12 N
Form the right hand palm rule, the direction of force is
according to negative z axis.
So, = - 0.60928 x 10-12 N eq.(102)
3 Fx = q Vy BZ sin θ
= 0.2198 x 107 m/s from eq.(79)
= -1 Tesla
sin θ = sin 90o = 1
• Fx = 1.6 x 10-19 x 0.2198 x 107 x 1 x 1 N
= 0.35168 x 10-12 N
Form the right hand palm rule, the direction of force is
according to positive x axis.
So, = 0.35168 x 10-12 N eq.(103)
See fig.(29)
Forces Acting On The Particle – Positron
Figure 29
4 Resultant force ( FR ) acting on the positron
FR2 = Fx2 + FY2 +FZ2
Fx = 0.35168 x 10-12 N from eq.(103)
F = Fy = Fz = 0.60928 x 10-12 N from eq.(101 and 102)
FR2 = Fx 2 + 2F2
FR2 = (0.35168 x 10-12)2 + 2(0.60928 x 10-12)2 N2
(0.1236788224 x10-24) +2(0.3712221184 x 10-24) N2
FR2 = (0.1236788224 x 10-24) + (0.7424442368x10-24) N2
FR2 = 0.8661230592 x 10-24 N2
FR = 0.9306 x 10-12 N eq.(104)
RADIUS OF THE CIRCULAR ORBIT FOLLOWED BY THE POSITRON
0.9306 x 10-12
R = 1.89039 x 10-5 m
R = 18.9039 x 10-6 m eq.(105)
CONCLUSION:
Confinement of positron
The positron is confined due to the forces acting on it. By seeing the directions of forces acting on the positron, we reach at the conclusion that the circular orbit followed by the confined positron lies in the plane made up of positive x – axis, negative y- axis and negative z – axis.
ELECTRON GUN: electron gun injects the electron into the the tokamak at point ‘F‘. The each electron is injected into the tokamak with 10 ev energy making angle 30o with the x-axis, 120o angle with the y-axis and 90o angle with the z-axis.
Injection Of The Electron
Figure 30
Annihilation of the positron
The injected electron reaches at point ‘F‘ and collides with the positron passing through it. The result of the collision is the annihilation of the positron and the electron and is the creation of a pair of gamma ray photons.
The positron annihilates with an electron and their mass Energy [me+ c2 + me- c2] and their kinetic energy [½ m e+
V2CN + ½ me- V2 injected] is carried away by the two gamma ray photons.
So, at point ‘ F‘ annihilation is as: e+ + e- 2 y photons
The produced gamma ray photons strike to the wall of the tokamak.
The positron (s) that do not collide with electron (s) continue follow the confined circular orbit until they annihilate with electron (s).
The injected electron (s) that reaches at point ‘F‘ but does not collide with any positron, due to presence of magnetic fields, undergo to a circular orbit.
R = mv2/ FR X. CONFINEMENT OF THE INJECTED ELECTRON
mv2 = 1.7592 x 10-17 J from eq.(100) a. VELOCITY OF THE ELECTRON
FR = 0.9306 x 10-12 J from eq.(104)
R = 1.7592 x 10-17 m
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V = 2E½ m/s
me-
Ee = 10 ev
me- = 9.1 x 10-31 J
V = 2x10x1.6x10-19 ½ m/s
9.1 x 10-31
V = 3.51648351648 x 1012 ½ m/s
V = 1.8752 x 106 m/s eq.(106)
COMPONENTS OF THE VELOCITY OF THE ELECTION
Each electron is injected into the tokamak making angle 30o with the x-axis, 120o angle with y-axis and 90o angle with the z-axis.
So, the components of the velocity of the electron at point ‘F‘ are –
= V cos α
V = 1.8752 x 106 m/s from eq.(106)
cos α = cos 30o =0.866
= 1.8752 x 106 x 0.866 m/s
= 1.6239 x 106 m/s eq.(107)
2 = V cos β
cos β = cos 120o = -0.5
= 1.8752 x 106 x (-0.5) m/s from eq.(106)
= -0.9376 x 106 m/s eq.(108)
3 = V cos y
cos y = cos 90o = 0
= 1.8752 x 106 x 0 m/s
= 0 m/s eq.(109)
c. FORCES ACTING ON THE PARTICLE –
ELECTRON
1 Fy = q Vx BZ sin θ
= 1.6239 x 106 m/s from eq.(107)
= - 1 Tesla from eq.(15)
sin θ = sin 90o = 1
q = 1.6 x 10-19 C
FY = 1.6 x 10-19 x 1.6239 x 106 x 1 x 1 N
2.5982 x 10-13 N
Form the right hand palm rule, the direction of force is
according to negative y-axis.
So, = - 2.5982 x 10-13 N eq.(110)
2 Fz = q Vx By sin θ
= 1 Tesla from eq.(16)
sin θ = sin 90o = 1
Fz = 1.6 x 10-19 x 1.6239 x 106 x 1 x 1 N
2.5982 x 10-13 N
Form the right hand palm rule, the direction of force is according to negative z- axis.
So, = - 2.5982 x 10-13 N eq.(111)
3 Fx = q Vy BZ sin θ
= -0.9376 x 106 m/s from eq.(108)
= -1 Tesla
sin θ = sin 90o = 1
• Fx = 1.6 x 10-19 x 0.9376 x 106 x 1 x 1 N
= 1.5001x 10-13 N
Form the right hand palm rule, the direction of force is
according to negative x- axis.
So, = - 1.5001x 10-13 N eq.(112)
RESULTANT FORCE (FR) ACTING ON THE ELECTRON
FR2 = Fx2 + FY2 +FZ2
Fx = 1.5001 x 10-13 N from eq.(112)
F = Fy = Fz = 2.5982 x 10-13 N from eq.(110,111)
FR2 = Fx2 + 2F2
FR2 = (1.5001 x 10-13)2 + 2(2.5982 x 10-13)2 N2
(2.25030001 x 10-26) + 2(6.75064324 x 10-26) N2
FR2 = (2.25030001 x 10-26) + (13.50128648x10-26) N2
FR2 = 15.75158649 x 10-26 N2
• FR = 3.9688 x 10-13 N eq.(113)
RADIUS OF THE CIRCULAR ORBIT FOLLOWED BY THE ELECTRON
= mv2/ FR ½ mv2 = 10 ev
mv2 = 2x10x1.6 x 10-19 J
=32x10-19 J
FR = 3.9688 x 10-13 N from eq.(113)
R = 32x10-19m
3. 9688 x 10 -13
R = 8.0628 x 10-6 m eq.(114)
CONCLUSION: The electron that does not collide with positron is not confined. By seeing the direction of forces acting on the electron we reach at the conclusion that the circular orbit to be followed by the electron lies in the plane made up of negative x-axis, negative y-axis and negative z-axis where the magnetic fields are not applied. so, In trying to follow a confined circular orbit , the electron starts its circular motion from point ‘F‘ and reaches in a region made up of negative x-axis , negative y-axis and negative z- axis where the magnetic fields are not applied.
So, as the electron get rid of the magnetic fields, it starts its linear motion leaving the circular motion.
Firstly, the electron starts circular motion from point ‘F‘ then it give up its circular motion and starts its linear motion towards the downward to strike the base wall of the tokamak.
For fusion reaction (3):
11H + 21 H 23 He + y rays
Formation of compound nucleus [23 He]:
Interaction Of Nuclei
The injected proton as reaches at point ‘F‘, it interacts [ experiences a repulsive force due to the confined deuteron passing through the point F ] with the confined deuteron at point F. the injected proton overcomes the electrostatic
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repulsive force and – a like two solid spheres join - the injected proton dissimilarly joins with the confined deuteron .
See fig (31)
where, [23 He] is a compound nucleus formed due to proton and deuteron fusion.
Interaction Of Nuclei
interaction (1) interaction (2)
Figure 31
Formation Of The Homogeneous Compound Nucleus
The constituents (quarks and gluons) of the dissimilarly joined nuclei (the proton and the deuteron) behave like a liquid and form a homogeneous compound nucleus having similarly distributed groups of quarks with similarly distributed surrounding gluons.
Thus within the homogeneous compound nucleus – each group of quarks is surrounded by the gluons in equal proportion. so, within the homogeneous compound nucleus there are 3 groups of quarks surrounded by the gluons.
See fig(32)
Homogeneous Compound Nucleus
Figure 32
where,
= velocity of the compound nucleus and
α = 300 [from eq(132)]
β = 600 [from eq(133)]
y = 900 [from eq(134)]
Formation of lobes within into the homogeneous compound nucleus or the transformation of the homogeneous compound nucleus into the heterogeneous compound nucleus:
The central group of quarks with its surrounding gluons to become a stable and the next higher nucleus (the hellion –
than the reactant one (the deuteron) includes the other two (nearby located) groups of quarks with their surrounding gluons and rearrange to form the ‘A‘ lobe of the heterogeneous compound nucleus.
While the remaining gluons [the gluons or the mass that is not included in the formation of the lobe ‘A‘] rearrange to form the ‘B‘ lobe of the heterogeneous compound nucleus .
Due to formation of two dissimilar lobes within into the homogeneous compound nucleus, the homogeneous compound nucleus transforms into the heterogeneous compound nucleus.
See fig(33)
Formation Of Lobes Within Into The Homogeneous Compound Nucleus
Figure 33
Where,
Inner side - lobe ‘A‘ is formed. [The is the helium – 3 nucleus is formed ]
Outer side - The remaining gluons [or the reduced mass] form the ‘B‘ lobe of the heterogeneous compound nucleus].
FINAL STAGE OF THE HETEROGENEOUS COMPOUND NUCLEUS
The remaining gluons [That compose the ‘B‘ lobe of the heterogeneous compound nucleus ] remains loosely bonded to the helium – 3 nucleus [ that compose the ‘A‘ lobe of the heterogeneous compound nucleus]. Thus, the heterogeneous compound nucleus, finally, becomes like a coconut into which the remaining gluons represent the outer shield while the helium – 3 nucleus represent the inner part of the coconut.
See fig.(34)
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Final Stage Of The Heterogeneous Compound Nucleus 2 = Pb cos y = 0 kgm/s from eq.(43)….. eq.(122)
Figure 34
XI. CALCULATIONS FOR THE COMPOUND NUCLEUS [32 HE]
JUST BEFORE FUSION, THE LOSS IN THE KINETIC ENERGY OF THE INJECTED PROTON
As the injected proton reaches at point F, the injected proton fuses with the confined deuteron (passing through the point F) to form the compound nucleus at point F.
Just before fusion, to overcome the electrostatic repulsive force, the injected proton loses its energy equal to 2.5 kev.
So, the kinetic energy of the proton just before fusion is [Eb]
Eb = [EP- Eloss ]
EP = 102.4 kev from eq. (4)
Eloss = EP-D = 2.5 kev from eq. (3)
Eb = [102.4 -2.5] kev
= 99.9 kev
= 0.0999 Mev eq. (115)
THE VELOCITY OF INJECTED PROTON JUST BEFORE FUSION IS [ VB ]
Vb = 0.4371 x 107 m/s from eq. (37) …. eq.(116)
THE COMPONENTS OF THE VELOCITY OF THE PROTON JUST BEFORE FUSION ARE
(1) = Vb cos α = 0.3785 x 107 m/s from eq.(38)… eq.(117)
(2) = Vb cos β = 0.2185 x 107 m/s from eq.(39)… eq.(118)
(3) = Vb cos y = 0 m/s from eq.(40)……. eq.(119)
COMPONENTS OF THE MOMENTUM OF THE INJECTED PROTON JUST BEFORE FUSION ARE
1 = Pb cos α = 0.6330 x 10-20 from eq.(41)… eq.(120)
2 = Pb cos β = 0.3654 x10-20kgm/s from eq.(42)..eq.(121)
COMPONENTS OF THE MOMENTUM OF THE COMPOUND NUCLEUS
The injected proton penetrates the confined proton. So, Just before fusion, there is a loss in kinetic energy of the injected proton but the kinetic energy of the confined deuteron remains same. So, the momentum of the confined deuteron remains same as with which it was produced.
X – component of the momentum of the compound nucleus =
X – component of X – component of
the momentum of + the momentum of
the confined the injected
deuteron at point Proton just before
‘F’ fusion at point ‘F’
= [1.2731 x 10-20] + [0.6330 x 10-20] kgm/s from eq.(81
and 120)
= 1.9061 x 10-20 kgm/s
= PCN cos α = 1.9061 x 10-20 kgm/s eq.(123)
Y– component of the momentum of the compound nucleus( ) =
y – component of y – component of
the momentum of + the momentum of
the confined the injected
deuteron at point Proton just before
‘F ‘ fusion at point ‘F’
• = [0.7348 x 10-20] + [0.3654 x 10-20] kgm/s from eq.(82 and 121)
= 1.1002 x 10-20 kgm/s
• = PCN cos β = 1.1002 x 10-20 kgm/s from eq.(124)
z – component of the momentum of the compound nucleus ( ) =
z– component of z – component of
the momentum of + the momentum of
the confined the injected
deuteron at point Proton just before
‘F’ fusion at point ‘F’
• = [ 0 ] + [ 0 ] kgm/s from eq.(83 and 122)
• = 0 kgm/s
• = PCN cos y = 0 kgm/s eq.(125)
MASS OF THE COMPOUND NUCLEUS
M = md + mp
= [3.3434 x 10-27] + [1.6726 x 10-27] kg
= 5.016 x 10-27 kg eq.(126)
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COMPONENTS OF THE VELOCITY OF THE COMPOUND NUCLEUS
= VCN cos α = / M = PCN cos α
M
PCN cos α = 1.9061 x 10-20 kgm/s from eq.(123)
M = 5.016 x 10-27 kg from eq.(126)
• = VCN cos α = 1.9061 x 10-20 kgm/s
5.016 x 10-27 kg
• = VCN cos α = 0.3800 x 107 m/s eq.(127)
= VCN cos β = = / M = P CN cos β
M
PCN cos β = 1.1002 x 10-20 kgm/s from eq.(124)
= VCN cos β = 1.1002 x 10-20 kgm/s 5.016 x 10-27 kg
• = VCN cos β = 0.2193 x 107 m/s eq.(128)
= VCN cos y = = /M = PCN cos y
M
PCN cos y = 0 kgm/s from eq.(125)
= 0kgm/s
M kg
• = VCN cos y = 0 m/s eq.(129)
h. VELOCITY OF THE COMPOUND NUCLEUS
(VCN)
V2 CN = Vx2 + Vy2 + Vz2
Vx = 0.38 x 107 m/s from eq.(127)
Vy = 0.2193 x 107 m/s from eq.(128)
Vz = 0 m/s from eq.(129)
V2CN = (0.38 x 107)2 + (0.2193 x 107)2 + (0)2 m2/s2
(0.1444 x 1014) + (0.04809249 x 1014) + 0 m2/s2
V2CN = 0.19249249 x 1014 m2/s2 eq.(130)
VCN = 0.4387 x 107 m/s eq.(131)
i. ANGLES THAT MAKE THE VELOCITY OF THE COMPOUND NUCLEUS (VCN) WITH POSITIVE X, Y, AND Z AXES AT POINT ‘ F ‘
with x – axis
cos α = /VCN = VCN cos α/ VCN
VCN cos α = 0.38 x 107 m/s from eq.(127)
VCN = 0.4387 x 107 m/s from eq.(131)
• cos α = 0.38 x 107 m/s = 0.8661
0. 7
4387 x 10 m/s
• α = 30o eq.(132) [cos 30o = 0.8660]
with y– axis
cos β = /VCN = VCN cos β / VCN
VCN cos β = 0.2193 x 107 m/s from eq.(128
cos β = 0.2193x 107 m/s
0.4387 x 107 m/s
Cos β = 0.4998
• β = 60o eq.(133) [cos 60o = 0.5]
with z– axis
cos y= /VCN = VCN cos y / VCN
VCN cos y = 0 m/s from eq.(129)
cos y = 0 m/s = 0
0.
4387 x 107 m/s
y = 90o eq.(134)
The splitting of the heterogeneous compound nucleus:
The remaining gluons are loosely bonded to the helium – 3 nucleus.
At the poles of the helium – 3 nucleus, the remaining gluons are lesser in amount than at the equator
So, during the rearrangement of the remaining gluons [or during the formation of the ‘B‘ lobe of the heterogeneous compound nucleus], the remaining gluons to be homogeneously distributed all around, rush from the equator to the poles.
In this way, the loosely bonded remaining gluons
separates from the helium – 3 nucleus and also divides itself into two parts giving us three particles – the first one is the one – half of the reduced mass, second one is the helium -3 nucleus and the third one is the one – half of the reduced mass.
Thus, the heterogeneous compound nucleus splits according to the lines parallel to the velocity of the compound nucleus into three particles – the first one is the one – half of the reduced mass ( m/2), the second one is the helium -3 nucleus and the third one is the another one – half of the reduced mass (Δm/2).
By the law of inertia, each particle that has separated from the compound nucleus has an inherited velocity ( ) equal to the velocity of the compound nucleus ( ).
So, for the conservation of momentum
= ( m/2 + mhe-3 +Δm/2) Where,
M = mass of the compound nucleus
velocity of the compound nucleus m/2 = one – half of the reduced mass
mhe-3 = mass of the helium -3 nucleus See fig (35) and (36)
The Splitting Of The Heterogeneous Compound Nucleus
splitting -1
Figure 35
The loosely bonded remaining gluons rush from equator to the poles, but before they reach at the equator, it
breaks up.
Thus the heterogeneous compound nucleus splits according to the lines parallel to the velocity of the
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compound nucleus ( ) into three particles – one is the half of the reduced mass (Δm/2 ) the second is the helium -3 nucleus and the third is the one – half of the reduced mass (Δm/2).
The Spitting Of The Heterogeneous Compound Nucleus
splitting -2
Figure 36
The heterogeneous compound nucleus splits into three particles – The one – half of the reduced mass, the helium – 3 nucleus and the one – half of the reduced mass.
XII. INHERITED VELOCITY OF THE PARTICLES
Each particles that has separated from the compound nucleus has an inherited velocity ( ) equal to the velocity of the compound nucleus ().
i. Inherited velocity ( ) the particle hellion – 3
Vinh = VCN = 0.4387 x 107 m/s from eq.(131)
Components of the inherited velocity of the hellion – 3
= Vinh cos α = VCN cos α = 0.38 x 107 m/s from eq. (127)……eq.( 135)
2 = Vinh cos β = VCN cos β = 0.2193 x 107 m/s from
eq.(128)…. eq.( 136)
3 = Vinh cos y = VCN cos y = 0 m/s from eq.(129)….
eq.(137)
For the each one – half of the reduced mass (Δm/2):
Vinh = VCN = 0.4387 x 107 m/s from eq.(131)
Propulsion of the particles a. REDUCED MASS
Δm = [md +mp] - [mHe-3]
md = 2.01355 amu
mp = 1.00727 amu
mHe-3 = 3.01493 amu
Δm = [2.01355+1.00727] – [3.01493] amu Δm = [3.02082] – [3.01493] amu
Δm = 0.00589 amu
Δm = 0.00589 x 1.6605 x 10-27 kg
Δm = 0.009780345 x 10-27 kg
i.e. Δm = 0.00589 amu = Δm = 0.009780345 x 10-27 kg eq.(138)
b. THE INHERITED KINETIC ENERGY OF THE TOTAL REDUCED MASS (ΔM)
Einh = ½ Δm V2 inh = ½ Δm V2 CN
Δm = 0.009780345 x 10-27 from eq.(138)
V2CN = 0.19249249 x 1014 m2/s2 from eq.(130)
Einh = ½ x 0.009780345 x 10-27 x 0.19249249 x 1014 J
Einh = 0.00094132148 x 10-13 J
Einh = 0.000588 Mev eq.(139)
RELEASED ENERGY (ER)
ER = Δmc 2
Δm = 0.00589 amu from eq.(138)
ER = 0.00589 x 931 Mev
ER = 5.48359 Mev eq.(140)
TOTAL ENERGY (E T)
E T = Einh + ER
E T = (0.000588) + (5.48359) Mev from eq.(139 and 140)
E T = 5.484178 Mev eq.(141)
e. PROPULSION OF THE HELIUM – 3 NUCLEUS
The each one – half of the reduced mass (Δm/2) converts into energy. so, the energy (E) carried by the produced pairs of gamma ray photons is –
E = ET/ 2
ET = 5.4841 Mev from eq.(141)
• E = 5.4841/2 Mev
• E = 2.7420 Mev eq.(142)
When a pair of gamma ray photon make a head – on collision with a nucleus, it imparts its extra energy to the nucleus by a pair production [that is producing a positron and an electron].
So, for pair production each pair of gamma ray photon carrying high energy must have an energy (1.02 Mev) equal to or greater than the sum of the energies – the energy equal to the rest mass of the positron (me+ c2) and the energy equal to the rest mass of electron (me- c2).
Number of pairs of gamma ray photons (N y)
When one – half of the reduced mass (Δm/2) converts into energy, the energy (E) carried by the pairs of gamma ray photons is 2.7420 Mev.
Each pair of gamma ray photon that carry a part of energy (E) must have an energy equal to or more than 1.02 Mev.
So,
Number of pairs of = Energy (E) produced due to Δm/2
gamma ray photos Energy that must carried by a pair
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of g.r. photon
E = 2.7420 Mev from eq.(142)
• Ny = 2.7420 Mev
1.02 Mev
Ny = 2.6882
Taking the whole digit, we may say that there are 2 pairs of gamma ray photons that carry the energy 2.7420 Mev.
Thus, there are the 4 pairs of gamma ray photons that carry the total energy (ET) equal to 5.4841 Mev
Energy carried by the each pair of gamma ray photon [Ey]
Energy carried by the each pair of gamma ray photon is equal to the energy (E) produced due to one – half of the reduced mass (Δm/2) divided by the total number of pairs of gamma ray photons that carry the energy (E).
Ey =energy (E) produced due to Δm/2
Total number of pairs of g.r. photons that carry energy (E)
E = 2.7420 Mev from eq.(142)
Ey = 2.7420 Mev 2
Ey = 1.3710 Mev
PROPULSION OF PARTICLE
As the reduced mass converts into energy, the total energy (ET) is carried away by the gamma ray photons.
Conservation Of Momentum
We know that the reduced mass has separated from the compound nucleus with an inherited velocity () equal to the velocity of the compound nucleus ( ).
So, for conservation of momentum, if we sum up the momenta of all the produced gamma ray photons, we will again get the total momentum equal to the reduced mass multiplied by the velocity of the compound nucleus (ΔmVCN).
Now, if we denote the momentum of a gamma ray photon by Py then,
The total momenta of all the produced four gamma ray photons will be equal to ΔmVCN.
Or
Py ( 1 ) + Py ( 2 ) + Py ( 3 + Py ( 4 ) = ΔmVCN
See fig ( 37,39,40)
We know that the inherited kinetic energy of the reduced mass is negligible. So, we can take the inherited momentum of the reduced mass (Δm) is equal to zero.
If we take the inherited momentum of the reduced mass (Δm ) is equal to zero then the sum of the momenta of all the produced 4 gamma ray photons will also be equal to zero.
That is –
Py ( 1 ) + Py ( 2 ) + Py ( 3 + Py ( 4 ) = 0
So, we reach at this conclusion that out of 4 gamma ray photons, the sum of momenta of 2 gamma ray photons is equal and opposite to the sum of momenta of rest 2 gamma ray photons.
Or we may say that there is a photon having equal and opposite momentum to the another photon.
See fig (38,39,40)
When one – half of the reduced mass (Δm/2) converts into energy, the one – half of the total energy (ET / 2) is carried away by the 2 pairs of gamma say photons.
The pair of gamma ray photon numbered as ‘1‘ travelling in 1st quadrant has equal and opposite momentum to the pair of gamma ray photon numbered as ‘3‘ travelling in the 3rd quadrant.
Similarly, the pair of gamma ray photon numbered as ‘2‘ travelling in the 2nd quadrant has equal and opposite momentum to the pair of gamma ray photon numbered as ‘4‘ travelling in the 4th quadrant.
That is,
Py ( 1 ) = - Py ( 3 )
Similarly
Py ( 2 ) = -Py ( 4 )
See fig (37,38,39,40)
The sum of the momenta of the pairs of gamma ray photons travelling in the I quadrant is equal and opposite to the sum of the momenta of the pairs of gamma ray photons travelling in the III quadrant.
Similarly, the sum of the momenta of the pairs of gamma ray photons travelling in the II quadrant is equal and opposite to the sum of the momenta of the pairs of gamma ray photons travelling in the IV quadrant.
Conclusion: Increased kinetic energy of helion -3.
For conservation of momentum there the three conditions arise -
All the four pairs of gamma ray photons strike to the helium -3 nucleus where a pair of gamma ray photon has equal and momentum to another. So, the net change in momentum of the helium -3 nucleus is zero. As the momentum of helion -3 is not increased (changed) the increament in the kinetic energy of the helion -3 is zero see fig (37,38)
All the four pairs of gamma ray photons move outward and no any pair strike to the nucleus. In this condition also, the change in momentum of the helium -3 nucleus is zero. And so the increased kinetic energy of the helion -3 is zero. see fig.(39)
Only two pairs strike to the nucleus with equal and opposite momentum. So that the net change
[increatment] in the momentum of the helium -3 is zero or we may say that the increased kinetic energy of the helion-3 is zero. see fig.(40)
For the 1st condition: All the four pairs of gamma ray photon strike to the helium -3 nucleus.
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Figure 37
Each pair of gamma ray photon travel towards the nucleus to make a head – on collision with the nucleus.
Each photon carry 1.3710 Mev energy.
The pair of gamma ray photon that has numbered as ‘1‘ travelling in the `I quadrant has equal and opposite momentum to the pair of gamma ray photon that has numbered as ‘3‘ travelling in the III quadrant. And same condition is applicable for the pairs of gamma ray photons numbered as ‘2‘ and ‘4‘
I.e Py ( 1 ) = -Py ( 3 ) and Py ( 2) = -Py ( 4 )
For the 1st condition: Collision between a pair of gamma ray photon and the helium -3 nucleus.
Figure 38
Each incident photon carry 1.3710 Mev.
Each incident photon make head – on collision with the helium – 3
The angle of incidence is equal to the angle of reflection
Where,
i1 = The incident photon numbered as ‘1‘ r1 = The reflected photon numbered as ‘1‘
The helium – 3 nucleus is energised by the 1.404 Mev energy.
Conclusion: Four pairs of gamma ray photon strike to the helium -3 nucleus where one pair has a momentum equal and opposite to the another. So, the net change in momentum of the helion -3 is zero. So, though, there is a loss in the energy of pairs of gamma ray photons, the net increased energy of the helion -3 is zero.
For the second condition: No any pair of gamma ray photon strike to the helium -3 nucleus.
Figure 39
Each gamma ray photon carry 1.3710 Mev
The gamma ray photon that is numbered as ‘1‘ travelling in the first quadrant has equal and opposite momentum to the photon numbered as ‘3‘ travelling in the III quadrant
Similarly 2nd and 4th photon also have equal and opposite momenta
Conclusion: No any gamma ray photon make a head –on collision with the helium -3 nucleus. so, the net change in momentum of helium -3 is zero. So, the increased kinetic
energy of the helium -3 nucleus is zero.
For 3rd condition: only two pairs of gamma ray photons strike to the helium-3 nucleus.
Figure 40
In the third condition, only half number of the produced gamma ray photons [that is equal to 2] strike to the
helium-3 nucleus.
While the other two pairs of gamma ray photons do not strike to the helium-3 nucleus.
Conclusion: two pairs of gamma ray photons strike to the helium-3 nucleus where each pair has equal and opposite momentum to the other one. So, the net increased momentum
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of the helion-3 is zero and hence the net increased kinetic energy of the helion-3 is zero.
INCREASED KINETIC ENERGY OF THE HELIUM -3 NUCLEUS
In each condition, either the gamma ray photons strike to the helium-3 nucleus or not, the net change in momentum of the helium-3 nucleus is zero. So, the net change [increament in the kinetic energy]. In energy of the helium-3 nucleus is zero.
Or we say, the increased kinetic energy of the helium-3 nucleus is zero i.e.
Einc = 0 eq.(143)
INCREASED VELOCITY (VINC) OF THE HELIUM-3 NUCLEUS
As the increased kinetic energy (Einc) of the helium-3 nucleus is equal to zero. So, the increased velocity (Vinc) of the helium-3 nucleus is also equal to zero.
that is –
Vinc = 0 eq.(144)
COMPONENTS OF THE INCREASE VELOCITY ( ) OF THE HELIUM-3 NUCLEUS
As , Vinc = 0
So, the components of the increased velocity are –
= Vinc Cos α = 0 m/s eq.(145)
= Vinc Cos β = 0 m/s eq.(146)
= Vinc Cos y = 0 m/s eq.(147)
Components of the final velocity () of the helium – 3 nucleus
According Inherited Increased Final velocity
to - Velocity ( Velocity ( ) = +
X – axis = 0.38x107m/s = 0 m/s = 0.38x107m/s
eq.(148)
y – axis = = 0 m/s = 0.2193x107m/s
0.2193x107m/s eq.(149)
z – axis = 0 m/s = 0 m/s = 0 m/s eq.(150)
from eq.135,136,137 from eq.145,146,147
respectively respectively
Table 3
Thus the final velocity () of the helion -3 is equal to the inherited velocity () of the particle.
= = = 0.4387 x 107 m/s from eq.(131)
Final kinetic energy of the helium – 3 nucleus
EK = ½ mhe-3 Vf2
Vf2 = 0.19249249 x1014 m2 /s2 from eq.(130)
mhe-3 = 5.00629 x 10-27 kg
Ek = ½ x 5.00629 x 10-27 x 0.19249249 x1014 J
Ek = 0.48183661388 x 10-13 J eq.(151)
Ek = 0.3011 Mev
mhe-3 Vf 2 = 2 Ek = 2x0.4818 x 10-13 Jfrom eq.(151)
mhe-3 Vf 2 = 0.9636 x 10-13 J eq.(152)
Forces acting on the helium – 3 Nucleus 1 Fy = q Vx B Z sin θ
= 0.38 x 107 m/s from eq.(148)
= - 1 Tesla from eq.(15)
= 2 x 1.6 x 10-19 C sin θ = sin 90o = 1
FY = 2 x1.6 x 10-19 x 0.38 x 107 x 1 x 1 N
1.216 x 10-12 N
Form the right hand palm rule, the direction of force is
according to negative y axis.
So,
= - 1.216 x 10-12 N eq.(153)
2 Fz = q Vx By sin θ
= 1 Tesla from eq.(16)
sin θ = sin 90o = 1
Fz = 2 x 1.6 x 10-19 x 0.38 x 107 x 1 x 1 N
= 1.216 x 10-12 N
Form the right hand palm rule, the direction of force is
according to negative z - axis.
So,
= -1.216 x 10-12 N eq.(154)
3 Fx = q Vy BZ sin θ
= 0.2193 x 107 m/s from eq.(149)
-1 Tesla
sin θ = sin 90o = 1
Fx = 2 x1.6 x 10-19 x 0.2193 x 107 x 1 x 1 N
0.70176 x 10-12 N
Form the right hand palm rule, the direction of force is according to positive x axis.
So,
= 0.70176 x 10-12 N eq.(155)
See fig (41)
Forces Acting On The Helium – 3 Nucleus
Figure 41
Resultant force ( FR ) acting on the helium – 3 nucleus:
FR2 = Fx2 + FY2 +FZ2
Fx = 0.70176 x 10-12 N from eq.(155)
F = Fy = Fz = 1.216 x 10-12 N from eq.(153,154)
FR2 = Fx 2 + 2F2
FR2 = (0.70176 x 10-12)2 + 2(1.216 x 10-12)2 N2
FR2 = (0.4924670976 x 10-24) + 2(1.478656 x 10-24) N2
FR2 = (0.4924670976 x 10-24) + (2.957312 x 10-24) N2
FR2 = 3.4497790976 x 10-24 N2
• FR = 1.8573 x 10-12 N eq.(156)
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See fig (41)
Radius of the circular orbit followed by the helion – 3
r = mv2 / FR
mv2 = 0.9636 x 10-13 J from eq.(152)
FR = 1.8573 x 10-12 N from eq.(156)
r = 0.9636 x 10-13 J
1.8573 x 10-12 N
R = 0.5188 x 10-1 m
r = 5.188 cm = 0.5188 x 10-2 m eq.(157)
See fig.(42)
The circular orbit to be followed by the helion-3
The line segment FC he-3 is the radius of the circular orbit to be followed by the helion-3 and is equal to 5.188 x 10-2 m.
Cos β = FR cos B / FR = / FR
= -1.216 x 10-12 from eq.(153)
Cos β = -1.216 x 10-12 N
1.8573 x 10-12 N
Cos β = -0.6547 eq.(159)
• β = 130.8 degree [cos ( 130.8) = - 0.6534]
3 with z- axis
Cos y = FR cos y / FR = / FR
= -1.216 x 10-12 from eq.(154)
Cos y = -1.216 x 10-12 N
1.8573 x 10-12 N
Cos y = -0.6547 eq.(160)
y = 130.8 degree [cos (130.8) = - 0.6534] See fig.(43)
Figure 42
The circular orbit to be followed by the hellion -3 lies in the plane made up of positive x-axis, negative y-axis and negative z-axis.
‘F‘ is the center of fusion or the point where the he-3 nucleus is produced.
CH-3 = center of the circular orbit to be followed by the hellion – 3.
FR = Resultant force acting on the hellion – 3.
By seeing the directions of forces [, ,] acting on the helium-3 nucleus [when the helium -3 nucleus is at point ‘F‘], we reach at the conclusira that the circular orbit to be followed by the helion-3 lines in the plane made up of positive x-axis, negative y-axis, and negative z- axis.
Angles that make the resultant force (FR) [acting on the
particle when the helium - 3 is at point ‘F‘] with respect to positive x, y and z-axes.
1 with x- axis
Cos α = FR cos α / FR = / FR
= 0.70176 x 10-12 from eq.(155)
FR =1.8573 x 10-12 from eq.(156)
Cos α = 0.70176 x 10-12 N
1.8573 x 10-12 N
• Cos α = 0.3778 eq.(158)
• α = 67.8 degree [cos ( 67.8 ) = 0.3778]
2 with y- axis
Figure 43
Angles that make the resultant force (FR) [acting on the helion – 3 when the hellion -3 is at point ‘F’] with respect
to positive x, y, and z- axes.
Where,
α = 67.8 degree from eq.(158)
β = 130.8 degree from eq.(159)
Y = 130.8 degree from eq.(160)
The cartesian coordinates of the points P1 (x1, y1, z1) and P2 (x2, y2, z2) located on the circumference of the circle obtained by the helium – 3 nucleus
1 cos α = x2 - x1
d
d = 2 x r
r = 5.188x10-2 m from eq.(157)
d = 2 x 5.188x10-2 m
d = 10.376 x 10-2 m
Cos α = 0.37 from eq.(158)
• x2 - x1 = d x cos α
• x2 - x1 = 10.376 x 10-2 x 0.37 m
• x2 - x1 = 3.8391 x 10-2 m
• x2 = 3.8391 x 10-2 m [x1 = 0] eq.(161)
2 cos β = y 2 - y 1
d
cos β = -0.65 from eq.(159)
y2 - y1 = d x cos β
y2 - y1 = 10.376 x10-2x (-0.65) m
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• y2 - y1 = -6.7444 x 10-2 m
• y2 = -6.7444x 10-2 m [y1, = 0] eq.(162)
3 cos y = Z2 - Z1
d
cos y = -0.65 from eq.(160)
• z2 - z1 = d x cos y
z2 - z1 = 10.376 x10-2x (-0.65) m
z2 - z1 = -6.7444 x 10-2 m
• z2 = -6.7444 x 10-2 m[z1, = 0] eq.(163)
See fig.(44)
The cartesian coordinates of the points P1 ( x1 , y1 , z1 )
and the P2 ( x2 , y2 , z2 ) located on the circumference of
the circle obtained by the hellion- 3:
Figure 44
The line P1P2 is the diameter of the circle.
Conclusion: At point ‘F‘ the proton fuses with the
deuteron to produce helium-3 nucleus. The produced helium-3 nucleus under the influence of the magnetic lines of force undergo to a circular orbit.
It starts its circular motion from point p1 (x1, y1, z1) [or the point ‘F‘] and reaches at point p2 (x2, y2, z2)
As the helium -3 nucleus reaches at point p2 (x2, y2, z2), it enters into the mouth of the horse pipe located at the point p2 (x2, y2, z2). So, the helium-3 nucleus attains gaseous state and then be extracted out from the tokamak with help of vacuum pump attached to the another end of the horse pipe.
Thus, the helium-3 nucleus is not confined.
For fusion reaction (4):
11H + 21 H +11H [ 43Li ] 23 He +11H + Energy Formation of compound nucleus [ 43Li ]:
1. Interaction of nuclei:
The injected proton reaches at point ‘F‘ and interacts [ experiences a repulsive force due to the confined deuteron and also due to the confined proton that are passing through the point F] with the confined deuteron as well as confined proton that are passing through the point ‘F‘. The injected proton overcomes the electrostatic repulsive force and dissimilarly joins with the confined deuteron.
Similarly, as the confined proton reaches at point ‘F‘, it interacts [experiences a repulsive force due to the injected proton reaching at point ‘F‘ and also due to the confined
deuteron passing through the point ‘F‘] with the injected proton as well as the confined deuteron at point ‘ F‘.
The confined proton overcomes the electrostatic repulsive force and dissimilarly joins with the confined deuteron.
So, at point ‘F’, two protons reach and dissimilarly join with the deuteron.
Thus, at point F, all the three nuclei dissimilarly join with each other.
See fig.(45 and 46)
Interaction Of Nuclei
interaction (1)
Figure 45
Here, to overcome the electrostatic repulsive force, the loss in kinetic energy of the confined proton is equal to the loss in kinetic energy of the injected proton.
Here, the two protons overcome the electrostatic repulsive force and join with deuteron as well as with each other. So, just before fusion, the loss in kinetic energy of the deuteron is taken as zero.
So, the confined deuteron passes through the point ‘F‘ with the same momentum with which it was produced.
Interaction Of Nuclei
interaction -2
Figure 46
All the three nuclei dissimilarly join with each other. Or both the protons dissimilarly join to the deuteron one.
2. Formation of the homogeneous compound nucleus
The constituents (quarks and gluons) of the dissimilarly joined nuclei (The Protons, the deuteron and the proton) behave like a liquid and form a homogeneous compound
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nucleus having similarly distributed groups of quarks with Formation Of Lobes Within Into The Homogeneous
similarly distributed surrounding gluons. Compound Nucleus
Thus within the homogeneous compound nucleus – each group of quarks is surrounded by the gluons in equal proportion. So, within the homogeneous compound nucleus there are 4 groups of quarks surrounded by the gluons.
See fig.(47)
The Homogeneous Compound Nucleus
Figure 47
Where,
= velocity of the compound nucleus
α = 30o from eq.(179)
β = 60o from eq.(180)
y = 90o from eq.(181)
The z-axis is perpendicular to both the axes ‘x‘ and ‘y‘ as well as to the velocity of the compound nucleus ().
Formation of lobes within into the homogeneous compound nucleus or the transformation of the homogeneous compound nucleus into the heterogeneous compound nucleus:
The central group of quarks with its surrounding gluons to become a stable and the next higher nucleus (the helion – 3) than the reactant one ( the deuteron) includes the other two (nearby located) groups of quarks with their surrounding gluons and rearrange to form the ‘A‘ lobe of the heterogeneous compound nucleus.
While the remaining groups of quarks to become a stable nucleus (the proton) includes its surrounding gluons or mass [out of available mass (or gluons) that is not included in the formation of the lobe ‘A‘] and rearrange to form the ‘B ‘ lobe of the heterogeneous compound nucleus.
Thus, Due to formation of two dissimilar lobes within into the homogeneous compound nucleus, the homogeneous compound nucleus transforms into the heterogeneous compound nucleus.
See fig.(48)
Figure 48
The greater one is the lobe ‘A‘ and the smaller one is the lobe ‘B‘ while the remaining space represents the remaining gluons.
Within into the homogeneous compound nucleus, the greater lobe [lobe ‘A‘] represents the helium – 3 nucleus while the smaller lobe [the lobe ‘B‘] represents the proton.
4. Final stage of the heterogeneous compound nucleus
The process of formation of lobes creates void (s) between the lobes. So, the remaining gluons [the mass that is not involved in the formation of any lobe] rearrange to fill the void (s) between the lobes and thus the remaining gluons form a node between the dissimilar lobes of the heterogeneous compound nucleus.
Thus, the reduced mass (or the remaining gluons) keeps both the dissimilar lobes of the heterogeneous compound nucleus joined them together
So, finally the heterogeneous compound nucleus becomes like a dumb – bell.
See fig.(49 and 50)
Final Stage Of The Heterogeneous Compound Nucleus
final stage (1)
Figure 49
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Final Stage Of The Heterogeneous Compound Nucleus
final stage (2)
Figure 50
Calculation for the compound nucleus [43 Li] For fusion reaction (4) is:
11H + 21 D +11H [ 43Li ] 23 He +Δm + 11H
1 The minimum kinetic energy (Em) required by a proton: In this case, the two protons reach at point ‘ F ‘ and fuse with the deuteron one so, there is a loss in kinetic energy of both the protons but the loss in kinetic energy of hte deuteron is zero. So, just before fusion, there is a same change in momentum of both the protons but the momentum of the confined deuteron remains same as with which it was produced (and then confined to). For the above described fusion reaction, a proton has to overcome the electrostatic
repulsive force exerted by the other two positive charges. So,
Em = Ep-3 he = Ep-p x Z22
Ep-p = 2.5 kev from eq.(2)
Z2 = 2
• Em = 2.5kev x 22
Em = 10.0 Kev eq.(164)
NOTE: Even though, the confined proton and the
confined deuteron have the different time periods, they may pass through the point ‘F’ at a same time and fuse. meanwhile, if the injected proton also reach at point ‘F’, then the above fusion reaction is likely to happen.
2 Just before fusion, the kinetic energy of the proton [either it is injected or confined]:
Just before fusion, the proton [either it is injected or confined] to overcome the electrostatic repulsive force, loses its energy equal to 10.0 Kev. from eq.(164)
So, just before fusion, the kinetic energy (Eb) of the proton is –
Eb = Einjected - Eloss
Einjected = EP =102.4 kev from eq.(4)
Eloss = Em = 10.0 kev from eq.(164)
Eb = [102.4 - Kev] – [10.0 kev]
Eb = 92.4 kev
Eb = 0.0924 Mev eq.(165)
3. Just before fusion, the momentum of the proton [either
it is injected or confined] is:
Pb = [2mp x Eb] ½
Eb = 0.0924 Mev from eq.(165)
=0.0924 x 1.6 x10-13 J
Pb = [2 x 1.6726 x 10-27 kg x 0.0924 x 1.6 x 10-13 J] ½
Pb = [0.494554368 x 10-40] ½ kgm/s
Pb = 0.7032 x 10-20 kgm/s eq.(166)
4. Just before fusion, the components of the momentum (Pb) of the proton [either it is injected or confined].
Each proton [either it is injected or confined
when reaches at point ‘F‘ makes angle 30o with the x-
axis, 60o angle with the y- axis and 90o angle with the z-axis. so,
Just before fusion, the x- component of momentum (pb)
of the proton is –
= Pb cos α
Pb = 0.7032 x 10-20 kgm/s from eq.(166)
Cos α = cos 30o = 0.8660
= 0.7032 x 10-20 x 0.866 kgm/s
• = 0.6089 x 10-20 kgm/s eq.(167)
Just before fusion, the y- component of momentum (Pb) of the proton is -
= Pb cos β
Cos β = cos 60o = 0.5
= 0.7032 x 10-20 x 0.5 kgm/s
= 0.3516 x 10-20 kgm/s eq.(168)
Just before fusion, the z - component of momentum (Pb) of the proton is -
= Pb cos y
Cos y = cos 90o = 0
= 0.7032 x 10-20 x 0 kgm/s
= 0 kgm/s eq.(169)
5. Final momentum of the compound nucleus (PCN):
According Just before Just before Just before Final
to - Fusion , the Fusion , the Fusion , the Momentum of
momentum momentum momentum the compound
of the of the () of the nucleus
injected confined confined ( )
proton deuteron proton
0 1 2 3 4 = 1+2+3
X – axis = = = = 2.4909x10-20
0.6089x10-20 1.2731x10-20 0.6089x10-20 kgm/s
kgm/s kgm/s kgm/s eq.(170)
y – axis = = = = 1.438x10-20
0.3516x10-20 0.7348x10-20 0.3516x10-20 kgm/s
kgm/s kgm/s kgm/s eq.(171)
z – axis = 0 kgm/s = 0 kgm/s = 0 kgm/s = 0 kgm/s
eq.(172)
from eq.(167,168,169) from eq.(81,82,83) from eq.(167,168,169)
respectively respectively respectively
Table 4
In this case there is a same loss in kinetic energy of both the protons (either it is injected or confined) but the loss in kinetic energy of the deuteron is zero. So, the momentum of the confined deuteron remains same as with which it was produced to.
Mass of the compound nucleus (M):
M = mp + md + mp
[1.6726 x 10-27 + 3.3434 x 10-27 + 1.6726 x 10-27] kg
= 6.6886 x 10-27 kg eq.(173)
7. The components of the velocity of the compound
nucleus ( ):
x – component of the velocity of the compound nucleus
• = VCN cos α = PCN cos α /M = MVCN cos α /M = /M
= 2.4909 x 10-20 kgm/s from eq.(170)
M = 6.6886 x 10-27 kgm/s from eq.(173)
• = 2.4909 x10-20 kgm/s = 0.3724 x 107 m/s eq.(174)
6.68860-27 kg
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y– component of the velocity of the compound nucleus:
= VCN cos β= PCN cos β / M = MVCN cos β/M = / M
-20 kgm/sfrom eq.(171)=1.438x10
M = 6.6886 x 10-27 kgm/s
= 1.438 x10-20 kgm/s = 0.2149 x 107 m/s eq.(175) 6.68860-27 kg
z - component of the velocity of the compound nucleus
= VCN cos y = MVCN cos y /M = PCN cos y / M = /M
= 0 kgm/s from eq.(172)
M = 6.6886 x 10-27 kgm/s
• = 0 . kgm/s = 0 m/s eq.(176)
6.6886 x 10-27 kg
8. Velocity of the compound nucleus (VCN)
V2CN = Vx2 + Vy2 + Vz2
Vx = 0.3724 x 107 m/s from eq.(174)
Vy = 0.2149 x 107 m/s from eq.(175)
Vz = 0 m/s from eq.(176)
V2CN = (0.3724 x 107)2 + (0.2149 x 107)2 + (0)2 m2/s2
(0.13868176 x 1014) +(0.04618201 x 1014) +0 m2/s2
• V2CN = 0.18486377 x 1014 m2/s2 eq.(177)
• VCN = 0.4299 x 107 m/s eq.(178)
Angles that make the velocity of the compound nucleus (VCN) with respect to positive x, y, and z axes.
with x – axis
cos α = VCN cos α/ VCN = /VCN
= 0.3724 x 107 m/s from eq.(174)
VCN = 0.4299 x 107 m/s from eq.(178)
cos α = 0.3724 x 107 m/s = 0.8662 0.4299 x 107 m/s
• α = 30o eq.(179)
2 with y– axis
cos β = VCN cos β / VCN = / VCN
= 0.2149 x 107 m/s from eq.(175)
cos β = 0.2149x 107 m/s = 0.4998
0.4299 x 107 m/s
• β = 60o eq.(180)
3 with z– axis
cos y= = VCN cos y / VCN = /VCN
= 0 m/s from eq.(176)
cos y = 0 m/s = 0
0.
4299 x 107 m/s
y = 90o eq.(181)
The splitting of heterogeneous compound nucleus:
The heterogeneous compound nucleus, due to its instability, splits according to the lines parallel to the direction of the velocity of the compound nucleus ( ) into three particles – the helium -3 nucleus, the proton and the
reduced mass (Δm)
Out of them, the two particles (helion – 3 and the proton) are stable while the third one (reduced mass) is unstable.
According to the law of inertia, each particle that has separated from the compound nucleus, has an inherited velocity ( ) equal to the velocity of the compound nucleus ( ).
So, for conservation of momentum
M = (mhe-3 + Δm + mp)
Where,
M = mass of the compound nucleus
= velocity of the compound nucleus
mhe-3 = mass of the helium- 3 nucleus mp = mass of the proton Δm = reduced mass
See fig.(51,52)
The Splitting Of The Heterogeneous Compound Nucleus
Splitting -1
Figure 51
The heterogeneous compound nucleus splits according to the lines parallel to the velocity of the compound nucleus ( ).
The Splitting Of The Heterogeneous Compound Nucleus
Splitting – 2
Figure 52
The heterogeneous compound nucleus splits into three particles –The helium-3 nucleus, the reduced mass (Δm) and the proton.
= inherited velocity of the particle with which it has separated from the compound nucleus.
Inherited velocity ( ) of the particles
Each particles that has separated from the compound nucleus has an inherited velocity ( ) equal to the velocity of the compound nucleus ( ).
Inherited Velocity Of The Helium – 3 Nucleus
Vinh = VCN = 0.4299 x 107 m/s from eq.(178)
Components of the inherited velocity of the helium – 3 nucleus
1 = Vinh cos α = VCN cos α = 0.3724 x 107 m/s from
eq.(174) eq.(182)
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2 = Vinh cos β = VCN cos β = 0.2149 x 107 m/s from eq.(175) eq.(183)
3 = Vinh cos y = VCN cos y = 0 m/s from eq.(176)
eq.(184)
Inherited Velocity Of The Proton
Vinh = VCN = 0.4299 x 107 m/s
Components of the inherited velocity of the proton
1 = Vinh cos α = VCN cos α = 0.3724 x 107 m/s from eq.(174) eq.(185)
2 = Vinh cos β = VCN cos β = 0.2149 x 107 m/s from eq.(175) eq.(186)
3 = Vinh cos y = VCN cos y = 0 m/s from eq.(176)
eq.(187)
Inherited Velocity (Vinh ) Of The Reduced Mass (Δm)
Vinh = VCN = 0.4299 x 107 m/s Propulsion Of The Particles
The reduced mass converts into energy and the total energy (ET) is divided between the produced particles [the helion-3 and the proton] according to their inverse mass ratio.
The total energy propel both the particles with equal and opposite momentum according to a ray (line) perpendicular to the direction (line) of the velocity of the compound nucleus see fig.(53)
1 Reduced mass is equal to the subtraction of the sum of the masses of the reactant nuclei and the sum of the masses of the product nuclei.
Δm = [mp+md + mp] - [mHe-3 + mp]
mHe-3 = 3.01493 amu
mp = 1.00727 amu
md = 2.01355 amu
Δm = [1.00727+2.01355+1.00727] – [3.01493 +1.00727]
amu
Δm = [4.02809] – [4.0222] amu
Δm = 0.00589 amu eq.(188)
Δm = 0.00589 x 1.6605 x 10-27 kg
Δm = 0.009780345 x 10-27 kg eq.(189)
1 Inherited kinetic energy (Einh) of the reduced mass.
Einh = ½ Δm V2inh = ½ Δm V2 CN
Δm = 0.009780345 x 10-27 kg from eq.(189)
V 2inh =V2 CN = 0.18486377 x 1014 m2/s2 from eq.(177)
Einh = ½ x 0.009780345 x 10-27 x 0.18486377 x 1014 J
Einh = 0.00090401572 x 10-13 J
Einh = 0.0005650 Mev eq.(190)
Released energy (ER) ER = Δmc 2
Δm = 0.00589 amu from eq.(188)
ER = 0.00589 x 931 Mev
ER = 5.48359 Mev eq.(191)
2 Total energy ( E T )
E T = Einh + ER
E T = (0.000565 Mev) + (5.48359 Mev) from eq.(190
and 191) resp.
E T = 5.484155 Mev eq.(192)
5. Increased kinetic energy of the particles
The total energy (ET) is divided between the particles according to their inverse mass ratio. So,
The increased kinetic energy (Einc) of the helion – 3 Einc = mp x ET
mhe-3 + mp
mp = 1.00727 amu
mhe-3 = 3.01493 amu
ET = 5.484155 Mev from eq.(192)
Einc = 1.00727 x 5.484155 Mev 3.01493+1.00727
Einc = 1.00727 x 5.484155 Mev 4.0222
Einc = 0.25042762667 x 5.484155 Mev
Einc = 1.373383 Mev eq.(193)
(ii) Increased kinetic energy of the proton
Einc = ET - [increased kinetic energy of the helion -3]
ET = 5.484155 Mev from eq.(192)
Einc of helion -3 = 1.373383 Mev from eq.(193)
Einc = [5.484155 Mev] - [1.373383 Mev]
Einc = 4.110772 Mev eq.(194)
6. Increased velocity (Vinc) of the helium – 3 nucleus
Increased velocity (Vinc) of the helium – 3 nucleus.
Vinc = [ 2Einc / mhe-3 ]1/2
Einc = 1.373383 Mev from eq.(193)
Einc = 1.373383x1.6x10-13 J
mhe-3 = 5.00629x10-27 kg
Vinc = 2x1.373383x1.6x10-13 J ½ 5.00629x10-27 kg
• Vinc = 4.3948256x10-13 ½ m/s
5.00629x10-27
Vinc = [0.87786077114 x 1014] ½ m/s
Vinc = 0.9369 x 107 m/s eq.(195)
Increased velocity (Vinc) of the proton.
Vinc = [2Einc / mp] ½
Einc = 4.110772 Mev from eq.(194)
Einc = 4.110772x1.6x10-13 J
• Vinc = 2x4.110772x1.6x10-13 J ½
1.6726x10-27 kg
Vinc = 13.1544704x10-13 ½ m/s 1.6726x10-27
Vinc = [7.86468396508 x 1014] ½ m/s
• Vinc = 2.8044 x 107 m/s eq.(196)
7 Angle of propulsion
1 As the reduced mass converts into energy , the total energy (ET) propel both the particles with equal and opposite momentum according to a ray (line) perpendicular to the direction ( line ) of the velocity of the compound nucleus ( ).
2 We know that when there a fusion process occurs, then we find the lighter nucleus in the forward direction [or in the direction of ion beam or in the direction of the velocity of the compound nucleus ( ).]
Now,
At point ‘ F ‘, as VCN makes 30o angle with x-axis, 60o angle with y-axis and 90o angle with z-axis.
4 So, the proton is propelled making 60o angle with x-axis, 150o angle with y-axis and 90o angle with z-axis.
While the helium -3 nucleus is propelled making 120o angle with x-axis, 30o angle with y-axis and 90o angle with z-axis.
See fig (53)
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Propulsion Of The Particles
According Inherited Increased Final velocity
to - Velocity () Velocity () () = (+ ()
X – axis = = - = -
0.3724x107 0.4684x107 0.096x107m/s
m/s m/s eq.(203)
y – axis = = =
0.2149x107 0.8113x107 1.0262x107m/s
m/s m/s eq.(204)
z – axis = 0 m/s = 0 m/s = 0 m/s
eq.(205)
from eq.(182,183,184) from eq.(197,198,199)
resp. resp.
Table 5
2 For Proton
According Inherited Increased Final velocity
to - Velocity () Velocity () () = () + ()
X – axis = = =
0.3724x107 1.4022x107m/s 1.7746x107m/s
m/s eq.(206)
Figure 53 y – axis = = - = -
0.2149x107 2.4286x107m/s 2.2137x107m/s
= The direction of the increased velocity ( ). m/s eq.(207)
The reduced mass converts into energy and propel both
z – axis = 0 = 0 m/s =0 m/s
the particles according to a ray (line) perpendicular to the
m/s eq.(208)
direction (line) of the velocity of the compound nucleus (
). from eq.(185,186,187) from eq.(200,201,202)
is perpendicular to z- axis. resp. resp.
8. Components of the increased velocity) of particles. Table 6
Components of the increased velocity ( ) of helium -3 10. Final velocity (Vf ) of the helium – 3 nucleus
nucleus Vf 2 = Vx2 + VY2 +VZ2
1 = Vinc cos α Vx= 0.096X107 m/s from eq.(203)
Vinc = 0.9369 x 107 m/s from eq.(195) Vy= 1.0262X107 m/s from eq.(204)
cos α = cos 120o = -0.5 Vz= 0 m/s use fig.(53) from eq.(205)
= 0.9369 x 107 x (- 0.5) m/s Vf 2 = (0.096X107)2 + (1.0262X107)2 +(0)2 m2/s2
= -0.4684 x 107 m/s eq.(197) Vf 2 = (0.009216X1014)+(1.05308644X1014)+0 m2/s2
2 = Vinc cos β Vf 2 =1.06230244X1014 m2/s2 eq.(209)
cos β = cos 30o = 0.8660 use fig.(53) Vf = 1.0306x107 m/s eq.(210)
= 0.9369 x 107 x 0.866 m/s Final kinetic energy (Ef) of the helium – 3 nucleus
= 0.8113 x 107 m/s eq.(198) Ef = ½ mhe-3 Vf2
3 = Vinc cos y mhe-3 = 5.00629X10-27 kg
Cos y = cos 90o = 0 use fig.(53) Vf 2= 1.06230244x1014 m2/s2 from eq.(209)
= 0.9369 x 107 x 0 m/s = 0 m/s eq.(199) Ef = ½ X 5.00629X10-27 X 1.06230244X1014 J
Components of the increased velocity () of protons = 2.65909704117X10-13 J
1 = Vinc cos α = 1.6619 Mev
Vinc = 2.8044 x 107 m/s from eq.(196) • mhe-3 Vf2 = 5.00629X10-27 x 1.06230244 x 1014 J
cos α = cos 60o = 0.5 use fig.(53) =5.3181X10-13 J eq.(211)
= 2.8044 x 107 x 0.5 m/s Forces acting on the helium – 3 nucleus
= 1.4022 x 107 m/s eq.(200) 1 Fy = q Vx BZ sin θ
2 = Vinc cos β q = 2 x 1.6 x 10-19 C
cos β = cos 150o =- cos 30o = -0.8660 use fig.(53) = -0.096 x 107 m/s from eq.(203)
= 2.8044 x 107 x (- 0.866) m/s = - 1 Tesla from eq.(15)
= -2.4286 x 107 m/s eq.(201) sin θ = sin 90o = 1
3 = Vinc cos y • FY = 2 x1.6 x 10-19 x 0.096 x 107 x 1 x 1 N
cos y = cos 90o = 0 use fig.(53) = 0.3072 x 10-12 N
= 2.8044 x 107 x 0 m/s = 0 m/s eq.(202) Form the right hand palm rule, the direction of force is
9. Components of the final velocity () of the particles according to positive y axis.
1 For helium – 3 nucleus So, = 0.3072 x 10-12 N eq.(212)
2 Fz = q Vx By sin θ
= 1 Tesla from eq.(16)
sin θ = sin 90o = 1
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Fz = 2 x1.6 x 10-19 x 0.096 x 107 x 1 x 1 N
= 0.3072 x 10-12 N
Form the right hand palm rule, the direction of force is according to positive z- axis. (that is upward)
So, = 0.3072 x 10-12 N eq.(213)
3 Fx = q Vy BZ sin θ
= 1.0262 x 107 m/s from eq.(204)
= -1 Tesla
sin θ = sin 90o = 1
Fx = 2 x 1.6 x 10-19 x 1.0262 x 107 x 1 x 1 N
3.2838 x 10-12 N
Form the right hand palm rule, the direction of force is
according to positive x- axis.
So, = 3.2838 x 10-12 N eq.(214)
See fig (54)
Forces Acting On The Helium – 3 Nucleus
Figure 54
4. Resultant force (FR) acting on the particle [when
See fig.(55)
The circular orbit to be followed by the helium-3 nucleus:
By seeing the directions of forces [ , ] acting on the helium-3 [when the helium-3 is at point ‘ F‘], we reach at the conclusion that the the confined circular orbit to be followed by the helium-3 nucleus lies in the region made up of positive x-axis, positive y-axis and the positive z-axis.
Where,
Che-3 = center of the circular orbit followed by hellion -3
= The direction of the resultant force acting on the hellion -3, when the hellion -3 is at point ‘F‘.
‘ F‘ is the point where helion-3 is produced.
Figure 55
Angles that make the resultant force () [acting on the particle when the particle is at point ‘F ‘] with positive x, y and z-axes respectively.
particle helium – 3 is at point ‘F‘].
FR2 = Fx2 + FY2 +FZ2
Fx = 3.2838 x 10-12 N from eq.(214)
F = Fy = Fz = 0.3072 x 10-12 N from eq.(212,213)
FR2 = Fx2 + 2F2
FR2 = (3.2838 x 10-12)2 + 2 (0.3072 x 10-12)2 N2
FR2 = (10.78334244 x 10-24 ) + 2 (0.09437184 x 10-24) N2
FR2 = (10.78334244 x 10-24) + (0.18874368 x 10-24) N2
FR2 = 10.97208612 x 10-24 N2
FR = 3.3124 x 10-12 N eq.(215)
See fig(54)
5. Radius of the circular orbit to be followed by the
helium – 3 nucleus:
r = mv2/ FR
mv2 = 5.3181 x 10-13 J from eq.(211)
FR = 3.3124 x 10-12 N from eq.(215)
• r = 5.3181 x 10-13 J
3.3124 x 10-12 N
• r = 1.60551 x 10-1 m
• r = 16.0551 x 10-2 m eq.(216)
For helium- 3 nucleus 1 with x- axis
Cos α = FR cos α / FR = / Fr
= 3.2838 x 10-12 N
Fr = 3.3124 x 10-12 N
Cos α = 3.2838 x 10-12 N
3.3124 x 10-12 N
Cos α = 0.9913
α = 7.4 degree 2 with y- axis
Cos β = FR cos β / FR = / Fr
= 0.3072 x 10-12 N
Cos β = 0.3072 x 10-12 N
3.3124 x 10-12 N
• Cos β = 0.0927
• β = 84.7 degree
3 with z- axis
Cos y = FR cos y / FR = / Fr
= 0.3072 x 10-12 N
Cos y = 0.3072 x 10-12 N
3.3124 x 10-12 N
• Cos y = 0.0927
from eq.(214)
from eq.(215)
eq.(217)
from eq.(212)
eq.(218)
from eq.(213)
eq.(219)
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y = 84.7 degree See fig.(56)
The resultant force ( ) makes angle α, β, y with positive
x, y, z –axes respectively.
Where,
α = 7.4 degree from eq.(217)
β = 84.7 degree from eq.(218)
y = 84.7 degree from eq.(219)
The cartesian coordinates of the points P1 (x1, y1, z1) and P2 (x2, y2, z2) located on the circumference of the circle to be followed by the helium-3.
• Where, the line P1 P2 is the diameter of the circle.
Unit of distance is in meter.
Figure 56
The cartesian coordinates of the points P1 (x1, y1, z1) and P2 (x2, y2, z2) located on the circumference of the circle obtained by the helium-3 nucleus.
cos α = x2 - x1
d
d = 2 x r
r = 16.0551x10-2 m from eq.(216)
2x16.0551x10-2 m
32.1102 x 10-2 m
Cos α = 0.9913 from eq.(217)
• x2 - x1 = d x cos α
• x2 - x1 = 32.1102 x10-2x0.9913 m
• x2 - x1 = 31.8308 x 10-2 m
• x2 = 31.8308 x 10-2 m [x1 = 0] eq.(220)
cos β = y2 - y1 d
cos β = 0.0927 from eq.(218)
• y2 - y1 = d x cos β
• y2 - y1 = 32.1102 x10-2x 0.0927 m
• y2 - y1 = 2.9766 x 10-2 m
• y2 = 2.9766x 10-2 m [y1 , = 0] eq.(221)
cos y = z2 - z1 d
cos y = 0.0927 from eq.(219)
• z2 - z1 = d x cos y
z2 - z1 = 32.1102 x10-2x 0.0927 m
• z2 - z1 = 2.9766 x 10-2 m
• z2 = 2.9766 x 10-2 m[z1 , = 0] eq.(222)
See fig.(57)
Figure 57
CONCLUSION:
Two protons [one is injected and the another is confined] and confined deuteron fuse to form an unstable homogeneous compound nucleus – the lithion -4.
The lithion -4 splits to produce two stable particles – one is helium -3 and the other is proton.
As soon as the helium-3 nucleus is produced at point ‘F‘, under the influence of magnetic lines of force, it undergo to a circular orbit.
It starts its circular motion from point p1 [or F(0,0,0,) ] and reaches at point p2 (x2, y2, z2).
As it reaches at point p2 (x2, y2, z2), in spite of completing its circle, it enters into the mouth of the horse pipe which is located at the point p2 (x2, y2, z2).
Then it attains the gaseous state in the horse pipe and thus be extracted out of the tokamak with the help of vacuum pumps attached to the another end of the horse pipe.
Confinement of the produced proton:
1. Final velocity of the produced proton
Vf2 = Vx2 + VY2 +VZ2
Vx= 1.7746X107 m/s from eq.(206)
Vy= 2.2137X107 m/s from eq.(207)
Vz= 0 m/s from eq.(208)
Vf2 = (1.7746X107)2 + (2.2137X107)+(0)2 m2/s2
Vf2=(3.14920516X1014)+(4.90046769X1014)+0m2/s2
2 = 8.04967285X1014 m2/s2eq.(223)Vf
Vf = 2.8371x107 m/s eq.(224)
2. Final kinetic energy (Ef) of the proton
E = ½ mP Vf2
mP = 1.6726X10-27 kg
Vf 2= 8.04967285x1014 m2/s2 from eq.(223)
= ½ X1.6726X10-27 X 8.04967285X1014 J
E= 6.73194140445X10-13 J
E=4.2074Mev
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Mp Vf2 = 1.6726X10-27 x 8.04967285 x 1014 J
-13 Jeq.(225)=13.4638X10
3. Forces acting on the proton
1 Fy = q Vx BZ sin θ
q = 1.6 x 10-19 C
= 1.7746 x 107 m/s from eq.(206)
= - 1 Tesla from eq.(15)
sin θ = sin 90o = 1
FY = 1.6 x 10-19 x 1.7746 x 107 x 1 x 1 N
2.8393 x 10-12 N
Form the right hand palm rule, the direction of force is
according to nagative y axis.
So, = -2.8393 x 10-12 N eq.(226)
2 Fz = q Vx By sin θ
= 1 Tesla from eq.(16)
sin θ = sin 90o = 1
Fz = 1.6 x 10-19 x 1.7746 x 107 x 1 x 1 N
2.8393 x 10-12 N
Form the right hand palm rule, the direction of force is according to negative z- axis. [That is downward].
So, = -2.8393 x 10-12 N eq.(227)
3 Fx = q Vy BZ sin θ
= -2.2137 x 107 m/s from eq.(207)
= -1 Tesla
sin θ = sin 90o = 1
Fx = 1.6 x 10-19 x 2.2137 x 107 x 1 x 1 N
3.5419 x 10-12 N
Form the right hand palm rule, the direction of force is
according to negative x- axis.
So, = -3.519 x 10-12 N eq.(228)
See fig.(58)
4. Resultant force (FR) acting on the proton [when the
proton is at point ‘F‘] .
FR2 = Fx2 + FY2 +FZ2
Fx = 3.5419 x 10-12 N from eq.(228)
F = Fy = Fz = 2.8393 x 10-12 N from eq.(226,227)
FR2 = Fx2 + 2F2
FR2 = (3.5419 x 10-12 )2 + 2 (2.8393 x 10-12 )2 N2
FR2 = (12.54505561 x 10-24) + 2 (8.06162449 x 10-24 ) N2
FR2 = (12.54505561 x 10-24) + (16.12324898 x 10-24) N2
FR2 = 28.66830459 x 10-24 N2
FR = 5.3542 x 10-12 N eq.(229)
5. Radius of the circular orbit to be followed by the
proton: = mv2/ FR
r
mv2 = 13.4638 x 10-13 J from eq.(225)
FR = 5.3542 x 10-12 N from eq.(229)
• r = 13.4638 x 10-13 J
5.3542 x 10-12 N
• r = 2.51462 x 10-1 m
• r = 25.1462 x 10-2 m eq.(230)
See fig.(58)
6. Conclusion: For proton that has produced
Conclusion: By seeing the directions of the forces [, ,] being applied on the proton, we reach at the conclusion that the confined circular orbit to be followed by the proton lies in the region made up of negative x – axis, negative y- axis and negative z – axis where the magnetic fields are not applied.
In trying to follow the confined circular orbit, the proton reaches in the region where the magnetic fields are not
applied. So as soon as the proton gets rid of the magnetic fields, it starts its linear motion.
So, the proton starts its circular motion from the point ‘F‘ and after travelling a negligible circular path, it give up the circular motion and then travel downward to strike to the base wall of the tokamak.
Thus the produced proton is not confined See fig.(58)
The confinement of the produced proton:
The produced proton is not confined:
Figure 58
By the directions of the components of the resultant force [, ,] acting on the proton, we have come to know that the circular orbit to be followed by the proton lies in a region that is made by negative x, negative y and negative z-axis where the magnetic fields are not applied.
So, the proton starts its circular motion from point ‘F‘ and reaches in a region where the magnetic fields are not applied. Thus, the proton gets rid of the magnetic fields and hence finally moves downward to strike to the base wall of the tokamak.
Thus, the produced proton is not confined.
The power produced
To calculate the heat energy produced, we will consider the main fusion reactions only.
1 11 H + 11 H 12 H + e+ + νe
[ Einjected] [ Einjected ] [ Einh ] [ Einh ] [0.4189 Mev]
2 11 H + 12 H 23 He + y rays + 5.4841 Mev
[ Einjected ] [ Einh ]
ignoring the inherited kinetic energy of positron and the energy carried away by the neutrino, the energy produced is –
E produced = 3 11 H 23 He + y rays + 5.4841 Mev eq.(231) Conclusion: 3 protons fuse to form a Helium-3 nucleus
and to produce 5.4841 Mev.
Total input energy
Each proton is injected with 0.1024 Mev energy. from eq.(4)
So, the total input energy carried by the 3 injected protons is –
Einput = 3x 0.1024 Mev
• E input = 0.3072 Mev eq.(232)
Net yield energy
Net yield = Eproduced - Einput
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Net yield = 5.4841 Mev - 0.3072 Mev from eq.(231,232)
resp.
Net yield = 5.1769 Mev eq.(233) Conclusion: The 3 protons fuse and give us the net yield
5.1769 Mev.
VBM fusion reactor and the power produced
The 3 protons fuse to yield 5.1769 Mev. From eq.(233) Or the 3 protons fuse to yield 5.1769 x 1.6 x 10-13 J
Then, if the 3 x 1019 protons fuse per second then the power produced is-
= 5.1769 x 1.6 x 10-13 x 3 x 1019 J/s 3
P = 8.2830 x 106 J/s
P = 8.2830 MW
Horse pipe and the extraction of the undesired charged particles:
A horse pope is a hollow cylindrical pipe that allows the gaseous atoms to pass through it.
The one end (mouth) of the horse pipe is located in the main tokamak at the point P2 (x2, y2, z 2) where the Confined charged particle (s) reaches. While the another end is located outside the main tokamak.
Now as the confined charged particle, to follow the circular orbit, reaches from point P1 (0, 0, 0) to point P2 (x2, y2, z2), it enters into the mouth of the horse pipe. So, the charged particle at point ‘P2‘, in spite of completing its circle, strike to the horse pipe. Thus the confined charged particle imparts its all the heat energy to the horse pipe which in turn transfers this heat energy to the tokamak.
The slowed charged particle gets free electron (s) and become a gaseous atom.
With the help of vacuum pumps, attached to the another end of horse pipe, we can get the undesired gaseous atoms (helion -3 or helium - 4 gas) on the another end of the horse pipe which is located at the outside of the main tokamak.
XIII. SUMMARY
Either the charged particles are confined or not confined, the cartesian coordinates of the charged particles and the extraction of the undesired charged particles (the helium -3) with the help of horse pipe:
The injected proton remains confined within into the tokamak.
The radius of the circular orbit followed by the confined proton is 3.498 x 10-2 m.
The Cartesian coordinates achieved by the confined
proton are –
P1 (0, 0, 0) and P2 (2.58 x 10-2 m, - 4.54 x 10-2 m, -4.54 x 10-2 m)
For fusion reaction
11 H + 11 H 12 H + e+ + νe
[ injected ] [ confined ]
a. FOR THE DEUTERON
The deuteron remains confined within into the tokamak.
The radius of the circular orbit followed by the confined deuteron is 6.943 x 10-2 m.
The Cartesian coordinates achieved by the confined deuteron are –
P1 (0, 0, 0) and P2 (5.13 x 10-2 m, - 9.02 x 10-2 m, -9.02 x 10-2 m)
FOR THE POSITRON
The positron annihilates with the free electron to produce a pair of gamma ray photon.
FOR THE NEUTRINO
The produced neutrino does not interact with any particle and pass out from the tokamak.
For fusion reaction
11 H + 12H 23 He + gamma rays
[ injected ] [ confined ]
FOR THE HELIUM – 3 NUCLEUS
The helium -3 nucleus is not confined.
The radius of the circular orbit to be followed by the helium -3 nucleus is 5.188x10-2 m.
The Cartesian coordinates achieved by the helium -3 nucleus are –
P1 (0, 0, 0) and P2 (3.8391 x 10-2 m, - 6.7444 x 10-2 m, - 6.7444 x 10-2 m)
The helium -3 nucleus starts its circular motion from
point P1 (x1, y1, z1) and reaches at point P2 (x2, y2, z2).
As the helium – 3 nucleus reaches at point P2 (x2, y2, z2), it enters into the mouth of the horse pipe and thus be extracted out of the tokamak.
Where,
P1 (x1, y1, z1) = P1 (0, 0, 0)
And P2 (x2, y2, z2) = P2 (3.8391 x 10-2 m, - 6.7444 x 10-2 m, -6.7444 x 10-2 m)
e. FOR THE GAMMA RAYS
The gamma rays heats the tokamak.
For the fusion reaction
11 H + 12H + 11 H 23 He + 11 H
[injected] [confined] [confined]
f. FOR THE HELIUM – 3 NUCLEUS
The helium -3 nucleus is not confined within into the tokamak.
The radius of the circular orbit to be followed by the helium – 3 nucleus is 16.0551 x 10-2 m.
The Cartesian coordinates achieved by the helium – 3 nucleus are –
P1 (0, 0, 0) and P2 (31.83 x 10-2 m, 2.97 x 10-2 m, 2.97 x 10-2 m)
The helium – 3 nucleus starts its circular motion from point P1 (x1, y1, z1) and reaches at point P2 (x2, y2, z2).
As the helium – 3 nucleus reaches at point
P2 (x2, y2, z2), it enters into the mouth of the horse pipe and thus be extracted out of the tokamak.
Where,
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P1 (x1, y1, z1) = P1 (0, 0, 0)
And P2 (x2, y2, z2) = P2 (31.83x 10-2 m, 2.97 x 10-2 m, 2.97 x 10-2 m)
g. FOR THE PROTON
The produced proton is not confined within into the tokamak.
The produced proton starts its circular motion form point ‘F‘ [or the point P1 (0, 0, 0)] and then reaches in a region where the magnetic fields are not applied. So, the proton give up its circular motion and then travel downward to strike to the base wall of the tokamak.
VBM fusion reactor and the lawson’s criterion:
VBM fusion reactor has a particular point – the center of
fusion (or the point ‘F‘) that governs all the fusion reactions. So, the VBM fusion reactor exceeds the value of Lawson ‘s criterion [ne te or ne T te] while in any other thermonuclear fusion reactor, to achieve the Lawson’s criterion is still a challenge.
Mode Of Output
Figure 59
The heat is transferred by a water – cooling loop from the tokamak to a heat exchanger to make steam.
The steam will drive electrical turbines to produce electricity.
The steam will be condensed back into water to absorb
more heat from the tokamak.
REFERENCES
Books of NCERT, New Delhi, India
ITER: The Journey to Nuclear Fusion
Nuclear fusion: www.splung.com
RF Linac: Introduction, proton linear accelerators
Cyclotron radiation: chapter 5
Charged particle in a magnetic field: http; farside.ph.utexas:edu
Magnetic fields due to currents: chapter 20
Portable neutron generators: http/www.sciner.com/ Neutron-generators-basics.htm
Helmholtz coil: http/en.wikipedia.org/wiki/Helmholtz-coil
Duoplasmatron: Taking a closer look at LHC:
Some Notes on SI Vs Cgs Units: By Jason Harlow
planes and direction cosines: Geospatial science
Nuclear Reaction: Wikipedia, the free encyclopedia
Helium -3: Fusion without radioactivity: Sci Forums.com
Positron: Wikipedia the free encyclopedia
Annihilation: Wikipedia the free encyclopedia
How stuff works “How nuclear fusion reactor work“
Proton – Proton chain reaction: Wikipedia the free encyclopedia
Isotopes of hydrogen: Wikipedia the free encyclopedia
Isotopes of helium: Wikipedia the free encyclopedia
Beta decay –wikipedia the free encyclopedia
Posted Date:0000-00-00 00:00:00
Posted By: BADRILAL MANMYA