CHAPTER 2 The nucleus and

2.1 The and its nucleus An atom is characterized by the total positive charge in its nucleus and the atom’s . The positive charge in the nucleus is Ze , where Z is the total number of in the nucleus and e is the charge of one . The number of protons in an atom, Z, is known as the and dictates which element an atom represents. The nucleus is also made up of N number of neutrally charged particles of similar mass as the protons. These are called . The combined number of protons and neutrons, Z+N , is called the number A. A specific nuclear species, or , is denoted by A Z Γ 2.1 where Γ represents the element’s symbol. The subscript Z is often dropped because it is redundant if the element’s symbol is also used. We will soon learn that a of protons and a mole of neutrons each have a mass of approximately 1 g, and therefore, the mass of a mole of Z+N should be very close to an integer 1. However, if we look at a , we will notice that an element’s atomic weight, which is the mass of one mole of its , is rarely close to an integer. For example, ’s atomic weight is 192.22 g/mole. The reason for this discrepancy is that an element’s number N can vary. with different neutron numbers but the same atomic number Z are called 191 193 . Thus, Iridium has two isotopes, 77 Ir and 77 Ir , which are characterized respectively by 114 neutrons and 116 neutrons; a mole of the former weighs 190.960580 g and the latter weighs 192.962917 g. In naturally occurring Iridium, 37.3% of the atoms are 191 Ir and the remaining atoms are 193 Ir. The atomic weights reported in the periodic table represent the average of each element.

2.2 The mass of an atom How is the mass of an atom determined? One way to determine the mass of an atom is to use a mass spectrometer. Figure 1 shows schematically how one type of mass spectrometer works. The mass spectrometer begins with an source, which produces a beam of ionized atoms or molecules (in Chapter 4, we will discuss the different types of ion sources). This beam of passes through a velocity sector, through which only ions with a particular velocity are sampled. The velocity sector makes use of an electric (E) and a ( B). The E and B fields are oriented such that the electric and magnetic on the ions are equal and opposite, that is, qE = -qv×B 2.2 where q is the charge of the ion. Thus, only ions having a velocity equal to E/B are permitted to continue on their flight paths as ions with any other velocity would be deflected to the walls of the mass spectrometer. Once ions have been selected for their velocities, they enter a magnetic sector, where the magnetic field is uniform and perpendicular to the path of the ions. The magnetic field exerts a qv×B acting

1 The mass of a proton is approximately 2000 times that of an , so contribute only a small part to the mass of an atom. Cin-Ty Lee


rs cto lle co



Ion source Velocity Sector Magnetic Sector

Modified from Krane 1988

Figure 1. One type of mass spectrometer used for separating isotopes.

perpendicular to the ion flight path. This force is balanced by the centripetal force mv 2/r , that is, mv 2/r = qv×B 2.3 where r is the radius of curvature of the ion flight path. Upon rearranging, Eq. 2.3 can be expressed in terms of r r = mv/qB 2.4 It can be seen from Eq. 2.4, that a magnetic sector separates ions based on their momentum mv . However, as q, B, and v are uniquely determined, each different mass appears at a different radius r. If we know the magnitude of the electric and magnetic fields precisely, then obtaining the mass of an atom is trivial. To be useful, the precision of our mass determinations must be on the order of one part in 10 6. Unfortunately, it is not possible to know the magnitude of the electric and magnetic fields to this precision. We can, however, measure relative mass differences very precisely by keeping the electric and magnetic fields constant and measuring the difference in radii of curvature, which we can do more precisely. By convention, the 12 C atom is taken to be exactly 12.000000 units on the atomic mass scale, against which all other masses are calibrated. The masses of other isotopes are determined using the “doublet” approach, which is best explained through an example. If we set up the mass + + spectrometer to collect the molecular ions, C 9H20 (nonane) and C 10 H8 (naphthalene), the difference ∆ in mass units on this scale is 0.09290032±0.00000012 units 2 1 12 ∆ = m(C9 H 20 ) − m(C10 H 8 ) =12 m( H ) − m( C) 2.5 where m(C 9H20 ) is the mass of C 9H20 and so forth. It follows that the mass of , 1H, is given by

2 We neglect molecular binding , which are negligible for this treatment. Cin-Ty Lee


1 m(1H ) = [m(12 C) + ∆] 12 1 = .1 00000000 + ∆ 2.6 12 = .1 00782503 u Using this approach, we can work through the periodic table and obtain the mass of each of interest. We can also determine the masses of isotopes and elementary particles by considering the energies of particles in nuclear reactions by assuming and mass are interchangeable ( E=mc 2, where E is energy, m is mass, and c is the speed of ). Regardless of the approach, all masses are calibrated against 12 C. On such a scale the proton is 1.00727647 u, the electron is 0.0005485803 u, and the neutron is 1.00866501 u. One mass unit is defined to be 1.660566×10 -24 g, such that one mole or 6.022045×10 23 atoms (Avogadro’s number) of 12 C weighs exactly 12.000000 g. Nuclide masses are often tabulated in terms of mass defects ( ∆=m - A), which denotes the difference between the actual mass m and the atomic A (Appendix Table 1). Once nuclide weights are known, the atomic weight of an element can be determined by measuring an element’s relative abundances. The atomic weight AW of an element is simply the weighted average of each nuclide

AW = ∑ i mi 2.7

where Xi is the atomic proportion of nuclide i and its mass mi. For most naturally occurring elements, the relative nuclide abundances are constant and their numbers have now been tabulated for convenience. As we will learn later, there are two situations when isotopic abundances in nature vary. For example, isotopic abundances can be fractionated by certain physical processes in nature or in the lab. Another situation occurs when a nuclide is radioactive or is the product of the decay of a radioactive nuclide. Both of these processes yield a variable atomic weight. For example, Pb has 4 naturally occurring isotopes 204 Pb, 206 Pb, 207 Pb, and 208 Pb. The latter three are decay products of 238 U, 235U, and 232 Th, and therefore the relative abundances of the Pb isotopes depends on how much radiogenic Pb has accumulated from U and Th. The non- constancy of the atomic weight of was recognized very early on and was one line of evidence suggesting that elements are made up of different isotopes.

2.3 Binding energies and mass defects The astute reader might wonder now that if we know the masses of each elementary particle (proton, neutron, and electron), then it seems trivial to calculate the masses of all the nuclides as long as we know how many protons, neutrons and electrons are in a particular nuclide. In fact, the mass of a nuclide determined by the “doublet method” discussed above is slightly smaller than the combined masses of the constituent elementary particles. We can use the case of 191 Ir as an example. Iridium-191 is made up of 77 protons, 77 electrons and 114 neutrons. Therefore, the combined mass of the elementary particles is given by

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M = Zm p + Zm e + Nm n

= 77 .1( 00727647 ) + 77 .0( 0005485803 ) +114 .1( 00866501 ) 2.8

=192 .590340 u

(where mp, m e, and mn refer to the masses of a proton, electron and neutron). The true mass of 191 Ir (e.g., m( A Γ ) ) is actually 190.960584 u, which is 1.62976 u lower than the combined mass of the elementary particles. The reason for this decrease in mass is that the fusion of these elementary particles into the nucleus is energetically favorable, that is, energy is released when protons and neutrons come together. As energy is equivalent to mass according to the well-known equation E = mc 2 2.9 where E is the energy, m is the mass, and c is the speed of light, the decrease in energy is equivalent to a decrease in mass. The unit conversion between mass and energy is c2=931.50 MeV/u . A The energy released by combining N neutrons and Z protons to make a nucleus Z Γ is called the (Appendix Table 1): A 2 E B = [Zm p + Nm n − (m ( Γ ) − Zm e )]c 2.10 The binding energy for electrons is small compared to the nuclear binding energies, which is why in Eq. 2.10, we subtract out the mass energy of the electrons. Eq. 2.10 can be simplified by recognizing that the most elementary nucleus is the Hydrogen nucleus, 1H:

9 Fe Mg Si O Ne 8 C

7 B 6 Be

5 Li


3 He Fusion Fission Average Binding Energy Per (MeV) Per Nucleon Energy Average Binding 2 0 4 8 12 16 20 24 30 60 90 120 150 180 210 240

Mass Number A

Figure 2. Binding energy per nucleon.

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1 A 2 EB = [Z( H ) + Nm n − m( Γ)]c 2.11 For 191 Ir, we find that its binding energy is 1478.77 MeV. Figure 2 shows the binding energy per nucleon (for 191 Ir this is 1478.77 MeV/191 = 7.7423 MeV) as a function of mass number, e.g. EB/A . Except for mass numbers below ~20 u, it can be seen that EB/A is remarkably constant, lying between 7 and 10 MeV per nucleon. This feature indicates that the binding energy increases approximately linearly 56 with mass number. Another feature of the EB/A curve is that there is maximum at Fe. This means that fusion of masses lower than mass number 56 or fission of masses greater than 56 will release energy. It is this property that we take advantage of when generating nuclear energy by fusion or fission. We will see in Section 2.4 that an understanding of binding energy is fundamental to understanding why certain elements undergo radioactive decay. A nuclide will energetically tend towards decay by a particular mode (α emission, β emission, or ) if its atomic mass is greater than the sum of the masses of the products formed by that decay mode. This means that nuclei above A>100 are unstable towards spontaneous fission into two nuclei of approximately the same mass and that nuclei above A>140 will tend to decay by α emission.

2.3.1 A physical explanation for the binding energy Figure 3 shows a plot of Z versus N for all stable nuclei. One of the most important features is that at Z<20, the number of protons and neutrons is approximately equal. For nuclides having Z>20 , there are more neutrons than protons as shown by the deviation of the mass curve towards the neutron-rich side of the Z=N line. The array represents the most energetically favorable nuclides, that is, the combination of neutrons and protons that yields the highest binding energy. This array is often called the “”. The excess of neutrons at high mass numbers can be qualitatively explained by the fact that the more protons one packs together in the nucleus, the higher the Coulombic repulsion (which lowers the binding energy) and therefore the more neutrons needed to physically increase the separation distance of the protons so that their Coulombic repulsion is decreased. In more detail, physicists have come up with a semi-empirical equation to predict the binding energy of a nuclide. The binding energy can be expressed in terms of A and Z as follows (A − 2Z)2 E (Z, A) = a A − a A 3/2 − a Z(Z − )1 A− 3/1 − a +δ 2.12 B v s c sym A We now discuss each term in sequence. The first term is an empirical relationship that describes the fact that the binding energy is approximately a linear function of the atomic mass number A. That EB ~ a vA indicates that each nucleon interacts with only its nearest neighbors and that each nucleon contributes roughly the same amount to the binding energy. The constant av must be on the order of 8 MeV per nucleon. The second term accounts for the fact that on the surface of the nucleus interact less with other nucleons. As the surface area is proportional to R2 and R ∝ A1/3 , the surface term is proportional to A2/3 (this assumes that the nucleons are rather incompressible as is implied by the fact that EB ~ a vA). The third term accounts for the Coulombic repulsion of the protons, which decreases the binding energy. The Coulombic repulsion is directly proportional to Z(Z-1) and inversely proportional to the separation distance R, or in other words, proportional to A-1/3 . The A-1/3 proportionality allows for the Coulombic repulsion Cin-Ty Lee


N = Z

Z Lawrence Berkeley National

N Figure 3. Chart of Nuclides in a plot of Z versus N.

energy to decrease as more neutrons are added and has its greatest effect for high mass numbers. The fourth term accounts for the fact that at low masses ( Z < ~20), there is a tendency for the number of protons and neutrons to be symmetric, e.g. A = 2Z . The ( A- 2Z )2 proportionality is to account for the fact that as A becomes greater than 2Z , the binding energy is reduced. The inverse proportionality with respect to A is to account for the decreasing importance of and the increasing importance of the Coulombic term as the mass number increases. Finally, the last term accounts for the fact that nuclides with even numbers of Z or N are energetically more stable as is exemplified by the observation that there are 167 stable nuclei in nature with even N and even Z, but only four stable nuclei in nature with both odd N and odd Z (2H, 6Li, 10 B, 14 N). This term thus accounts for the pairing force, which accounts for the tendency of like nucleons to pair up. -3/4 For even Z and N, the pairing energy is usually expressed as + apA , for odd Z and N the -3/4 term is -apA , and for odd A the pairing term is zero. The constants av, a s, a c, a sym , and ap must be chosen to best fit the data in Figure X. One choice of constants is av=15.5 MeV, a s = 16.8 MeV, a c = 0.72 MeV, a sym = 23 MeV, and ap = 34 MeV. Using Eq. 2.12 for EB, we can rearrange Eq. 2.12 to yield the semi-empirical mass formula A 1 2 m(Z Γ) = Zm ( H ) + Nm n − EB (Z, A /) c Eq. 2.13 For constant A, Eq. 2.13 represents a parabola of m versus Z. On Figure X, nuclides of constant A are said to lie on an , which represent lines having a slope of negative one. A cross-section at constant A through the nuclide mass chart will therefore look like

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a valley, hence the trend of naturally occurring nuclei is called the “valley of stability”. For constant A, we can calculate the minimum mass or the most energetically favorable A nuclide configuration by letting (∂m( Z Γ /) ∂Z) A = 0 : 1 − 3/1 [mn − m( H)]+ ac A + 4asym Z min = − 3/1 −1 2.14 2ac A + 8asym A

Since ac = 0.72 MeV and asym = 23 MeV , the first two terms in the numerator are negligible and thus, A 1 Z min ≈ 3/2 2.15 2 1+ A ac /( 4asym )

It can be seen from Eq. 2.15 that for small A, Zmin is approximately equal to A/2 . For heavy nuclides, Z/A ~ 0.4 as expected from observations of the heavy stable nuclei.

2.4 Radioactive decay If we look at the chart of nuclides, which is a plot of naturally occurring nuclides by atomic number (Z) versus (N), we see the resulting array defines the region of greatest stability, often referred to as the “valley of stability”. For nuclides of low atomic mass, greatest stability is achieved when Z ~ N. However, as atomic mass increases, the stable Z/N ratio increases to ~1.5, that is, a larger number of neutrons are needed in order to maintain stability. This can be explained by the semi-empirical mass formula (Eq. 2.12). In words, we can say that as the number of protons increases in the nucleus, extra neutrons (e.g., N>Z) are needed to offset the increase in Coulombic

A Z+1 N-1 β−

A Z N β+ E.C Z . y c a A e d Z-1 α N+1

A-4 Z-2 N-2


Figure 4. Z versus N diagram showing how Z, A and N change during different decay schemes.

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repulsion caused by the larger number of protons. It follows that nuclides, which lie away from the “valley of stability”, will be unstable, and have a tendency to decay towards the “valley of stability”. In other words, a nuclide will energetically tend towards decay by a particular mode (e.g., α emission, β emission, or spontaneous fission; Fig. 4) if its atomic mass is greater than the sum of the masses of the products formed by that decay mode. It turns out that nuclei above A>100 are unstable towards spontaneous fission into two nuclei of approximately the same mass and that nuclei above A>140 will tend to decay by α emission. Below, we describe various types of radioactive decays.

2.4.1 At masses above ~A>140, many nuclides may decay by alpha emission because of the Coulombic repulsion between the increased number of protons being forced into the nucleus. Alpha particles are made up of 2 protons and 2 neutrons (Z=2, N=2) and therefore have the nuclear make-up of the 4He. Alpha decay can be expressed as follows: A A−4 4 Z Γ→Z −2 Γ+2 He + Q Eq. 2.16 where Q is the energy released by the decay process. The question arises as to why a prefers to decay by alpha emission rather than by emission of a larger A particle. Assuming the parent radionuclide Z Γ is at rest, requires that the energy released by alpha decay is 2 Q = (mA − mA−4 − mα )c Eq. 2.17 It turns out that the is one of the most tightly bound nuclides, having a low mass defect (or low binding energy). It can be seen from Eq. 2.17 that the smaller the mass defect of the emitted particles, the greater the release of energy Q and the more favorable the decay reaction is. The alpha particle, 4He, has a mass defect of +0.002603, which is smaller than the mass defects of other low mass nuclides, such as 1H, 2H, 3H and 3He, which means that emission of an alpha particle is an energetically more favorable process than emission by any of these other nuclides. We leave you to verify this in the Problem Set. In general, the greater Q of alpha decay is for a given nuclide, the more likely it is to decay by alpha decay. Thus, it turns out that if we look at the half-lives of alpha decay for various , the half-life decreases with increasing Q. There also appears to be a difference between atomic masses with even numbers of neutrons and protons from atomic masses with odd numbers of neutrons and/or protons. For a given Z and Q, the alpha decay half-lives of the latter are much longer than those of even-even nuclides. Geologically relevant alpha decay systems include the of 238 U to 206 Pb, 235 U to 207 Pb, and 147 Sm to 143 Nd.

Alpha recoil energies We now consider the distribution of energy caused by alpha decay. When a radionuclide decays to generate an alpha particle and a product nuclide, energy Q is released. This energy is taken up as due to the recoil of the product particles. Assuming that the radionuclide was at rest, the following relationship must hold 1 1 Q = m v 2 + m v 2 2.18 2 P P 2 α α

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where the two terms on the right hand side represent the kinetic energies of the product nuclide and the alpha particle (where v represents velocity). Because linear momentum must also be conserved, the following relationship also holds

mPvP = mα vα 2.19 Combining Eqs. 2.19 and 2.18, we arrive at

1 2  m   α  Q = mα vα 1+  2.20 2  mP  which shows that the velocity of the alpha particle is uniquely determined by the release energy Q, its own mass and the mass of the product nuclide. Because the mass of the alpha particle is very small, Eq. 2.20 also shows that most of the is taken up by the alpha particle. Typically, the alpha particle accounts for ~98% of the decay energy. The recoil distance of an alpha particle is therefore much larger than the product nuclide. In many instances, an alpha particle has enough energy to rip apart the lattices of its host material, leaving a short path of destruction, called an “alpha recoil track”. Recoil tracks can be seen in geologic materials that have high concentrations of radionuclides, such as 238 U and 235 U, and that have had sufficient time to accumulate noticeable amounts of alpha decay products. Very old are often so full of alpha recoil tracks that they serve as transport pathways for the diffusive loss of U or its decay products.

2.4.2 Beta decays and In addition to alpha decay, nuclides can undergo either negative ( β−) , positive beta decay ( β+) , or orbital electron capture (E.C.). These decay processes are as follows: Negative beta decay n → p + e− Positive beta decay p → n + e+ Electron capture p + e− → n In words, negative beta decay results in the conversion of a neutron to a proton and an electron (the negative ), positive beta decay results in the conversion of a proton to a neutron and a (positive beta particle), and electron capture results from the capture of an orbital electron by a proton in the nucleus to form a neutron. One of the dilemmas facing nuclear physicists during the early stages of beta decay research was that the measured kinetic energy of the beta particle and the daughter nuclide did not add up to the energy that should be released by the decay. The beta particles were in fact found to have a spectrum of kinetic energies, ranging from nearly zero to a maximum energy that was equal to the energy difference between the initial and final states. This “missing” energy appears to violate conservation of energy. Physicists have hypothesized that this “missing” energy is taken up by a second particle, which has a neutral charge. For negative beta decay, this second particle is called the antineutrino ν . For positive beta decay, the second particle is called the ν . Therefore, the complete decay process for negative beta decay is A A − Z Γ→Z +1 Γ + β +ν + Qβ − 2.21

where Qβ − is the mass difference between the initial and final states A A 2 Qβ − = [m(Z Γ) − m( Z +1 Γ)] c 2.22

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Qβ − represents the kinetic energies shared by the negative beta particle and the neutrino. Negative beta decay results in an increase in proton number at the expense of a neutron. Some geologically relevant negative beta decay systems include the decays of 87 Rb to 87 Sr, 187 Re to 187 Os, and 176 Lu to 176 Hf. The decay equation for is as follows A A + Z Γ→Z −1 Γ + β +ν + Qβ + 2.23 In this case, the proton number decreases by one and the neutron number increases by one. An example of positron emission is the decay of 26 Al to 26 Mg, the former a very short-lived radionuclide found only as a or as a nucleosynthetic product of . In the electron capture decay process, an electron reacts with a proton in the nucleus to generate a neutron. Thus, the neutron number increases at the expense of a proton, and for this reason, electron capture decay looks similar to positive beta decay A A Z Γ→Z −1Γ 2.24 The electron that is captured into the nucleus usually comes from the lowest energy levels because such electrons are closest to the nucleus and hence have the highest probability of being captured. Once such an electron is captured, an electron vacancy appears which is then filled by an electron from a higher . The transferring of a higher energy electron to a lower energy level results in the release of characteristic X-rays. The decay of 40 K to 40 Ar is an example of an electron capture decay relevant to geologists. In all of the above decays, the nucleus can be left in an . This excited nuclide can then decay to a lower energy level. This process results in the release of characteristic gamma rays (or ).

2.4.3 Spontaneous Fission Recall from our discussion of the binding energy (Fig. 2) that 56 Fe has the highest binding energy. Nuclides with mass numbers above 56 have lower binding energies and therefore it is energetically favorable for very high mass particles to separate into two nuclides with subequal masses. The release in energy caused by fission decay has been harnessed in nuclear reactors. Some nuclides, such as 235 U can be induced to fission if they are bombarded by neutrons. The fission products are not exactly equal in mass and are not always the same. There appears to be a spectrum of fission product masses. A possible neutron-induced fission reaction is 235 U + n = 93 Rb + 141 Cs + 2n where the product nuclides invariably fall on the neutron-rich side of the valley of stability and therefore undergo negative beta decay.

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Useful units Speed of light c 2.99792458 x 10 8 m/s Charge of electron e 1.602189 x 10 -19 C 23 -1 Avogadro’s Number NA 6.022045 x 10 mole

Particle rest masses Mass units ( u) MeV/c 2 Electron 5.485803 x 10 -4 0.511003 Proton 1.00727647 938.28 Neutron 1.00866501 939.573 Alpha 4.00150618 3727.409

Conversion Factors 1 eV = 1.602189 x 10 -19 J 1 u = 931.502 MeV/c 2 = 1.660566 x 10 -27 kg 1 barn = 10 -28 m2 = 10 -24 cm 2 1 Ci = 3.7 x 10 10 decays/s

References and useful websites Faure, G., 1986, Principles of Isotope Geology, John Wiley and Sons, NY, 589 pp. Classic text on the application of isotope chemistry to geology. Krane, K. S., 1988, Introductory , John Wiley and Sons, NY, 845 pp. Comprehensive treatment of and physics at the moderately advanced level. (http://www2.bnl.gov/ton/ ) – Lawrence Livermore National Laboratory website. Table of isotopes (http://ie.lbl.gov/toi.html ) – Lawrence Livermore National Laboratory website. Web of Elements (http://www.webelements.com/ ) – comprehensive source of elemental properties, electron configurations, ionization potentials, naturally occurring isotopes, binding energies, etc.

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