ALine Posted December 4, 2020 Posted December 4, 2020 (edited) I remember reading somewhere about neutrons decaying after about 10-15 minutes of being free from any atomic interaction. This got me thinking that if you were to somehow reflect neutrons in a concentrated source then you would be able to form neutron fusion due to the neutrons decaying into protons. And because they are past the coulombic barrier they would be able to undergo fusion due to the weak force. This feels wrong however I am unsure why. Please correct me on where I am messing up. Thank you for your time and patience in answering my question. Edited December 4, 2020 by ALine change of title
swansont Posted December 4, 2020 Posted December 4, 2020 A pair of Neutrons will not form a bound state. There is no “neutron fusion”
ALine Posted December 4, 2020 Author Posted December 4, 2020 thank you for the correction, apologies for using the term "neutron fusion." What I meant by that was neutrons decaying into protons in the same region forming fusion. It was only quicker for me to use that bunched together term in order to give an assigned name to it. I will try and be more careful next time with my wording. Also I will do some reading up on bound states to get a better grasp on the concept. Thank you for the quick reply and your time for answering this question swansont.
swansont Posted December 4, 2020 Posted December 4, 2020 I don’t know what process you are envisioning. Quote neutrons decaying into protons in the same region forming fusion. Forming fusion? What is undergoing fusion?
ALine Posted December 4, 2020 Author Posted December 4, 2020 ahhh, blasted words. you have failed me for the last time. 0) At t = 0, 2 neutrons (n1, n2) are formed by some means and placed inside a container (c) using some means. 1) At t = 1, 2 different neutron particles (n1, n2) are brought near one another while inside of c by some means, separated by a certain distance (xf). 2) At t = td, 1 neutron particle (n) will decay into 1 proton particle (p). 3) when 2 proton particles (p1, p2) are within a certain distance from one another, that distance being (xf), they may fuse. 4) when 2 proton particles (p1, p2) fuse, the resultant will be 1 deuterium atom, (d) 1 positron (e+) , and 1 neutrino (neu). 5) assuming that n1 and n2 are brought near one another to a distance xf as initially stated and a certain amount of time td has passed while n1 and n2 are inside c, both n1 and n2 will decay into p1 and p2 while remaining a distance xf away from one another. Because they are at a distance (xf) away from each other and time td has been reached they may fuse, forming 1 (d), 1 (e+) and 1 (neu).
Bufofrog Posted December 4, 2020 Posted December 4, 2020 8 minutes ago, ALine said: 3) when 2 proton particles (p1, p2) are within a certain distance from one another, that distance being (xf), they may fuse. 4) when 2 proton particles (p1, p2) fuse, the resultant will be 1 deuterium atom, (d) 1 positron (e+) , and 1 neutrino (neu). They won't get close to each other and fuse at anything remotely close to room temperature. 1
swansont Posted December 4, 2020 Posted December 4, 2020 9 minutes ago, ALine said: ahhh, blasted words. you have failed me for the last time. 0) At t = 0, 2 neutrons (n1, n2) are formed by some means and placed inside a container (c) using some means. 1) At t = 1, 2 different neutron particles (n1, n2) are brought near one another while inside of c by some means, separated by a certain distance (xf). 2) At t = td, 1 neutron particle (n) will decay into 1 proton particle (p). 3) when 2 proton particles (p1, p2) are within a certain distance from one another, that distance being (xf), they may fuse. 4) when 2 proton particles (p1, p2) fuse, the resultant will be 1 deuterium atom, (d) 1 positron (e+) , and 1 neutrino (neu). 5) assuming that n1 and n2 are brought near one another to a distance xf as initially stated and a certain amount of time td has passed while n1 and n2 are inside c, both n1 and n2 will decay into p1 and p2 while remaining a distance xf away from one another. Because they are at a distance (xf) away from each other and time td has been reached they may fuse, forming 1 (d), 1 (e+) and 1 (neu). Why bother with steps 1 and 2? Protons are easy to come by. The reason we don’t bother with this is that p-p fusion is really unlikely. The average proton in the core of the Sun waits 9 billion years before it successfully fuses with another proton. https://en.wikipedia.org/wiki/Proton–proton_chain_reaction It’s only because there are a lot of protons in the sun’s core that there’s significant reaction rate, and that’s only possible because the large mass lets you confine them 1
ALine Posted December 4, 2020 Author Posted December 4, 2020 ok then gotcha, thank you for the assist.
MigL Posted December 5, 2020 Posted December 5, 2020 I think his intention is to get neutrons close enough together, as opposed to protons which have coulombic potential to overcome, such that when they decay, the resultant protons are close enough already to fuse. Not sure if neutrons in such close proximity are not equivalent to a bound state ( if at all possible ), in which case there would be no decay. 2
ALine Posted December 8, 2020 Author Posted December 8, 2020 On 12/4/2020 at 11:04 PM, MigL said: I think his intention is to get neutrons close enough together, as opposed to protons which have coulombic potential to overcome, such that when they decay, the resultant protons are close enough already to fuse. Not sure if neutrons in such close proximity are not equivalent to a bound state ( if at all possible ), in which case there would be no decay. that was the plan so no cold fusion? darn need to go back to the drawing board.
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