Elementary Particle Physics Atomic Physics The physics of the electronic, extra-nuclear structure of atoms Nuclear Physics The physics of the atomic nucleus, believed to be constituted of neutrons and protons Elementary Particle Physics The physics of quarks and gluons, believed to be the constituents of protons and neutrons, and of leptons and gauge bosons and…who knows what else! Quarks, gluons, leptons, and gauge bosons are believed to have no substructure.

Isotopes: same Z 40Ca, 42Ca, 44Ca often, ‘isotope’ used instead of ‘nuclide’ isotopes have same Z, so same number of electrons => same chemistry use radioactive isotope in place of stable one – can monitor decay for tracer studies Isotones: same N 40Ca, 42Ti, 44Cr Isobars: same A 42Ca, 42Ti, 42Cr Isodiaphors: same neutron excess 42Ca, 46Ti, 50Cr isodiaphors isotopes isobars isotones Z

Classification of Nuclides Stable nuclei: 264; 16O Primary natural radionuclides: 26; very long half-lives; 238U with t1/2 = 4.47 x 109 y Secondary natural radionuclides: 38; 226Ra t1/2 = 1600 y decay of 238U Induced natural radionuclides: 10; cosmic rays; 3H t1/2 = 12.3 y; 14N(n,t)12C Artificial radionuclides: 2-4000, 60Co, 137Cs…

Chart of Nuclei

4-1 The naturally occurring nuclei

Line of Stability 1. Apart from three or four exceptions, the naturally occurring elements up to lead are stable and lie on or near a line (the line of stability) in the Z, N plane. 2. These elements have a neutron number N which is equal to Z, Z+1,or Z+2 in nuclei up to A = 35 (except for 1H and 3 He, for which N =Z –1) but which thereafter increases faster than Z until in lead N ≈ 1.5 Z.

What we would like to know. 1. We would like to understand why the stable nuclei have this property and what happens if nuclei are produced in which N is greater or less than the stable optimum. 2. We would also like to understand why it is that for A > 209, there are no stable nuclei.

Binding Energy

Learning Objectives To define binding energy. To define mass defect. To know which are the most stable nuclei. Explain why energy is released in nuclear fission and nuclear fusion.

3 quarks baryons mp = 1.6726 x 10-27 kg = 938.26 MeV = 1.007276 u mn = 1.6749 x 10-27 kg = 939.55 MeV = 1.008665 u Charge: 0 Charge: e

The Atomic Mass Unit The atomic mass unit (u) is far more convenient to use with nuclear masses.  It uses carbon-12 as a reference and is defined as: 1 atomic mass unit (u) = 1.661 × 10-27 kg. Exactly 1/12th the mass of a carbon 12 atom.

The table shows particle masses in atomic mass units. Note that the numbers are expressed to a large number of significant figures as the changes are quite subtle. Particle Mass (u) Electron 0.000549 Neutron 1.008665 Proton 1.007276 Hydrogen atom (1p+ + 1e-) 1.007825 Helium atom (2p+ + 2n + 2e-) 4.002603 α particle (2p+ + 2n) 4.001505

Be Careful! Remember to distinguish between the atomic mass and the nuclear mass. The atomic mass is the mass of an atom complete with its electrons. The nuclear mass is the mass of the nucleus alone.  To get the nuclear mass we need to take away the mass of the electrons.

Binding Energy If you want to remove a nucleon from the nucleus of an atom, then you have to do work to overcome the strong nuclear force. Definition from specification book:- The removed nucleons gain potential energy. The binding energy of a nucleus is the work that must be done to separate a nucleus into its constituent neutrons and protons.

4-2 The nuclear binding energy For the study of nuclei the nuclear mass (M) is a very important quantity. It is related to measurements of both binding energy (B) and separation energy (S).

Binding Energy II If we run that process in reverse, then the binding energy can also be defined as the energy released when a nucleus is assembled from its constituent nucleons. This means that the mass of the nucleus is less than the mass of the separate nucleons because energy has been released…

Mass Defect Definition from the specification book:- The mass defect Δm of a nucleus is defined as the difference between the mass of the separated nucleons and the mass of the nucleus.

Calculating Mass Defect For a nucleus of an isotope ZAX, composed of Z protons and (A-Z) neutrons with mass MNUC, the mass defect, Δm, is given by:-

Question 1 What is the mass defect of a helium atom?

Answer 1 What is the mass defect of a helium atom? The atomic mass of a helium atom is 4.002603 u Therefore mass defect, Δm=0.030377 u Particle Mass (u) Number Total (u) Proton 1.007276 2 2.014552 Neutron 1.008665 2.017330 Electron 0.000549 0.001098 Total 4.032980

Calculating Binding Energy The mass defect Δm exists because energy is released when the constituent nucleons bind together to form a nucleus. The energy released is equal to the binding energy of the nucleus:- Note that Δm must be in kg to get energy in J.

Question 2 What is the binding energy of the helium atom whose mass defect is 0.030377 u?   Express your answer in MeV. m = 0.030377 u × 1.661 × 10-27 kg = 5.046 × 10-27 kg E = mc2 = 5.046 × 10-27 kg × (3 × 108)2 = 4.541 × 10-12 J E = 4.541 × 10-12 J = 28.38 × 106 eV= 28.38 MeV Note: 1 u = 931.3 MeV

Binding Energy Per Nucleon If we know the binding energy in a nucleus, and the number of nucleons, we can work out the binding energy per nucleon, which is the work done needed to remove each nucleon.  The higher the binding energy per nucleon, the more stable is the nucleus.  For helium (4He) the binding energy per nucleon is: Binding energy per nucleon = 28.38 MeV/4 = 7.1 MeV

Question 3 What is the mass defect in atomic mass units (u) and in kilograms for the lithium nucleus which has 7 nucleons, and a proton number of 3?  What is the binding energy in J and eV?  What is the binding energy per nucleon in eV?  The nuclear mass = 7.014353 u.  

Answer 3 Add them together to get 7.056488 u Now take away the nuclear mass from the number above to get the mass deficit. 7.056488 u - 7.014353 u = 0.042135 u   Now convert to kilograms: 1 u = 1.661 ´ 10-27 kg 0.042135 u × 1.661 ´ 10-27 kg  = 6.9986235 × 10-29 kg     Now use E = mc2 to work out the binding energy: E = 6.9986235 × 10-29 kg × (3 × 108 m/s)2 =  6.3 × 10-12 J In electron volts, this is 6.3 × 10-12 J ÷ 1.6 × 10-19 eV/J = 3.9 × 107 eV = 39 MeV.   There are 7 nucleons so the binding energy per nucleon = 3.9 × 107 eV ÷ 7 = 5.6 × 106 eV  

Question 4 What is the mass defect in atomic mass units (u) and in kilograms for the copper nucleus which has 63 nucleons, and a proton number of 29?  What is the binding energy in J and eV?  What is the binding energy per nucleon in eV?  The nuclear mass = 62.91367 u.

Answer 4 Number of protons = 29; number of neutrons = 63 – 29 = 34 Mass of protons = 29 ´ 1.007276 = 29.211004 u  Mass of neutrons = 34 ´ 1.008665 = 34.29461 u Total mass = 29.211004 u + 34.29461 u = 63.505614 u Mass defect = 63.505614 u – 62.91367 u = 0.591944 u Mass defect in kg = 0.591944 ´ 1.661 ´ 10-27 = 9.83218 ´ 10-28 kg Binding energy = mc2 = 9.83218 ´ 10-28 kg ´ (3 ´ 108 m/s)2 = 8.85 ´ 10-11 J Binding energy in eV = 8.85 ´ 10-11 J ¸ 1.6 ´ 10-19 J/eV = 5.53 ´ 108 eV Binding energy per nucleon =  5.53 ´ 108 eV ¸ 63 = 8.78 ´ 106 eV

Binding Energy per Nucleon We can plot a graph of binding energy per nucleon against nucleon number.

From this graph we can see that The maximum value is about binding energy of 8.7 MeV per nucleon and iron is the most stable nuclide. Helium has a particularly high value of binding energy per nucleon, much higher than the light isotopes of hydrogen. There is a trend for nuclides of nucleon numbers in multiples of 4 to be particularly stable (i.e. have a high binding energy). The largest nuclides tend to be less stable, with slightly lower binding energies per nucleon.

The Most Stable Nucleus Iron has the highest binding energy per nucleon so is the most stable nucleus.  If we look at large nuclei (greater than iron), we find that the further to the right (greater nucleon number) the less stable the nuclei.  This is because the binding energy per nucleon is getting less. The explanation for this observation lies in that the strong nuclear force that binds the nucleus together has a very limited range, and there is a limit to the number of nucleons that can be crammed into a particular space. 

Nuclear Fission A large unstable nucleus splits into two fragments which are more stable than the original nucleus. The binding energy per nucleon increases in this process and energy is released. The change in binding energy per nucleon is about 0.5 MeV in a fission reaction.

Nuclear Fusion Small nuclei fuse together to form a larger nucleus. The product nucleus has a higher binding per nucleon as long as A is no greater than ~50. The change in binding energy per nucleon can be more than 10 times greater in a fusion reaction than a fission reaction.

The average binding energy per nucleon versus mass number A Bave = B/A

1. The saturation property observed in the figure is the manifestation of short range characteristics of nuclear force. 2. The short range nuclear interaction can be studied by examining data collected from the (p,p) and (n,p) scatterings as well as from the binding energy of deuteron.

4He 8Be 12C 16O 24Mg Bave = B/A

(1). The separation energy of a neutron Sn Separation energy (S) (1). The separation energy of a neutron Sn (4) (5)

(2). The separation energy of a proton Sp Separation energy (S) (2). The separation energy of a proton Sp (6) (3). The separation energy of a α-particle Sα (7)

Summary Atomic Mass Unit: 1/12th the mass of a carbon atom Mass defect: Difference between the mass of nucleons separately and together within a nucleus. Difference between the two sides of a nuclear interaction equation. Energy worked out by E = mc2.                                                     Binding Energy: Energy equivalent of the mass defect in a nucleus.  Binding energy per nucleon increases in more stable nuclei. Fission Splitting of a nucleus.  Rarely spontaneous.  Occurs after the nucleus has been tickled with a neutron Fusion  Joining together of two light nuclei to make a heavier nucleus.

finer features of nuclear forces are ignored, 4-3 The liquid drop model 1. A detailed theory of nuclear binding, based on highly sophisticated mathematical techniques and physical concepts, has been developed by Brueckner and coworkers (1954-1961). 2. A much cruder model exists in which the finer features of nuclear forces are ignored, but the strong inter-nucleon attraction is stressed. It was derived by von Weizsäcker (1935) on the basis of the liquid-drop analogy for nuclear matter, suggested by Bohr.

Bave = B/A The mass density of nuclear matter is approximately constant throughout most of the periodic table. Over a large part of the periodic table the binding energy per nucleon is roughly constant. These two properties of nuclear matter are very similar to the properties of a drop of liquid, namely constant binding energy per molecule, apart from surface tension effect, and constant density for incompressible liquids.

The essential assumptions of the liquid drop model: 1. A spherical nucleus consists of incompressible matter so that R ~ A1/3. 2. The nuclear force is identical for every nucleon and in particular does not depend on whether it is a neutron or a proton. Vpn = Vpp= Vnn (V denotes the nuclear potential) 3. The nuclear force saturates.

The binding energy of a nucleus Definition: (9) From the liquid drop model ̶ Weizsäcker’s formula (10) Carl Friedrich von Weizsäcker, 1993 A German physicist (1912-2007)

(10) is the “volume term” which accounts for the binding energy of all the nucleons as if every one were entirely surrounded by other nucleons. is the “surface term” which corrects the volume energy term for the fact that not all the nucleons are surrounded by other nucleons but lie in or near the surface. Nucleons in the surface region are not attracted as much as those in the interior of a nucleus. A term proportional to the number of nucleons in the surface region must be subtracted from the volume term.

(10) is the “Coulomb term” which gives the contribution to the energy of the nucleus due to the mechanical potential energy of the nucleus charge. Assume a charged sphere of radius r has been built up, as shown in the figure (a). The additional work required to add a layer of thickness dr to the sphere can be calculated by assuming the charge (4/3)πr3ρ of the original sphere is concentrated at the center of the shell [see figure (b)]. The electrical potential energy of the nucleus is therefore where and

(10) Three terms that were discussed previously are in a sense classical. The following terms that are to be discussed are quantum mechanical. (1) the asymmetry term (2) the paring term These include (3) the shell effect correction term is the “asymmetry term” which accounts for the fact that if all other factors were equal, the most strongly bound nucleus of a given A is that closest to having Z = N.

Different system energies due to asymmetric configurations (10) The Pauli exclusion principle states that no two fermions can occupy exactly the same quantum state. At a given energy level, there are only finitely many quantum states available for particles. Different system energies due to asymmetric configurations

1. If Z = N, then both wells are filled to the same level (the Fermi level). 2. If we move one step up away from that situation, say in the direction of N > Z (or Z > N), then one proton must be changed into a neutron. All other things being equal (including equal proton and neutron mass), this state has energy ΔE greater than the initial state, where ΔE is the level spacing at the Fermi level. 3. A second step in the same direction causes the energy excess to become 2ΔE. 4. A next step means moving a proton up three rungs as it changes from proton to neutron and the excess becomes 5ΔE.

requires an energy of ~ (N – Z)2ΔE/8. Cumulative effect 4. Therefore to change from N – Z = 0 to N > Z, with A = N + Z held constant, requires an energy of ~ (N – Z)2ΔE/8. 5. 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. 6. The energy levels of a particle in a potential well have a spacing inversely proportional to the well volume, thus we put ΔE ~ A-1. (11) This is the asymmetry term.

(10) is the “pairing term” which accounts for the fact that a pair of like nucleons is more strongly bound than is a pair of unlike nucleons. 1. For odd A nuclei (Z even, N odd or Z odd, N even) →δ = 0. 2. For A even there are two cases; (a). Z odd, N odd (oo) → – δ (b). Z even, N even (ee) → + δ (12)

(10) is the term accounts for the nuclear shell effect when Z or N is some magic number. This term is much less important than other terms. Therefore this term is not included in most of the applications. A favorable set of values for the coefficients: aV = 15.560 MeV aS = 17.230MeV aC = 0.6970 MeV aA = 23.385 MeV aP = 12.000 MeV (13)

(10)

(15) (16) 4-4 Mass parabolas and the stability line If we make some rearrangement the Weizsäcker’s formula can be written as (14) For a fixed mass number A, this is the equation of parabola with respect to the variable Z. We may differentiate the equation (14) and find the root (Z0, usually not an integer) of the following equation (15). Z0 is the optimum nuclear proton number for a fixed mass number A. The nuclear system with a specified mass number A is the most stable with proton number Z0. (15) (16)

(16) From the equation (15) the most stable nuclear systems of various mass numbers A are determined by the value of Z0. By using the relation A = Z0 + N we are able to plot stability lines on the N-Z plot. This follows exactly the shape of the empirical stability line in the figure. From expression (16), we can recognize that the deviation of the stability line from N = Z or Z = A/2 is caused by the competition between the Coulomb energy, which favors Z0 < A/2, and the asymmetry energy which favors Z0 = A/2.

beta (electron) decay takes place from Z to Z+1 For odd-A isobars, δ = 0, and equation (14) gives a single parabola, which is shown in the figure (a) for a typical case. We will see later that if M(A,Z) > M(A, Z+1) beta (electron) decay takes place from Z to Z+1 M(A,Z) > M(A, Z-1) electron capture and perhaps positron decay takes place from Z to Z - 1 (15) It is clear from the figure (a) that for odd-A nuclides there can be only one stable isobar. (14)

For even-A isobars, two parabolas are generated by the equation (14), differing in mass by 2δ. A typical case is given in the figure (b). Depending on the curvature of the parabolas and the separation 2δ, there can be several stable even-even isobars. Figure (b) shows that for certain odd-odd nuclides both conditions (15) are met so that electron and positron decay from the identical nuclide are possible and do indeed occur. (14)

Three types of β-decay