Why is fusion stronger than fission?

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  • #26
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There is no explanation as to "why" any of the fundamental forces exist. They simply do.
[I know this thread is a mix of 6-year-old and month-old messages, but I just spotted it....]

There is a reasonably straightforward explanation for the shape of the nuclear binding energy curve. The Bethe-Weizsäcker mass formula accounts for most of the contributions to atomic mass. However, it does not account for "shell" effects, i.e. the existence of magic numbers that create islands of stability in the chart of the nuclides.... See

http://en.wikipedia.org/wiki/Semi-empirical_mass_formula" [Broken]

BBB
 
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  • #27
Morbius
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[I know this thread is a mix of 6-year-old and month-old messages, but I just spotted it....]

There is a reasonably straightforward explanation for the shape of the nuclear binding energy curve. The Bethe-Weizsäcker mass formula accounts for most of the contributions to atomic mass. However, it does not account for "shell" effects, i.e. the existence of magic numbers that create islands of stability in the chart of the nuclides.... See

http://en.wikipedia.org/wiki/Semi-empirical_mass_formula" [Broken]

BBB
BBB,

Yes - see post #23.

Greg
 
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  • #28
jim hardy
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The easiest way to paint a word picture for answer to this question is to turn the binding energy curve upside down and use a hill analogy.

Hydrogen would then be high on left side of the curve and transuranics high on right(both hilltops), and iron at the valley in middle.

Moving left to right on the curve is fusion, opposite direction is fission.

Next imagine yourself at one or the other end of the curve with a skateboard made of that material(hydrogen or uranium or trans-U), ready to ride it down that hill toward iron.
It's clear that progressing downhill toward iron from either end changes binding energy(potential) into kinetic. Once you've arrived at iron you cannot get any more energy out by either fusion or fission. To move past iron you must put energy back in.

Judging by steepness of hills, which direction is the more thrilling ride?

old jim
 
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  • #29
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a minor clarification if I might.
Simply put - In fusion you overcome electrostatic repulsion to release strong nuclear interaction's energy. In fission you overcome strong nuclear force that is holding nucleus together to release electrostatic energy of the nucleus. The energy that you get from fission comes from the fission fragments repelling each other. The energy that you get from fusion comes from nuclei snapping to each other after you brought them close. For the bigger nuclei than iron, the electrostatic energy is larger than strong nuclear interaction's energy.
There are many different fusion reactions that have different yields.
And also - in a practical fusion reactor, or a nuclear fusion bomb, high energy neutrons from fusion may be used to fission U-238 (which is not fissile with the neutrons from fission); AFAIK in a typical bomb half of the yield comes from such fission.
 
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  • #30
Morbius
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AFAIK in a typical bomb half of the yield comes from such fission.
Dmytry,

I can't tell you what the breakdown is for a typical weapon, because that would be
telling you something about how the weapon works. However, I can point you to
public information on the Sedan nuclear test:

http://en.wikipedia.org/wiki/Sedan_(nuclear_test)

Sedan was a thermonuclear device with a fission yield less than 30% and a fusion yield about 70%

Greg
 
  • #31
Morbius or Greg thank you for your clarification. I am just a student from college but I am very curious of things, which has led me to read and learn things way ahead of the current classes I am taking. This Message is just to answer: No! It is not too hard to understand that the amount of energy released is directly dependent on the particular reaction applied and not whether is fusion or fission. If I stated this incorrectly please let me know. Many of the posts here were confusing and contradicting and I found yours very useful and I admire your knowledge and passion. I am looking forward to learn more from this forum, again I am just a student so I will not contribute much, instead will absorb as much as I can.
 
  • #32
Drakkith
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Morbius or Greg thank you for your clarification. I am just a student from college but I am very curious of things, which has led me to read and learn things way ahead of the current classes I am taking. This Message is just to answer: No! It is not too hard to understand that the amount of energy released is directly dependent on the particular reaction applied and not whether is fusion or fission. If I stated this incorrectly please let me know. Many of the posts here were confusing and contradicting and I found yours very useful and I admire your knowledge and passion. I am looking forward to learn more from this forum, again I am just a student so I will not contribute much, instead will absorb as much as I can.
Of course! The energy released depends entirely on fuel used and the particular reaction that happens. The energy released from each type varies drastically with different fuels.
 

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