What temperature would provide a mean kinetic energy of 0.5 MeV?

What temperature would provide a mean kinetic energy of 0.5 MeV? By comparison, the temperature of the surface of the sun 6000 K.

Interior Zones of the Sun 6,000 K Convective Zone < 0.01 g/cm3 500,000 K Radiative Zone < 0.01-10 g/cm3 8,000,000 K Core < 10-160 g/cm3 15,000,000 K

In simple 1-dimensional case an exponential decay connects the wave functions either side of the barrier V E III II I where

The Nuclear pp cycle 4 protons  4He + 6g+ 2e + 2p 26.7 MeV

The effective kinetic energy spectrum of nuclei in a gas (at thermal equilibrium) is given by the high energy part of the Maxwell-Boltzmann distribution

vz And for each value of energy E = ½ mv2 = ½ m(vx2 + vy2 + vz2 ) vy Notice for any fixed E, m this defines a sphere of velocity points all which give the same kinetic energy. vx The number of “states” accessible by that energy are within the infinitesimal volume (a shell a thickness dv on that sphere). dV = 4v2dv

The probability distribution Maxwell Boltzmann distribution With a root mean square speed of

While the cross-section will have an energy dependence dominated by the barrier penetration probability where the details follow…

In simple 1-dimensional case an exponential decay connects the wave functions either side of the barrier V E x = r1 x = r2 III II I probability of tunneling to here

Where this time we’re tunnelingin with an energy from r2 where: minimum energy to reach the barrier E r2 R which we can just write as

hence with E=Ta becomes then with the substitutions:

Performing the integral yields: and for R « r2 the term in the square brackets reduces to

with E = ½ mv2 can also write as

which I will abbreviate as Cross sections for a number of fusion reactions

The probability of a fusion event at kinetic energy E is proportional to the product of these two functions. N(E) P(E) (E) kinetic energy E Em

which has a maximum where and at that maximum:

Since we can evaluate  , for example, for the case of 2 protons: from which we can see: with kT in keV

with kT in keV If T ~ 106K, kT ~ 0.1 keV for T ~ 108K, kT ~ 10 keV This factor of 100 change in the temperature leads to a change in the fusion rate > 1010!!!

There are actually twodifferent sequences of nuclear reactions which lead to the conversion of protons into helium nuclei.

I. The proton-proton cycle The sun 1st makes deuterium through the weak (slow) process: Q=0.42 MeV then Q=5.49 MeV 2 passes through both of the above steps then can allow Q=12.86 MeV This last step won’t happen until the first two steps have built up sufficient quantities of tritium that the last step even becomes possible. 2(Q1+Q2)+Q3=24.68 MeV plus two positrons whose annihilation brings an extra 4mec2 = 40.511 MeV

II. The CNO cycle Q=1.95 MeV Q=1.20 MeV Q=7.55 MeV Q=7.34 MeV Q=1.68 MeV Q=4.96 MeV carbon, nitrogen and oxygen are only catalysts

Interior Zones of the Sun 6,000 K Convective Zone < 0.01 g/cm3 500,000 K Radiative Zone < 0.01-10 g/cm3 8,000,000 K Core < 10-160 g/cm3 15,000,000 K

The 1st generation of stars (following the big bang) have no C or N. The only route for hydrogen burning was through the p-p chain.  T17 In later generations the relative importance of the two processes depends upon temperature. Rate of energy production  T4 p-p chain sun CNO cycle Shown are curves for solar densities 105 kg m-3 for protons and 103kg m-3 for 12C.

The heat generated by these fusion reactions raises the temperature of the core of the star. The pressure of this "black body" radiation is sufficient to counteract gravitational collapse. However once the hydrogen in the central region is exhausted gravitational collapse resumes. The temperature will rise as gravitational potential energy converts to kinetic energy of the nuclei.

At 108 K helium burning starts fusing: Q=-91.9 keV Q=190 keV Note: these reactions are reversible. 8Be exists as a resonance decaying with  10-16 sec. Its formation requires 91.9 keV kinetic energy shared between the initial states. At T= 108 K the fraction of helium nuclei meeting this threshold is given by the Boltzmann factor e-91.9/kT ~ 2.2  10-5 (with kT= 8.6 keV, the mean thermal energy).

the small branching ratio for the -decay makes it only 4 x 10-4 as likely as a return to the initial state: Once stable 12C has been produced, further  absorption can occur through Q=7.16 MeV Q=4.73 MeV Q=9.31 MeV

As the helium supply in the core is exhausted further collapse leads to even higher temperatures. At ~5 x 108-109 Kcarbon and oxygen fusion can take place. Q=13.9 MeV Q=16.5 MeV and others, which yield protons, neutrons or helium nuclei

During the silicon burning phase (2109 K) elements up to iron are finally produced. Even at these temperatures the Coulomb barrier remains too high to allow direct formation: Instead it is done in an equilibrium process of successive alpha particle absorptions balanced against photo-disintegration: At these temperatures the thermal photons have an average energy of 170 keV and their absorption can easily lead to the break up of nuclei.

Then absorption of the 4He by other 28Si nuclei eventually leads to the build up of 56Ni etc.

Beyond iron, nickel and cobalt there are no more exothermic fusion reactions possible. Heavier elements cannot be built by this process.

What temperature would provide a mean kinetic energy of 0.5 MeV?