I am curious: I read lots of places where proton/boron 11 fusion generates 3 alphas. Why wouldn't it produce very stable carbon 12? I can understand if the nucleus is unstable after being rammed by a proton, but that implies to me that if the capture is just above the repulsive force that the nucleus would settle down rapidly into stable carbon 12. If the stability is determined by the impact of the proton, is there any research (theoretical or otherwise) on what kind of impact would be necessary to give a high probability that the nucleus would split into the alphas?
Thanks!
Why p + 11B != 12C ?
Re: Why p + 11B != 12C ?
I can tell you why.. this reaction proceeds at 96 kv and up.
The temperature is in the 10's of millions of degrees.
The carbon can never form due to the fact that the alpha particles have more than enough energy to tunnel away.
(Don't hit me it's quantum physic's fault)
In nature the carbon atom is more energetic in total energy than the 3 alphas...in our universe ma nature tends to lowest energy.
We have a whole field of study called thermodynamics to to explain it. The entropy is less for the 3 alphas than the single
carbon atom. Less energy for more entities.... we call it entropic balance....heat death for short.
Fusion is Fun!
Larry Leins
Fusion Tech
The temperature is in the 10's of millions of degrees.
The carbon can never form due to the fact that the alpha particles have more than enough energy to tunnel away.
(Don't hit me it's quantum physic's fault)
In nature the carbon atom is more energetic in total energy than the 3 alphas...in our universe ma nature tends to lowest energy.
We have a whole field of study called thermodynamics to to explain it. The entropy is less for the 3 alphas than the single
carbon atom. Less energy for more entities.... we call it entropic balance....heat death for short.
Fusion is Fun!
Larry Leins
Fusion Tech
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JohnCuthbert
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Re: Why p + 11B != 12C ?
Are you sure you have that the right way round?
S=K ln W
where W is the number of microstates. Surely there are more possible arrrangements of 3 He nuclei than 1 C.
(Though I have to say I'm not very good at thermodynamics)
S=K ln W
where W is the number of microstates. Surely there are more possible arrrangements of 3 He nuclei than 1 C.
(Though I have to say I'm not very good at thermodynamics)
Re: Why p + 11B != 12C ?
If carbon is more energetic than the alphas, why doesn't it spontaneously decay? I could understand if there was a period of instability just after the fusion event and some fraction of the newly formed carbon would split into the alphas, but the idea that all carbon nuclei are less stable than the alphas doesn't make sense to me.
Re: Why p + 11B != 12C ?
Its the exact same type of scenario with D-D fusion. He-4 is stable, but with the reaction D+D->He4+Gamma, the fraction of D-D reactions where this occurs is around .001% (i dont know the exact but it is fractions of a percent) with He3+n and T+p being the usual reaction products.
Re: Why p + 11B != 12C ?
I think it basically boils down to the fact that you're walloping the Boron so hard that it gets blown to bits. If you could *gradually* place a proton into a boron nucleus, you might get carbon 12. But you can't.
Think of it this way: The Boron nucleus is sitting in the bottom of a hole that's surrounded by a tall hill. If you don't throw the proton hard enough to reach the top of the hill, it never has the chance to fall into the hole. (Ok, a very slight chance, due to tunneling.) If it does make it to the top of the hole, falling down the hole provides enough energy that the resulting carbon 12 nucleus flies apart.
Think of it this way: The Boron nucleus is sitting in the bottom of a hole that's surrounded by a tall hill. If you don't throw the proton hard enough to reach the top of the hill, it never has the chance to fall into the hole. (Ok, a very slight chance, due to tunneling.) If it does make it to the top of the hole, falling down the hole provides enough energy that the resulting carbon 12 nucleus flies apart.