Understanding Nuclear Fusion Cross Section Diagrams

Click For Summary
SUMMARY

This discussion focuses on the interpretation of nuclear fusion cross section diagrams, specifically regarding deuterium fusion at high velocities. The fusion cross section at 107 keV is stated to be 8e-30 m², suggesting that deuterium atoms must approach within 1.60 femtometers for fusion to occur. However, the simulation indicates that due to Coulomb repulsion, the closest approach is 65 femtometers, raising questions about the role of tunneling and electron interactions in fusion processes. The consensus is that tunneling significantly contributes to fusion probabilities, which are not accounted for in the initial simulation.

PREREQUISITES
  • Understanding of nuclear fusion principles
  • Familiarity with Coulomb's law and its implications in particle interactions
  • Knowledge of quantum tunneling effects in nuclear reactions
  • Basic proficiency in programming particle simulations
NEXT STEPS
  • Research quantum tunneling in nuclear fusion reactions
  • Explore the role of electron interactions in fusion processes
  • Learn about thermal averages in reaction cross sections
  • Investigate advanced particle simulation techniques for nuclear reactions
USEFUL FOR

Physicists, particle simulation developers, and researchers interested in nuclear fusion processes and their computational modeling.

BrandonBerchtold
Messages
46
Reaction score
6
TL;DR
How are nuclear fusion cross section diagrams supposed to be interpreted?
I'm working on programming a particle simulation that visually shows the nuclear fusion reaction rate of deuterium at different densities and temperatures, but I'm having trouble understanding exactly how nuclear fusion cross section diagrams are supposed to be interpreted. (The simulation assumes the particles are just deuterium atoms with one elementary positive charge each, and the only force acting on them is the coulomb repulsion force.)

Particularly, if I simulate (in 1D) one deuterium atom moving towards another stationary deuterium atom at 2 million meters per second (which corresponds to about 107 keV), the closest the atoms get during the collision before moving apart again is 65 femtometers.

According to the attached diagram, the fusion cross section at 107 keV is 8e-30 m^2, which I believe implies that if one deuterium atom gets within the radius of a sphere with this cross sectional area of the other deuterium atom, the atoms should fuse. However, an area of 8e-30 m^2 requires the atoms to get within 1.60 femtometers of one another, which is much closer than the coulomb repulsion allows the particles to get in my simulation. I am pretty confident in my math and numerical integration of the forces acting on the particles in my simulation so I doubt that they are at fault. How can the fusion cross section radius be smaller than the closest distance coulomb repulsion allows the particles to get during a collision?

I understand that tunneling is the major pathway by which fusion reactions occur but I assumed tunneling is factored into creating the fusion cross section diagrams. Is this incorrect?

Alternatively, I ignored all effects from the electrons (which I assume is probably a poor assumption since they likely play a non negligible role in the fusion process), so is that possibly to blame?
 

Attachments

  • Fusion Cross Section.png
    Fusion Cross Section.png
    11.7 KB · Views: 323
  • Like
Likes   Reactions: Drakkith
Physics news on Phys.org
You're given the total reaction cross section. The reactions per second per volume, ##f##, is then given by

##f = \frac{N^2}{2}\langle \sigma v\rangle##

where ##\langle \sigma v\rangle## is the thermal average of the particle speed, ##v##, times the cross section at that energy, ##\sigma##. ##N## is the particle number per volume.
 
  • Like
Likes   Reactions: BrandonBerchtold
BrandonBerchtold said:
I understand that tunneling is the major pathway by which fusion reactions occur but I assumed tunneling is factored into creating the fusion cross section diagrams.
Correct, but it is not considered in your simulation, that's why you get a difference.
If the protons approach each other to 70 fm then you have a small chance of fusion happening due to tunneling. The chance is small, but the impact parameter can be much larger than 2 fm. An integral would produce the cross section you have.

There is an additional effect you didn't consider yet: What you use is the average kinetic energy. The fusion probability increases rapidly with increasing energy. Fusion processes mainly happen in the tail of the kinetic energy distribution, where both deuterium nuclei have above average energy.
 
  • Like
Likes   Reactions: BrandonBerchtold

Similar threads

  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 29 ·
Replies
29
Views
5K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 19 ·
Replies
19
Views
5K
  • · Replies 3 ·
Replies
3
Views
4K