C h a p t e r 6

Nuclides composite particles of nucleons

A nuclide is a type of atoms whose nuclei have specific numbers of protons and neutrons (nucleons). The standard model considers up and down quarks as basic components of nuclei, but nucleons (protons and neutrons) are convenient units. On the other hand, nuclides are energy states in the form of masses. Stable states remain unchanged for an indefinite period whereas unstable ones undergo radioactive decay. For example, the energy equivalents of protons, deuterons and 238U are 938 MeV, 1.88 GeV and 212 GeV respectively. For convenience, we discuss nuclides in terms of nucleons.

Discoveries of protons and neutrons infer the existence of isotopes. Since isotopes refer to atoms of the same element with different number of neutron, the term nuclide is more appropriate when referring to a type of atom. Nuclear properties are specific for nuclides, but not necessarily for chemical elements.

The number of protons is the same as the atomic number, Z, and the mass number, A, is the number of nucleons in the nucleus. Thus the number of neutrons, N, is A - Z. Any nuclide is an isotope of an element, and the symbol of the element is used to represent the nuclide, but the mass number is given as a pre-superscript, such as 16O, where 16 is the mass number. For clarity and convenience, the atomic number is given as a post-superscript, 16O8.

Stable nuclides exist for an indefinite period of time, and they are the constituents of ordinary material. Unstable nuclides emit subatomic particles, with 4, n, p being the most common. Few undergo nuclear fission. However, radioactive nuclides with long half-lives are also present in nature.

Discovery consists in seeing what everyone else has seen and thinking what no one else has thought.

177

A nuclide as an isotope of element E.

AEZ

Stable NuclidesStable nuclides remain unchanged for an indefinite period, and they are not radioactive. Of the natural elements on Earth, only elements with atomic number less than 83 have stable isotopes, except technetium (Tc, Z = 43) and promethium (Pm, Z = 61). Only 81 elements have at least one stable isotope. However, there are 280 stable nuclides, and many elements have more than one stable isotope. The composition of isotopes in an element is an important piece of information in many technologies. Elements with more than 92 protons are all man-made, as are technetium (Tc) and promethium (Pm), because they have no stable isotopes. Elements with atomic number greater than 83 have no stable isotopes, but they are decay products from uranium and thorium, and they are constantly produce, but at the same time they decay and convert eventually to Bi or Pb.

There are many factors affecting the stability of nuclides. In this section, we describe some common factors of stable nuclides. We can only describe one at a time, but all factors concertedly affect the stability of nuclides.

Numbers of Protons and NeutronsProtons are charged but neutrons are not. Although they may not exist as individual nucleons in the nucleus, their numbers are obvious properties. From accounting point of view, we consider them as individuals.

How do numbers of protons and neutrons affect the stability?What is the role of neutrons in the atomic nuclei?What are the ratios of neutron numbers to proton numbers for stable nuclides?Do the ratios vary systematically?

We have examined some light nuclides in an Chapter 5 when we introduced the Nuclide Chart as a means to organize nuclear information. For convenience, we divide the large Nuclide Chart into three sections: light-weight nuclides with atomic number less than 20 (Z < 20), medium-weight nuclides with 20 < Z < 50, and heavy- weight nuclides with Z > 50.

Light-weight nuclides are given in the next page. Some general observations regarding their stability are given below:

178

Only 1H and 3He have more protons than neutrons in their nuclei. Natural hydrogen contains 0.0015 % of the isotope D, and only a trace of 3He is present in natural helium (mostly 4He). All other nuclides have either equal numbers of protons and neutrons or more neutrons than protons. The heaviest nuclide with equal numbers of protons and neutrons is 40Ca20, (20 being a magic number).

Most elements with odd atomic numbers, Z, have only one stable isotope. The number of stable isotopes generally increases as Z increases. Nuclides with odd numbers of neutrons have only one stable isotone (nuclides with the same number of neutrons), and the number of stable isotones also increases as the number of neutrons, N, increases.

Four nuclides 2D1, 6Li3, 10B5, and 14N7 have equal and odd numbers of protons and neutrons. For these light stable nuclides, the ratio N/Z ranges from 1 (D) to 1.22 (40Ar), ignoring H for which N/Z = 0. All but a few stable nuclides have more neutrons than protons. Uncharged neutrons probably moderate the proton-proton repulsion, making heavy nuclide stable.

A chart of stable nuclides with 20 < Z < 70 is given on the next page. For convenience in our discussion, nuclides with atomic numbers range between 20 (Ca) and 50 (Sn), two magic numbers called medium-weight nuclides.

Z Stable Nuclides| (Magic numbers and double magic-number nuclides are in bold) (to be continued)21 Sc20 . . . . . . . . . . . . . . . . . . . Ca Ca Ca Ca Ca Ca19 K K K18 Ar Ar Ar .17 Cl Cl16 S S S S15 . . . . . . . . . . . . . . P14 Si Si Si . .13 Al12 Mg Mg Mg . . .11 Na10 . . . . . . . . . . Ne Ne Ne 9 F . . . 8 O O O 7 N N 6 C C . . . . 5 . . . . B B 4 Be . . . . 3 Li Li 2 He He . . . . . 1 P D N 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27

179

A line with equal numbers of protons and neutrons is marked by “+” signs. For stable nuclides, the ratio N/I increases as Z (or N) increases: N/I = 1.14 for Sc (Z = 21), 1.27 for Nb (Z = 41), 1.41 for Sb (Z = 51), and 1.43 for Tb (Z = 65). The effect of this increase can be seen from the deviations of the locations of stable nuclides from the line marked by “+”.

Many stable isotopes appear on lines for Z = 8, 20, 28, 50, and 82 (see list below). Similarly, unusual numbers of stable isotones also appear on vertical lines with N = 8, 20, 28, 50, and 82. The existence of many stable nuclides with these numbers of protons or neutrons supports calling them magic numbers.

For the medium-weight nuclides, 1 (40Ca) < N/Z < 1.48 (124Sn50). However, 1 < N/Z < 1.40 for the six stable isotopes of calcium alone: 40Ca20, 42Ca20, 43Ca20, 44Ca20, 46Ca20, and 48Ca20. There are ten (10) stable isotopes of tin (112, 114, 115, 116, 117, 118, 119, 120, 122, & 124Sn50), for which 1.24 < N/Z < 1.48.

180

181

Stable Nuclides (Continue)

Z The Line with equal numbers of p and n is marked by “+”. (Double magic-number nuclides and magic numbers are in bold)74 W 73 Ta To be continued in Table form72 Hf X XXXX71 Lu XX70 Yb . . . . . . . . . . . . . . . . . . . . . . . . + . . . . . . . . . . X XXXXX X69 Tm X68 Er X X XXX X67 Ho X66 Dy X X XXXXX65 Tb . . . . . . . . . . . . . . . . . . . . . + . . . . . . . . . . X64 Gd X XXXXX X .63 Eu X X62 Sm X XXXX XX .61 Pm - - -60 Nd . . . . . . . . . . . . . . . . . . + . . . . . . . XXXXX X X .59 Pr X58 Ce X X X X . .57 La XX56 Ba X X XXXXX . .55 Cs . . . . . . . . . . . . . . . + . . . . . X54 Xe X X XXXXX X X . .53 I X52 Te X XXXXX X X . . .51 Sb X X 50 Sn . . . . . . . . . . . . . + . . . X XXXXXXX X X . . .49 In X X48 Cd X X XXXXX X . . . .47 Ag X X46 Pd X XXX X X . . . .45 Rh . . . . . . . . . . +. . X44 Ru X XXXXX X . . . .43 Tc - - -42 Mo X XXXXX X . . . . .41 Nb X40 Zr . . . . . . . . + . . . XXX X X . . . . .39 Y X38 Sr X XXX . . . . .37 Rb X X 36 Kr X X XX X . . . . .35 Br . . . . . + . . X X34 Se XXXX X X . . . . .33 As X32 Ge X XXX X . . . . . .31 Ga X X30 Zn . . . + . X XXX X . . . . . .29 Cu X X28 Ni X XXX X . . . . . . .27 Co X26 Fe X XXX . . . . . . .25 Mn + X24 Cr X XXX . . . . . . .23 v XX22 Ti XXXXX. . . . . . . .21 Sc X 20 Ca X X 2 2 3 4 5 6 7 8 8 9 10 0123456789012345678901234567890123456789012345678901234567890123456789012345678901234567

Because we run out of room in the graph above, the stable isotopes of elements with Z > 70 are listed below with their mass numbers. Number of neutrons (N) can be evaluated by subtracting the atomic number (Z) from the mass number (A), N = A - Z.

In the table below, the abundance in percentage, %, of the isotopes are given in the parentheses (% omitted). Gold (Au) and bismuth (Bi) are two elements that have only one stable isotope, the percentages of 197Au and 209Bi are 100%.

Mass Numbers of Heavy Stable Elements (70 < Z)

Z Symbol Mass (abundance or half life)

71 Lu 175 (97.4), 176 (2.6)72 Hf 174 (0.163), 176 (5.21), 177 (18.56), 178 (27.1), 179 (13.75), 180 (35.22) 73 Ta 180 (0.0123), 181 (99.9877)74 W 180 (0.135), 182 (26.4), 183 (14.4), 184 (30.6), 186 (28.4)75 Re 185 (37.07), 187 (62.93)76 Os 184 (0.018), 186 (1.59), 187 (1.64), 188 (13.3), 189 (16.1), 190 (26.4), 192 (41.0)77 Ir 191 (38.5), 193 (61.5)78 Pt 190 (0.0127), 192 (0.78), 194 (32.9), 195 (33.8), 196 (25.2), 198 (7.19)79 Au 197 (100)80 Hg 196 (0.146), 198 (10.02), 199 (16.84), 200(23.13), 201(13.22), 202(29.8), 204(6.85)81 Tl 203 (29.5), 205 (70.5)82 Pb 204 (1.4), 206 (25.1), 207 (21.7), 208 (52.3)83 Bi 209 (100)

90 Th 232 (100% half life 1.4x1010 y)

92 U 235 (0.720, half life 7.04x108 y), 238 (99.276, half life 4.5x109 y)

For more information about nuclides, consult the Handbook of Chemistry and Physics by CRC Press, and the Handbook of Isotopes by CRC Press. More information about the Nuclide Chart is also available from the following web sites:

http://www.fysik.lu.se/~nsr/isoexpl/isoexpl.htmhttp://www.dne.bnl.gov/~burrows/usndn/usndn.htmlhttp://csa5.lbl.gov/~fchu/isoexpl/man73.htm

The contents in these web sites include the following. Description of the US Nuclear Data Network Center for Nuclear Information Technology, San Jose State

University

182

Idaho National Engineering Laboratory Isotopes Project, Ernest Orlando Lawrence Berkeley National

Laboratory Lund Nuclear Data Service, University of Lund, Sweden Tandem Accelerator Laboratory, McMaster University, Canada National Nuclear Data Center, Brookhaven National Laboratory Nuclear Data Evaluation Project, Triangle Universities Nuclear

Laboratory Nuclear Data Project, Oak Ridge National Laboratory Permanent Mass-chain Evaluation Responsibilities International Nuclear Structure and Decay Data Network (NSDD) Glossary of nuclear data evaluation and WWW jargon

Skill Building Questions

1. How does the ratio N/Z vary as Z or N increases?

2. What are the stable nuclides that have equal numbers of protons and neutrons. Which of these have odd atomic numbers?

3. Which medium-weight elements do not have stable isotopes? What are their atomic numbers? How many stable isotones are there with N = 19, 31, 35, 39, 61, and 89?

4. Can you calculate the atomic weight of say lead (Pb) from the abundance given to each stable isotope in the table of heavy stable nuclide, why or why not? If not, what more information is required?

Pairing of NucleonsSince free nucleons have ½ spin, they obey Pauli's exclusion principle by allowing two protons or neutrons each with opposite spin to occupy a quantum state (if they are nucleons in a nucleus). There is a preference for having pairs of protons or neutrons, and it is known as pairing of nucleons.

How does pairing of protons and neutrons affect the stability of nuclides?What evidence suggests pairing of nucleons contributes to the stability of nuclide?

183

The numbers of stable nuclides due to even or odd numbers of protons and neutrons seem to suggest support the preference theory.

There are 166 stable nuclides with both Z and N even, 57 with even Z odd N, and 53 with odd Z even N. The only 4 light stable nuclides with both Z and N being odd are 2D1, 6Li3, 10B5, and 14N7. Note that H, Li, and B have two stable isotopes, which belong to the odd-even type. The distribution of stable nuclides due to odd and even number of Z and N can be interpreted as due to pairing of nucleons. We have mentioned that technetium (Z = 43) and promethium (Z = 61) have no stable isotopes. There are no stable isotones for N = 19, 31, 35, 39, 61, and 89.

The distinction between stable and long-lived nuclides is vague, but we choose to list the stable ones only. The effect of pairing also affects the abundance of the isotopes in elements, as well as the abundance of a nuclide on a planet, galactic or universal scale.

Pairing of nucleons also affects the decay or transmutation of unstable nuclides, particularly transmutation of isobars with even mass numbers A. For this type of nuclide, a beta decay converts an even-even nuclide into an odd-odd nuclide.

Skill Building Questions

1. Which medium elements do not have stable isotopes? What are their atomic numbers? How many stable isotones are there with N = 19, 31, 35, 39, 61, and 89?

2. How many stable isotones are there with N = 2, 8, 20, 50, and 82?

Magic Numbers of NucleonsMagic numbers 2, 8, 20, 28, 50, 82, and 126 have been mentioned in previous chapters in connection with the shell model of nuclear structures.

How does the distribution of stable nuclides support a number to be a magic number?How do magic numbers affect the stability of a nuclide?

Effect of Paring Nucleons

Z N No. of stable nuclideseveneven 166evenodd 57oddeven 53odd odd 4

184

One of the supports for the shell model comes from the fact that there are more stable isotopes with magic numbers of proton and more stable isotones with magic number of neutrons when compared to nuclides of similar mass numbers.

The emission of 4He2 (or particle) nuclei in the form of radioactive decay supports 2 as a magic number. Oxygen is a very abundant element, with 16O8 being the most abundant isotope (99.759%), plus 0.037% of 17O and 0.204% of 18O. There are 2 isotones (15N, 16O) with N = 8, plus the long-lived 14C. These data suggest 8 as a magic number.

There are 5 stable isotones with N = 20, but 0 with N = 19, 1 with N = 21, and 3 each for N = 18 and 22. There are 6 stable Ca isotopes (Z = 20); the abundances of Ca isotopes are: 40Ca 96.97%, 42Ca 0.647%, 43Ca 0.145%, 44Ca 2.06%, 46Ca 0.0033%, 48Ca 0.185%. The light and heavy isotopes of Ca have 20 and 28 neutrons. There are also 5 Ni (Z = 28) isotopes and 4 isotones with 28 neutrons. These data reinforce the effect of pairing of protons, and pairing of neutrons and the magic numbers 20 and 28. Note that 4He2, 16O8, 40Ca20, 48Ca20, and 208Pb82 are double-magic-number nuclides, because they have magic number of protons and neutrons.

Numbers of stable isotopes and isotones for magic numbers 50 and 82 are very convincing. Furthermore, three families of radioactive series decay to a stable lead isotope (Z = 82) after many steps of transmutations.

The heaviest stable nuclide 209Bi83 has 126 neutrons; so has its isotone 208Pb82. You may think that two stable isotones are not impressive, but no heavier nuclides are stable.

So far, elements with Z = 112 have been artificially made. One of the objectives for people who synthesize heavy nuclides is to test the large magic numbers. Glenn T. Seaborg, a Nobel laureate in Chemistry (1951) believed that element 126 may have a half-life long enough to be observed. However, this element has not been observed yet.

Skill Building Questions

1. What are magic numbers?What are the reasons for them to be magic numbers?

2. Give five double-magic number nuclides.

185

Abundance (%) of calcium isotopes

40Ca 96.9742Ca 0.64743Ca 0.14544Ca 2.0646Ca 0.0033

Abundances of ElementsThe abundance of an element or nuclide is its amount in a system.

What is the most abundant element in the universe?How about the solar system and on the planet Earth?How are they estimated and based on what evidences?

Is the abundance of a nuclide related to its stability?Aside from stability, what else affects the abundance?

The sun has 99.9% of the mass of the solar system. Hydrogen atoms contribute 72%, and helium 4He 26% to all atoms in the Sun according to reliable estimates from spectroscopic, density, and meteorite studies. Light nuclides H and He are the major components of the Sun, where nuclear reactions convert H into He. Since stars are by far more massive than planets, the most abundant element in the universe is also hydrogen, followed by helium.

Taking as a whole, the most abundant element of the planet Earth is iron, which is the major component of the earth (molten) core. Additional evidence comes from the many iron meteorites, which are considered debris from outer space. However, the most abundant element of the Earth crust is oxygen in terms of number of atoms, but silicon is the most abundant element by mass. Their physical and chemical properties made them accumulate in different systems.

The atomic abundance (abundance based on the number of atoms) of elements are made from mineral and meteorites analyses. For example, the Earth crust is mostly silicate, SiO2, and the most abundant elements in the crust are O8 and Si14. Taking other data into consideration, oxygen is the most abundant element of the inner solar system, of which the Earth is a major part. The abundance of oxygen is given as 1 (log 1 = 0) for reference. Magnesium, silicon, and iron are the next most abundant elements, but their abundances are less than 0.5. Abundances of Ni, Ca, S, Al and Na are less than 0.1 whereas those of N and F are 0.0001 as shown in the diagram below.

The abundance of elements is one of the most important parameter in order to understand a planet, a satellite, or any body in the universe. Thus, space explorations include the probe to find the abundances of elements in the moon and mars.

186

The abundances of elements with even atomic numbers are usually higher than those with odd atomic numbers of comparable masses. The relative high abundances of O, Ca, Fe, Sn and Pb are due to the magic numbers 8, 20, 28, 50, and 82. Many factors influence the stability of nuclides and their abundances.

A general notion is that stable nuclides are abundant on a planetary or galactic scale. However, other factors such as the process of their synthesis and sources of nuclides for their productions also affect their abundances. The history and formation of the Sun, the Earth, and its satellite are different, and these differences give rise to their composition differences. The abundance, however, is partially related to their stability.

The group of elements C6, N7, O8, Ne10, Mg12, Si14, S16, Ar18, and Ca20 with even atomic numbers are relatively more abundant than the group with odd atomic numbers: F9, Na11, Al13, P15, and Cl17. The abundances of Li3, Be4, and B5 are very low. The abundance decreases from Ca20 on as the atomic numbers increase, but there is a relatively high peak at iron, Fe26, and nickel Ni28. Thus, the abundance strongly supports the theory that pairing of nucleons is an important factor for their stability.

Skill Building Questions

1. What are the most abundant elements in the Earth crust and why?

2. How do the magic numbers affect the abundances of elements or nuclides?

187

Mass and Stability of NuclidesUnstable nuclides under go radioactive decay or fission, and they are radioactive nuclides. Some long-life unstable nuclides occur naturally, but many of them have been made artificially, and their properties well studied. Their making (synthesizing) involves nuclear reactions, which will be discussed in the next Chapter. We are only concerned with mass or energy regarding their stability in this section.

Aside from the stable isotopes of 12C and 13C, radioactive carbon isotopes with mass number 9, 10, 11, 14, 15 and 16 have been made and identified. Their half-lives are 127 ms, 19.3 s, 20.3 m, 5730 y, 2.45 s, and 0.75 s respectively. Light carbon isotopes 9C, 10C, and 11C decay by emitting positrons or electron captures whereas heavy carbon isotopes 14C, 15C and 16C are beta

emitters. Their masses hold the key for their stability. Radioactivity is a process by which unstable nuclides convert to stable ones.

The Binding EnergyRadioactivity or nuclear decays are spontaneous and exothermic reactions. The decay energy is the difference in total energy content before and after the decay. Energy and mass are equivalent, and the relative mass is the key to stability.

Is the mass of a nuclide the sum of masses of its components?Why or why not? How is the mass of a nuclide related to its stability?

Masses of nuclides have been carefully measured, and the mass of a nuclide is usually less than the sum of masses of components. This suggests that the formation of a nuclide from its component is an exothermic reaction. For convenience in our discussion, a special name is given to the energy released when a nuclide is made from its components.

The binding energy (BE) of a nuclide is the energy released when the atom is synthesized from the appropriate numbers of hydrogen atoms and neutrons. Hydrogen atoms and neutrons are convenient components, because they provide appropriate number of electrons and nucleons. The binding energies of hydrogen atoms and neutrons are thus zero in this definition. The concept of BE applies to both stable

188

Half life9C127 ms,10C 19.3 s11C 20.3 m12C stable13C stable14C 5730 y15C 2.45 s

and radioactive nuclides, and the definition given above can be represented by a hypothetical equation:

Z H + N n = AEZ + BE

where H, n, and AEZ represent the energy equivalent of hydrogen, neutron, and the nuclide AEZ. In terms of masses, we have

Z mH + N mn = mE + BE

where mH, mn, and mE are masses of H, n, and the nuclide AEZ respectively. Thus, BE can be evaluated using:

BE = Z mH + N mn - mE.

The binding energy is thus the minimum energy required in order to decompose nuclide AEZ into Z H and N n, i.e., AEZ + BE = Z H + N n. The more the binding energy, the more stable is the nuclide.

To evaluate BE, we need to know the mass of a hydrogen atom (1.007825 amu, more than the mass of a proton), the mass of a neutron (1.008665 amu) and the mass of the nuclide AEZ. For example, the binding energies of 3He (mass 3.016031) and 4He (mass 4.0260) are:

BE (3He) = (2 x 1.007825 + 1.008665 - 3.01603) 931.481 MeV = 7.72 MeV

BE (4He) = (2 x 1.007825 + 2 x 1.008665 - 4.00260) 931.481 MeV

= 28.30 MeV

The estimates here show that formation of 4He releases much more energy than the formation of 3He, 20 MeV per atom more. Although 4He contains only one neutron more than 3He does.

For comparison purpose, we can look at the amount of energy released when one nucleon is added to a nuclide. The binding energy of a nucleon (BEn, or BEp) is the energy released when a nucleon is added to a nuclide. From the above estimates, the binding energy of a neutron for 3He is 20.58 MeV. As another example, the binding energies of proton and neutron for 234U are calculated below. For these calculations, the masses of 234U92, 235U, and 235Np93 are 234.040946, 235.043924 and 235.044056 amu respectively. These values are looked up from data Table of nuclides. By definition, the binding energy of a neutron, BEn in 234U is the amount of energy released in the hypothetical reaction:

189

234U + n 235U + BEn. Thus,BEn = (234.040946 + 1.008665 - 235.043924) amu = 0.00566 amu x 931.5 MeV / amu = 5.30 MeV

The binding energy of proton (BEp) when it is added to 234U to make a 235Np atom (234U + p 235Np + BEp) is,

BEn = (234.040946 + 1.008665 - 235.044056) amu = 0.00555 amu x 931.5 MeV / amu = 5.17 MeV

The binding energy for a group of nucleons such as an particle (BE) is the negative energy of decay Edecay. For example, we can write the equation for the decay of 212Po84 as

212Po84 = 208Pb82 + 4He + Edecay

But for the reverse reaction, we have 208Pb82 + 4He = 212Po84 + BE.

Thus, BE = – Edecay

Note, however, BE is the energy of binding an particle in the nuclide 212Po84, not the binding energy of 4He from 2 H and 2 n, neither is it the binding energy of 212Po84. By the way, the particle energy from 212Po is 6.09 is MeV (looked up from a table).

Skill Building Questions

1. Evaluate the binding energy of 12C (mass = 12.00000), 14N (mass = 14.003074) and 16O (mass = 15.994915). Discuss your results.

2. Discuss the binding energies of 16O, 17O (mass = 16.999131), and 18O (mass = 17.999160).

3. Calculate the binding energies of a proton and a neutron to 4He. You need to look up the masses of 5He and 5Li.

4. Is it possible for 235Np to be a proton emitter and 235U to be a neutron emitter? Is the binding energy of an alpha particle in an alpha emitter positive or negative?

The Average Binding EnergyThe average binding energy, Eab, of a nuclide is its binding energy (BE) divided by the number of nucleon in it (or mass number A), Eab, =

190

BE/A. The average binding energy is also called the packing fraction. This term is given to reflect the tighter the nucleons pack in a nuclide, the more energy is released.

How does binding energy, BE, vary as the mass number A increases?How does the average binding energy vary as A increase?Which nuclide has the largest average binding energy, 16O, 56Fe or 235U?

Comparison of BE provides an indication about relative stability for similar nuclides, but BEs of different nuclides are not based on the same numbers of H atoms and neutrons.

As we evaluate the average binding energy BEav of some of the nuclides, pay attention to the variation of binding energy as A increases:

BEav (3He2) = 7.72 MeV / 3 = 2.57 MeV / nucleonBEav (4He2) = 28.3 MeV / 4 = 7.08 MeV / nucleonBEav (16O8) = 127.6 MeV / 16 = 7.98 MeV / nucleonBEav (56Fe26) = 492.3 MeV / 56 = 8.79 MeV / nucleon BEav (54Fe26) = 471.76 MeV / 54 = 8.74 MeV / nucleonBEav (208Pb82) = 1636.44 MeV / 208 = 7.87 MeV / nucleonBEav (238U92) = 1801.7. MeV / 238 = 7.57 MeV / nucleon

In general, the binding energy BE increases as A increases. For example, the binding energy for 16O, 56Fe (mass = 55.934939 amu), 208Pb82 (mass = 207.976627) and 238U (mass = 238.050784) are 127.6, 492.3, 1636 and 1801.7 MeV respectively. There is a steady increase as A increases.

There is a dramatic increase of BEav from 3He to 4He. Note that 4He, 16O and 208Pb are double-magic-number nuclides, but their BEavs are lower than those of the two iron isotopes. In fact BEavs for the two iron isotopes are the highest among the group, indicating a very high stability of iron in terms of BEav. There is a very slight difference between BEavs of the two iron isotopes, but only 54Fe contains a magic number (28) of neutrons.

191

A general trend of the variation of average binding energy as a function of the mass number is sketched here. As the atomic number increases, the average binding energy increases, reaching a peak at Fe, and then it decreases gradually. From the average binding energy point of view, nuclides with mass number around 58 are the most stable.

Since the average binding energy for Fe is the highest, synthesis of Fe from hydrogen atoms and neutrons will release the most energy per nucleon compared to all other nuclides. Thus, energy will be released when light nuclides such as H and D combines to form He, or even when He combine to form C, O etc. Combining light nuclides to form heavy ones is called nuclear fusion and it usually is accompanied by a release of (fusion) energy.

When a heavy nuclide split up, it is called nuclear fission. Fission also releases energy. The release of energy in fusion and fission is evident from the sketch.

The BEav is an indicator of energy frozen per nucleon in a nuclide. The more average energy is released when a nuclide is synthesized, the less energy is frozen per nucleon in a nuclide. Thus, BEav is a parameter for the stability of a nuclide, stable and unstable.

Skill Building Questions

1. Calculate the average binding energies of 11B (Mass, 11.00931), 30Si (29.97376), 52Cr (51.9405), 100Ru (99.9030), 150Sm (149.9170), and 208Pb (207.9766).

2. Use a spreadsheet to evaluate the average binding energy of the above nuclides and those given in this section. Use the spreadsheet to plot or sketch the negative values of BEav as a function of A.

Mass ExcessesThe mass of 12C is defined as 12 atomic mass units (amu) exactly, and thus the average mass of a nucleon in 12C is exactly 1 amu. The

192

difference between the mass of a nuclide and its mass number, A, is called the mass excess (ME),

ME = mass - A.

For example, the mass excesses of H (1.007825 amu) and neutron (1.008665 amu) are 0.007825 and 0.008665 amu respectively. Ironically, mass excess is also called mass deficiency because many nuclides have negative mass excesses. Like binding energy and average binding energy, mass excess is a convenient and useful criterion regarding a nuclide's stability.

How is the mass excess (ME) evaluated?What is the ME for 12C?What is the relationship between ME and BEav?Sketch ME as a function of A, and use it to describe fusion and fission energy.

The mass excess of 3He (mass = 3.01603) is 0.01603 amu. This quantity is easily calculated for nuclides whose average nucleon masses are more than 1 amu. For these nuclides, the mass excesses are positive. The mass of 54Fe is 53.939612 amu for 54 nucleons, and the mass excess is (53.939612 - 54.000000 =) -0.060388 amu, a negative value. In fact, MEs are negative for most nuclides except the light ones (H, He, … C) and heavy ones (Ra, Th, U, … etc).

In general, the variation pattern of negative average binding energy (- BEav), and mass excess (ME) are the same, but there is no direct way of converting ME into BEav and vice versa. A comparison is given in a table form below:

Comparison Among ME. Eab.

Nuclide Mass/amu ME/amu -Eab /amu BE/amu

H 1.0078250.007825 0 0n 1.0086650.008665 0 0

3He 3.01603 0.01603 -0.00276 0.008284He 4.00260 0.00260 -0.0076 0.030412C 12.0000000 -0.00825 0.0989416O 15.994915-0.005085 -0.00857 0.1369

40Ca 39.96259 -0.03741 -0.00917 0.3669 54Fe 53.939612-0.060388 -0.00938 0.506556Fe 55.934939-0.065061 -0.00944 0.52851

208Pb82207.976627-0.023373 -0.00845 1.757238U92238.0507840.050784 -0.00813 1.934

193

The binding energies of H atoms and neutrons are zero, 0. The more H atoms and/or neutrons are combined, the more energy is released. Thus, the BE increases as mass number A increases. The average binding energies (BEav) are calculated by treating H atoms and neutrons in a similar manner.

The mass of 12C is a zero point of reference for ME. Since the mass of H atoms and neutrons are different, ME values cannot be converted to BE values directly, but the two parameters vary in a similar manner. Both ME and BEav are good indicators for the stability of a nuclide. Because ME values are easily evaluated, they are widely used. For example, a Table of Nuclides may choose to list ME for unstable nuclides instead of masses. In such a list, however, the unit MeV is often used.

The variation of ME as a function of mass number A is sketched here. The MEs for light nuclides oscillate rapidly due to the effect of nucleon pairing and magic numbers of nucleons. However, the variation for medium and heavy nuclides are not as dramatic. In general, the ME decreases as A increases from 1 to 56, but it increases as A increases from 56 to 208. The ME values indicate that nuclides with A = ~56 are more stable than light or heavy nuclides. Thus, combining light nuclides such as H, D, 3He, into heavier nuclides 12C, 20Ca etc. in a nuclear fusion reaction will release energy, and splitting heavy nuclides such as 235U in a nuclear fission reaction into lighter nuclides such as Sr, La, Xe etc. will also release energy. The variation of ME as a function of A is an important piece of information concerning nuclear energy.

Another application of ME is to use it to calculate the energy of radioactive decay. For example, the ME’s of 40Sc21 and 40Ca20 from a handbook are -20.527 and -34.847 MeV respectively. Thus the energy of the decay process

40Sc21 40Ca20 + + or 40Sc21 + e– 40Ca20

isEdecay = -20.527 - (-34.847) = 14.32 MeV

194

Of course, 1.02 MeV (2 times the rest mass of an electron) has to be spent in producing the positron-electron pair in positron decay, but not in electron capture. This example illustrates the fact that the ME not only vary as mass number A changes, it also vary as the atomic number Z changes for isobars. Thus, a plot of ME versus Z and A shall produce a 3-dimension plot. The three parameters are BE, A, and Z. We shall discuss the variation of ME as Z changes for the isobars in the next section.

Skill Developing Questions

1. The masses are given for the following nuclides. Calculate the mass excesses in MeV:2D 2.0140 amu 6Li 6.015121 amu9Be 9.012182 10B 10.01293711B 11.009305 13C 13.00335514C 14.003241 14N 14.003074

2. Use the ME to calculate the decay energy for the decay of 14C 14N7

+ –?

3. The ME’s of 40Sc21 and 40Ca20 from a handbook are -20.527 and -34.847 MeV respectively. What are the masses of these two nuclides?

4. What is the difference between ME and BEav of a nuclide? Discuss the advantages and disadvantages of their usage as indicators.

Mass Excesses of IsobarsIsobars have the same number of nucleons, but different numbers of protons and neutrons. Their mass excesses are usually different. Thus, the mass excess among isobars varies as atomic number, Z, changes.

How does mass excess vary as a function of atomic number Z?Why is the variation of mass excess interesting for isobars?What decay will decrease or increase the atomic number, Z?What is the relationship between the mass excess the decay energy?

Energy is the driving force for change, and a change usually results in a system containing less energy. Isobars convert to each other by the beta decay process, which change the atomic number giving a more stable nuclide. Thus, the variation of mass excess as a function of Z is interesting, because it shed some light about the beta decay process.

195

If the mass number is odd, all isobars are even-odd or odd-even type in numbers of protons and neutrons, whatever Z is. These nuclides do not have a special contribution to ME due to pairing of protons or neutrons, because in either type, there is one unpaired nucleon.

As an example, the mass excesses of isobars with an arbitrary mass number of 123 are given below:

In49 Sn50 Sb51 Te52 I-53 Xe54 Cs55 Ba56

-0.0896 -0.0943 -0.0958 -0.0967-0.0944 -0.0915 -0.0870-0.0808 amu

Note that all the mass excesses are negative. These values are usually available from a handbook or from a table of nuclides. However, the units may be in MeV or keV.

For these isobars, the two with the lowest mass excesses, Sb and Te, are stable. For a comparison, the mass excesses are plotted as a function of Z as shown here. The further away a nuclide is from its stable isobars, the higher (less negative) are their mass excesses. If we connect all these points to form a curve, its shape looks like a hyperbola. This meaning that ME varies as a function of Z2.

Since the relative mass excesses of the isobars are directly related to their masses, their locations on the diagram represent their thermodynamic stability. The lower the position, the more stable is the isobar.

196

Although mass excess cannot be converted to binding energy directly, the two quantities are similar, especially if –BE is used. For the isobars with mass number 57, the BEs and - BEs are listed and plotted here:

Isobars of Mass number 57

Cr24 Mn25 Fe26 Co27 Ni28

Mass 56.9434 56.9383 56.9354 56.936356.3980 /amu

BE 0.53031 0.53462 0.53667 0.534930.53240 /amu

-BE -494 -498 -500 -498 -496/MeV

For a small mass number of 57, the number of isobars listed in handbooks is limited. Isobars with atomic number outside the 24 - 29 range are too unstable to be observed.

For mass numbers 57 and 123, the effect of nucleon pairing is not present, because all these isobars have either the protons or neutrons paired, but not both. The effect of nucleon pairing is important for isobars with even mass numbers, because the beta decay process convert a even-even (paired) nuclides into odd-odd ones, which have two unpaired nucleons. Isobars with even mass numbers are discussed next.

Isobars with 120 nucleons have two stable nuclides 120Sn50 (mass = 119.902200 amu, mass excess = -0.097800 amu = –91.1 MeV) and 120Te52 (mass = 119.904048 amu, mass excess = 0.095952 amu). The former has a magic number of protons.

197

The atomic numbers Z, element symbols (Isobar), mass differences (Mass) [compared to 120Sn50, which is taken as zero] decay energies (Edecay), decay modes, and half lives, are given below:

Z 47 48 49 50 51 52 53 54 55 56Isobar Ag Cd In Sn Sb Te I Xe Cs BaMass/MeV 15.32 7.135.3 0 1.51 1.72 7.12 9.07 17.3

22.6Edecay/MeV 8.2 1.80 5.3 stable 2.68 stable 5.4 2.0 8.22

5.3Decay mode – – – – +,EC – +,EC +,EC +,EC

+, ECHalf life 1.2 s 50 s 44 s 15.9 m – 1.35 h 40 m 64 s

32 s

The decay energy and mass differences are consistent for isobars which undergo – decay, but for those undergo + and EC decays, there is a little discrepancy, perhaps due to production of positrons.

The mass differences (Mass) can be converted to mass excesses by subtracting 91.1 MeV, which is the mass excess of 120Sn (ME = (Mass - 91.1) MeV). For simplicity, the mass differences are plotted against the atomic number Z here. From the graph, the - decay series Ag (- -) Cd (- -) In (- -) Sn, reduces the energy of a nuclide via several

Relative Mass Differences (MeV) forIsobars with Mass Number 120.

Ba22

20

18 Cs16 Ag14

12

10 Xe 8 Cd I 6 In 4

2 Sb Te 0 Sn 47 48 49 50 51 52 53 54 55 56

198

isobars in steps. So does the decay series by + emission or electron capture that convert Ba via Cs, Xe, I, to Te in five steps.

The stable and even-even type isobar 120Te52 contains slightly higher energy than the radioactive and odd-odd type nuclide 120Sb51, which is converted to the most stable isobar 120Sn50, which has the lowest mass (or energy). The two decay modes + and EC of 120Sb51 occur at 41 % and 59 % respectively.

To see the effect of nucleon pairing, we connect the even-even type and the odd-odd isobars by lines. These lines form two hyperbolas.

There is a difference in Mass for odd-odd type nuclides and the even-even type nuclides due to the effect of nucleon pairing. The even-even and odd-odd nuclides appear to belong to two separate classes. The gap between these two types of nuclides depends on the mass number A. An average over many nuclides has given a difference of 250/A MeV. For A = 120, the gap is 1 MeV, and this value seems reasonable from the two curves in the plot.

The variation of binding energy for mass 57 represents typical isobars with odd number nuclons, and the variation of binding energy for mass 120 reflects the importance of nucleon pairing. In both examples, there are two stable nuclides, but more often, there is only one stable nuclide for a series of isobars. From these graphs, we gain a little more insight to the energy for the beta decay process.

Skill Building Questions

1. List atomic numbers Z, element symbols (Isobar), mass excess (ME) decay energies (Edecay), decay modes, and half lives for isobars with mass number A = 140 and 141 respectively. (Please find the data from a handbook)

2. Plot the - BE of the nuclides given above versus atomic number.

3. Use a spreadsheet to plot the variation of ME as functions of both Z and A for a selected group of nuclides, for example Z = 1 to 20 or for Z=21 to 40.

A Semi-empirical Equation for Binding EnergyBinding energy (BE) varies in a systematic manner as a function of mass number A and atomic number Z.

Can the systematic variation be represented by a formula?What a theoretical model can be proposed for variations of BE?

199

The variation of BE is complicated but somewhat systematic and a model for calculating BE has been proposed by Seeger in 1961. He used a semi-empirical approach. The equation is based on some theoretical consideration, but mostly based on empirical data (the masses of nuclides). His theoretical consideration is based on the liquid-drop model, in which the nucleus volume is directly proportional to A. Its radius is proportional to the cubic root of A (A1/3) and its surface area is proportional to the square of its radius A2/3. The function BE as a function of A and Z is,

BE(A,Z) = 14.1A - 13A2/3 - - 20 2 2( )A ZA + Eo.

For nuclides with A 80, the equation gives reasonable results (in MeV). The rational for various terms are described below:

1. The 1st term (14.1 A) shows BE being proportional to the number of nucleons A. The more nucleons that pack into a nucleus, the more energy is released. Thus, BE is proportional to A.

2. The 2nd (negative) term (- 13A2/3) indicates an increase in surface area causes a decrease in stability. The factor A2/3 reflects the square of the radius (A1/3).

3. The 3rd (negative) term represents the instability caused by

protons due to Coulomb interaction, which is proportional to Z2, but inversely proportional to the radius.

4. The 4th (negative) term 20 2 2( )A ZA attributes to a favorable

number-of-neutron to number-of-proton (N/Z) ratio for a nucleus of mass A, and the instability is proportional to (A - 2 Z)2 but inversely proportional to A.

5. The last term (Eo) is a constant reflecting stability due to pairing of nucleons, Eo = 0 for odd-even and even-odd nuclides, = -125/A for even-even nuclides, and = +125/A for odd-odd nuclides.

All isobars have the same value A, and their binding energy is thus a function of Z and Z2. For isobars with an odd mass number, the nuclides are either odd-even or even-odd type, and the last term is zero. This equation suggests that a plot of BE for a set of isobars as a function of Z is a parabola with its vertex at the stable isotope. For isobars with an odd number of nucleons, there are usually one or two

200

nuclides that contain the least amount of energy. These are stable isobars.

For isobars with even mass numbers, the equation suggests two parabolas, one corresponding to the odd-odd type, and one to the even-even type due to the different values of the last term E0.

Skill Building Questions

1. Use the semi-empirical equation to calculate the binding energies for the isobars with mass number 120 giving earlier. Compare the values with those listed earlier and comment on your results.

2. Apply the semi-empirical equation to calculate the binding energy for Cr24, Mn25, Fe26, Co27, and Ni28 for mass number 57, and discuss your results.

201

Exercises1. What are the lightest and heaviest stable nuclides? Which element

has the largest number of isotopes? What elements with atomic number less than 83 are absent and why? What are some of the common features of these absent nuclides? What roughly is the number of stable nuclides?

2. What are the ratios of number-of-neutrons to number-of-protons ratio (N/Z) for the stable nuclides: 1H, 14N, 35Cl, 56Fe, 80Br, 100Rh, 150Sm, 200Hg, 209Bi? Describe the variation of N/Z ratios as the mass number increases.

3. The heaviest stable nuclide is 209Bi (Z = 83). How many neutrons are there in the nucleus? What is its N/Z ratio? (Ans. 126 n and 1.52)

4. Give the four stable nuclides that have equal and odd numbers of protons and neutrons?

5. What evidences support the conclusion that nuclides with even numbers of protons and neutrons are more stable? What theory can you give to explain this fact?

6. List a summary of the considered factors that affect the stability of nuclides.

7. List some features about the stable nuclides. For example, explain why there are six stable nuclides for the element hafnium but only two each for lutetium and tantalum?

8. Is there a relationship between abundance of nuclides and their stability? If so, what evidence can you give? If not, provide some arguments.

9. Enumerate the magic numbers. Why are they called magic numbers? Give some double magic-number nuclides? Comment on the fact that 209Bi (Z = 83) is stable, whereas 209Pb (Z = 82) is a positron emitter.

10. What are the most abundant nuclides in the solar system and in the planet earth? How are the estimates made? Comment on your answer.

11. Calculate the binding energies (BE) and average binding energies of 56Fe (Z = 26, mass = 55.9349 amu), and 238U (Z = 92, mass = 238.0508). (Ans. BE = -492.3 MeV and -1801.7 MeV respectively.)

202

12. Use a spreadsheet to evaluate the average binding energy of the above nuclides and those given in this section. Use the spreadsheet to plot or sketch the negative values of Eab as a function of A (one of the Skill Building Questions).

13. The disintegration energies for isobars with mass number 133 are given below. Make a plot of energy content as a function of Z for these isobars from the disintegration energy (in MeV). Te (52), 2.4; I (53), 1.8; Xe (54), 0.43; Cs (55), stable; Ba (56), 0.489; La (57), 1.2l Ce(58), 1.8. Do the points fall on a parabola?

14. Calculate disintegration energies of 133Ba and 133La using the function of binding energy?

15. Plot the binding energy of isobars with mass number 114 as a function of Z. Which nuclide is stable? (Check the properties of nuclides from a handbook please.)

16. Discuss the stability of isobars in terms of binding energy (BE). In particular, how does BE vary as a function of the atomic number Z and mass number A? What decay process is responsible for the transmutation of isobars? How does a function for the calculation of BE takes the effect of even or odd number of protons and neutrons into account?

Further Reading and Work CitedHarvey, B.G. (1969), Introduction to nuclear physics and chemistry, 2nd Ed. Prentice Hall.

Kuroda, P.K. (1982), The origin of the chemical elements and the oklo phenomenon, Springer-Verlag.

Lederer, C.M., Hollander, J.M. and Perlman, I. (1986) Table of isotopes 8th Ed.

Noddack, I. and Noddack, W. (1930), Naturwissenschaften 18, 757.

Pagel, B. (1965), New Scientist, 26, 103.

Seeger, P.A. (1961) Nuclear Physics, 25, 1

Suess, H.E. and Urey, H.C. (1956), Rev. Mod. Phys. 28, 53.

Useful Web Sites203

The following web sites contain very handy data about nuclides. http://www.nndc.bnl.gov/from that, choose the Nuclear Wallet Cards to go to http://www.nndc.bnl.gov/wallet/walletcontent.shtmlOn it is a list of atomic numbers, such as 0-n 1-H 2-He 3-Li 4-Be 5-B 6-C 7-N 8-O 9-F 10-Ne 11-Na 12-Mg 13-Al 14-Si 15-P 16-S 17-Cl 18-Ar 19-K 20-Ca 21-Sc 22-Ti 23-V 24-Cr 25-Mn 26-Fe 27-Co 28-Ni 29-CuEach of these items is hyper linked to another file which contains data on all isotopes of the elements. Access these data is easier than looking up from a handbook.

If you click the NuDat, you will go to the web site http://www.nndc.bnl.gov/nndc/nudat/and there are various choices of nuclear data.

204