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How to prove N=1.5Z for heavy nuclei?

  1. Nov 16, 2009 #1
    Hi, there.

    I was trying to prove a relation about the most stable situtation for a nucleus. According to the semiempirical mass formula, the less mass a nuclei has, the most binding energy has and, of course, the most stable is. I followed the "liquid drop model" (Weizsacker) for the binding energy, and I found where nuclear mass becomes minimum. That happens for Z=A/2 (N=Z). But that's true ONLY for small nuclei (A<=20).

    How can I prove that N=1.5Z for heavy nuclei?
     
  2. jcsd
  3. Nov 16, 2009 #2
    with the other terms of the Weizsacker formula... o_O
     
  4. Nov 16, 2009 #3
    What do you mean? Can you be more specific. I want a mathematical relation which proves that N=1.5Z for heavy nuclei.
     
  5. Nov 16, 2009 #4
  6. Nov 17, 2009 #5

    Vanadium 50

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    You won't be able to prove this, as it is not strictly true - it is an approximation. It comes about because 82 and 126 are nuclear magic numbers, and 126/82 is 1.53...
     
  7. Nov 17, 2009 #6

    arivero

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    I assume you are referring to [tex]Z \approx {1\over 2} {A\over 1 + A^{2/3} {a_C\over 4 a_A}}[/tex] in that page, are you? And surely this is the way the OP has tried too. As you say, All the trick is hidden in the value of {a_C\over 4 a_A} so that for small A, the unity term dominates. On the other hand, a_C and a_A are empirical, so the point that they are forced by the magic number could be argued. Two different models of the nuclei, shell and empirical.
     
    Last edited: Nov 17, 2009
  8. Nov 17, 2009 #7

    arivero

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    I wonder if we could use an argument directly from the breaking of SU(2) isospin. Of course when nuclear isospin is restored, Z=A/2 and N=Z. It is interesting that the breaking drives to a situation where N=2A, has a sort of topology scent. Can be coincidental, as it has been said, due to the doubly magic number of the Pb area.

    The equality of the distances to the driplines is preserved after the breaking of isospin. Again, it can be coincidental, but it is puzzling. http://arxiv.org/abs/nucl-th/0407118

    The driplines are given by
    [tex]M[Z,Z+N]-M[Z-1,Z+N-1]=m_p[/tex]
    [tex]M[Z,Z+N]-M[Z,Z+N-1]=m_n[/tex]

    when isospin is restored, the functions [itex]Z_p(N), Z_n(N)[/itex] are symmetric respect to the Z=N axis. Can we recover the stability axis from this pair of functions? In the symmetric case, this axis is a geometric axis, really, so the question that arises is which formula to use to define a "symmetry axis" for two generic functions when Z and N is a explicit coordinate system (Coming from physics, protons and neutrons do exist. Without an explicit coordinate system we should use some variational equation)

    It is noticeable that solving for X in the pair of "forward fitting" equations
    [tex]Z=Z_p(N-X), Z=Z_n(N)+X[/tex]
    recovers very well the stability line [itex]Z(N)[/itex] not only for Z=N but for the real, isospin broken, case.
     
    Last edited: Nov 17, 2009
  9. Nov 20, 2009 #8

    bcrowell

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    This is incorrect. The reason that the ratio is approximately 1.5 for heavy nuclei is simply the bulk properties of nuclear matter. It has nothing to do with magic numbers.

    There are many doubly-magic nuclei. Some are closer to the line of stability than others. 208Pb happens to be both doubly magic and close to the line of stability. 100Sn (N=Z=50) is also doubly magic, but is far from the line of stability.
     
    Last edited: Nov 20, 2009
  10. Nov 20, 2009 #9

    bcrowell

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    The constant aC simply depends on the Coulomb constant and the radius of the nucleon. It has nothing to do with magic numbers. The significance of the aA term is explained in the WP article. It also has nothing to do with magic numbers, although it is quantum-mechanical.
     
  11. Nov 20, 2009 #10

    bcrowell

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    It has nothing to do with topology or magic numbers. Yes, you can describe it in terms of the breaking of isospin symmetry, but that's a very fancy way of expressing a very simple fact, which is simply that protons repel one another. In a heavy nucleus, you get greater stability by having more neutrons, so that the protons can be farther apart.
     
  12. Nov 20, 2009 #11

    bcrowell

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    To try to clear up some of the confusion in this thread, here is a simple way to understand the division betwen microscopic and macroscopic that's going on in this problem. Imagine that the single-particle energy levels available to the neutrons and protons were all spaced in a fairly uniform way. The density of states will increase gradually as you go up in energy, but suppose that there are no shell gaps, i.e., no magic numbers. (This is not even terribly unrealistic; many nuclei are nonspherical, and in the nonspherical potential, the shell gaps are not very pronounced.) In this situation, you will still have a quantum-mechanical system, with discrete energy levels, quantized angular momentum, etc. You will still have the Pauli exclusion principle, and in particular nothing changes in the explanation of the asymmetry term given at http://en.wikipedia.org/wiki/Semi-empirical_mass_formula#Asymmetry_term . The line of stability will still be determined by the compromise between the asymmetry energy (which arises from the exclusion principle and the existence of two types of nucleons) and the Coulomb energy (which is completely classical).

    To formalize this concept, you can use the Strutinski shell-correction method.

    The shape of the line of stability has nothing to do with magic numbers.
     
  13. Nov 23, 2009 #12

    arivero

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    But the whole point of the thread is that the mass formula does not predict 1.5. If one puts Z= k A, N= (1-k) A in
    [tex]E_{B} = a_{V} A - a_{S} A^{2/3} - a_{C} \frac{Z(Z-1)}{A^{1/3}} - a_{A} \frac{(A - 2Z)^{2}}{A}[/tex]
    and then asks for the optimum value of k in the limit where A goes to infinity, then the coulomb term dominates, and the prediction is Z=0. On the contrary, if the Coulomb term does not dominate, the prediction should be Z=0.5 A and thus N=Z. Then the original poster was asking for some way to explain N=1.5 Z
     
  14. Nov 23, 2009 #13
    Watch This:
    binding_energy.gif
    If you go to infinite A you cannot Have a finite fraction of charge hoping to bind the system. In fact infinite Nuclear matter is almost pure neutron matter in the Neutron Stars (also known as Pulsar).

    You have to calculate for a specific region of interest, when you define "heavy nuclei"... I dunno, you can tell heavier then Pb to U...
    You must try a definite A that you think heavy, you cannot do a limit to the infinite matter with the Weizsacker formula...
     
  15. Nov 23, 2009 #14

    bcrowell

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    N=1.5Z is not a limiting result as A approaches infinity. N=1.5Z is simply a description of the highest mass numbers that have been observed.

    Raghnar is correct. The only thing I would caution about is that a neutron star is a qualitatively different system than a nucleus, since it's bound by gravity.

    Another thing to keep in mind here is that the parameters of the liquid drop energy are fits to data. They were fitted to data with A=0 to about 250, and N/Z only as far from stability as we have data on. The liquid drop model with standard parameters cannot be accurately extrapolated beyond the region to which the parameters were fitted.
     
  16. Nov 24, 2009 #15
    You're right, thanks for the puntualization.
     
  17. Nov 24, 2009 #16

    arivero

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    Actually, and lets forget about Weizsacker, is there any way to understand the decrease of binding energy per nucleon as we increase A? Suppose the nucleus were only neutrons, the binding being nuclear strong force. Should them fissionate, or kept together?
     
  18. Nov 25, 2009 #17

    bcrowell

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    We do observe bound systems like helium-8, which is stable against neutron emission, but decays by beta decay. Helium-10, on the other hand, decays by neutron emission.

    The reason systems with very high N/Z decay by neutron emission is basically a particle-in-a-box argument, plus the fact that neutrons are fermions. You have an attractive potential well, but it only has a finite number of bound states. Once you fill all those bound states, the only states left are the unbound continuum states.
     
  19. Nov 25, 2009 #18

    arivero

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    Hey, thanks, now I got it. I was thinking that the fact of adding a neutron was going to do the well deeper, enough for a new neutron to fit it, and so on. Probably with a lighter pion it could work in that way, but ultimately the short range of the nuclear force stops the process. Of course we can fill the well twice, one with neutrons and one with -less, because of repulsion- protons. But even so, the process of filling and "digging" the well eventually stops.

    Is there some drawing of the theoretical neutron emission line supposing that beta decay is forbbidden? Of course, the neutron decays beta because/when the available level in the proton sector is at lower energy that the current level of the neutron. When the proton sector is full, there is no more places to decay into, but we can keep adding neutrons until the well fills. What is funny is that the protons are contributing to digg the well, so were no protons, we could add less neutrons.
     
  20. Nov 25, 2009 #19

    bcrowell

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    This is called the neutron drip line, and you can find theoretical papers that try to predict it. It doesn't matter whether beta decay is forbidden, because the half-life for beta decay is on the order of milliseconds or something, while the half-life for neutron emission is on the order of the time it takes a neutron to travel across the nucleus.

    The drip lines are very hard to predict accurately, because the models are all based on fits to the available data, and the available data are mostly for nuclei close to stability. Extrapolating them to extremes of N/Z is unreliable. As an example, nobody really knows for sure whether the dineutron (N=2, Z=0) is bound or not. Extensive experimental searches for dineutrons haven't ever turned one up, so we suspect it's not bound.
     
  21. Nov 26, 2009 #20

    arivero

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    I had read of it, but forgotten it. It is fascinating because, I guess, most models of the nuclear force would bound it. As you said, it is about having a potential well and for every hole up to four particles (spin and isospin) can be put into.

    Having no dineutron implies that the exchange of spin 1 mesons overrrules the exchange of spin 0 mesons. Amusing.
     
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