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Number of neutrons & stability?

  1. Oct 31, 2012 #1
    Hi everyone. I'm kind of new here, please be nice.

    So I was reading up some physics and came across the fact that if the no. of neutrons is too high compared to the no. of protons, the nucleus becomes unstable. I can understand why this happens with protons, due to the coloumbic repulsion between the charged protons, but I don't understand why this happens with neutrons. I tried googling it and the best explanation I got was "because of the nuclear force". I know what the nuclear/strong force is (well, at a high school level), which is exactly why I'm asking this question.

    Neutrons, without any charge, is a mean to hold the nucleus together despite the strong electric force, by its nuclear force. Nuclear force holds the protons together, because in very small distances, smaller than the diameter of the nucleons, it overpowers the electric force, thus acts like cement linking the protons together.

    But, let's say we have a fairly large but stable nucleus, like iron, and we throw in a few neutrons. The neutrons should not affect the nucleus in a negative way, as the only force it can interact with the neutron with, is the strong force, which is attractive. (Well, it's repulsive for very, very small distances, but all this does is keep the nucleons at a stable distance from each other, and makes sure they aren't a "continuous mass") It'd be just like adding more glue.

    But, clearly this isn't the case. It'd be great if someone could help me out here, by either proving my logic wrong, or giving me more information on how the nuclear force works.

    I hate high school physics sometimes. It just makes you assume stuff without explaining how each components works. And then, later, in a test, it asks you to "explain". It's almost like asking me to write an essay on how E=mc^2 is derived, when the only thing they've given me is the equation.

    Sorry about my complaint/rambling, at the end there :P Thanks
     
  2. jcsd
  3. Oct 31, 2012 #2
    I am interested to see a good explanation of this as well. As I understand it the strong force is always attractive and is proportional to the inverse of distance between colour charges. Forces between nucleons (protons and neutrons) is a result of the colour charges in each nucleon having uneven distribution in each.
    At no point is it repulsive.

    I think the decay of unstable nucleii has more to do with the weak nuclear force than the strong one, but I can't explain it any further than that. I do not expect the explanation will be a simple one but there are some pretty clever people here who I am sure can let us know more.
     
  4. Oct 31, 2012 #3

    tiny-tim

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    welcome to pf!

    hi almondsize! welcome to pf! :smile:
    neutrons aren't glue, they require glue

    the more neutrons, the more glue is required

    in stability terms, a nucleus likes full shells, in the same way that electrons like full shells …

    put too many electrons in, and some will want to break away …

    it's the same with neutrons! :wink:

    see http://en.wikipedia.org/wiki/Nuclear_shell_model and http://en.wikipedia.org/wiki/Island_of_stability for some detail
     
  5. Oct 31, 2012 #4
    Thanks tiny-tim.

    I guess from that I can refine the question.

    As protons and neutrons have identical spin, what is preventing a neutron from occupying a shell/state that is reserved for a proton?
     
  6. Nov 1, 2012 #5
    Thanks for the reply everyone. I guess that kind of made sense, so the nucleus does have shells. And some numbers of neutrons are more stable than other numbers of neutrons, and there are excited states for the nucleus. There are "magic numbers" for the no. of neutrons and protons.

    But there are several things that makes the electron shell anology somewhat irrelevant. First of all, the no. of electrons must equal the no. of protons, due to the electrostatic force. The charge of the electrons cancel out the chrage of the protons in the nucleus, so there is, somwhat, of a relationship between the no. of protons and number of electrons. The electrons also have orbital However, for neutrons, there is no relationship directly linking the protons to the neutrons. The strong force is essentially always attractive, almost like gravity. to the extent of my knowledge, there is no one to one relationship between the proton and the neutron.

    Looking through, I found that the p-n strong force is stronger than n-n or p-p strong force, and I thought the answer to my question might have something to do with this. But I don't know. Either way, this is getting quite complicated. I guess it shows how much I still have to learn :P
     
  7. Nov 1, 2012 #6

    tiny-tim

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    Hi Justin! :smile:
    What a refined question! o:)

    The protons and neutrons have separate shells, and more or less ignore each other :wink:

    http://en.wikipedia.org/wiki/Nuclear_shell_model: [Broken] Note that the shells exist for both protons and neutrons individually, so that we can speak of "magic nuclei" where one nucleon type is at a magic number, and "doubly magic nuclei", where both are …​

    (here's a short thread about the history: https://www.physicsforums.com/showthread.php?t=545853)
     
    Last edited by a moderator: May 6, 2017
  8. Nov 1, 2012 #7
    Thanks again although I must say that is a slightly unsatisfying answer.

    I re-read that wiki page a few times and while I do not pretend to understand everything there, I was drawn to the Interacting Boson Model which wiki says very little about.

    Is the IBM II a generally accepted model? Meaning proton and neutrons can pair up and become exempt from the exclusion principle?

    While it still does not entirely explain why p & n are treated differently I presume there are other quantum properties involved that identify them as different particles aside from spin.
     
  9. Nov 2, 2012 #8
    Adding more neutrons wouldn't make nucleus fall apart, true. But soon enough you will reach a situation when you have (26 protons + 33 neutrons), and you have a problem: it has more energy than (27 protons + 32 neutrons), and thus "wants" to convert to that state. Hence, you'd get beta-active Fe-59.
     
  10. Nov 2, 2012 #9
    I'm not exactly sure what the OP's question was, but I'll guess it has something to do with the differential ratio of neutrons to protons and the stability of the nucleus. The short answer is that it is a mystery and researchers are working on it, but, there does seem to be some consistent pattern to how these ratios scale. In general, the evidence seems to suggest that as the nucleus gets bigger, it needs a greater ratio of neutrons to protons to provide stability. Check this link:http://www.youtube.com/watch?v=H8Yd2T9MQBU&list=EC06C3C4E3F84C6A24&index=14&feature=plcp
     
  11. Nov 3, 2012 #10
    I think it is too much to call it a mystery. There is too much known about it.

    As for the nature of neutron drip line - the simplest case is helium 5.

    Understand that a neutron, like any other microscopic particle, has to have zero point energy. Which means it can only be caught in a potential well which is sufficiently deep and wide to have space for the state the neutron can occupy. If you do have attraction, but it is too weak, then the neutron has no bound states.

    The electric monopole attraction, with its 1/r2 bottomless pit and 1/r2 endless tails always has infinitely many states. But other potential well shapes, like attraction of electric monopole to an induced dipole or the short range nuclear force have a finite number of bound states, and that finite number can be small or zero.

    After you have the alpha particle with 2 protons and 2 neutrons filling all the 1s orbitals, further nucleons can only go to higher orbitals. But the problem is, the strong force attraction of He-4 to the extra neutron is sufficiently weak that the extra neutron has no bound states. It merely scatters off the nucleus.

    If you add 2 neutrons simultaneously then the weak attraction of neutrons to each other is sufficient to make He-6 a bound state. He-7 is unbound again - neutrons after 4 have to find another higher level, and He-6 is not enough to bind them. He-8 again is bound - but no higher He isotopes.

    He-6 is a bound state, because the neutrons are weakly bound to the alpha. But, in fact, a proton would be far better bound. So He-6 undergoes rapid beta decay to Li-6, where it is 1 proton and 1 neutron on the outer shell.

    Again, for lithium (3 protons) you will reach neutron drip line at Lithium 10. 3 protons can bind stably 3 or 4 neutrons, 5 or 6 neutrons would be bound but they would be better off decaying into 4th proton, 7th neutron is at a shell which is too high to be bound.

    Et cetera. A number of protons can only bind a finite number of neutrons by strong force. Further neutrons are forbidden from going to lower states, which are full, while the higher orbitals are unbound.
     
  12. Nov 3, 2012 #11
    Well, Snorkack, your post looks like a lot of hand waving to me because your ostensibly sophisticated but actually myopic analysis doesn't address the salient question of why, for small nuclei, the ratio of protons to neutrons is 1:1, but for larger nuclei it scales up linearly to 1:5.
     
  13. Nov 3, 2012 #12

    Vanadium 50

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    DiracPool, a little politeness goes a long way.

    It's not true that "the ratio of protons to neutrons is 1:1, but for larger nuclei it scales up linearly to 1:5." The scaling is only approximately linear and there are gaps: e.g. Cl-35 and Cl-37 are both stable; Cl-36 is not. Pb-204, 206, 207 and 208 are stable. Pb-205 is not. So there isn't an exact answer.

    There is an approximate answer, and snorkack gave it.
     
  14. Nov 3, 2012 #13
    Huh, Ok, I stand corrected and apologize for the impoliteness.:redface:
     
  15. Nov 3, 2012 #14
    And I apologize for not covering some parts of the questions because I was thoroughly explaining some others.

    It is very important - though related - matter as to whether a nucleus is unstable as in flying apart and dripping neutrons or, although bound, unstable as in undergoing beta decay.

    A free neutron is radioactive. It undergoes beta decay. The neutron is so much more massive than proton that the excess is enough for the 511 keV mass of electron and 782 keV are left over for matters like kinetic energy of electron, energy of antineutrino and recoil of the proton.

    A neutron in deuteron is stable. It is bound by 2 MeV 225 keV - more than the 782 keV of free neutron decay energy. And since a diproton is unbound, breaking a deuteron costs energy no matter whether it is broken into neutron and proton, two protons or two neutrons. Instead, you can have the opposite process - 2 protons and one electron combine spontaneously into a deuteron and release the energy, 1 MeV 443 keV, to a neutrino (minus recoil). But this process has a high and wide energy barrier - the electrostatic repulsion of protons - so the tunnelling is slow except at high temperatures or densities.

    Now, the binding energy of a triton is 8 MeV 482 keV.

    The binding energy of a neutron in a triton is 6 MeV 257 keV. Far bigger than in deuteron! Obviously the neutron in deuteron cannot decay into a free proton.

    But while diproton is unbound, 2 protons and 1 neutron is a bound state, like 1 proton and 2 neutrons is.

    So a neutron in tritium might decay into a bound proton. Which it actually does.

    Yet the decay energy of tritium is, guess what, 18,6 keV. Far smaller than that of free neutron!

    Although the nuclear forces of a neutron and a proton can bind either second proton or second neutron, the binding of second neutron is stronger, by 763 keV. This energy is due to electrostatic repulsion of proton, which neutron does not experience. If a He-3 nucleus encounters a free neutron, it may replace a proton with the neutron, releasing 763 keV (which goes to proton energy and triton recoil).

    While either a proton or a neutron added to alpha particle would be unbound, after there are 3 of both neutrons and protons, adding fourth of either is bound. But the electrostatic repulsion to proton gets even stronger. Lithium 7 is stable - beryllium 7 actually captures electron to turn into Li-7. In even bigger nuclei, that repulsion effect gets stronger. Boron 9, like helium 5, is actually unbound - beryllium 9 is not only bound but stable! Carbon 11, nitrogen 13 and oxygen 15 all have the energy to not only capture electrons, but actually emit positrons, only to turn proton into a less repulsive neutron.
     
  16. Nov 6, 2012 #15
    Beginning with Li (Z=3) to N (Z=7), the stable nuclei are those with N=Z or N=Z+1, with exception of Be-8 which is unstable not to beta decay but to alpha decay or spontaneous fission. It actually follows that the stable N/Z decreases - Li-7 is stable, C-14 is not. The reason, obviously, is that the lone excess neutron in Li-7 is more stable than the lone excess proton in Be-7, whereas the 2 excess neutrons in C-14 have the option to turn into 1 proton and 1 neutron.

    However, paired nucleons do have an energy advantage. After N, the energy penalty from deviating from the 1/1 ratio decreases to the extent that it is overwhelmed by the pairing energy. O-18 is the first stable nucleus with N=Z+2.

    Under Mattauch rule, from 2 isobars which differ by 1 proton, only one is stable. There are just 2 exceptions: Sn 123 and Sb 123 are both stable, and so are Ta-180, Hf-180 and W-180.

    From O (Z=8) to P (Z=15) every even Z has 3 stable isotopes (N=Z, Z+1, Z+2) and every odd Z has one stable isotope (N=Z+1).
     
  17. Nov 6, 2012 #16
    I think it's also possible to look at the quark content of the neutrons and proton; if they are really confined such that you have to pick one or the other using Pauli's exclusion principle, you can have different colors of quarks which validates the statement of how proton and neutron can occupy the same state.
     
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