3 ms·

Perhaps the same reason neutrons bound in atoms don't decay.

by iwwr 15y ago

pavel_lishin 15y ago
I wish he hadn't glossed over this. What is it that keeps them stable in atoms? Edit: Wikipedia isn't very helpful, I may just be too dumb to understand this paragraph: "When bound inside of a nucleus, the instability of a single neutron to beta decay is balanced against the instability that would be acquired by the nucleus as a whole if an additional proton were to participate in repulsive interactions with the other protons that are already present in the nucleus. As such, although free neutrons are unstable, bound neutrons are not necessarily so. The same reasoning explains why protons, which are stable in empty space, may transform into neutrons when bound inside of a nucleus."
- Zaak 15y ago
  The basic idea is that a neutron all by itself has a lower energy state available to it, which is to decay into a proton, an electron, and an antineutrino. On the other hand, in the nucleus of a stable atom, a neutron decaying would result in a higher energy state for the whole nucleus. Does that help?
- neutronicus 15y ago
  The difference can be explained by potential energy. A neutron decaying in empty space causes a proton and an electron to wink into existence on level ground. A neutron decaying near a proton causes a proton to wink into existence on top of a cliff (the proton's electromagnetic field) and an electron to wink into existence at the bottom of a trench (ditto). As far as what critical line is crossed, I can't really say. In fact, I'm not even sure if QCD / Electroweak Theory are yet able to calculate that.
- nagrom 15y ago
  In some senses, protons exert an "electrostatic pressure" on each other. The bound neutron cannot become a proton, because of this electrostatic pressure applied by its companion protons - if it were to become a proton, it would need to get over the hurdle of that "pressure". In empty space, this "pressure" doesn't exist, so the neutron can easily slide into a proton-electron-antineutrino state. As an analogy, when you're hot (say temperature t1) you take your jacket off. However, if you are in a really cramped room with a lot of other people, you'll need to become a lot warmer (say temperature t2) to go through the hassle of removing your jacket and elbowing your neighbour in the eye. If the room is air-conditioned so that you only ever get to temperature t1.5 (where t1 < t1.5 < t2), you'll not take your jacket off, even though you would if you were in a free state.

Perhaps the same reason neutrons bound in atoms don't decay.

by iwwr 15y ago

Perhaps the same reason neutrons bound in atoms don't decay.

I wish he hadn't glossed over this. What is it that keeps them stable in atoms? Edit: Wikipedia isn't very helpful, I may just be too dumb to understand this paragraph: "When bound inside of a nucleus, the instability of a single neutron to beta decay is balanced against the instability that would be acquired by the nucleus as a whole if an additional proton were to participate in repulsive interactions with the other protons that are already present in the nucleus. As such, although free neutrons are unstable, bound neutrons are not necessarily so. The same reasoning explains why protons, which are stable in empty space, may transform into neutrons when bound inside of a nucleus."

Zaak 15y ago
The basic idea is that a neutron all by itself has a lower energy state available to it, which is to decay into a proton, an electron, and an antineutrino. On the other hand, in the nucleus of a stable atom, a neutron decaying would result in a higher energy state for the whole nucleus. Does that help?
neutronicus 15y ago
The difference can be explained by potential energy. A neutron decaying in empty space causes a proton and an electron to wink into existence on level ground. A neutron decaying near a proton causes a proton to wink into existence on top of a cliff (the proton's electromagnetic field) and an electron to wink into existence at the bottom of a trench (ditto). As far as what critical line is crossed, I can't really say. In fact, I'm not even sure if QCD / Electroweak Theory are yet able to calculate that.
nagrom 15y ago
In some senses, protons exert an "electrostatic pressure" on each other. The bound neutron cannot become a proton, because of this electrostatic pressure applied by its companion protons - if it were to become a proton, it would need to get over the hurdle of that "pressure". In empty space, this "pressure" doesn't exist, so the neutron can easily slide into a proton-electron-antineutrino state. As an analogy, when you're hot (say temperature t1) you take your jacket off. However, if you are in a really cramped room with a lot of other people, you'll need to become a lot warmer (say temperature t2) to go through the hassle of removing your jacket and elbowing your neighbour in the eye. If the room is air-conditioned so that you only ever get to temperature t1.5 (where t1 < t1.5 < t2), you'll not take your jacket off, even though you would if you were in a free state.