344. Thermonuclear fusion. Coulomb barrier and reaction rates
Save This Post as PDFThis post is the next in the series devoted to the discussion of our main energy source in the 22 century – thermonuclear fusion 
Today let us talk a bit about reaction rates. Somehow, it is accustomed that we estimate these rates in terms of the maximal effective cross-section of the reaction. Here are some important and most common reactions that happen in Sun (as well as their cross-sections):
| reaction | energy released |  ,  |
energy of incoming particle, corr. to |
 |
2.2 MeV |  barn |
|
 |
5.5 MeV |  barn |
|
 |
19.7 MeV | barn |
|
 |
4.0 MeV | 0.16 barn | 2.0 MeV |
 |
3.3 MeV | 0.09 barn | 1.0 MeV |
 |
24.0 MeV | ||
 |
17.6 MeV | 5.0 barn | 0.13 MeV |
 |
11.3 MeV | 0.10 barn | 1.0 MeV |
Note that reactions involving light particles like p, d (
nucleus) and t (
nucleus) are rather low. The last column basically shows how much energy you need to pump into the system in orde to start nuclear reaction – recall that 1 eV is about 10000 K.
The picture below shows how the effective cross-sections for different reaction behave with increasing the projectile energy:
How to calculate the total cross-section of a nuclear reaction? There are two factorized contributions into . First, you need to overcome the Coulomb barrier (estimation of the associated probability is an easy task for any person who is familiar with quantum mechanics). Second, the probability that nuclear transformation actually happens should also be taken into account – this one is much harder to estimate, so I postpone discussion of this contribution till tomorrow.
The probability to overcome the Coulomb barrier is estimated as follows. The height of the barrier is given by
,
where 
and 
are electric charges of nuclei. Even for smallest 
and 
possible (equal to 1 as in the reaction d+d), this height is about 200 keV.
On the other hand, the temperature of plasma in the center of a star (like our Sun) is about 
K, which corresponds to 
keV. So, we have to conclude that in stars Coulomb barrier is overcomed because of the quantum tunnelling
When the energies of particles participating in the reaction are much lower than the height of the Coulomb barrier, the probability of tunnelling is given by Gamov exponent (first instanton discovered, whether you want it or not)
,
where 
and 
.
Interestingly, in contemporary experiments this simple expression ceases to describe physics properly. The reason is the presence of nuclei in the beam with energies higher or comparable with the height of the Coulomb barrier.
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Dear Dmitry, is the last reaction, “t+4″, supposed to be “t+t”? Although I admit, colliding a tritium against the number four would be more exciting. You would surely create “pi” and 0.86 of pseudo-GeV would be left.
t indeed, fixed!
Thanks,
Dmitry.
Very nice. But what about fusion of heavy nuclei?
Thank you
Dear Elbasher,
You mean fission, not fusion, I guess, don’t you?
Cheers,
Dmitry.