PHY–309 L. Solutions for homework set # 10. Non-textbook problem #II from the previous homework set: (a) Let’s start with the isotope naming conventions. A complete name of an isotope has form A 1 16 35 63 ZEℓ, for example 1H, O, 17Cl, 29Cu,... (1) Here ‘El’ is the chemical symbol for the element — ‘H’ for hydrogen, ‘O’ for oxygen, ‘Cl’ for chlorine,‘Cu’ for copper, etc. In addition, to the left of the chemical symbol there are two numbers describing the nucleus of the isotope. The upper left number A is the net number of nucleons — i.e., protons and neutrons, — while the lower left number Z is the number of protons. The number of neutrons in the nucleus is not written, but it can be easily obtained as the difference N = A − Z. 63 In particular, 29Cu is a copper isotope whose nuclei have 63 protons and neutrons, 65 including 29 protons; the remaining 63 − 29 = 34 nucleons are neutrons. Likewise, 29Cu is another copper isotope whose nuclei have 65 neutrons and protons, including 29 protons and 65 − 29 = 36 neutrons. The number of protons is the nucleus controls its electric charge Qnucleus = N × 0 + Z × (+e) = +Ze (2) and hence the number of electrons E in a neutral atom: Qatom = Qnucleus + E × (−e)=0 =⇒ E = Z. (3) This is the only number that matters for chemistry, so all isotopes with the same Z belong to the same chemical element. Conversely, all isotopes of the same chemical element have the same proton number Z. In fact, the serial number of an element in the Periodic Table — usually called the atomic number — is precisely the proton number Z. 1 Thus, all isotopes of hydrogen have one proton in the nucleus and one electron in the neutral atom, all isotopes of oxygen have 8 protons in the nucleus and 8 electrons in the neutral atom, etc. In particular, all isotopes of copper have 29 protons in the nucleus and 29 electrons in the neutral atom. Since the proton number Z is implicit in the chemical element symbol, it is usually 63 63 omitted from the isotope’s name. Thus, instead of 29Cu, people write simply Cu since 1 1 any copper isotope has proton number Z = 29. Likewise, H means 1H since any hydrogen 16 16 isotope has Z = 1, O means 8H since any oxygen isotope has Z = 8, etc., etc. Please remember this rule for your next homework. If an isotope’s name is written down without a left subscript and you need to find its proton number, look up the chemical element in the Periodic Table: The atomic number of the element — it’s sequential number in the Table — is the proton number of any isotope of that element. (b) Let’s start with atomic weights of pure isotopes. All atoms of a pure isotope have exactly the same mass binding energy Matom = Z × (Mproton + M )+ (N = A − Z) × Mneutron − . (4) electron c2 To a very good approximation — three or four significant figures — this mass is proportional to the net number of nucleons A = Z + N, and the atomic mass unit is chosen such that Matom ≈ A amu. (5) In other words, the atomic weight of a pure isotope is very close to the net number A of protons and neutrons in the nucleus. 63 In particular, the copper isotope 29Cu has atomic weight 62.929, 597 ≈ 63.0, while the 65 isotope 29Cu has atomic weight 64.927, 789 ≈ 65.0. (The high-precision atomic weights are taken from the NIST web page.) But the natural sources (ores, etc.) for many chemical elements provide mixtures of different isotopes, and since the isotopes of the same element have similar chemical properties, they don’t get separated when the element is refined from its ore or participates in chemical 2 reactions. Although there are ways to separate isotopes in a lab, or even on the industrial scale, this is very difficult and expensive, so it’s done only when the element is used for its nuclear properties (for example, uranium). Thus, most chemical elements and compounds you can find on Earth do not contain pure isotopes but rather natural mixtures of several ⋆ isotopes. In particular, all copper on this planet that hasn’t been through a nuclear lab is 63 65 † a mixture of two isotopes: about 69% of Cu and 31% of Cu. As far as chemistry is concerned, the atomic weight of an element with multiple isotopes is the weighted average of the isotopes’ atomic weights, weighted by their relative abundances, µ(element) = µ(isotope)×fraction(isotope) ≈ A(isotope)×fraction(isotope). X X isotopes isotopes (6) In particular, 63 63 65 65 µ(copper) = µ Cu × fraction Cu + µ Cu × fraction Cu ≈ 63 × 0.69 + 65 × 0.31 (7) = 63.62 ≈ 63.6. To see how this works, note that to a chemist, the atomic weight of an element — or a molecular weight of a compound — is simply the mass (in grams) of 1 mol of the substance. And 1 mol is the the Avogadro’s number 6.02 · 1023 of atoms (or molecules), regardless of their isotopes. Indeed, one mol of copper — 6.02 · 1023 copper atoms — is what combines 1 with 2 mol of oxygen O2 to make copper oxide CuO, or with 1 mol of sulfuric acid to make copper sulfate, etc., etc., and it does not make any difference what isotopes do those copper atoms belong to, as long as they are all copper atoms and there are 6.02 · 1023 of them. On the other hand, the net mass of those 6.02·1023 copper atoms depends on the isotope 63 65 mixture. One mol of natural copper comprises 0.69 mols of Cu and 0.31 mols of Cu, so ⋆ Some elements — beryllium, fluorine, sodium, aluminum, phosphorus, scandium, manganese, cobalt, arsenic, yttrium, niobium, rhodium, iodine, cesium, gold, bismuth, and some lantanids — have only one stable isotope. But all the other elements are present on Earth as isotope mixtures. † Actually, there are small differences between isotope ratios of elements coming from different sources. 63 For example, copper from different mines can have from 68.98% to 69.38% of Cu. 3 its net mass is 63 65 M(1 mol of natural Cu) = M 0.69 mol of Cu + M 0.31 mol of Cu 63 65 =0.69 × µ Cu g +0.31 × µ Cu g (8) ≈ 0.69 × 63g + 0.31 × 65 g ≈ 63.6 g. To a chemist, thus mass (in grams) of 1 mol of natural copper is its atomic weight, thus µ(natural copper) ≈ 63.6. (9) Textbook problem SP.2 at the end of chapter 19: (a) The chemical symbols in the thorium decay chain stand for the following elements: Th is thorium, Z = 90; Ra is radium, Z = 88; Ac is actinium, Z = 89; Rn is radon, Z = 86; Po is polonium, Z = 84; Pb is lead, Z = 82; Bi is bismuth, z = 83. (b) The α and β decays have different effects on the atomic number Z: In an α decay it decreases by 2, while in a β decay, it increases by 1. Thus, comparing the atomic numbers of a parent isotope and its decay product will immediately tells us if the decay in question is α or β. In particular, in the thorium decay chain in question, in the decays Th → Ra, Ra → Rn, Rn → Po, Po → Pb (10) the atomic number decreases by 2, so these are α decays, while in the decays Ra → Ac, Ac → Th, Pb → Bi, Bi → Po (11) the atomic number increases by 1, so these are β decays. 4 (c–d) Now that we know which decay is α and which is β, we may determine the mass numbers A of all the isotopes involved by following a simple rule: In an α decay A decreases by 4, while in a β decay A does not change. Thus: 232 228 4 (α) 90Th → 88Ra + 2He, 228 228 − (β) 88Ra → 89Ac + e +ν, ¯ 228 228 − (β) 89Ac → 90Th + e +ν, ¯ 228 224 4 (α) 90Th → 88Ra + 2He, 224 220 4 (α) 88Ra → 86Rn + 2He, 220 216 4 (12) (α) 86Rn → 84Po + 2He, 216 212 4 (α) 84Po → 82Pb + 2He, 212 212 − (β) 82Pb → 83Bi + e +ν, ¯ 212 212 − (β) 83Bi → 84Po + e +ν, ¯ 212 208 4 (α) 84Po → 82Pb + 2He. And the whole decay chain can be summarized as 232 α 228 β 228 β 228 α 224 α 220 α 90Th → 88Ra → 89Ac → 90Th → 88Ra → 86Rn → (13) α 216 α 216 α 212 β 212 β 212 α 208 → 84Po → 84Po → 82Pb → 83Bi → 84Po → 82Pb. There is a Walter Fendt applet showing this decay chain step by step. It also shows several 238 206 235 207 other decay chains, in particular 92U →···→ 82Pb and 92U →···→ 82Pb. Textbook problem Q.13 at the end of chapter 19: NO. During a half-life time, one half of the radioactive atoms decay while the other half survive. But the surviving atoms have no memory, and the probability of each atom decaying during some subsequent time does not depend on on how long it has been existing before that. Thus, during the second half-life period, one half of the atoms that have survived the first half-life time will decay, and the other half will survive again. 1 In terms of the original number N0 of atoms, 2 N0 atoms decay during the first half-life 1 1 1 1 time. Out of the remaining 2 N0 atoms, 2 × 2 N0 = 4 N0 decay during the second half-life 5 1 time, while the remaining 4 N0 atoms survive again.