Liquid Drop Model From the definition of Binding Energy we can write the mass of a nucleus X Z

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1 Our first model of nuclei. The motivation is to describe the masses and binding energy of nuclei. It is called the Liquid Drop Model because nuclei are assumed to behave in a similar way to a liquid (at least to first order). The molecules in a liquid are held together by Van der Waals force that is only between near neighbors. We have already seen that the nuclear force is a short range force that appears to act only between neighboring nucleons.

2 From the definition of Binding Energy we can write the mass of a nucleus X Z as, M( Z, ) c To first order we can write B Zm c p ( Z) m c /3 1/3 ( Z, ) av as acz / The first term = Volume Term This is due to the short range attraction between nucleons. If we ignore the fact that the nucleus has a surface, each nucleon is attracted to the same number of nearest neighbors. i.e. BE n B( Z, )

3 But: some nucleons are on the surface they have fewer neighbors. Therefore: BE will be reduced by an amount number of nucleons on the surface. Therefore: the second term = Surface Term surface area r and since r 1/3 /3 i.e. surface term = a S /3.

4 The Coulomb repulsion between the protons in the nucleus also reduces the Binding Energy. The Coulomb potential energy will be Z and 1/x x = average distance between proton pairs x r 1/3 So the third term = Coulomb Term = a C Z / 1/3 ctually the PE of one proton is 1 So PE for a nucleus would be Z(Z 1) but we will simplify to Z ( e )( Z 1) e

5 There are other terms in the Liquid Drop Model. These additional terms are quantum mechanical in origin. The symmetry Term. Imagine that the neutrons and protons are in a potential well (the nucleus). We will assume that the well has approximately equally spaced energy levels. Let the spacing be E. Neutrons and protons are fermions (spin ½ particles). So the Pauli exclusion principle allows only p and n in each level. E

6 For a nucleus with a given, the lowest energy will be when N = Z. N Z (Z N = 0) If we change a neutron to a proton (e.g. via decay) we increase the energy by 1E from Z N = 0 case. (Now Z N = )

7 N Z If we change another neutron to a proton we increase the energy by E for a total change of E from Z N = 0 case. (Now Z N = 4) nother neutron to a proton we increase the energy by 3E for a total change of 5E from Z N = 0 case. (Now Z N = 6)

8 Continuing this, we can find the energy shift for a change in Z N or N Z. ( Z N or N Z Total Energy Shift (E) We can approximate this energy increase by N Z) ( N 8 Z) E

9 It has been found that E decreases with like 1. (This can be justified by considering the phase space available for fermions but it is not worth the effort for this phenomenological formula.) Therefore we can write the asymmetry term as a ( N Z) ( Z) a (Some authors use (Z/) instead of (Z). In that case a will be different.)

10 Pairing Term. This term reflects the observation that the nucleus is in a lower energy state when neutrons are all paired with other neutrons. We find a similar thing for protons. e.g. plot of the Neutron Separation Energy for Barium isotopes = energy needed to extract a neutron from a nucleus. Neutron separation energy N Ba N N

11 Neutron Separation Energies

12 Proton Separation Energies

13 We find that the energy of the nucleus is changed by an amount pair = N even Z even = 0 odd (N or Z even) = + N odd Z odd i.e. Nuclei with both neutrons and protons paired are in a lower energy state than nuclei with just the neutrons or protons paired, and these are in a lower energy state than if neither are paired. The best fit to data is with: # of Stable Nuclei a P 1/

14 Finally we have the: Semi-Empirical Mass Formula. even - even nuclei / odd nuclei 0 odd - odd nuclei / / ) ( / ), ( 1/ 1/ 1/3 /3 a a Z a Z a a a Z B P P C S V Best Fit Parameters a V = MeV a S = 17.3 MeV a C = MeV a = 3.85 MeV a P = 1.0 MeV ), ( ) ( ), ( Z B m c Z c Zm c Z M n H

15 The value of the model is that we can predict the masses of nuclei that are far from stability, where masses are hard to measure. e.g. Nucleosynthesis models are highly dependent on knowing such masses. Many refinements, beyond what we have done, have been added to the model to improve accuracy. The liquid drop model has several shortcomings. e.g. it is not quantum mechanical assumes nuclei are spherical symmetric (which disagrees with the fact that nuclei have non-zero quadruple moments more later.) Nevertheless, the model does help us understand nuclear stability.

16 Odd Nuclei (Paring term is zero.) e.g. = 101 isobars. = decay by either EC or + Liquid Drop Model Conclusion: Odd nuclei have only one stable nucleus for a given. (Stable, unless it can decay by emission.)

17 Even Nuclei (Paring term alternates.) e.g. = 100 isobars. = decay by either EC or + Conclusions: o-o nuclei can always decay. (There are 4 exceptions mass formula not accurate enough to predict.) o-o nuclei e-e nuclei e-e nuclei often have two (or more) stable isobars.

18 The semi-empirical mass formula does a reasonable job of describing the general observations of nuclear masses. e.g. Binding energy per nucleon for stable nuclei This indicates that our model of the nucleons being bound in the nucleus by a Nuclear Force with a short range is valid at least to first order. mass formula measured for stable nuclei