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Structure of Atomic Nuclei

Introduction

In our investigation of the electronic properties (chemical, physical, etc.) of , the nucleus is regarded as a positively charged point that embodies the great majority of the of the . The interaction between the nucleus and the atomic holds the atom together. The structure of the nucleus plays almost no role in atomic properties because nuclear excitation energies are several orders of magnitude higher than atomic or electronic excitation energies. 1 Hence, in normal atomic interactions (e.g., chemical reactions), the nucleus is always in the . However, we do know, based upon experiments conducted by Rutherford, Chadwick, and others, that the nucleus does have structure. Our question is: how are the structure of the nucleus and the properties of the nucleus related?

Nuclear Constituents

Nuclei are made up of (charge + e) and (no charge). is unique in that its nucleus has a single and no neutrons. The sum of the number of protons ( Z) and the number of neutrons ( number N) is called the ( A) of the nuclear species or : Z+N = A. Protons and neutrons are collectively called . Each A nuclide is symbolized by its chemical symbol (X) along with the values of A and Z: Z X.

4 8 14 Examples: 2 He , 4 Be, 6 C. A [The symbol in the textbook is ZNX , but this is redundant.]

Isotopes are that have the same Z, but different N (and therefore different A). Since the number of electrons in a neutral atom (i.e., Z) determines the chemical behavior of an element, have the same chemical symbol. 4 3 12 14 Examples: 2 He , 2 He ; 6 C, 6 C.

Isobars are nuclides that have the same mass number A, but different Z (and therefore different N). 2 14 14 Examples: 7 N , 6 C

Nuclear Sizes and Shapes

Like atoms, nuclei have no precisely defined size because of the extended of the wave functions of nuclear constituents. However, one can define an “average” radius. From scattering experiments, we know that nuclear radii have values in the range 2 – 8 fermi (10 -15 m).

1 MeV’s versus eV’s 2 Isotones have the same N: 7 B, 6 Li. 4 3

1 Experiments also show that nuclei are for the most part spherical. Departures from sphericity are usually slight (either elongation or flattening).

Another interesting feature is that the density of the nucleons, i.e., the number of the nucleons per unit volume, is virtually independent of the mass number of the nucleus. (See diagram above, taken from Krane’s Modern .) Hence, to good approximation, A = constant, 4 3 3 π R i.e., 1/3 RRA= 0 , (Empirical) (9.1) where R0 is a constant of proportionality. The value of R0 must be determined experimentally. Its value depends on how the nuclear radius R is defined. One such definition is the (from the center of the nucleus) at which the falls to one-half of its value at the center of the nucleus. Whatever the definition for R, R0 usually lies in the range 1.0 – 1.5 fermis. Because the nucleus occupies such a small volume and contains virtually all the mass of the atom, the mass densities of nuclei are immense. 12 ≈ Example: 6 C; R 2.7 fm, A = 12 12 u ρ = = 2.4 ×10 17 kg/m 3! 4 15 3 3 π (2.7× 10 m)

3 3 For comparison purposes, ρwater = 10 kg/m

To measure nuclear sizes, one can do scattering experiments in which charged nucleons 3 are scattered from the nucleus to be investigated. Departures from the Rutherford indicate that the nucleus is being impinged upon. 4 Why is there a deviation from the Rutherford formula when the nucleus is approached closely? The deviation suggests that when nucleons are in close proximity, i.e., within ~2 fermis of each other, another interaction dominates the electromagnetic Coulomb interaction. This interaction is the strong nuclear interaction .

3 E.g., protons or alpha (which contain protons) 4 Distance of closest approach < nuclear radius

2 The Strong Nuclear

It is easy to see that another interaction between U(r) nucleons (apart from the Coulomb interaction) must exist inside the nucleus. The Coulomb interaction causes a repulsive interaction among the protons in the nucleus. Since neutrons have no charge, they are unaffected by the Coulomb interaction. Why then does a nucleus remain bound? The reason is that for 1 2 -nucleon of up to about 2 fm, the strong dominates the Coulomb force. r (fm) At typical inter-nucleon distances in the nucleus, the strong force is attractive and therefore binds the -100 MeV nuclear constituents together. [Note that the attractive gravitational force is about 36 orders of magnitude too weak to do the job.] Experiments have shown that the strong force has the following properties (see sketch of nucleon-nucleon vs. separation distance above):

1. It is extremely short-ranged: For nucleon-nucleon separations r > ~ 3 fm, the strong force is negligibly small. 5

2. For r ≤ 1 fm, the strong force is strongly repulsive. The constancy of nucleon densities is evidence for this repulsive interaction. (If the interaction were purely attractive, nuclear densities at the centers of large nuclei would be significantly larger than those for small nuclei.)

3. The strong force acts equally between any pair of nucleons. Thus, the strong interactions for n- n, n-p, and p-p are the same for a given nuclear state. The strong nuclear force is therefore charge-independent.

4. The affects only certain types of particles called (e.g., neutrons, protons, ). Non-hadrons e.g., electrons, are not acted upon by the strong nuclear force. Thus, when high-energy electrons are used to bombard the nucleus, the characteristics of the scattering are not influenced by the strong force. [ diffraction is a useful tool for determining nuclear sizes.]

[Show Fig. 9.3 p. 235, Krane.]

Nuclear and Binding Energies

We saw in chapter 3 that the mass of a nucleus is less than the total mass of its constituent nucleons because energy is liberated when individual nucleons bind together to form a nucleus. This energy, called the , decreases the mass of the nucleus relative to that of the separated nucleons, according to mass-energy equivalence. Thus,

5 The constancy of nuclear density supports this, as it indicates that each nucleon only interacts with its nearest neighbors and not with all other nucleons.

3 E = (Nm + Zm − m )c2 , (9.2) b n p A X,nuc Z where m is the mass of the A nucleus. Since it is the masses of atoms that are usually A X,nuc Z X Z A measured, we add the mass of Z electrons to the protons and to the Z X nucleus to obtain: 2 Eb=(). Nm n + Zm1 − m A c (9.3) 1H Z X,atom Note that we have neglected the binding energy of the electrons in the atom under consideration. However, since nuclear binding energies are several orders of magnitude greater than electronic binding energies, it is an excellent approximation to neglect electronic binding energies.

56 Example: Find the binding energy and the binding energy per nucleon for 26 Fe.

Solution: Z = 26, N = 30. mn =1.0086654 u.

m1 = 1.007825 u, and m56 = 55.934939 u 1 H 26 Fe,atom Eb = (30 ×1.008665 u + 26 ×1.007825 u – 55.934939 u) × 931.5 MeV / u = 492.3 MeV, where we have used the equality c2 = 931.5 MeV / u. 56 [Note that binding energy for innermost electrons in 26 Fe is tens of keV.]

56 Binding energy per nucleon for 26 Fe = Eb /A = 492.3 MeV / 56 = 8.791 MeV/nucleon.

238 A similar calculation for 92 U gives Eb = 1802 MeV and Eb / A = 7.571 MeV per nucleon. 238 Hence, 92 U has the greater binding energy as would be expected, but it takes more energy per 56 nucleon to break up the 26 Fe nucleus.

The diagram [p. 237 Krane or TZ&D] shows the trend in binding energy per nucleon for all the 1 elements. The lightest nuclei have the smallest binding energies per nucleon ( 1 H has a binding energy of zero). This is because the number of nearest-neighbors is smaller than the "saturation" value, i.e., the maximum number consistent with the short range of the strong force. The binding 56 energy per nucleon rapidly climbs, peaks at 26 Fe, and then slowly declines.

The binding energy per nucleon chart explains the release of energy in and . If a very massive nucleus is split into two fragments having comparable masses, then each fragment will have a larger Eb/A value than the original. Thus, energy is released in this process. The process of splitting a massive nucleus with attendant release of energy is called nuclear fission . To more clearly see that energy is released, we write the : X → C + D (X splits into C and D). Let us consider the binding energies. Total energy is conserved, so

−EEEb,X,C,D= − b− b + energy. (9.4) Note that the binding-energy terms are negative (the binding energy Eb itself is defined to be positive) because the nuclei are bound states , i.e., energy must be supplied to separate the nuclei into individual nucleons having no kinetic or potential energy. We must have positive energy (energy released) on the right side of Eq. (9.4) because the binding energy per nucleon of X is

4 smaller than that of C and also smaller than that of D. Since the number of nucleons in X is equal to the sum of the number of nucleons in C and the number nucleons in D, it follows that

EEEb,X,C,D< b+ b , so, −EEEb,X,C,D> − b− b .

This can also be seen the point of view of rest energy (or rest mass). A greater binding energy per nucleon implies a smaller rest energy per nucleon inside the nucleus. Thus, in the fission process, the total rest energy of the fragments is less than the rest energy of the original nucleus. The rest energy lost appears as energy released.

We note that is very rare because there is an energy barrier to the fission process. (The repulsive Coulomb potential energy of the fragments causes a peak in the energy of the fragments relative to the original nucleus.) Hence, fission is usually initiated by bombarding the heavy nucleus with slow neutrons. The heavy nucleus will absorb a neutron, forming a new nucleus in an . The excitation energy is great enough to exceed the fission barrier, therefore fission ensues.

If two very nuclei are fused together, the binding energy per nucleon of the new nucleus is greater than each of those of the original two nuclei. Hence, energy is released in this process also. The fusing of light nuclei with attendant energy release is called nuclear fusion . Nuclear fission and fusion are of great practical importance. Fission is now used to provide nuclear energy for power plants and engines. Both fission and fusion have been used in nuclear weapons. Fusion is the source of energy for the , and research continues to find an economical method of using fusion for commercial power generation.

Nuclear Stability

We have seen that the strong force binds the nucleus into a stable unit despite the electrostatic repulsion of the protons. About 260 stable nuclei exist; 6 however, there are many more that are unstable. Unstable nuclei spontaneously disintegrate or decay to form other nuclei. This process of spontaneous decay is called radioactivity . Unstable nuclei will undergo successive decays until a stable nucleus results. Why are some nuclei stable and others are not? The chart of the stable nuclei gives us some interesting clues.

1. Stable nuclei (especially light ones) tend to have N = Z. (Due to the symmetry effect.) 2. For the more massive nuclei, N > Z. 3. There are no stable nuclei for Z > 83. 4. The overwhelming majority of stable nuclei have both N and Z even (~150). Approximately equal numbers (~50) have odd-even nuclei [ N odd, Z even or N even, Z odd]. A very small number of stable nuclei (4) have both N and Z odd. (Due to the pairing effect.)

Qualitative Classical Explanation for Properties of Stable Nuclei

We explain with a classical argument items (2) and (3) in the list above. The more massive nuclei have a larger number of protons and therefore there is a larger electrostatic repulsive

6 Show plot of N vs. Z pg. 241 Krane; Table 12.1 pg. 297 TZ&D.

5 potential energy tending to break the nucleus apart. Now, the electromagnetic force is long- ranged, while the strong force is very short-ranged. Hence, in larger nuclei, some protons would tend to be more widely separated, leading to a weakening of the strong force and the consequent dominance of the electrostatic repulsive force. To maintain stability, the number of neutrons exceeds the number of protons so that even though the protons are farther apart on average, each proton is close to more than one neutrons so that the strong force between neutrons and protons keeps the nucleus bound and stable. Evidently, if the nucleus gets too large, the electrostatic repulsion between the protons will to an unstable nucleus.

Quantum-Mechanical Explanation

Symmetry Effect (Why N ≈ Z for Light Nuclei) To fair approximation, the potential energy of a nucleon in a nucleus due to the strong interaction can be modeled as a square well. 7 For protons, the electrostatic repulsion increases the potential energy, raising the proton “well” relative to the neutron well. Let us first consider the situation for light nuclei, where the Coulomb repulsion is less important. To simplify the discussion, we assume (as seen in the diagram) that no two single-nucleon states have the same energy. 8 (This is not true in general because of degeneracy, i.e., different states with the same energy.) p n The Pauli Principle is at the heart of the symmetry effect. The Pauli 12 Principle says no two identical nucleons can occupy the same quantum 6 C state. (The Pauli Principle applies to all particles called that have “half-integral ,” where spin is one of the quantities that specifies a .) Thus, in our simplified model, each can have at most four nucleons – two protons and two neutrons (the arrows indicate the value of the spin – “up” or “down”). 9

12 12 Consider the energy levels for 6 C (for which N = Z) and an isobar 5 B(Z < N) as shown below. 12 12 Note that the ground state energy of 6 Cis lower than that of 5 B. A similar conclusion applies 12 for 7 N for which Z > N. For any set of isobars of light nuclei, one finds that the nucleus that has N = Z has the lowest ground-state energy and therefore tends to be stable. Consider a 15 hypothetical 6 C nucleus. The nucleus would be unstable because it could lower its total energy 15 by transforming the highest-energy neutron to a proton, resulting in a 7 N nucleus. This nucleus is stable because the uppermost occupied neutron and proton levels are nearly coincident, so it is not energetically favorable for any more transformations of nucleons to take place.

7 Figures 12.1 – 12.3, pp. 155-157, Six Ideas 8 Apart from spin. 9 Energy levels for neutrons well and proton well are the same because of charge independence of strong force.

6 p n p n p n 12 12 B 12 6 C 5 7 N Why N > Z for Heavy Nuclei For larger nuclei, the Coulomb repulsive energy of the protons is significant and cannot be neglected. It raises the energies of the protons relative to the neutrons. As seen in the diagram, stability is most probable when N > Z, since this must be the case when the energy difference between the uppermost protons and neutrons is smallest (so it will not be energetically favorable to transform a neutron to a proton or 44 vice versa). Consider the 22 Tinucleus. This nucleus is unstable (even though N = Z) because of the Coulomb repulsion. 44 [Show decay of 22 Ti.]

p

n Semi-empirical Binding Energy Formula ( Drop Model)

The Semi-empirical Binding Energy Formula (first written down by C. F. von Weizsacker) is a formula that accurately predicts the binding energies of all nuclei, with the exception of the very lightest ones (H, He, Li).

The formula consists of a sum of five terms, each of which depends on the mass number and/or the atomic number of the nucleus. The formula is ZZN2()− 2 ε E= a A − a A2/3 − a − a + a . (9.5) b vol surf CoulA1/3 symA pair A1/ 2

The values of the “ a” parameters are determined by using the formula to fit experimental data on binding energies. Let us interpret each of the terms in the formula.

The first term, called the volume term , is the most dominant term in the formula. It expresses the fact that for all but the least massive nuclei, the binding energy per nucleon is approximately constant. (This is due to the fact that each nucleon interacts with only a fixed number of nearest neighbors due to the short range of the strong nuclear force.) Therefore, this contribution to the

7 binding energy is proportional to A. This is analogous to the fact that the latent heat of vaporization of a drop of liquid is proportional to its mass.

The second term, called the term , is due to the fact that nucleons near the surface of the nucleus have fewer neighbors than those in the interior. This effect tends to decrease the binding energy so the term is negative. Since the density of nucleons is constant within the nucleus (it is assumed that the nuclear density drops discontinuously to zero at the surface), the contribution due to the surface term is proportional to the surface area of the nucleus. This is because the number of surface nucleons is proportional to the surface area. Now, for a spherical nucleus, 2 1/3 2 2 2/3 2/3 surface area = 4πRRARA= 4 π (0 ) = 4 π 0 . Hence the surface term is proportional to A .

The third term is the Coulomb term , and is due to the Coulomb repulsion between the protons in the nucleus. This repulsion decreases the binding energy, so the term is negative. The electrostatic potential energy of a uniform spherical charge distribution of radius R and total 3 2 1/3 charge Ze can be calculated fairly easily. The result is UCoul = 5 k()/ Ze R . Since R ∝ A , we 2 1/3 see that UZACoul ∝ / . Since nuclei are not exactly spherical, the expression for UCoul given above is not precisely correct.

The next term in the formula is the symmetry term , which is due to the fact the nuclei with Z = N tend to have the greatest binding energy in a set of isobars. As we previously saw, this effect is 2 attributable to the Pauli Exclusion Principle. The term −asym ()/ Z − N A captures the symmetry effect since any departure from Z = N decreases the binding energy. Further, the factor of A in the denominator expresses the fact that changes in binding energy for a given value of | Z–N| are smaller for larger nuclei because the spacing between energy levels is smaller.

The last term is the pairing term , which accounts for the fact that nuclei in which both N and Z are even numbers, are especially stable, i.e., have especially large binding energies. Also, nuclei with both Z and N odd tend to be unstable. This effect is due to quantum mechanical considerations involving the angular of the nucleus. The pairing term is expressed as 1/ 2 ∆Eb = ε a pair / A , where apair is a constant to be obtained from fitting the experimental data and ε is a parameter that reflects the greater stability of even-even nuclei: ε = 1 when N and Z are both even; ε = 0 when N or Z is even (but not both); ε = –1 when N and Z are both odd. 10

Fitting the formula with the experimental data yields the parameter values avol = 15.75 MeV, asurf = 17.8 MeV, aCoul = 0.711 MeV, a sym = 23.7 MeV, apair = 11.2 MeV.

[Show semi-empirical binding energy formula fit to binding-energy data.]

Note that the binding energy formula can be used to obtain the rest masses of nuclei: 2 mnuc= Zm p + Nm n − ( E b / c ). (9.6) The mass of the neutral atom is

10 Fig 16.14 ,TZ&D, p. 552

8 2 matom= Zm1 + Nm n − ( E b / c ). (9.7) 1 H 56 Example: find the mass of 26 Fe atom using the semiempirical binding energy formula

Solution: Z = 26, N = 30 This is an even-even nucleus, so ε = +1. (0.711 MeV)(26)2 (23.7 MeV)(26− 30)2 ( + 1)(11.2 MeV) E =(15.75 MeV)(56) − (17.8 MeV)(56)2/ 3 − − + b (56)1/ 3 56 561/2

= 491 MeV. (From a previous example, we obtained Eb = 492 MeV.)

2 1 u  m56 =(26)(1.007825 u) + (30)(1.008665 u) − (491 MeV /c )2  26 Fe,atom 931.5 MeV / c  = 55.9363 u. The accepted value (Appendix D) is 55.9349 u.

Shell Model and Magic Numbers

We saw that the majority of stable nuclei had both Z and N even. Furthermore, the experimental data show that when N = 2, 8, 20, 28, 50, 82, or 126, the nuclei are particularly stable. These "magic numbers" also apply to the atomic number Z, except for Z = 126 which does not exist. Nuclei are even more stable of both N and Z are magic numbers. 11

These magic numbers remind one of the closed-shell configurations of electrons in atoms. These configurations occur for atoms with Z = 2, 10, 18, 36, 54, 86. These are the inert or noble , which are particularly inactive chemically, i.e., they are very stable against . The magic numbers are in fact due to a similar shell structure for nuclei. Maria Goeppert-Mayer and J. Jensen shared the 1963 Physics Nobel Prize for their independent development of the shell model of the nucleus. To fully understand the shell model, we need the concept of the quantization of , which we have not dealt with as yet. Hence, we shall not go into further detail about the shell model.

11 Trasp. T&Z fig 12.15 p. 316

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