What is the energy needed for Deuterium-Tritium fusion to occur?

  • Thread starter Thread starter PhyS194
  • Start date Start date
  • Tags Tags
    Energy Fusion
Click For Summary
SUMMARY

The energy required for Deuterium-Tritium fusion involves overcoming the Coulomb barrier, calculated at 7.11 x 10-14 J. The kinetic energy of the deuteron must match this barrier to initiate the fusion reaction, resulting in a required velocity of 6.5199 x 106 m/s. Momentum is calculated as 2.18 x 10-20 N s. The total energy can be derived using the equation Etot = sqrt((p*c2)2 + (m*c2)2), although initial calculations may yield values around 10-3 J, indicating a need for further verification.

PREREQUISITES
  • Understanding of Coulomb barrier in nuclear physics
  • Familiarity with kinetic energy calculations (Ekin = m*v2/2)
  • Knowledge of momentum (p = m*v)
  • Basic grasp of relativistic energy equations (Etot = sqrt((p*c2)2 + (m*c2)2))
NEXT STEPS
  • Research the implications of the Coulomb barrier on fusion reactions
  • Study the relationship between kinetic and potential energy in particle collisions
  • Explore advanced momentum calculations in nuclear physics
  • Investigate the role of exothermic reactions in Deuterium-Tritium fusion
USEFUL FOR

Students and researchers in nuclear physics, particularly those focusing on fusion energy, as well as educators teaching advanced concepts in particle physics and energy calculations.

PhyS194

Homework Statement


How much total energy, kinetic energy and momentum is needed for Deuterium-Tritium fusion to happen when deuterium is accelerated and tritium is the target(v=0)?
Which exothermic reactions are caused by the fusion?

Homework Equations

The Attempt at a Solution


I've calculated the height of the coulomb barrier to 7.11*10-14
Since the height of the coulomb barrier is the energy needed for the two nucleiis to get close enough to each other for the reaction to start, the kinecting energy of the moving deutron should be as large as the height of the coulomb barrier? In that case, the kinetic energy needed is also 7.11*10-14
Using m=2.0141u = 2.0141*1.660*10-27
Using Ekin=m*v2/2, v =6.5199*106 m/s
Then the momentum can be calculated as p=m*v = 2.18*10-20 N s
Does this appear to be correct? What about the total energy?

I tried to use Etot=sqrt((p*c^2)^2+(m*c^2)^2) but getting something in the range of 10-3, is this correct then maybe?

best regards
 
Physics news on Phys.org
Can you convert the whole kinetic energy to potential energy? What would that imply for the motion of the particles at the point of closest approach, and does that look right?

You should work with units everywhere, that helps spotting errors.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
7K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 13 ·
Replies
13
Views
6K