PHY401: Nuclear and Particle Physics

PHY401: Nuclear andParticle PhysicsLecture 9,Thursday, September 10, 2020

Dr. Anosh JosephIISER Mohali

The Fermi Gas Model -Continued

PHY401: Nuclear and Particle Physics Dr. Anosh Joseph, IISER Mohali

The Fermi Gas Model - Continued

Using Fermi momentum we can estimate Fermienergy.

Fermi energy depends on the density of states.

It is different for protons and neutrons.



The reason is that in general there is a differentnumber of particles in the proton and neutronpotential well.

For protons, we have the Fermi energy

EpF =

pp2F

2mp=

( hc)2

2r20mpc2

(9π4

) 23(

ZA

) 23

(1)



For neutrons, the Fermi energy is

EnF =

pn2F

2mn=

( hc)2

2r20mnc2

(9π4

) 23((A − Z)

A

) 23

(2)

ForZA

∼(A − Z)

A∼

12

(3)

andr0 = 1.2 fm, (4)

we have Fermi energy

EF ∼ 33 MeV. (5)



The difference B ′ between the top of the well and theFermi level is constant for most nuclei.

It is just the binding energy per nucleon B/A ≈ 8 MeV.

Depth of potential and Fermi energy are, to a goodextent, independent of mass number A

V0 = EF + B ′ ≈ (33 + 8) MeV = 41 MeV. (6)



EF

Bosons Fermions

Figure: Bosons and fermions in the potential well.



Figure: Sketch of the proton and neutron potentials andstates in the Fermi gas model.



Occupied states

Unoccupied states

Fermi surface

nz

ny

nx

rn

Figure: Occupied and unoccupied states in the Fermi gasmodel.



We get the same depth for protons and neutrons sincewe assumed the equal number for both.

If the number of protons and neutrons are differentinside the nucleus we get different depths for potentialwells for each species.



The potential depth difference comes out from thedependence of the Fermi energy on the number ofprotons or neutrons.

The difference in depths is a consequence of theCoulomb repulsion between the protons.



In stable nuclei the Fermi level is at the same energyfor protons and neutrons.

In such a case there is no energy gain fromtransforming one type of nucleons into another by βdecay.

In unstable nuclei the Fermi level is different forprotons and neutrons, this opens a path to transformnucleons from one well to the other through β decay.



The β decay proceeds until Fermi levels in both wellsare equal.


Average Momentum of a Nucleon

Average momentum of a nucleon depends on thedensity of momentum states which is the number ofmomentum states dn for a particle with momentum pand p + dp.

We can calculate this number using a similar trick towhat we used to calculate the number of states up tothe Fermi momentum.

Let us consider the space defined by the quantumnumbers (nx , ny, nz).



We argued that there are two momentum states perunit volume in this space.

Particles with momentum between p and p + dp definea spherical shell with radius

R =pLπ h

(7)

and thickness

dR =dpLπ h

(8)



Leads to the following volume in this space

V = 4πR2dR = 4π(

Lπ h

)3

p2dp =4Vπ2 h3 p2dp (9)

We need to find the number of states....

Remember the condition that for nx > 0, ny > 0, nz > 0we need to take 1/8 of the volume V...

We also need to multiply V by 2 to account for twospin state (up or down) of the nucleon



The density of states is

dn = 2× 18× V =

14

4Vπ2 h3 p2dp =

Vπ2 h3 p2dp. (10)

This gives

dndp

=Vπ2 h3 p2 (11)



Now we can compute the average momentum of anucleon.

We have

〈p〉 =

∫pF0 p dn

dp p2dp∫pF0

dndp p2dp

=

∫pF0 p3dp∫pF0 p2dp

=34

pF ≈ 188 MeV. (12)

We need to compute the average energy.



Note that average energy is not the energycorresponding to average momentum.

To calculate average energy, need to know the densityof energy states, which is the number of states for aparticle with energy between E and dE.

Can get this number from the density of themomentum states

dn =Vπ2 h3 p2dp (13)



E =p2

2m=⇒ p2 = 2mE =⇒ pdp = mdE (14)

dp =mdE√2mE

=

√m2

dE√E

(15)

dn =√

2Vπ2 h3 m

32√

EdE (16)



Thus the average kinetic energy of a nucleon is

〈E〉 =∫EF

0 E dndE dE∫EF

0dndE dE

=

∫EF0 E

32 dE∫EF

0 E12 dE

=35

EF ≈ 20 MeV. (17)

Note: energy which corresponds to the averagemomentum is

E =〈p〉2

2m=

916

p2F

2m=

916

EF ≈ 18.6 MeV. (18)



Total kinetic energy of the nucleus is

Ekin(N , Z) = N〈En〉+ Z〈Ep〉

=3

10m

[N(pn

F )2 + Z(pp

F )2]

(19)

Upon using

N =V (pn

F )3

3π2 h3 (20)



Z =V (pp

F )3

3π2 h3 (21)

and

V =43πR3 =

43πr3

0A (22)



we get

Ekin(N , Z) =3

10m

h2

r20

(9π4

)2/3 N5/3 + Z5/3

A2/3 . (23)

Note that we have again assumed that the radii of theproton and the neutron potential wells are the same.



This average kinetic energy has, for fixed massnumber A but for varying N (or equivalently Z ), aminimum at N = Z .

Hence the binding energy shrinks for N 6= Z .

Expand the above equation in the difference N − Z . Weget

Ekin(N , Z) =3

10m

h2

r20

(9π8

)2/3(A +

59(N − Z)2

Z+ · · ·

).

(24)



This gives us the functional dependence upon theneutron surplus.

First term contributes to the volume term in the massformula.

Second term describes the correction which resultsfrom having N 6= Z .



This so-called asymmetry energy grows as the squareof the neutron surplus and the binding energy shrinksaccordingly.

We thus see that the simple Fermi gas model, wherenucleons move freely in an averaged-out potential, canalready render the volume and asymmetry terms inthe semi-empirical mass formula plausible.

The first term corresponds to the volume term in theliquid drop model while the last one corresponds tothe symmetry energy term in the liquid drop model.

Note: We will see liquid drop model soon.



Therefore, based on the Fermi model...

... we can conclude that the symmetry energy term inthe liquid drop model is a quantum mechanical effect...

... related to the way fermions occupy allowed statesin the proton/neutron potential well.


Global Properties of Nuclei

We can use the particles produced by radioactivedecay to bombard other elements in order to study theconstituents of the latter.

This experimental ansatz is the basis of modernnuclear and particle physics.

The development of ion sources and massspectrographs now permitted the investigation of theforces binding the nuclear constituents, i.e., theproton and the neutron.



These forces were evidently much stronger than theelectromagnetic forces holding the atom together,...

... since atomic nuclei could only be broken up bybombarding them with highly energetic α-particles.

The binding energy of a system gives informationabout its binding and stability.

This energy is the difference between the mass of asystem and the sum of the masses of its constituents.



It turns out that for nuclei this difference is close to1% of the nuclear mass.

This phenomenon, historically called the mass defect,was one of the first experimental proofs of themass-energy relation E = mc2.

The mass defect is of fundamental importance in thestudy of strongly interacting bound systems.


Nuclides

Nuclide: A distinct kind of atom or nucleuscharacterized by a specific number of protons andneutrons.

The atomic number: The atomic number Z gives thenumber of protons in the nucleus.

The charge of the nucleus is, therefore, Q = Ze, wheree is the elementary charge e = 1.6× 10−19 C.


Nuclides

In a neutral atom, there are Z electrons, whichbalance the charge of the nucleus, in the electroncloud.

The atomic number of a given nucleus determines itschemical properties.

The mass number: In addition to the Z protons, Nneutrons are found in the nucleus.


Nuclides

The mass number A gives the number of nucleons inthe nucleus, where A = Z + N .

Different combinations of Z and N (or Z and A) arecalled nuclides.


Nuclides

1. Nuclides with the same mass number A are calledisobars.

2. Nuclides with the same atomic number Z arecalled isotopes.

3. Nuclides with the same neutron number N arecalled isotones.


Nuclides

The binding energy B is usually determined fromatomic masses, since they can be measured to aconsiderably higher precision than nuclear masses.

We have

B(Z , A) =[Z M(1H) + (A − Z)Mn − M(A, Z)

]c2. (25)


Nuclides

Here M(1H) = MP + me is the mass of the hydrogenatom.

Mn is the mass of the neutron.

M(A, Z) is the mass of an atom with Z electrons whosenucleus contains A nucleons.


Nuclides

The rest masses of these particles are

Mp = 938.272 MeV/c2, (26)

Mn = 939.565 MeV/c2, (27)

me = 0.511 MeV/c2. (28)


Nuclides

In nuclear physics, nuclides are denoted by AX , Xbeing the chemical symbol of the element.

For example, the stable carbon isotopes are labelled12C and 13C ; while the radioactive carbon isotopefrequently used for isotopic dating is labelled 14C.

Sometimes the notations AZ X or A

Z XN are used,whereby the atomic number Z and possibly theneutron number N are explicitly added.


Parametrization of Binding Energies

Apart from the lightest elements, the binding energyper nucleon for most nuclei is about 8 MeV.

Depending only weakly on the mass number, it can bedescribed with the help of just a few parameters.

The parametrization of nuclear masses as a functionof A and Z , which is known as the Weizsacker formulaor the semi-empirical mass formula, was firstintroduced in 1935.



The mass of an atom with Z protons and N neutronsis given by the following phenomenological formula:

M(A, Z) = NMn + ZMp + Zme − avA + asA2/3

+acZ2

A1/3 + aa(N − Z)2

4A+

δ

A1/2 , (29)

with N = A − Z .



The exact values of the parameters av, as, ac, aa and δdepend on the range of masses for which they areoptimized.



One possible set of parameters is given below:

av = 15.67 MeV/c2, (30)

as = 17.23 MeV/c2, (31)

ac = 0.714 MeV/c2, (32)

aa = 93.15 MeV/c2, (33)

δ =

−δ0 for even Z and N (even-even nuclei)

0 for odd A (odd-even nuclei)

+δ0 for odd Z and N (odd-odd nuclei)

(34)

whereδ0 = 11.2 MeV/c2. (35)



The individual terms can be interpreted as follows:

Volume term, av: This term, which dominates thebinding energy, is proportional to the number ofnucleons.

Each nucleon in the interior of a (large) nucleuscontributes an energy of about 16 MeV.



From this we deduce that the nuclear force has ashort range, corresponding approximately to thedistance between two nucleons.

This phenomenon is called saturation.

Due to saturation, the central density of nucleons isthe same for all nuclei, with few exceptions.



The average inter-nucleon distance in the nucleus isabout 1.8 fm.

Surface term as: For nucleons at the surface of thenucleus, which are surrounded by fewer nucleons, thebinding energy is reduced.

This contribution is proportional to the surface area ofthe nucleus (R2 or A2/3).



Coulomb term ac: The electrical repulsive force actingbetween the protons in the nucleus further reducesthe binding energy. This is approximately proportionalto Z2/A1/3.

Asymmetry term aa: Heavier nuclei accumulatemore and more neutrons, to partly compensate for theincreasing Coulomb repulsion by increasing thenuclear force. This creates an asymmetry in thenumber of neutrons and protons.

The dependence of the nuclear force on the surplus ofneutrons is described by the asymmetry term(N − Z)2/4A.



This shows that the symmetry decreases as thenuclear mass increases.

Pairing term ap: A systematic study of nuclearmasses shows that nuclei are more stable when theyhave an even number of protons and/or neutrons.

This observation is interpreted as a coupling ofprotons and neutrons in pairs.



The pairing energy depends on the mass number, asthe overlap of the wave functions of these nucleons issmaller in larger nuclei.

Empirically this is described by the term δ

A12.

The Weizsacker formula is often mentioned inconnection with the liquid drop model.



In fact, the formula is based on some propertiesknown from liquid drops: constant density,short-range forces, saturation, deformability, andsurface tension.

An essential difference, however, is found in the meanfree path of the particles.

For molecules in liquid drops, this is far smaller thanthe size of the drop; but for nucleons in the nucleus, itis large.



Therefore, the nucleus has to be treated as a quantumliquid, and not as a classical one.

At low excitation energies, the nucleus may be evenmore simply described as a Fermi gas...

That is, as a system of free particles only weaklyinteracting with each other.


References

I B. Povh, K. Rith, C. Scholz and F. Zetsche, Particlesand Nuclei: An Introduction to the Physical Concepts,Springer, 6th edition (2008).


End


Documents

PHY401: Nuclear and Particle Physics