I've read that free neutrons can decay into hydrogen, but it's rare because the energy from the decay usually sends the electrons away, unable to bind with the protons. But if trillions of free neutrons formed and then decayed around the same time in reasonably close proximity, could most of the decay result in hydrogen formation?
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$\begingroup$ Neutrons decay into a proton, an electron and an anti-electron neutrino. The decay energy is around 780keV if I am not mistaken and that means that the proton and the electron will not form a neutral hydrogen atom unless they can thermalize in the surrounding matter. $\endgroup$– FlatterMannCommented Jun 30 at 16:01
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1$\begingroup$ @FlatterMann you are wrong. I was also thinking like you did, and was so very shocked to find out the truth. $\endgroup$– naturallyInconsistentCommented Jun 30 at 16:05
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$\begingroup$ @naturallyInconsistent How do I get rid of the excess energy in the final products? From kinematics we mostly end up with a proton and an electron that are flying apart rapidly, unless most of the energy is in the anti-electron neutrino? Is that what happens? Maybe my intuition is wrong, then. $\endgroup$– FlatterMannCommented Jun 30 at 16:08
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1$\begingroup$ @FlatterMann physics.stackexchange.com/a/784220/364064 $\endgroup$– naturallyInconsistentCommented Jun 30 at 16:14
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1$\begingroup$ Related: What would happen to a room full of neutrons? $\endgroup$– PM 2RingCommented Jun 30 at 19:31
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2 Answers
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Let's just crank it out with:
$$ n \rightarrow H + \bar{\nu_e} $$
I'm going to set $m_{\nu}=0$, for ...reasons.
We have:
$$ p^{\mu}_n = (m_n, 0) $$
going to:
$$ p^{\mu}_H = (E', p') $$ $$ p^{\mu}_{\bar\nu_e} = (p', -p') $$
which is trivial in the space part, the time component says:
$$ E' + p' = m_n $$
or
$$ p'^2 + M_H^2 = (m_n-p')^2 = m_n^2 + p'^2 - 2m_np'$$
or:
$$ p' = \frac{m_n^2 - M_H^2}{2m_n} = 0.78145\,{\rm MeV}$$
while
$$ \Delta E = m_n-M_H = 0.78178\,{\rm MeV} = 1.00042 p'$$
which I think means there is not a lot of phase space for that neutrino.
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$\begingroup$ This has nothing to do with the question. The OP explicitly stated that it is rare for the free neutron to decay into bound hydrogen. That part is not going to change just by having a gas of free neutrons. It stands to reason that other mechanisms are intended. $\endgroup$ Commented Jul 1 at 0:47
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Definitely, almost obviously.
Most free neutron decay would send the electron out, with only a rare subset having bound hydrogen.
However, at large enough numbers, then an electron that had already been flung out, could lose energy by Bremsstrahlung near a second proton, and then be slowed down enough to be captured by a third proton.
answered Jun 30 at 16:08
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$\begingroup$ Where does that magic second, third etc. proton come from? $\endgroup$ Commented Jun 30 at 16:10
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1$\begingroup$ The OP said that there is a lot of neutrons; they all decay, and will make the protons $\endgroup$ Commented Jun 30 at 16:13
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$\begingroup$ Trillions of neutrons is not a lot of hydrogen in the final state. Avogadro's number (6e23/mol) is very, very large compared to trillions and high energy electrons require quite a bit of matter to be thermalized. $\endgroup$ Commented Jun 30 at 16:16
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$\begingroup$ @FlatterMann that kind of nitpicking is not of any use. The OP had not stated how many trillions; the word is thus more generally applicable as large multiples of Avogadro's number would do. $\endgroup$ Commented Jun 30 at 16:21
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1$\begingroup$ I wasn't nitpicking. The OP stated trillions. Not to mention that the thermal energy of the recoiling hydrogen atoms will heat the gas to a plasma state anyway, even based on your calculation. So even if we assume opacity, we still end up with a plasma rather than atomic hydrogen. $\endgroup$ Commented Jun 30 at 16:25