2-Liquid Drop Model

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LIQUID DROP MODEL

- A non-rotating drop of liquid in the absence of gravitational or other external fields adjusts its

shape to minimize its energy. That shape is spherical and it minimizes the positive surface

tension energy.

- If the liquid is incompressible then the drop's density is constant, independent of radius R and R is given , where n is the number of molecules in the drop.

- Let each molecule (except one in or near the surface) be bound in the drop with energy a;

this is the energy required to remove the molecule from the inside of the drop and is due to

the forces that can exist between molecules,

- Typically these forces are negligible at large separations, can become attractive at separations

comparable to the molecular size and become strongly repulsive at closer separations.

Volume energy. When an assembly of nucleons of the same size is packed together into the

smallest volume, each interior nucleon has a certain number of other nucleons in contact with it.

So, this nuclear energy is proportional to number of particle n in the volume. Equal to

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Surface energy. A nucleon at the surface of a nucleus interacts with fewer other nucleons than one in the interior of the nucleus and hence its binding energy is less. This surface energy term takes that into account and is therefore negative and is proportional to the surface area. Equal to

(T is the surface tension) or

- Therefore the binding energy B of the drop:

where β contains all the constants of the surface term.

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- If the drop carries an electric charge Q, there is an extra term due to mechanical potential

energy of the charge distribution.

- If the charge is distributed uniformly in the surface, then that Coulomb energy is

- If it s distributed uniformly through the drop it is .

Coulomb Energy . The electric repulsion between each pair of protons in a nucleus contributes toward decreasing its binding energy.

- This energy decreases the binding energy, which becomes

,

where γ contains all the Coulomb effects.

- We now examine the analogy with nuclei. We assume that

(1) the nucleus is spherical;

(2) the nucleons in the nucleus behave like the molecules in a drop - that is there is a

short-range attractive force holding the nucleons together, and a shorter-range

repulsive force which stops the nucleons collapsing into one another,

(3) the nuclear density is constant.

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https://en.wikipedia.org/wiki/Coulomb's_law


With these assumptions we can write down a formula for the nuclear binding energy B(Z,A) by simple analogy, changing , and ,

where for nuclear case; for volume term,

for surface term,

for the Coulomb term.

- First we note another important assumption: the nuclear binding of neutrons is identical

to that for the protons.

- Therefore imagine two potential wells each with an associated set of energy levels that

are identical, one for the protons and one for the neutrons. These levels fill accordingly to

the Pauli exclusion principle because both neutron and proton are fermions (spin of half

integer).

- If Z=N, then both wells are filled to the same level (the Fermi level).

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Fig. 4.3

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- If we move one step away from that situation, say in the direction of N>Z, then one proton must be changed into a neutron (see Fig. 4.3).

- This state has energy ΔE greater than the initial state, where ΔE is the level spacing at the Fermi level.

A second step in the same direction causes the energy excess to become 2xΔE.

- A next step means moving a proton up 3 rungs as it changes from

proton to neutron and the excess becomes 3ΔE.

- For 2 protons move to become 2 neutrons the excess energy

become 2x3ΔE.


In fact, listing in units of ΔE, as each step (which changes proton to neutron) is made, we find the changes require energy in units of ΔE

1, 1, 3, 3, 5, 5, 7, ...

so that the cumulative effect is

1, 2, 5, 8, 13, 18, 25, 32, …… unit ΔE

for N-Z = 2, 4, 6, 8, 10, 12, 14, 16, … ,

Therefore to change from N-Z= 0 to N > Z, with A = N+Z held constant, requires an energy of

.

- This is independent of whether it is N or Z that becomes larger and it means that, if all other things are equal, nuclei with Z= N have less energy and are therefore more strongly bound than a nucleus with Z≠ N.

- Thus we must add a term which reduces the binding energy when Z≠N. Since the energy

levels of a particle in a potential well have a spacing inversely proportional to the well

volume, we can put ΔE A-1.

- Therefore we include a term which reduces the binding energy for nuclei for which Z≠N. This is the asymmetry term:

to be added to the binding energy formula.

- The pairing term. It reflects the fact that it is found experimentally that 2 protons or 2 neutrons

are always more strongly bound than 1 proton and 1 neutron. That is, like nucleons 'pair'.

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For odd A nuclei (Z even, N odd (eo) or Z odd, N even (oe)) this term is taken to be zero.

For even A nuclei there are two cases;

(1) Z odd, N odd (oo),

(2) Z even, N even (ee).

The binding energy will be greater for case 2 than for 1 so we add into the binding energy

formula a quantity δ(Z,A) for case 2 and subtract it for case 1. Bohr and Mottleson (1969) show

that it has the form

, MeV,

fits the rather scattered data with a precision better than 1 MeV for all A > 20.

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Before putting the formula for binding energy together we note that there is one refinement that is sometimes made.

The charge on the nucleus is carried in discrete units, one on each proton. The charge on the

proton does not interact with itself (or if the proton constituents do interact, that energy is already

included in the proton mass). It is therefore sensible to replace Z2 appropriate to a continuous

charge distribution by Z(Z-1) which is appropriate to this discreteness of the nuclear charges.

However, we do not do that: the reason is that the apparently best set of coefficients av, etc. has

been determined using the formula with Z2. In addition, the final precision of the formula is

probably not sufficient to allow improvements at this level to be discerned.

So putting all our terms together, we have

The values of the coefficients have to be found by fitting to the binding energy data for medium and heavy nuclei. The light nuclei (A<20) are not included as there is no smooth curve of binding energy against A or Z due to the effects of shell closures.

The fit is not perfect because these effects persist throughout the periodic table and because some nuclei are not spherical.

We have written the whole formula for the nuclear mass (Table 4.1) but as rest mass energy—hence the c2 attached to the real masses. Also shown is a favoured set of values for the coefficients.

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Fig. 4.5 shows how the various contribution (Except the pairing term) change with A throughout the periodic table.

Fig 4.5

What is surprising is that this formula is good from A≈20 to the end of the periodic table with a precision beter than 1.5 present on the binding energy. This is shown in the case of the odd A nuclei in Fig. 46.

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Fig 4.6

The binding energy as a function of A for odd-A nuclei from 15-259.

The solid points are prediction of semi empirical mass formula as given in Table 4.1

The open circles are measured values

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2-Liquid Drop Model