Chemistry 201: General Chemistry II - Lecture Dr. Namphol Sinkaset Chapter 21 Study Guide Concepts 1. There are several modes of radioactive decay: (1) alpha (α) decay, (2) beta (β) decay, (3) gamma (γ) ray emission, (4) positron emission, (5) electron capture. 2. Recall that specific isotopes are represented as A ZX, where A is the mass number, Z is the atomic number, and X is the chemical symbol. 3. In nuclear chemistry, the term nuclide is used to refer to a particular isotope of an element. 4. In nuclear chemistry, a proton is represented as 1 1p, a neutron is represented as 1 0n, and an electron is represented as 0 1 e. 5. When a nuclide undergoes alpha decay, it sheds an α particle, represented by 4 2He. 6. In a nuclear equation, a parent nuclide decays into a daughter nuclide. 7. Nuclear equations are balanced by ensuring the sum of mass numbers and the sum of atomic numbers on both sides are equal. 8. Alpha particles are the most massive particles emitted by nuclei. Alpha radiation has the highest ionizing power, but the lowest penetrating power. 9. When a nuclide undergoes beta decay, it emits a beta particle. A beta particle is an electron. 10. Through beta decay, a neutron becomes a proton in the nucleus. 11. Beta particles are less ionizing but more penetrating than alpha particles. 12. Gamma ray emission involves emission of high-energy photons, not particles. Gamma rays are a type of electromagnetic radiation. 13. Nuclides emit gamma rays when they are too energetic. 14. Gamma rays have low ionizing power, but very high penetrating power. 15. Positron emission involves emission of the electron s antiparticle, known as the positron. The positron is represented as 0 1e. 16. Through positron emission, a proton becomes a neutron in the nucleus. 1
17. A nuclide can pull in an electron from an inner orbital in a process known as electron capture. 18. In electron capture, the electron combines with a proton to form a neutron. 19. Nuclides undergo radioactive decay to become more stable. 20. The strong force binds protons and neutrons together, but it only works within very short distances. 21. Stability of the nucleus is a balance between +/+ repulsions between protons and the strong force attraction. 22. Neutrons increase strong force attractions without increasing +/+ repulsions. 23. The valley of stability shows which N/Z ratios result in stable nuclei. 24. Just like electrons, nucleons occupy energy levels within the nucleus. Also like electrons, certain numbers of nucleons are stable. 25. Unique stability is obtained when N or Z equals 2, 8, 20, 28, 50, 82 and when N equals 126. These are known as magic numbers. 26. An unstable nuclide will undergo radioactive decay in order to reach the valley of stability. The decays a nuclide undergoes is known as a decay series. 27. Radioactivity is all around us. It exists due to the constant production of radioactive nuclides through various decay series. 28. Radioactive processes follow first order kinetics. 29. Radioactivity can be used to estimate the age of objects. 30. In radiocarbon dating, the rate of carbon-14 decay in an object is compared to the rate of carbon-14 decay in the atmosphere. 31. In nuclear fission, large nuclides are split into smaller nuclides, releasing a large amount of energy. 32. A nuclear chain reaction requires a critical mass of material. 33. In nuclear reactions, it s possible for mass to be converted into energy. 34. Some important relationships: 1 amu = 931.5 MeV (mega electron volts). 1 MeV = 1.602 10 13 J. 35. Stable nuclei have masses that are less than their components. The difference in mass is known as the mass defect. 36. The mass defect is used in Einstein s equation to calculate the nuclear binding energy. This is the energy needed to break up a nucleus into its component nucleons. 2
37. When the binding energy is divided by the number of nucleons, a sense of the nucleus s stability can be calculated. The most stable nuclides have mass numbers around 60. 38. In nuclear fusion, smaller nuclides combine into more stable nuclides and release a great amount of energy. High temperatures are needed to over come the +/+ repulsions in nuclear fusion. 39. There are three categories of radiation effects: (1) acute radiation damage, (2) increased cancer risk, and (3) genetic defects. 40. Radiation exposure can be measured in curies (Ci), grays (Gy), rads, or rems. Equations 1. ln Nt N 0 = kt (1st order equation for radioactive decay) 2. ln Rate t Rate 0 = kt (1st order equation for radiometric dating) 3. E = mc 2 (Mass/energy equation) Representative Problems 8. Write balanced nuclear equations for the following: (a) alpha decay of 234 92U, (b) electron capture by neptunium-232, (c) positron emission by 12 7N. A relatively simple problem in which we must write correct, balanced nuclear equations. The key is understanding the terminology associated with the types radioactive processes. In (a), uranium-234 is undergoing alpha decay. In alpha decay, the alpha particle appears as a product along with the daughter nuclide. We just have to make sure the Z and A sum to the same number on both sides and that we identify the correct daughter elemental symbol based on Z. 234 92U 230 90Th + 4 2He In (b), we have electron capture. In this process, an electron from an inner orbital is pulled in by the nucleus. Thus, the electron can be thought of as a reactant, appearing on the parent side. We have to get the Z number of neptunium in order to create the proper nuclear equation. Looking at the periodic table, we see that Z = 93. Again, we must ensure that Z and A sum to the same number on both sides. 232 93Np + 0 1 e 232 92U In (c), we have positron emission. The positron is the anti-matter equivalent of an electron. Since the positron is being emitted, it will appear on the daughter side of the nuclear equation. 12 7 N 12 6C + 0 1e 3
48. Volcanism was much more common in the distant past than today, and in some cases, specific events can be dated. A volcanic eruption melts a large area of rock, and all gases are expelled. After cooling, 40 18Ar accumulates from the ongoing decay of 40 19K in the rock (t 1/2 = 1.25 10 9 yr). When a piece of rock is analyzed, it is found to contain 1.38 mmol of 40 K and 1.14 mmol of 40 Ar. How long ago did the rock cool? This problem demonstrates the usefulness of radioactivity in determining dates in geology. Although not entirely necessary to solve this problem, we begin by writing the nuclear equation for the process. 40 19K 40 18Ar + 0 1e Notice however, that the following nuclear reaction also satisfies what is described. 40 19K + 1 0 e 40 18Ar Thus, argon-40 could derive from either positron emission or electron capture. Without additional information, there s no way to know which process occurs. From other sources, it is found that electron capture is the actual process that occurs. In any event, we are told that a rock has 1.38 mmol of 40 K and 1.14 mmol of 40 Ar. Since the 40 Ar used to be 40 K, there must have been 2.52 mmol of 40 K when the rock was formed. We will be using the first-order integrated rate law to calculate how old the rock is, but we first must get the value of k. t 1/2 = ln 2 k 1.25 10 9 yr = ln 2 k k = ln 2 1.25 10 9 yr k = 5.54 51774 10 10 yr 1 Now we can use the first-order integrated rate law for radioactive decay. ln N t N 0 = kt 1.38 mmol ln 2.52 mmol = (5.54 51774 10 10 yr 1 )t t = 1.08 5944342 10 9 yr t = 1.09 billion years 4
81. The isotopic mass of 210 86Rn is 209.989669 amu. When this nuclide decays by electron capture, it emits 2.368 MeV. What is the isotopic mass of the resulting nuclide? Recall that nuclides undergo radioactive decay in order to become more stable. As such, it makes sense that energy is being released in a nuclear reaction. In this problem radon-210 captures an electron according to the nuclear equation below. 210 86Rn + 0 1 e 210 85At Derived from Einstein s E = mc 2, we have the relationship 1 amu = 931.5 MeV. The energy that is lost came from a loss of mass from the parent nuclide. We calculate the mass that corresponds to an energy of 2.368 MeV. 2.368 MeV 1 amu 931.5 MeV = 0.002542 1363392 amu The mass of the daughter nuclide, 210 85At, must be this much less than the mass of the parent nuclide, 210 86Rn. mass 210 85At = (mass 210 86Rn) (mass lost in form of energy) = 209.989669 amu 0.002542 1363392 amu = 209.987126 8636608 amu = 209.987127 amu 5