This document discusses nuclear fusion and fission, focusing on their respective reactions and energy outputs. It explains the conditions for nuclear fusion, particularly in stars like the sun, and highlights various reactions involving deuterium and helium isotopes. Additionally, it addresses the challenges and potential of fusion as a power source, including neutron supply issues and energy partitioning in the reaction products.

1. NUCLEAR FUSION REACTION
Semester Fall 2020
Date ( Dec-1st-2020)
Submitted by : Syed Hammad Ali ( 19012510-085)
Program : MSc-III / D
Course Title : Plasma Physics
Course Code : PHY-451
Department of Physics
University of Gujrat

2. NUCLEAR REACTION:
1. There are two types of reactions:
Nuclear Fission and Nuclear Fusion.
2. Nuclear reaction:
Hard to produce
Powerful
Use neutrons

3. Nuclear Fission process
• Heavy nucleus is bombarded with slow
moving neutrons and two nuclei produced
with more neutrons and high amount of
energy is released.
• Considering U-235 as a heavy element. A
slow moving neutron is set to strike on it
and resulting two lighter nuclei(Ba-144
and Kr-89) produced along with 3 moving
neutrons and high amount of energy
which can be denoted as Q.

4. Nuclear Fusion
Nuclear fusion is when two small, light
nuclei join together to make one heavy
nucleus.
Fusion reactions occur in stars where two
hydrogen nuclei fuse together under high
temperatures and pressure to form a
nucleus of a helium isotope.
There are a number of different nuclear
fusion reactions happening in the Sun.

6. Light Nuclei In Nuclear Reaction:
Light element Fission:
Bombardment of a deuterium nucleus 𝟏
𝟐
𝑯
with a neutron. The relevant reaction can
be written as:
0
1
𝑛 + 1
2
𝐻 → 1
1
𝐻 + 2(0
1
𝑛) + 𝐸
Reaction leads to the desired neutron
multiplication. The energy released as
calculated from the nuclear data show that
E = -2.23 MeV.
Negative sign indicates that energy is not
actually released but must be supplied as
an input to make the reaction take place.
Clearly this would be unacceptable as a
power source.
Light element Fusion:
Hypothetical example again assume that
a neutron collides with a deuterium
nucleus. The resulting
nuclear fusion reaction can be written
as:
0
1
𝑛 + 1
2
𝐻 → 2
3
𝐻𝑒 + 𝑒−
+ 𝐸
In this case the nuclear data show that E
= +6.27 MeV. The fusion reaction is
energetically favorable for power
production.

7. Fusion Is Un-Acceptable For Power Source:
• The reaction consumes neutrons.
• Since there are no readily available sources of neutrons, this reaction is not self-
sustainable.
• Therefore, it too is unacceptable as a practical power source.
How then can light element fusion reactions be
initiated?
• Replace the neutron with another light element; that is, generate a nuclear reaction
by having two light elements bombard each other, for instance two colliding
deuterium nuclei.
• The advantage of this idea is that the lack of a chain reaction is easily overcome by
simply providing a continuous supply of deuterium, which, unlike a neutron supply,
is readily and inexpensively available.

8. Disadvantage Is That For Two Deuterium Atoms:
• Deuterium is an isotope of Hydrogen atom which is proton.
• Proton and proton always repels.
• To undergo a nuclear reaction, their nuclei must be in very close proximity
to each other, typically within a nuclear diameter.
• At these close distances, the inter-particle Coulomb potential produces a
strong repulsive force between the two positively charged nuclei, which
diverts the particle orbits and greatly reduces the likelihood of a nuclear
reaction.

9. If Neglecting Issues:
• If these issues are not considered and attention is focused solely
on the nuclear energy production of various fusion reactions without
regard for how easy or difficult it may be to produce these reactions.
• Studies of the nuclear properties of light element fusion indicate that
three such reactions may be advantageous for the production of
nuclear energy. These involve deuterium, tritium, and helium-3, an
isotope of helium.

10. Fusion In Sun:
Inside the core of sun there is a
small nuclei.
Electrostatic force rip apart the
nucleus because of repulsion of
proton and proton.
The strong nuclear force holds
the nucleus.

11. The sun is much dense part so protons in it so protons collide in
it.
But somehow two protons collide and due to beta decay one
proton converts into neutron.
The Hydrogen-2 forms and cause explosion with release of
energy.
Hydrogen-2 continuously move and came near to another proton
and become Helium-3 with release of tons of energy.
The process continues and inside more helium-3 produces.

12. When two helium-3 nuclei
came close to each other they
came into a better combination
and results in formation of one
single helium-4 and results in
release of energy.
The 1st step is slowest because
we have to wait for occurrence
of beta decay.
The 3rd step, when two helium-
3 collides a helioum-4 nuclei
occurs with release of two
protons.

13. The D–D reaction:
• Nuclear interaction of two deuterium nuclei.
• This is the most desirable reaction in the sense of a virtually unlimited
supply of inexpensive fuel, easily extracted from the ocean.
• The D–D reaction actually has two branches, each occurring with an
approximately equal likelihood.
• The relevant reactions are as follows:

14. • In terms of energy content the two reactions produce 0.82 and 1.01
MeV per nucleon respectively. Macroscopically this is equivalent to 78
× 106and 96 ×106 MJ/kg of deuterium, typical of nuclear energy
yields.

15. The D–𝑯𝒆𝟑
reaction:
• It requires helium-3 as a component of the fuel and there are no natural supplies
of this isotope on earth.
• The reaction is also difficult to achieve, but less so than for D–D.
• The reaction is worth discussing since the end products are all charged particles.
• The reaction is:
• The energy released per reaction is impressive, even by nuclear standards. The
18.3 MeV corresponds to 3.66 MeV per nucleon, which is macroscopically
equivalent to 351 ×106 MJ/kg of the combined D–He3 fuel.

16. The D–T reaction:
• It can be written as:
• D–T reactions produce large numbers of neutrons and require a
supply of tritium in order to be capable of continuous operation, but
there is no natural tritium on earth.
• The tritium is radioactive with a half-life of 12.26 years.
• This corresponds to 3.52 MeV per nucleon and is macroscopically
equivalent to 338 × 106MJ/kg.

17. The one outstanding problem is the tritium supply:
• The solution is to breed tritium in the blanket surrounding the region
of D–T fusion reactions.
• The chemical element that is most favorable for breeding tritium is
lithium.
• The nuclear reactions of primary interest are:

18. • Both reactions produce tritium although the first reaction generates
energy while the second one consumes energy.
• Also natural lithium comprises 7.4% 3
6
𝐿𝑖 and 92.6% 3
7
𝐿𝑖. Even though
there is a much larger fraction of 3
7
𝐿𝑖, nuclear data show that the 3
6
𝐿𝑖
reaction is much easier to initiate and as a result it is this reaction that
dominates in the breeding of tritium.
• For present purposes one should assume that the issues have been
satisfactorily resolved. Consequently, breeding T from 3
6
𝐿𝑖solves the
problem of sustaining the tritium supply, assuming adequate supplies of
lithium are available.

19. Energy partition in fusion reactions:
• The end products which forms after fusion reaction explicit large amount of
energy appear in the form of kinetic energy.
• More importantly for the D–T reaction where one end product is electrically
charged and the other is not.
• The distribution can be easily determined by making use of the well-satisfied
assumption that the energy and momentum of each end product far surpasses that
of the initial fusing nuclei.
• Suppose we have two sub-scripts of end products namely,1 and 2. The initially
fusing particles are at rest. The conservation of energy and momentum relations
before and after the fusion reaction involve only the end products and thus have
the forms.

20. • The end products have the form given below,
⇒
1
2
𝑚1𝑣1
2
+
1
2
𝑚2𝑣2
2
= 𝐸
⇒ 𝑚1𝑣1 + 𝑚2𝑣2= 0
After solving for 𝑣1 and 𝑣2 we got below equations,
⇒
1
2
𝑚1𝑣1
2
=
𝑚2
𝑚1+𝑚2
𝐸
⇒
1
2
𝑚2𝑣2
2
=
𝑚1
𝑚1+𝑚2
𝐸
Energy partition in fusion reactions:

21. Energy partition in fusion reactions:
• As we see from above equations the kinetic energies are apportioned inversely
with the mass, either we can say lighter particle carries most of the energy.
• For example the end product for D-T reaction where E = 17.6 MeV consist of
alpha particle and neutron 𝑚α / 𝑚n = 4.
• Thus the kinetic energy of the alpha particle is equal to (1/5) E = 3.5 MeV,
while
• that of the neutron is equal to (4/5)E = 14.1 MeV. The neutron energy is four
times larger
• than that of the alpha particle.
• Rewrite the D–T fusion reaction in the slightly more convenient form
D + T → α(3.5 MeV) + n(14.1 MeV).
• This reaction that dominates the world’s fusion research program.

22. The binding energy curve and why it has the shape it does:
The nuclear reactions are initiated for
• Heavy metals called fission reaction.
• Light element called fusion reaction.
• The nuclear reaction does not start for intermediate elements.
This explanation is resulted from two observations.
• From binding energy vs atomic mass graph.
• The shape of binding energy curve.

23. The binding energy curve and why it has the shape it does:
• Graph explains binding forces of nuclei of heavy and light elements are weaker
than intermediate elements.
• Shape arises due to geometric competition between strong nuclear forces which
are short range and long range but weak coulombic forces.