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CHAPTER 2 the Nucleus and Radioactive Decay

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1 CHEMISTRY OF THE EARTH CHAPTER 2 The nucleus and radioactive decay 2.1 The atom and its nucleus An atom is characterized by the total positive charge in its nucleus and the atom’s mass. The positive charge in the nucleus is Ze , where Z is the total number of protons in the nucleus and e is the charge of one proton. The number of protons in an atom, Z, is known as the atomic number 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 neutrons . The combined number of protons and neutrons, Z+N , is called the atomic mass number A. A specific nuclear species, or nuclide , 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 mole 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 periodic table, we will notice that an element’s atomic weight, which is the mass of one mole of its atoms, is rarely close to an integer. For example, Iridium’s atomic weight is 192.22 g/mole. The reason for this discrepancy is that an element’s neutron number N can vary. Nuclides with different neutron numbers but the same atomic number Z are called 191 193 isotopes . 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 masses 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 ion 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 ions 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 magnetic field ( B). The E and B fields are oriented such that the electric and magnetic forces 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 force qv×B acting 1 The mass of a proton is approximately 2000 times that of an electron, so electrons contribute only a small part to the mass of an atom. Cin-Ty Lee 2 CHEMISTRY OF THE EARTH rs cto lle co E B B 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 Hydrogen, 1H, is given by 2 We neglect molecular binding energies, which are negligible for this treatment. Cin-Ty Lee 3 CHEMISTRY OF THE EARTH 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 isotope of interest. We can also determine the masses of isotopes and elementary particles by considering the energies of particles in nuclear reactions by assuming energy and mass are interchangeable ( E=mc 2, where E is energy, m is mass, and c is the speed of light). 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 mass number 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 = ∑ X 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 lead 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 Cin-Ty Lee 4 CHEMISTRY OF THE EARTH 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 .
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