Recent posts: D. Cubero. Thermal equilibrium in special relativity
Powered by MaxBlogPress 

344. Thermonuclear fusion. Coulomb barrier and reaction rates

Print This Post Print This Post   Save This Post as PDF                                


This 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 \sigma_{\rm max}, E<1{\rm MeV} energy of incoming particle, corr. to \sigma_{\rm max}
p+p\to{}d+e^++\nu 2.2 MeV 10^{-23} barn
p+d\to{}{}^3He+\gamma 5.5 MeV 10^{-6} barn
p+t\to{}^4He+\gamma 19.7 MeV 10^{-6} barn
d+d\to{}t+p 4.0 MeV 0.16 barn 2.0 MeV
d+d\to{}^3He+n 3.3 MeV 0.09 barn 1.0 MeV
d+d\to{}^3He+\gamma 24.0 MeV
d+t\to{}^4He+n 17.6 MeV 5.0 barn 0.13 MeV
t+t\to{}^4He+2n 11.3 MeV 0.10 barn 1.0 MeV

Note that reactions involving light particles like p, d ({}^2He nucleus) and t ({}^3He 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:

Effective cross-section for nuclear reactions

How to calculate the total cross-section of a nuclear reaction? There are two factorized contributions into \sigma_{\rm max}. 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

E\sim\frac{Z_1Z_2e^2}{R},

where Z_1e and Z_2e are electric charges of nuclei. Even for smallest Z_1 and Z_2 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 10^7\div{}10^8 K, which corresponds to 1\div{}10 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)

\exp\left(-\frac{2\pi{}Z_1Z_2e^2}{\hbar{}v}\right),

where v=\sqrt{2E/\mu} and \mu=m_1m_2/(m_1+m_2).

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.

  • Digg
  • Reddit
  • StumbleUpon
  • Technorati

If you liked the post, please kindly consider to leave a comment, subscribe to the RSS feed or get new posts sent directly to your Inbox. If you want to chat with me in real time, you can find me on Twitter. The posts below are probably related to the subject of this one:

379. Thermonuclear fusion: list of posts
346. Thermonuclear fusion. Nuclear reaction rates – second part
355. Introduction into thermonuclear reactors
361. NEQNET: first two weeks of April
358. Thermonuclear reactors. Inertial confinement

RSS feed | Trackback URI

6 Comments »

Comment by Lubos Motl
2009-04-09 16:29:28

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.

Comment by Dmitry
2009-04-09 16:39:50

t indeed, fixed!

Thanks,
Dmitry.

 
 
Comment by Elbasher
2009-04-24 19:25:44

Very nice. But what about fusion of heavy nuclei?

Thank you

Comment by Dmitry
2009-04-24 21:22:55

Dear Elbasher,

You mean fission, not fusion, I guess, don’t you?

Cheers,
Dmitry.

 
 
Please, enter your name (required)
e-mail (required - never shown publicly)
URI
or login via Facebook by clicking the button below
Your comment (smaller size | larger size)
For LaTeX in your comment, please use tags [tex] and [/tex]. Also, you may use the following HTML tags: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <strike> <strong> .

« Back to text comment
or subscribe me to comments RSS feed

Trackback responses to this post