Help with Understanding D-T Fusion for Class Project

In summary, Tom needed help with his class project on fusion, binding energy, and reactors. He wanted to understand the mathematical explanation for D-T fusion and why his calculations did not match the online output energy of 17.6 MeV. After taking into account initial energies, relativistic masses, and the Coulomb barrier, he still couldn't find the discrepancy. Upon further discussion, it was discovered that Tom was using incorrect values for atomic masses, which led to the 0.5 MeV difference in output energy. It was also noted that Tom should be careful with significant figures and consider the speed of particles in a fusion reactor as a possible factor.
  • #1
toml42
6
0
I need a bit of help here, as part of a class project i was trying to explain how fusion, binding energy, reactors etc work. I always like to go a bit more indepth than the syllabus requires, and i wanted an explanation of how D-T fusion worked mathematically, to clearly show how the binding energy becomes output energy - however, whenever i do the calculations, i don't get the same net energy output as I find online, 17.6 MeV. I always end up with about 17.1.
I tried to compensate for the initial energies, relativistic masses and the coulomb barrier, but I'm still not a huge deal closer. Can anyone give me any clues as to what i may be missing? If nothing is obvious i'll add a brief of summary of the calculations i did.
Thanks, Tom.
 
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  • #2
Well there is no way that we can see what misstake you made if you not show us your work.
 
  • #3
Ok, I'll give a brief summary, i was just wondering if I'd missed something face slappingly obvious. this is before i started trying to correct it with extra bits, they didn't help a huge amount.

BEFORE
Deutrium 2.013004633
Tritium 3.01550062
Total 5.028505253

AFTER
Helium 4.0015
Neutron 1.008664917
Total 5.010164917

Missing amu 0.018340336
In kg 3.04548E-29
E = MC2
energy 2.73714E-12

MeV 17.08
 
  • #4
He-4 atomic mass is 4.00260323 amu, there is your error.
 
  • #5
i used the mass of an alpha particle, which explains the difference, because the electrons shouldn't be involved in a nuclear reaction should they?
 
  • #6
then you must remove the electrons from the deuterim and tritium also. Remember that what is listes are ofte the ATOMIC masses. So one must be careful...and cosistent.

Yes you are right, in reality (in a fusion reactor) only nuclei takes part in the reaction.

Also: Post course work / homework questions in appropirate home work forum, this forum is not for those kind of questions which can be read here: https://www.physicsforums.com/showthread.php?t=171607
 
  • #7
I did remove the electrons from them too, sorry, i should have made that clearer. I didn't think to post it in the homework area because it isn't part of my syllabus, this is more of an outside concern really.
 
  • #8
ok in that case:

nuclear masses before=
5.02905362 amu

Nuclear masses after=
5.010170986 amu

gives 17.58906 MeV
 
  • #9
Right, thanks, that's close enough, and when i use those values it works with my calculations. my values for atomic mass must have been slightly off... for future reference, where do you go to find yours?
Thanks for your help, Tom.
 
  • #11
Thanks so much for your help, that website is really usefull! just out of interest though, can you think of anything that might account for the slight discrepancy we're still getting?

My calculations say that if the particles were traveling at about 2.5 million metres per second at the start of the reaction that would be enough extra energy to account for it... that sounds like a reasonable enough speed for high energy particles in a 100,000,000 K fusion reactor... am i on the right lines here maybe?
 
Last edited:
  • #12
1. The source you have ARE already listing nuclear masses and you substract the electron mass again..

2. Lack of accurateness, use ALL significant figures that you can. 17MeV is a very very small unit (3% of electron mass, and one electron is 1/1800 of a nucleon), so be careful!

3. The source you used have listed totaly wrong numbers for atomic masses.

With the numbers I used, I got 17.6MeV, which is what the answer tells you.
 

1. What is D-T fusion?

D-T fusion, or deuterium-tritium fusion, is a nuclear reaction in which two types of hydrogen isotopes, deuterium and tritium, fuse together to create a helium atom and release a large amount of energy.

2. How does D-T fusion work?

In D-T fusion, the nuclei of deuterium and tritium atoms are brought close together at extremely high temperatures and pressures. This causes them to overcome their natural repulsion and fuse, releasing energy in the form of heat and light.

3. What are the potential benefits of D-T fusion?

D-T fusion has the potential to provide a nearly limitless source of clean energy, as the fuel used (deuterium and tritium) is widely available in seawater. It also produces minimal radioactive waste compared to other forms of nuclear energy.

4. What are the challenges of achieving D-T fusion?

The main challenge in achieving D-T fusion is creating and sustaining the extremely high temperatures and pressures needed for the reaction to occur. This requires advanced technology and significant energy input, which is currently a barrier to making D-T fusion a viable energy source.

5. Are there any current efforts or projects focused on D-T fusion?

Yes, there are several international projects and research initiatives aimed at developing and harnessing D-T fusion as a source of energy. These include the International Thermonuclear Experimental Reactor (ITER) in France and the Joint European Torus (JET) in the United Kingdom.

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