** Nuclear
Fusion Reactions**:

In
the Sun, for example, two distinct nuclear fusion processes occur: 1) the
proton-proton cycle, and 2) the carbon-nitrogen cycle. In the first of these (the easier one because
it has fewer reactions):^{ }

^{1}**H + ^{1}H + e^{-} **

**®**

**+**

^{2 }H**n**

**+ 1.44 MeV**

^{2 }**D** + ^{1}H
**®**** ^{3
}He + **

**g**

**+ 5.49 MeV**

^{3 }**He + ^{3 }He **

**®**

^{4 }He +^{1}H +^{1}H + 12.85 MeVThe top line shows two protons fusing to yield deuterium (heavy hydrogen) with a positron and neutrino (n) emitted, along with 1.44 MeV of energy. Empirical evidence of this reaction is obtained from gallium detectors, of the neutrinos given off, which are within 1-2% of what theoretical models predict.[1] In the second fusion reaction, the deuterium combines with a proton to give the isotope helium 3, along with a gamma ray (g) and 5.49 MeV energy. In the final fusion, two helium-3 nuclei combine to yield one helium-4 nucleus, along with two protons, and 12. 85 MeV energy. Note that the two ending product protons commence the cycle anew, so that the generation of nuclear energy is ongoing.

The
ending quantities on the right sides of each part of the cycle denote the Q of
the reaction for that part. Let us check
the Q for the first and simplest part. We know the hydrogen mass *=
1.007825 u and for deuterium we have (from atomic tables): : *^{2 }D = * = 2.01410
u. Then:*

Q
= [ 2(1.007825 u) – 2.01410 u] c^{2}

Q
= [ 2.01565 u – 2.015941u] c^{2}

Q = [2.01565 – 2.01410] 931.5 MeV/u

Q
= [0.00155] 931.5 MeV/u = *1.44 MeV*

The
effect of ongoing fusion reactions such as this, means that the central core of the Sun becomes
heavier and heavier, as more and more helium is produced. This despite the fact
that the Sun as a whole is losing an amount of mass of roughly 4 x 10^{6}
metric tones per second

**Insight
Problem**:

If
the atomic mass for helium 3 (** ^{3 }He**)
is equal to 3.01603 u, then verify the other Q-values for the last two parts of
the proton-proton cycle. A simplified,
compressed “net reaction”:

^{1}H
+ ^{1}H +^{1}H + ^{1}H **®**** ^{4
}He + Energy**

Is sometimes used to evaluate the total energy released in the proton-proton cycle. Compute this energy and compare to the value obtained for the total energy released in the earlier example. Can you account for the difference?

** Nuclear
Fusion Reactions in the Aging Sun**:

At some stage, when nearly the entire solar core is helium a new helium fusion phase will be ushered in (at higher temperature), such that the following reaction series, known as the ‘triple alpha’ process, kicks in:

^{4}**He + ^{4}He **

**®**

^{8}Be +**g**

**(- 95 keV)**

^{8}**Be + ^{4}He **

**®**

^{12}C +**g**

**+ 7.4 MeV**

Here,
the two alpha particles (helium nuclei) first fuse to give unstable beryllium
and a gamma ray (**g****)**,
with 95 keV energy *absorbed*. Then the
beryllium fuses with a helium-4 to give carbon–12 plus a gamma ray and 7.4 MeV
energy given off.

In this way a new cycle commences, leading to a heavier molecular weight core. Each successive burning phase, however, is less efficient than its predecessor, as can be seen by comparing the energy given off in the triple alpha process to the energy given off in the proton-proton cycle. The key thing to bear in mind in terms of a stable phase (i.e. ‘Main sequence’) star like the Sun is that it is in pressure-gravity balance. The outer gas pressure balances the weight of its overlying layers. Any condition likely to disrupt this balance is therefore of paramount interest.

The stable lifetime of the Sun depends on how long before it consumes ninety percent of the hydrogen in its core. Theoretical investigations using data from nuclear reaction rates and cross sections suggest the Sun’s Main Sequence lifetime at 8-10 billion years. Since it already has spent 4.5 billion of those years, there are anywhere from 3.5 to 5.5 billion years remaining.

Once
the triple-alpha process gets underway and the energy balance declines, the Sun
will have to compensate for the lost energy to sustain any kind of balance. Thus,
the Sun’s core must contract and convert gravitational potential energy into
thermal energy. Meanwhile, ignition of
hydrogen burning in the Sun’s outer layers will create radiation pressure that
forces the outer layers to expands. The Sun will then become a “*Red Giant*”
and its new larger surface will be expected to engulf all the planets up to and
including Mars.

** Example
Problem**:

If
the atomic mass of beryllium 8 (** ^{8}Be**)
= 8.00531u, verify that the first part of the triple-alpha fusion process is
endothermic and has the value given.

*Solution:*

Q
= [ 2(4.00260 u) – 8.00531 u] c^{2}

Q
= [ - 0.00011] 931.5 MeV = 0.102465 MeV = - 102.4 keV

Of
course, not taken into account here is the gamma ray (**g**) which also comes off.
Hence we will have:

(-102.4
keV) + (E (**g****)) =
-95.7 keV**

So that:

E
(**g****) =
hc/ ****l**** =
6.7 keV **

Is
the missing energy of the gamma ray photon, with the difference factored in
yielding 95.7 keV.

*Aside: The Problem of the Coulomb Barrier in Solar Fusion**'The Structure and Evolution' of the Stars*’ (Dover, 1958).

**once every 14 billion (14 x 10**. Since the universe itself is only 13.8 billion years old this means it could never occur unless another factor was present to enable it.

^{9}) yearsIn order for thermonuclear fusion to be realized, the Coulomb barrier must be overcome. Fortunately quantum mechanics allows for a certain non-vanishing probability that a particle (say proton) of kinetic energy K, can overcome a barrier of energy V ("

*barrier potential*"), via the process of "

**".**

*quantum tunneling*Note that tunneling is a general feature of low mass systems, such as single proton (H) states.

Consider a deBroglie (matter) wave arising from a single proton (p+) of form:

U(x) ~ sin(kx)

Where x is the particle's linear displacement (e.g. in 1-D) and k, the wave number vector(k= 2π/l), where l denotes the wavelength.

Though the associated kinetic energy

**K < V**(the barrier "height") the wavefunction is *non-zero* within the barrier, e.g.

U(x

_{b})~ exp(- cx)

So, visualizing this behavior as shown below:

** **W*"tunneling*" through to the right side where it may have wave function, U(x) ~sin (kx + φ), where φ denotes a phase angle.

Note that if the barrier is not too much higher than the incident particle energy, and if the mass is small, then tunneling is significant.

It's important here to point out that the penetration of the barrier is a direct result of the wave nature of matter. In effect, this wave nature - which is uniquely quantum mechanical in origin- allows a higher energy barrier to be penetrated by a lower energy particle, something totally without parallel in classical, Newtonian physics!

Even given tunneling, an "offset" is required to reduce the low penetration probability , since clearly the Sun and other stars are shining by fusion.

This 'offset' arrives via enormously high density of protons, e.g. in the core, which: i) increases the probability enormously, since so many more protons are in extremely close proximity, and (ii)enhances temperatures to the point they can be sustained, and continue - thereby building up other fusion reactions to finish the initial one.

*actually penetrate a higher potential energy barrier*, say of energy V(x) > K. Note that the penetration of the barrier is a direct result of the wave nature of matter! (The matter wave form changes in the process of transmission through the barrier, say from an

**exp(-ikx)**function to a

**sin (kx +**

**f)**where f denotes phase angle). In effect, this wave nature - which is uniquely quantum mechanical in origin- allows a higher energy barrier to be penetrated by a lower energy particle, something totally without parallel in classical, Newtonian physics! Note that if the barrier is not too much higher than the incident energy, and if the mass is small, then tunneling is significant. It was insights such as this that paved the way to apprehending how much subtler nature was than hitherto realized, and how many more technological advances could be achieved when the wave nature of matter was factored into designs.

**The Quantum Treatment of the Deuteron.**

_{n}m

_{p}/ (m

_{n}+ m

_{p})

^{2}d/dr (r

^{2}dR/dr) + 2m’/

**ħ**

**[E – V] R = 0**

^{2 }^{2}U/dr

^{2}+ 2m’/ ħ

^{2 }[E – V] U = 0

_{o}and V = 0 outside the well. The diagram below shows how we are treating the deuteron in terms of the function V(r).

_{I}for r

__<__r

_{o}and

_{II}for r

__>__r

_{o}

**Inside the potential well:**

**d**

^{2}**u**

_{I}**/dr**

^{2}+ 2m’/**ħ**

^{2 }**[E + V**

_{o}] u_{I}= 0**a**=

^{2}**2m’/**

**ħ**

^{2 }**[E + V**

_{o}]**so that:**

**d**

^{2}**u**

_{I}**/dr**

^{2}+ a^{2 }u_{I}= 0

**For which we can show:**

_{I }= A cos (ar) + B sin(ar)

_{I }= B sin(ar)

**d**

^{2}**u**

_{II }**/dr**

^{2}+ 2m’/**ħ**

^{2 }**[E] u**

_{II}= 0**b**=

^{2}**2m’/**

**ħ**

^{2 }**[- E ]**

**Since the total energy of the neutron is negative, i.e. being bound to the proton. Then:**

**d**

^{2}**u**

_{II }**/dr**

^{2}- b^{2}u_{II}= 0**u**

_{II }**= C exp (-br) + D (exp(br))**

**u**

_{II }**= C exp (-br)**

**For continuity at r = r**

_{o}and u_{I}= u_{II }:**Thence:**

**du**

_{I}/dr = d**u**

_{II }**/dr**

**Þ**

**aB cos (a r**

_{o}) = -bC e^{-b r o}**a B cos (a r**

_{o}) / -b C e^{-b r o}**tan (a r**=

_{o})**- a/b**

**Quantum Numbers for Deuteron**:

__+__s2 = 0 or 1. The orbital angular momentum quantum number, L (describing the motion in space of the proton and neutron relative to each other) can assume the values L= 0, 1, 2 (i.e. S, P, and D states).

** Suggested Problems**:

1) Calculate the wavelength of the gamma ray photon
(in nm) which would be needed to balance the endothermic part of the triple
–alpha fusion equation. (Recall here that 1 eV = 1.6 x 10 ^{-19} J)

2) Verify the second part of the triple-alpha fusion reaction, especially the Q-value. Account for any differences in energy released by reference to the gamma ray photon coming off and specifically, give the wavelength of this photon required to validate the Q.

3) The luminosity or power of the Sun is measured to
be L = 3.9 x 10^{26} watts. Use
this to estimate the mass (in kilograms) of the Sun that is converted into
energy every second. State any assumptions made and reasoning.

4) a) Show that u _{I } = A cos (ar) + B sin(ar)

^{2}u

_{I}/dr

^{2}+ a

^{2 }u

_{I}= 0.

__cosine solution__needs to be discarded.

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