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):
1H + 1H + e- ® 2 H + n + 1.44 MeV
2 D + 1H ® 3 He + g + 5.49 MeV
3 He + 3 He ® 4 He + 1H + 1H + 12.85 MeV
The 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. 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] c2
Q = [ 2.01565 u – 2.015941u] c2
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 106 metric tones per second
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”:
1H + 1H +1H + 1H ® 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:
4He + 4He ® 8Be + g (- 95 keV)
8Be + 4He ® 12C + 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.
If the atomic mass of beryllium 8 (8Be) = 8.00531u, verify that the first part of the triple-alpha fusion process is endothermic and has the value given.
Q = [ 2(4.00260 u) – 8.00531 u] c2
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
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.
In 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(xb)~ exp(- cx)
So, visualizing this behavior as shown below:
With the "barrier" at height V, so we can visualize the particle of lesser energy K, moving from the left side of the E-axis "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.
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 1026 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)
 See, e.g. Physics Today: Reports, April, 1995, p. 19.
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