Contents 1 the Fermi Gas Model

PHY401 - Nuclear and Particle Physics

Monsoon Semester 2020 Dr. Anosh Joseph, IISER Mohali

LECTURE 09

Thursday, September 10, 2020 (Note: This is an online lecture due to COVID-19 interruption.)

Contents

1 The Fermi Gas Model - Continued 1 1.1 The Fermi Energy ...... 1 1.2 Average Momentum of a Nucleon ...... 2 1.3 Average Kinetic Energy of a Nucleon ...... 3

2 Global Properties of Nuclei 5 2.1 Nuclides ...... 5 2.2 Semi-empirical Mass Formula ...... 6

1 The Fermi Gas Model - Continued

1.1 The Fermi Energy

Upon using the Fermi momentum we can estimate the Fermi energy. Fermi energy depends on the density of states. It is diﬀerent for protons and neutrons. The reason is that in general there is a diﬀerent number of particles in the proton and neutron potential well. For protons, we have the Fermi energy

2 2 p2 2 3 3 p p F (~c) 9π Z EF = = 2 2 . (1) 2mp 2r0mpc 4 A

For neutrons, the Fermi energy is

2 2 n2 2 3 3 n p F (~c) 9π (A − Z) EF = = 2 2 . (2) 2mn 2r0mnc 4 A PHY401 - Nuclear and Particle Physics Monsoon Semester 2020

When Z (A − Z) 1 ∼ ∼ (3) A A 2 and

r0 = 1.2 fm, (4) we have the Fermi energy

EF ∼ 33 MeV. (5)

The diﬀerence B0 between the top of the well and the Fermi level is constant for most nuclei. It is just the binding energy per nucleon B/A ≈ 8 MeV. The depth of the potential and the Fermi energy are to a good extent independent of the mass number A. We have 0 V0 = EF + B ≈ (33 + 8) MeV = 41 MeV. (6)

We get the same depth for protons and neutrons since we assumed the equal number for both. When the number of protons and neutrons in a nuclei are we will get different depths for the potential wells for each species. The potential depth difference comes out from the dependence of the Fermi energy on the number of protons or neutrons. The difference in depths is a consequence of the Coulomb repulsion between the protons. Note the difference between the depth of the potential well and the Fermi level (energy of the highest occupied state). In stable nuclei the Fermi level is at the same energy for protons and neutrons. In such a case there is no energy gain from transforming one type of nucleons into another through β decay. In unstable nuclei the Fermi level is different for protons and neutrons, this opens a path to transform nucleons from one well to the other through β decay. The β decay proceeds until Fermi levels in both wells are equal.

1.2 Average Momentum of a Nucleon

The average momentum of a nucleon depends on the density of momentum states, which is the number of momentum states dn for a particle with momentum p and p + dp. We can calculate this number using a similar trick to what we used to calculate the number of states up to the Fermi momentum.

Let us consider the space deﬁned by the quantum numbers (nx, ny, nz). We argued that there are two momentum states per unit volume in this space. Particles with momentum between p and p + dp deﬁne a spherical shell with radius

pL R = , (7) π~

2 / 8 PHY401 - Nuclear and Particle Physics Monsoon Semester 2020 and thickness dpL dR = , (8) π~ in this space, with volume

L 3 4V V = 4πR2dR = 4π p2dp = p2dp. (9) π~ π2~3

To get the number of states we need to remember that the condition for nx > 0, ny > 0, nz > 0 requires that we take 1/8 of the volume V, and that we also need to multiply V by 2 to account for two spins state of the nucleon The density of states is

1 1 4V V dn = 2 × × V = p2dp = p2dp. (10) 8 4 π2~3 π2~3 . This gives dn V = p2. (11) dp π2~3 Now we can compute the average momentum of a nucleon. We have p R F p dn p2dp R pF 3 0 dp 0 p dp 3 hpi = = = pF ≈ 188 MeV. (12) R pF dn 2 R pF p2dp 4 0 dp p dp 0

1.3 Average Kinetic Energy of a Nucleon

We need to compute the average energy. Note that the average energy is not the energy for the average momentum. To calculate the average energy we need to know the density of energy state, which is the number of states for a particle with energy between E and dE. We can get this number from the density of the momentum states

V dn = p2dp. (13) π2~3

p2 E = =⇒ p2 = 2mE =⇒ pdp = mdE. (14) 2m

mdE rm dE dp = √ = √ . (15) 2mE 2 E

√ V 3 √ dn = 2 m 2 EdE. (16) π2~3

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Thus the average kinetic energy of a nucleon is

3 EF EF R dn R 2 0 E dE dE 0 E dE 3 hEi = = 1 = EF ≈ 20 MeV. (17) EF dn EF R R 2 5 0 dE dE 0 E dE

Note that the energy which corresponds to the average momentum is

hpi2 9 p2 9 E = = F = E ≈ 18.6 MeV. (18) 2m 16 2m 16 F

The total kinetic energy of the nucleus is

Ekin(N,Z) = NhEni + ZhEpi 3 = N(pn )2 + Z(pp )2 . (19) 10m F F

Upon using V (pn )3 N = F , (20) 3π2~3

V (pp )3 Z = F , (21) 3π2~3 and

4 4 V = πR3 = πr3A, (22) 3 3 0 we get 3 2 9π 2/3 N 5/3 + Z5/3 E (N,Z) = ~ . (23) kin 2 2/3 10m r0 4 A Note that we have again assumed that the radii of the proton and the neutron potential wells are the same. This average kinetic energy has, for ﬁxed mass number A but for varying N (or equivalently Z), a minimum at N = Z. Hence the binding energy shrinks for N 6= Z. If we expand the above equation in the diﬀerence N − Z we obtain

2 2/3 2 3 ~ 9π 5 (N − Z) Ekin(N,Z) = 2 A + + ··· . (24) 10m r0 8 9 Z

This gives us the functional dependence upon the neutron surplus. The ﬁrst term contributes to the volume term in the mass formula. The second term describes the correction which results from having N 6= Z. This so-called asymmetry energy grows as the square of the neutron surplus and the binding energy shrinks accordingly.

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We thus see that the simple Fermi gas model, where nucleons move freely in an averaged-out potential, can already render the volume and asymmetry terms in the semi-empirical mass formula plausible. The ﬁrst term corresponds to the volume term in the liquid drop model while the last one corresponds to the asymmetry energy term in the liquid drop model. Therefore, based on the Fermi model we can conclude that the asymmetry energy term in the liquid drop model is a quantum mechanical eﬀect related to the way fermion occupy allowed states in the proton/neutron potential well.

2 Global Properties of Nuclei

It now became possible to use the particles produced by radioactive decay to bombard other elements in order to study the constituents of the latter. This experimental ansatz is the basis of modern nuclear and particle physics. The development of ion sources and mass spectrographs now permitted the investigation of the forces binding the nuclear constituents, i.e., the proton and the neutron. These forces were evidently much stronger than the electromagnetic forces holding the atom together, since atomic nuclei could only be broken up by bombarding them with highly energetic α-particles. The binding energy of a system gives information about its binding and stability. This energy is the difference between the mass of a system and the sum of the masses of its constituents. It turns out that for nuclei this difference is close to 1% of the nuclear mass. This phenomenon, historically called the mass defect, was one of the first experimental proofs of the mass-energy relation E = mc2. The mass defect is of fundamental importance in the study of strongly interacting bound systems.

2.1 Nuclides

A nuclide is a distinct kind of atom or nucleus characterized by a speciﬁc number of protons and neutrons. The atomic number: The atomic number Z gives the number of protons in the nucleus. The charge of the nucleus is, therefore, Q = Ze, where e is the elementary charge e = 1.6×10−19 C. In a neutral atom, there are Z electrons, which balance the charge of the nucleus, in the electron cloud. The atomic number of a given nucleus determines its chemical properties. The mass number: In addition to the Z protons, N neutrons are found in the nucleus. The mass number A gives the number of nucleons in the nucleus, where A = Z + N. Diﬀerent combinations of Z and N (or Z and A) are called nuclides.

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1. Nuclides with the same mass number A are called isobars.

2. Nuclides with the same atomic number Z are called isotopes.

3. Nuclides with the same neutron number N are called isotones.

The binding energy B is usually determined from atomic masses, since they can be measured to a considerably higher precision than nuclear masses. We have 1 2 B(Z,A) = ZM( H) + (A − Z)Mn − M(A, Z) c . (25)

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

Mn is the mass of the neutron. M(A, Z) is the mass of an atom with Z electrons whose nucleus contains A nucleons. The rest masses of these particles are

2 Mp = 938.272 MeV/c , (26) 2 Mn = 939.565 MeV/c , (27) 2 me = 0.511 MeV/c . (28)

In nuclear physics, nuclides are denoted by AX, X being the chemical symbol of the element. For example, the stable carbon isotopes are labelled 12C and 13C; while the radioactive carbon isotope frequently used for isotopic dating is labelled 14C. A A Sometimes the notations Z X or Z XN are used, whereby the atomic number Z and possibly the neutron number N are explicitly added.

2.2 Semi-empirical Mass Formula

Apart from the lightest elements, the binding energy per nucleon for most nuclei is about 8 MeV. Depending only weakly on the mass number, it can be described with the help of just a few parameters. The parametrization of nuclear masses as a function of A and Z, which is known as the Weiz- sacker formula or the semi-empirical mass formula, was ﬁrst introduced in 1935. The mass of an atom with Z protons and N neutrons is given by the following phenomenological formula

2/3 M(A, Z) = NMn + ZMp + Zme − avA + asA Z2 (N − Z)2 δ +a + a + , (29) c A1/3 a 4A A1/2 with N = A − Z.

The exact values of the parameters av, as, ac, aa and δ depend on the range of masses for which they are optimized.

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One possible set of parameters is given below:

2 av = 15.67 MeV/c , (30) 2 as = 17.23 MeV/c , (31) 2 ac = 0.714 MeV/c , (32) 2 aa = 93.15 MeV/c , (33)  −δ0 for even Z and N (even-even nuclei)  δ = 0 for odd A (odd-even nuclei) (34)  +δ0 for odd Z and N (odd-odd nuclei) where 2 δ0 = 11.2 MeV/c . (35)

The individual terms can be interpreted as follows:

Volume term, av: This term, which dominates the binding energy, is proportional to the number of nucleons. Each nucleon in the interior of a (large) nucleus contributes an energy of about 16 MeV. From this we deduce that the nuclear force has a short range, corresponding approximately to the distance between two nucleons. This phenomenon is called saturation. Due to saturation, the central density of nucleons is the same for all nuclei, with few exceptions. The average inter-nucleon distance in the nucleus is about 1.8 fm.

Surface term as: For nucleons at the surface of the nucleus, which are surrounded by fewer nucleons, the binding energy is reduced. This contribution is proportional to the surface area of the nucleus (R2 or A2/3).

Coulomb term ac: The electrical repulsive force acting between the protons in the nucleus further reduces the binding energy. This is approximately proportional to Z2/A1/3.

Asymmetry term aa: Heavier nuclei accumulate more and more neutrons, to partly compensate for the increasing Coulomb repulsion by increasing the nuclear force. This creates an asymmetry in the number of neutrons and protons. The dependence of the nuclear force on the surplus of neutrons is described by the asymmetry term (N − Z)2/4A. This shows that the symmetry decreases as the nuclear mass increases.

Pairing term ap: A systematic study of nuclear masses shows that nuclei are more stable when they have an even number of protons and/or neutrons. This observation is interpreted as a coupling of protons and neutrons in pairs. The pairing energy depends on the mass number, as the overlap of the wave functions of these nucleons is smaller in larger nuclei. Empirically this is described by the term δA−1/2.

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The Weizsacker formula is often mentioned in connection with the liquid drop model. In fact, the formula is based on some properties known from liquid drops: constant density, short-range forces, saturation, deformability and surface tension. An essential diﬀerence, however, is found in the mean free path of the particles. For molecules in liquid drops, this is far smaller than the size of the drop; but for nucleons in the nucleus, it is large. Therefore, the nucleus has to be treated as a quantum liquid, and not as a classical one. At low excitation energies, the nucleus may be even more simply described as a Fermi gas; i.e., as a system of free particles only weakly interacting with each other.

References

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

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