Why are neutrons in the nucleus stable vs their free counterparts?

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Discussion Overview

The discussion revolves around the stability of neutrons within atomic nuclei compared to free neutrons, particularly focusing on the conditions under which a neutron can decay and the role of binding energy in this process. It touches on theoretical aspects of nuclear physics and the implications of neutron decay in different contexts.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions why free neutrons decay in about 15 minutes while neutrons in the nucleus appear stable.
  • Another participant suggests that if a neutron decays within a nucleus, it would result in a nucleus with one more proton and one less neutron, depending on the energy available for the decay.
  • A participant reiterates the previous point about decay and questions whether the energy required for beta decay is the same for free neutrons and those within a nucleus.
  • One participant provides an example using the 7Li nucleus, explaining that the mass difference between the initial and final nuclei determines whether decay can occur.
  • Another participant states that a neutron in a nucleus derives the energy required for decay from the difference in binding energy between the initial and final states of the nucleus.
  • A participant seeks clarification on whether the mass from the newly created proton affects the nucleus's stability or if they are misunderstanding the concept.
  • One participant explains that the mass of the final nucleus and the emitted electron must be less than that of the original nucleus, mentioning the complexities of nuclear forces and suggesting a simplified model for understanding binding energy.
  • A later reply acknowledges the role of binding energy in the stability of free neutrons, noting that a free neutron has a binding energy of zero, making other states more favorable.

Areas of Agreement / Disagreement

Participants express varying views on the mechanisms behind neutron stability and decay, with no consensus reached on the specifics of how binding energy influences these processes.

Contextual Notes

The discussion includes assumptions about energy requirements for decay, the role of binding energy, and the complexities of nuclear forces, which remain unresolved.

iced199
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If the free neutron decays in only 15 minutes, why are neutrons in the nucleus attached to protons stable? This has always bewildered me.


Just a side question, are 'tetraneutrons' stable or even possible?
 
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Because if a neutron inside a nucleus decayed, it would leave a nucleus having one more proton and one less neutron. Whether this can actually happen depends on whether there is enough energy available, i.e. whether the mass of the first nucleus exceeds the mass of the second.
 
Bill_K said:
Because if a neutron inside a nucleus decayed, it would leave a nucleus having one more proton and one less neutron. Whether this can actually happen depends on whether there is enough energy available, i.e. whether the mass of the first nucleus exceeds the mass of the second.

But still, surely a free neutron requires the same amount of energy to undergo Beta decay?
 
In order for a decay process to happen there must be at least as much rest mass in the original nucleus as in the end products. Here's an example:

The 7Li nucleus has a mass of 7.016003 amu. If one of its neutrons beta-decayed, the result would be a 7Be nucleus, which has one less neutron and one more proton. But 7Be has a mass of 7.016926 amu, which is more. So there's not enough energy for this process to happen.
 
In effect, a neutron in a nucleus gets the energy required for the decay, from the difference in binding energy between the initial and final nuclei.
 
So is the extra mass from the energy of the newly created proton opposing the nucleus or am i completely missing this?
 
The sum of the masses of the final nucleus and the electron has to be less than the mass of the original nucleus. Changing a neutron into a proton does increase the overall "internal electrical repulsion" of the nucleus, which would tend to increase the mass, but that's far from the whole story because of the strong nuclear force. Nuclear structure is a complicated subject. For a simplified model, look up the "semi-empirical binding energy formula." (Keep in mind that binding energy is negative.)
 
jtbell said:
In effect, a neutron in a nucleus gets the energy required for the decay, from the difference in binding energy between the initial and final nuclei.

Thanks I understand now, didn't think about binding energies but a free neutron has a binding energy of it's 'system' - (which is just itself) of 0 so therefore any other state has a higher binding energy and so is preferred?
 

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