ASTR 498N                     Lecture 13                           Nuclear burning I

 

(online at www.astro.umd.edu/~drabin/)

 

                Hot enough for ya?

               

 

Nuclear binding energy

 

An atomic nucleus composed of Z protons and N neutrons has a smaller mass, m(Z,N), than the total mass of its constituent nucleons.  This mass defect corresponds to a binding energy

 

 

 

A more significant quantity for the discussion of nuclear fusion is the binding energy per nucleon, , which is proportional to the fractional mass defect of the nucleus relative to its constituent nucleons.

 

The figure shows the shape of Q/A as a function of A with some nuclei identified.  Note the rapid rise with atomic weight among the light elements and the broad maximum for heavier elements with a peak value of about 8.8 MeV/nucleon for ⁵⁶Fe (“the iron peak”).   

 

From this figure follow some important potential properties of nuclear fusion and fission in stars (“potential” because we can’t yet say what processes actually occur).

 

1.     If two nuclei and their fusion product all lie to the left of the iron peak, the fusion product has a higher binding energy per nucleon than the original nuclei and, as the total number of nucleons has not changed, the reaction must release energy.

2.     If a succession of fusion reactions occurs in which the atomic weight increases gradually (a few nucleons at a time), further energy release from fusion becomes impossible as the iron peak is approached.

3.     If a nucleus and its two fission products all lie to the right of the iron peak, the newly formed nuclei both have a higher binding energy per nucleon than the initial nucleus and energy is released by the fission.  [For actual fission reactions, one of the products may lie to the left of the peak, in which case we compare Q’s rather than Q/A’s.]

4.     Because the curve is much flatter for nuclei heavier than iron than nuclei significantly lighter than iron, the maximum energy release from fission reactions is much less than the typical release of fusion reactions.  Combined with the predominance by mass of light elements, this indicates that fusion is by far the more important nuclear energy source in stars.

 

Consider the following net transformations and the energy they release:

 

                    H → ⁴He    releasing 6.3 × 10¹⁴ J kg^-1

                    H → ⁵⁶Fe    releasing 7.6 × 10¹⁴ J kg^-1

 

The second transformation represents essentially the maximum energy release that can be obtained from fusion processes.  Since the rest mass energy of 1 kg is 9 × 10¹⁶ J, this maximum release is just under 1% of the rest mass energy.  Note that the usual first step in fusionhydrogen to heliumalready releases over 80% of all the fusion energy available.

 

 

The strong interaction

 

The strong attractive interaction between baryons is a short-range force, effective only for separations of ~1 fm or less (1 fm = 10^-15 m).   Within that range, the strong interaction is much stronger than the electrostatic interaction, hence the name.

 

The existence of stable nuclei depends on a balance between the strong and electrostatic forces.  All the protons in a nucleus repel each other mutually because the Coulomb force is long-range.  On the other hand, a proton can only be attracted to a nearby proton or neutron via the short-range strong force.  Therefore, as the atomic number increases, the relative proportion of neutrons in a stable nucleus must increase in order to keep the protons from flying apart.

 

The weak interaction

 

The weak interaction is also a short-range force.  Where it is effective, it is weaker than the electrostatic interaction, hence the name.  Here are examples of weak interactions:

 

                     

 

The neutrinos in these reactions are all electron neutrinos.  Weak interactions involving muons and muon neutrinos are not significant in stellar evolution.

 

That the weak interaction will play a role in stellar nuclear fusion is apparent from the simple fact that ⁴He has two neutrons, so the creation of a single helium nucleus (alpha particle) from hydrogen must involve the conversion of two protons into neutrons:

 

                     

 

 The neutrinos, although nearly massless, carry energy.  However, because they interact solely through the weak interaction, almost all the neutrinos produced leave the star without any further interaction.  In particular, they do not heat the stellar gas and thereby contribute to its luminosity.  Conventionally, neutrino luminosity is subtracted from the total energy generation rate to arrive at the rate to be used in the stellar structure equation:  .

 

 

 Nuclear reactions and notation

 

Two notations are in common use (often free intermixed in the same paper).  They are illustrated here for three reactions from the proton-proton chain:

 

                                                

 

                                                     

 

                                

 

It is often physically appropriate to consider a reaction such as  to occur in two steps:  the creation of a compound nucleus which, although short-lived, has enough time to reach an internal equilibrium before choosing one of two (or more) decay modes:

 

                     

 

The relative frequency of these decay modes is given by the branching ratio.

 

Allowed nuclear reactions must respect conservation laws, including baryon number, lepton number, charge, angular momentum, and nuclear symmetries (parity and isospin).

 

 

The conundrum of stellar nuclear fusion

 

… is that the most obvious path is blocked in two ways:

 

1.     For the central temperature we expect in a star such as the Sun, the average classical distance of closest approach of two light nuclei, as set by the repulsive Coulomb force, is larger than the distance (~ 1 fm) at which the attractive strong force becomes effective.   Nuclei far in the high-energy tail of the Boltzmann distribution can get close enough to each other, but there are not nearly enough of them.  In other words, apparently most stars are too cold to initiate nuclear burning.

2.     All the compound nuclei created from two-particle combinations of hydrogen and helium are unstable; that is, the reactions

                              

almost always decay back the other way before they do anything else.   

The first barrier is penetrated (literally) by a quantum mechanical effect considered below.  As for the second barrier, to paraphrase Einstein:  everything should be as simple as it can be, but not simpler.

 

 

The Coulomb barrier

 

Consider two nuclei, Z_A and Z_B, approaching each other from infinity, head-on, with kinetic energy E.   Classically, as the particles approach, their relative speed decreases as kinetic energy is converted into potential energy by the repulsive Coulomb force.  When the potential energy equals E, the particles will momentarily come to rest and then bounce back.  This distance of closest approach, r_c, is related to E by

 

           

 

We know that the strong force is only effective for r_c  1 fm, so, thinking of r_c as the independent variable, the height E_C of the Coulomb barrier is, numerically,

 

                     

 

Compare this with the mean thermal energy of a particle at 10⁷ K, of order 1 keV.  The relative frequency of nuclei with thermal energy ~1 MeV is given by the Boltzmann factor, exp(E/kT) or exp(1000).  The Coulomb barrier appears to be insurmountable for the range of central temperatures demanded by hydrostatic equilibrium.

 

If the collision is not head-on, conservation of angular momentum adds an effective centrifugal potential to the potential V(r) governing the radial motion of the particles: 

 

                     

 

That is, the barrier only becomes higher.

 

 

Barrier penetration

 

Idealize the Coulomb barrier as shown in the figure.  In the center-of-momentum frame, the time-independent Schrödinger equation is

 

 

 

where μ is the reduced mass.  The probability that the two nuclei are separated by a distance between r and r + dr is

 

                     

 

Outside the Coulomb barrier, the wave function oscillates sinusoidally.    Within the barrier, V(r) = E_C ,  E_C  E > 0,  and

 

                     

 

(defining χ).  In a central potential V(r), the Schrödinger equation separates into radial and angular equations.  For a head-on approach (no angular momentum), the radial wave function u(r) satisfies an equation of the same form as the one above, .  Remembering that the radial part of the Laplacian in spherical coordinates is

 

                             [ not  ]

 

you can verify that

 

                     

 

satisfies the radial equation.  [The general solution also includes a term representing the wave reflected from the inner boundary r_N, but we ignore it for the purpose of a simple estimate.]  The probability that the nuclei penetrate the Coulomb barrier is roughly given by the relative probability densities in shells at the inner and outer radii,

 

                     

 

Now the Coulomb potential doesn’t look much like a rectangular barrier, but we can approximate it by a sequence of narrow rectangles of variable height (the approach familiar from elementary integral calculus).  The constant barrier height E_C is replaced by the Coulomb potential and the constant χ is replaced by

 

                     

 

The probability of penetrating a single “sub-barrier” at radius r_i is approximately .  To the extent that the penetration of each sub-barrier is an independent probabilistic event, the probability of penetrating the sequence of barriers is

 

                     

 

The integral  can be evaluated by making the substitution , with the result that the net probability of penetrating the Coulomb barrier is

 

                     

 

where the Gamow energy is defined as

 

                     

 

and

                     

 

is the fine structure constant.

 

 

Let’s compare this with our earlier estimate that only exp(1000) of protons at 10⁷ K have enough thermal energy to overcome the barrier classically.  For the fusion of two protons, the Gamow energy is 493 keV.  At 10⁷ K, the typical thermal energy kT is about 1 keV, so the probability of quantum penetration for typical protons is roughly   small, but much larger than the classical estimate.  Protons of higher-than-average kinetic energy will have an even better chance to fuse.

 

At this point you can pick up the development in O&C at Eq. (10.38), having semi-quantitatively derived the exponential dependence of the fusion cross section that O&C introduce by magic.  The cross section is usually written in the form

 

                     

 

The exponential term represents the probability of barrier penetration.  The E¹ factor is motivated by the expectation that the “intrinsic” nuclear cross section will be roughly proportional to the area of a circle with a radius of one de Broglie wavelength,

 

                     

 

The remaining factor, S(E), must be calculated quantum mechanically for each specific nuclear reaction but can ordinarily be expected to vary with energy much more slowly than the full cross section.  However, if the collision energy happens to be very close to an excited internal energy level of the nucleus,  S(E) will peak sharply at that energy.  A resonance of this kind is crucial to the triple-alpha process of helium burning.

 

For non-resonant reactions,  the energy dependence of the fusion probability, and consequently the rate of energy generation, is mainly determined by the product of an exponentially rising barrier penetration factor and an exponentially falling Boltzmann factor, as illustrated below.