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Transcript
Decay Mechanisms 1. Alpha Decay An alpha particle is a helium-4 nucleus. This is a very stable entity and alpha emission was, historically, the first decay process to be studied in detail. Almost all naturally occurring alpha emitters are heavy elements with Z > 82. This is the only way a heavy nucleus can reduce its nucleon number, A. The laws of conservation of charge and of nucleons require that for alpha decay, A Z P→ A− 4 Z −2 D + 24 He + Q 3.1 where Q is the energy release in the decay, the disintegration energy. An example is 212 83 Bi → Tl + 24 He 208 81 If the radioactive parent is taken to be initially at rest, the conservation laws of energy and of linear momentum yield MPc2 = (MD + Mα)c2 + KD + Kα 3.2 MDvD = Mαvα 3.3 where the M’s are the atomic rest masses of the parent, daughter, and helium atom, and the K’s and v’s are the kinetic energies and velocities of the daughter and α particle. Since the kinetic energies KD and Kα can never be negative, α decay is energetically possible only if MP > M D + M α If this inequality is not satisfied, α decay simply cannot occur. In other words, decay is energetically possible only for Q > O. Q = KD + Kα = (MP – MD - Mα)c2 3.4 In an α decay, it’s the energy of the α particle, Kα, that is usually measured, for example by measuring its radius of curvature in a magnetic field. So, how is Kα related to Q? Squaring the masses in the kinetic energy equation above (3.2), and multiplying each side by ½ gives: MD(½MDvD2) = Mα(½Mαvα2) 1 MDKD = MαKα In atomic mass units, the daughter and the α masses are approximately A – 4 and 4 u, respectively. Then the equation above becomes (A – 4)KD = 4Kα But 4 ⎞ ⎛ Q = Kα + KD = Kα ⎜1 − ⎟ A− 4⎠ ⎝ and therefore Kα = A−4 Q A 3.5 Since most alpha emitters are those with large A (i.e. A >> 4), Kα is only slightly less than Q; nearly all the energy released in the decay is carried away as kinetic energy by the light particle. The equation above also shows that in two-particle emission from an initially unstable nucleus at rest, the α particle emerges with a precisely defined energy; since Q has a precise value, so does Kα. This does not mean that all alpha emitters emit only one alpha energy. Some alpha decays leave the daughter nucleus in one of a number of discrete excited states as well as the ground state. This means that the emitter can emit a group of alpha particles all at discrete energies. Over 300 α emitters have been identified. The emitted α particles have discrete energies from about 4 to 10 MeV, a factor of 2, but half-lives ranging from 10-6 to 1017 s, a factor of 1023. Short-lived α emitters have the highest energies, and the contrary, as indicated by the examples in the table below. α emitter 238 92 U Kα(MeV) 4.19 T ½ (s) 1.42 x 1017 λ (s-1) 4.9 x 10-18 212 83 Bi 6.09 3.64 X 103 1.90 X 10-4 215 85 At 8.00 10-4 104 Hazard: Alpha particles have a very limited range, and can’t even penetrate the outer, dead layer of skin. They therefore generally pose no direct external hazard to the body. However, ingestion or inhalation of alpha-emitters can lead to damage to internal organs, such as the inhalation of the daughters of radon gas. We’ll be discussing this in more detail later in the course. 2 Decay Schemes: A decay scheme is a graphical or schematic representation of nuclear transformations showing decay mode, energy transitions, and abundance (branching ratios). It consists of a qualitative plot of energy (vertical axis) versus atomic number (horizontal axis). Transitions leading to an increase in Z are indicated by an arrow slanting down and to the right. Those leading to a decrease in Z are indicated by a downward arrow to the left. Gamma emission, which results in no change in z, is indicated by a vertical wavy line. Below is the decay scheme for radium-226 which decays by alpha decay to radon-222 by one of two paths. One path is 94.4 % probable and leads directly to the ground state of the daughter and leads to the emission of a 4.785 MeV alpha particle. The other pay is 5.5% abundant and leads to an excited state of the daughter nucleus via the emission of a 4.602 MeV alpha particle. The excited state of radon drops down to the ground state via emission of a gamma ray within about 10-14 seconds. The gamma ray energy is equal to the 4.785 – 4.602 = 0.186 MeV. [Note that gamma emission occurs only 3.3 % of the time, indicating that 2.2 % of the time, the energy of excitation is lost via another process. This is internal conversion, described on page 12.] Beta (β) Decay Beta decay can be defined as the radioactive decay process in which the charge of the nucleus is changed without any change in the number of nucleons. There are three types of beta decay: β− decay, β+ decay, and electron capture. 2. β− Decay Nuclides with neutron excess lie to the right of the line of stability on the Z vs. N plot and need to convert a neutron to a proton via the β− decay process. Again we will invoke conservation of electric charge and conservation of nucleons in describing the decay of the parent nucleus P into the daughter D: 3 A Z P→ A Z +1 D + 0 −1 0 e + 0ν 3.6 The symbol ν represents the neutrino, and ν represents its antiparticle, the antineutrino. Neutrinos (and antineutrinos) are chargeless particles with little or no mass. They interact extremely weakly with nuclei and are consequently very difficult to detect. An example of beta minus decay is: 60 27 60 Co→ 28 Ni + 0 −1 0 e + 0ν The cobalt nucleus has too many neutrons to be stable. The unstable cobalt nucleus decays to a lower state by changing one of its nucleons from a neutron to a proton. This changes the nucleus to a Ni nucleus, and, to conserve electric charge, one unit of negative charge must be created. Since an electron cannot exist within the nucleus, the created electron, or beta particle, must be emitted from the decaying nucleus. What are the requirements for β− decay? We simply need the mass of the parent to exceed the mass of the daughter, MP > MD. Let’s look at how that comes about. Mass-energy conservation requires that the rest mass (energy) of the parent nucleus, MP – Zme, exceed the rest masses (energies) of the daughter nucleus, MD – (Z +1)me, and the electron me, where MP and MD are the neutral atomic masses of the parent and daughter. Any excess energy Q, that is, energy released in the decay, appears as kinetic energy of the particles emerging from the decay. Therefore, MP – Zme = [MD – (Z +1)me] + me + Mp = MD + Q c2 Q c2 3.7 Thus β− decay is possible whenever MP > MD. Further, whenever β− decay can occur it , although the probability may be small and the half-life extremely long. So, where does the disintegration energy go? (i.e. the energy released in going from the less stable (60Co) to the more stable (60Ni) entity. The energy is shared between the recoil 60Ni nucleus, the beta particle, and the antineutrino. Thus, beta particles are emitted in a decay into three bodies which can share energy and momentum in a continuum of ways, resulting in a spectrum of possible beta particle energies. [This is unlike alpha particles which are emitted in a decay into two bodies which must share energy and momentum in a unique way; this gives rise to discrete alpha energies.] A typical beta-particle spectrum is shown below. Note that the maximum beta partilce energy, Kmax, is observed when the anti-neutrino receives no kinetic energy and, since the 4 mass of the daughter nucleus is so large, it receives negligible kinetic energy, and the entire Q will be given to the beta particle. It was the appearance of the beta particle spectrum that led Wolfgang Pauli to postulate the existence of neutrinos and antineutrinos. If they didn’t exist, beta emission would occur at discrete energies only (like alphas). The neutrino avoids what would otherwise be a violation of the principles of the conservation of momentum, angular momentum and energy. Further, without the neutrino the daughter nucleus and β− particle would move in opposite directions, contrary to observation. Below is the decay scheme for cobalt-60 which decays by β- decay, leading to the 2.505 MeV excited state of nickel-60. This 60Ni* nucleus immediately decays be a two-photon gamma cascade. 5 3. β+ Decay Nuclides that are excessively proton rich lie to the left of the stability line on the Z vs. N plot and can decay by positive electron (positron) emission. The nuclide attempts to gain stability by increasing the N/Z ratio by conversion of a neutron to a proton. P→ Z −A1 D + +10 e + ν Let’s now look at the energetics involved: Again we want to include all electron masses: A Z MP – Zme = [MD – (Z-1)me] + me + Q c2 We no longer see all electron masses canceling on both sides of the equation. Instead we see that β+ is energetically possible only if the mass of the parent atom exceeds the mass of the daughter atom by at least two electron masses (2 x .000549 u or its energy equivalent, 1.02 MeV). MP > MD + Zme MP = MD +2me + Q c2 3.8 Additional radiations are observed with positron decay. The positron, the anti-matter equivalent of the electron, has only a limited lifetime. A positron will slow down in matter and then combine with an atomic electron in the material (making ‘positroniuim’). The positron-electron pair then annihilates, giving rise to two photons, each having energy 0.511 MeV and travelling in opposite directions. This phenomenon (which we’ll investigate in more detail later in the course) is the basis of Positron Emission Tomography (PET imaging). 4. Electron Capture Some nuclides are too neutron rich for stability, but positron emission is not possible because the masses of the parent and daughter nuclides differ by less than 1.02 MeV. In this situation the nuclide can undergo the process of electron capture which has the same effect on the nucleus as β+ decay (reducing Z by one but leaving A unchanged) but has no emissions from the nucleus. The general relation for electron capture is written: 0 −1 e + ZA P→ Z −A1 D + ν In electron capture, an orbital electron is captured by the parent nucleus, and the products are the daughter nucleus and a neutrino. Mass energy is conserved when the energy Q released in the decay equals the sum of the rest masses entering the reaction minus the sum of the rest masses leaving the reaction. Therefore, 6 me + (MP – Zme) = [MD – (Z – 1)me] + MP = MD + Q c2 Q c2 3.9 Thus, electron capture is possible if MP > MD. In an electron-capture decay, energy is released. Where does it go? Unlike β+ and β− decay, electron capture produces only two particles. By momentum conservation, the neutrino and the daughter nucleus must move in opposite directions with the same momentum magnitude; the sum of their kinetic energies is the disintegration energy Q. Because only two particles appear in electron capture, each has a precisely defined energy. The neutrino’s rest mass is (probably) zero; therefore, almost all the energy is carried by a virtually unobservable neutrino, and the nucleus recoils with an energy of only a few electron volts. Nevertheless, some very delicate experiments have confirmed that the recoiling nuclei are monoenergetic and their energy is precisely the amount required to satisfy the laws of momentum and energy conservation. Without the accompanying neutrino in electron capture, this decay process is completely inexplicable. The most common electron capture process is one in which the nucleus captures a K shell electron. This is because the K shell electrons are closest to the nucleus; the electron cloud can even overlap the nucleus. Thus, although capture of electrons from other shells is possible, electron capture is sometimes referred to as “K-capture”. Shown below is the decay scheme for 22-sodium. The energy levels are drawn relative to the ground state of 22Ne as having zero energy. This nuclide decays by both positron emission and electron capture. In (a) we see the positron decay illustrated, showing that the daughter nucleus, neon-22, is left in an excited state. This energy of excitation is given off immediately in the form of a gamma ray. Positron emission is 89.8 % possible and the accompanying gamma ray is given off with every positron decay. In (b) we see the electron capture decay illustrated. This process occurs 10.2 % of the time and leaves the neon daughter in the same excited state. In (c) the complete decay scheme for 22Na is shown. The starting EC level is 2mc2 higher than the starting level for positron decay. [You should ask yourself why an energy difference of two electron masses is shown.] 7 Electron capture always leaves the atom with an orbital vacancy in an inner shell. These inner shells are higher energy shells than those farther from the nucleus and so the atom will rearrange its electron structure so as to get to a lower energy state. It will do this by allowing an electron in an outer (lower energy) shell to drop down into the inner (high energy) shell. When this happens the difference in energy is given off as electromagnetic radiation, i.e. a photon. This photon (as you already know) is called fluorescence. So, while there are no detectable radiations emitted by the nucleus during electron capture (the neutrino is really not detectable), there is atomic radiation detected. Note 8 however, that since the electron capture results in a new nuclide and the orbital rearrangement occurs in that nuclide, the fluorescent photons are representative of the daughter, not the parent. When the K-shell vacancy is filled by an electron from the L shell, the M shell, and so on, the photons are called Kα, Kβ etc. Similarly, if an L-shell vacancy is filled the photons are designated Lα, Lβ etc. The diagram below illustrates the phenomenon of fluorescence as well as the competing process of Auger electron emission. Auger Electrons There is another process that competes with fluorescence. That is, an atom in which an L-shell electron makes a transition to fill a K-shell electron does not always emit a photon, especially if the element is low Z. A different transition can occur in which an L electron is ejected from the atom, thereby leaving two vacancies in the L shell. The ejected electron is called an Auger electron. The fluorescence yield of an element is defined as the number of photons emitted per vacancy. Fluorescence yield varies from essentially zero for the low Z elements to almost 1.0 for high Z, as shown in the plot below. Thus, Auger electron emission is favored over photon emission for light elements. Note that orbital vacancies can arise from a number of processes. We’ve seen the consequences of vacancies caused by electrons hitting the atom and knocking out orbital 9 electrons (cathode ray tubes) and vacancies cause by electron capture. They can also be caused by internal conversion (see below) or by ejection of electrons by photoelectron absorption of a photon ( a topic to be covered later in the course). Note that emission of an Auger electron increases the number of vacancies in the atomic shells by one unit. Often we see Auger cascades in relatively heavy atoms as inner-shell vacancies are successively filled by the Auger process with simultaneous ejection of the more loosely bound atomic electrons. An original, singly charged ion with one innershell vacancy can thus be converted into a highly charged ion by an Auger cascade. This phenomenon is being studied in radiation research and therapy. Auger emitters can be incorporated into DNA and other biological molecules. For example, 125I decays by electron capture. The ensuing cascade can release some 20 electrons, depositing a large amount of energy (~ 1 keV) within a few nanometers. A highly charged 125Te ion is left behind. A number of biological effects can be produced, such as DNA strand breaks, chromatid aberrations, mutations, bacteriophage inactivation, and cell killing. 5. Gamma Decay All stable nuclides are ordinarily in their lowest, or ground, states. If such nuclei are excited and gain energy by photon or particle bombardment, they may exist in any one of several excited, quantized energy states. Excited nuclei can emit their energy excitation as a photon. A photon emitted from the nucleus in an excited state is called a gamma ray. [This distinguishes such photons from ‘x-rays’ which is the term used to refer to photons emitted from the atomic structure. Two examples of ‘x-rays’ are the fluorescent radiations (“characteristic x-rays”) observed when inner vacancies are filled, and the photons produced when an electron undergoes a change in acceleration in the field of a nucleus (bremsstrahlung – see notes from Lecture 1 page 5). 0.492 MeV _________________________________ 0.472 MeV _________________________________ 0.327 MeV _________________________________ γ γ γ γ γ 0.040 MeV _________________________________ 0 _________________________________ 10 Alpha or beta decay often leaves the daughter nucleus in an excited state instead of the ground state. This is often represented by an asterisk beside the symbol of the nuclide, e.g. AZ* Z → AZ + γ A * Gamma decay is very rapid, occurring with a half-life on the order of 10-14 seconds. A few gamma decaying nuclides have much longer half-lives, for example greater than 10-6s and thus relatively easy to measure. These nuclides are given the name “isomers”. An isomer is not chemically distinguishable from the lower-energy nucleus into which it slowly decays. 6. Internal Conversion A process that competes with gamma emission from an excited nucleus is internal conversion. In this process the nucleus transfers its energy of excitation to one of the inner atomic electrons. This electron, originally bound to the atom with binding energy Eb, leaves the atom with a kinetic energy, Ke, equal to: Ke = E*- Eb where E*is the energy of excitation of the nucleus. The internal conversion coefficient α for a nuclear transition is defined as the ratio of the number of conversion electrons Ne and the number of competing gamma photons Nγ for that transition: Ne α= . Nγ The conversion coefficient α increase as Z3, the cube of the atomic number, and decreases with E*. Internal conversion is thus prevalent in heavy nuclei, especially in the decay of low-lying excited states (small E*). Gamma decay predominates in light nuclei. 11 This concludes our discussion of decay mechanisms. You need to make sure you understand both what a decay scheme is showing you, and how to construct one given emission and energy data. Example: work out the decay scheme for 137Cs which undergoes β- decay using the following information: Q = 1.174 MeV βmax: 1.174 MeC (5%) βmax: 0.512 MeC (95%) γ : 0.662 (85%, 137mBa), Ba X-rays e- : 0.624, 0.656 Beta minus decay of cesium-137 leads to a nucleus with the same A but with a Z increased by one, therefore the daughter is Barium-137. From the radiation emission information given, we can see that 5 % of the time, beta decay leads directly to the ground state of the barium nucleus since the maximum beta energy is the same as the Q for this decay. We can illustrate this on a decay scheme by drawing a line downward to the barium daughter, and slanting to the right, indicating an increase in Z. 95 % of the time beta decay occurs via a different path. This beta has a lower maximum energy, indicating the decay leads to an excited state of the daughter. The energy of the daughter in this case is 1.174 - 0.512 MeV=0.662 MeV. A photon of this energy occurs with 85% frequency. Therefore, internal conversion occurs in 95-85 = 10% of the decays, giving rise to conversion electrons, e-. The conversion electron energies reflect the excitation energy, 0.662 MeV, minus the binding energies of two different orbital shells, most likely the K, and L shells. Characteristic Ba X-rays are emitted following the inner-shell vacancies created in the atom by internal conversion. The plot below shows the spectrum of electron energies from cthe decay of cesium-137. 12