Radioactive Decay Mechanisms (cont.) Beta (β) Decay: 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: ? βPP ? decay ? β + decay ? electron capture β ? decay Nuclides with excess neutrons need to convert a neutron to a proton to move closer to the line of stability. neutron → proton + negative electron + antineutrino ν 0 0 0 11 ++→ ?+ eDP A Z A Z Conservation of charge and nucleons must be observed. The symbol ν represents the neutrino, and ν represents its antiparticle, the antineutrino. Example: ν 0 0 0 1 60 28 60 27 ++→ ? eNiCo ? 60 Co has too many neutrons to be stable. ? One of the 60 Co neutrons is converted to a proton. ? An electron is created in the nucleus then ejected from the nucleus. 1 β ? decay energetics Mass-energy conservation: ν 0 0 0 11 ++→ ?+ eDP A Z A Z B ? The β ? decay e - leaves the nucleus. ? Another e - is required to take up an orbital position to balance the new charge (Z+1). ? There is no net gain or loss of electron masses in this reaction. ? No allowance needs to be made in the energy-mass calculations for e - shifts. Equivalent (from a mass-balance point of view) to M P = M D + Q, or Q = M P - M D [or Q β- = ? P - ? D ] For the example: QeNiCo +++→ ? ν 0 0 0 1 60 28 60 27 Use ? values from App. D Q = -61.651 – (- 64.471) = 2.82 MeV For β ? decay to be possible, M P > M D (i.e., Q = M P – M D ) Any excess energy Q will be shared by the 3 decay products. 2 β ? decay requirements (or what is a neutrino??) “Neutrinos, they are very small/ They have no charge and have no mass / And do not interact at all” “Cosmic Gall” John Updike, 1960 Early 1900s: Observations of beta decay puzzled investigators. ? They thought they were studying a 2-body decay process. ? This should produce electrons of a fixed energy. ? Same exact same β ? reaction produced electrons with variable energy. ? It appeared as if energy were being destroyed. ? This violated Conservation of Energy. ? The ejected electron and the recoil nucleus did not move apart on a straight line. ? This violated Conservation of Momentum. Enter the neutrino: “I have done a terrible thing. I have postulated a particle that cannot be detected.” Wolfgang Pauli, 1930 ? To explain the apparent violations of momentum and energy, Pauli “invented” the neutrino, a particle of no charge and little or no mass that would carry away energy and momentum. ? Actually he proposed the name “neutron” (the neutron we know was discovered in 1932) ? Enrico Fermi proposed the name “neutrino: little neutral one” In 1956 the neutrino was first detected. ? Elaborate experimental detectors are required. ? Neutrinos have a very small, but non-zero, mass. ? Implications? ? Neutrinos are everywhere: fusion in the sun, reactors produce copious amounts of neutrinos. ? Billions of neutrinos pass through your body every second. ? Only a few interactions expected in a light-year thickness of lead!!! ? Neutrino mass is a big issue in cosmology. 3 How is the energy, Q, distributed? Image removed. Fig. 3.5 in Turner J. E. Atoms, Radiation, and Radiation Protection, 2P ndP ed. New York: Wiley-Interscience, 1995. β ? decay produces three products, the daughter nucleus, the beta particle and the antineutrino. The daughter nucleus, because of its large mass, receives negligible energy. νβ EEQ += ? The energy is shared between the beta particle and the antineutrino. Depending on the orientation, the particles can have any energy between 0 and Q. Thus, the beta particle energy spectrum is continuous 0 ≤ E β- ≤ 0 The average beta particle energy ~ Q/3 Beta decay scheme Image removed. Fig. 3.5 in [Turner]. 60 Co decay details from Turner, Appendix D β ? 0.318 max (99.92%) 1.491 max (0.08%) γ 1.173 (99.98%) 1.332 (99.90%) 4 β + Decay Nuclides that are excessively proton rich can decay by positive electron (positron) emission. The nuclide attempts to gain stability by increasing the N/Z ratio by conversion of a proton to a neutron. proton → neutron + positive electron + neutrino ν 0 0 0 11 ++→ +? eDP A Z A Z β + Decay Energetics ? β + ejected from the nucleus: loss of 1 m e ? The daughter Z-1 nucleus must shed an orbital electron to balance charge: loss of 1 m e Net result is: M P → M D + 2 m e + Q Note: two electron masses must be accounted for in the mass-energy calculations. β + decay 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 0.000549 AMU or its energy equivalent, 1.02 MeV). M P > M D + 2m e M P = M D +2m e + Q Q = M p - M D – 2m e [or Q β+ = ? P - ? D – 2m e ] Example: QNeNa +++→ νβ 0 0 0 1 22 10 22 11 From App. D: Q = -5.182 – (-8.025) – 1.022 = 1.821 MeV 5 Positron Imaging in Nuclear Medicine β + is a positron, the antiparticle of the electron. Positron-electron annihilation releases two 0.511 MeV photons traveling in opposite directions. This is the basis of positron emission tomography (PET). 6 Electron Capture ? Nuclide too proton rich for stability ? Positron emission (β + ) decay not possible o (M P – M D < 1.022 MeV) Electron capture has the same effect on the nucleus as β + decay. ν 0 01 0 1 +→+ ? DPe A Z A Z proton + electron → neutron + neutrino ? An orbital electron is captured by the parent nucleus. ? Energy “spent” overcoming the electron binding energy, E B ? Products are the daughter nucleus and the neutrino. QNeNae ++→+ ? ν 0 0 22 10 22 11 Electron capture energetics: ? Mass-energy is conserved ? Compare the masses on both sides of the arrow. (m e - E B ) + M P = M D + Q M P - E B = M D + Q Q =M P - M D - E B [or Q EC = ? P - ? D – E B ] Electron capture is only possible if ? P - ? D > E B . Electron capture produces only two reaction products (unlike β - and β + decay). M D and the neutrino share the energy, move in opposite directions with the same momentum. Because of the mass difference, the (virtually undetectable) neutrino carries away almost all of the energy. 7 Na 22 11 can decay by both EC and β + pathways Image removed. Fig. 3.11 in [Turner]. Q β+ = ? P - ? D – 2m e = 1.821 MeV Q EC = ? P - ? D – E B = 2.843 MeV Gamma Emission ? Spectrum is discrete, characteristic of element ? Many decay modes emit gamma rays from excited nuclear states ? Gamma emission with no change in Z or A = isomeric transition Metastable states Excited nuclear states usually decay in ~ < 10 -10 sec Excited nuclear states with longer lifetimes termed metastable ν 0 0 99 43 99 42 +→ TcMo m ? halflife of 99m Tc = 6 hrs ? the 0.14 MeV gamma emission is used in nuclear medicine imaging 8 Internal Conversion ? Energy of excited nuclear state transferred to orbital electron ? Electron is ejected from the atom (K or L shell) IC is an alternative to gamma emission from the nucleus. IC coefficient = γ N N e [ratio of branching] E e = E* - E B Energy of the ejected electron = excitation energy – electron binding energy IC coefficient: increases as Z increases decreases as E* increases IC prevalent in heavy nuclei from low-lying excited nuclear states Summary of Decay Energetics Image removed. Table 3.1 in [Turner]. 9 Decay Scheme Exercise K 40 19 β - (89%) β - : 1.312 (max) EC (1%) γ: 1.461 (11%) Ar x rays 10