(A) Let's Start with T

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. And even if you wait a third half-life 1 time, you would still have 8 N0 atoms surviving the experience.

Textbook problem E.9 at the end of chapter 19: When some radioactive atoms decay with half-life time T , the number of surviving atoms depend on time t as

−t/T N(t) = N0 × 2 (14) where N0 is the original number of atoms at t = 0.

In this problem, we know that during t = 18 days, the number of atoms has dropped to

1/8 of the original number N0, whatever that was. In terms of eq. (14), this means

N(t) 1 = (15) N0 8

and hence N(t) 1 2−t/T = = = 2−3 (16) N0 8

Solving this equation for the half-life time T , we ﬁnd

t t 18 days =3 =⇒ T = = = 6 days. (17) T 3 3

Textbook problem Q.16 at the end of chapter 19: A quick look at the nuclide chart shows that heavy nuclei has larger neutron-to-proton ratios than the light nuclei. Speciﬁcally, the heavy nuclei have N/Z ≈ 1.5 to 1.6, while the light nuclei have N/Z ≈ 1. In the middle of the nuclide chart (80

The physical reason for this behavior is a combination of two potential energies: the negative (binding) energy of the strong nuclear forces, and the positive (un-binding) energy of the electrostatic repulsion between the protons. The strong forces favor equal numbers of protons and neutrons while the electrostatic forces ‘dislike’ the protons. Since the strong

6 forces saturate, while the electrostatic forces do not, the relative importance of the electrostatic potential energy increases with the number A of nucleons. And that’s why the N/Z ratio of the stable nuclei increases with A: For the light nuclei, the binding energy is dom- inated by the strong forces, so the most-bound — and hence stable — nuclei have N ≈ Z; but for heavier nuclei, the electrostatic energy becomes important and shifts the maximum of the binding energy to the higher N/Z ratios.

When a heavy nucleus with N/Z ≈ 1.55 ﬁssions into two fragments, both fragments have similar neutron-to-proton ratios. Since the fragments are smaller than the parent nucleus, such rations make them too neutron-rich and proton-poor for their sizes. This means that the daughter nuclei can further increase their binding energies by turning some neutrons into protons. Consequently, the daughter nuclei are unstable: they β-decay into nuclides with the same net numbers of nucleons A but lower N/Z ratios.

For example, consider a ﬁssion reaction:

235 100 132 92U + n → 40Zr + 52Te + 4n. (18)

Here the parent uranium-235 nucleus has N/Z = 143/92 = 1.554, and although is it spits out a few neutrons during the ﬁssion, the daughter nuclei have N/Z = 60/40 = 1.5 for zircon-100 and N/Z = 80/52= 1.538 for tellurium-132. Such N/Z ratios are way too high for the middle-of-the-chart nuclei, so they can increase their binding energies via β decay. Consequently, both daughter nuclei go through chains of β decays until the ﬁnal decay products become stable:

100 β 100 β 100 β 100 β 100 Zr → Nb → Mo → Tc → Ru, 40 41 42 43 44 (19) 132 β 132 β 132 52Te → 53I → 54Xe.

There are many ways for a parent uranium-235 nucleus to fission into two daughter nuclei — the reaction (18) is just an example. But in all such reaction, one or both daughter nuclei are too neutron-rich and proton-poor for their sizes, so they undergo chains of β-decays into the final fission products. In a nuclear reactor, such decays provide about 20% of the net thermal power.

7 Textbook problem E.11 at the end of chapter 19: Fission reactions do not involve weak interactions, so neutrons do not change into protons and vice verse. Therefore, the net number for protons and the net number of neutrons do not change during the reaction. In the reaction

235 130 A n + 92U → 50Sn + Z X+4n (20)

we start with 92 protons in the uranium nucleus, so the fission products must also contain 92 protons altogether. Fifty of these protons go to the tin-130 nucleus, so the remaining 92 − 50 = 42 protons must go to the other nucleus. Likewise, we start with 235 − 92 = 143 neutrons in the uranium nucleus, plus one additional neutron that makes it fission, so the fission products must have 143 + 1 = 144 neutrons between them. The tin-130 nucleus have 130 − 50 = 80 neutrons, plus 4 more neutrons fly free; the remaining 144 − 80 − 4 = 60 neutrons must be contained in the second daughter nucleus.

Altogether, the second daughter nucleus has 102 nucleons: 42 protons and 60 neutrons. Chemically, it belongs to element #42, which is molybdenum. Thus, the second daughter 102 nucleus is 42Mo and the ﬁssion reaction is

235 130 102 n + 92U → 50Sn + 42Mo + 4n. (21)

The new non-textbook problem: (a) The net mass of three initial deuterons is

3M(D) = 3 × 2.014, 102amu = 6.042, 306 amu. (22)

The net mass of the fusion products is

M(products) = M(α) + M(p) + M(n) = 4.001, 506amu + 1.007, 276amu + 1.008, 665 amu (23) = 6.017.447 amu.

8 The lost mass is the diﬀerence

∆M = M(3D) − M(α + p + n) = 0.024, 859amu; (24)

it’s converted into kinetic energy of the fusion products.

(b) According to the relativity theory, mass and energy can be converted into each other according to

E ↔ ∆M × c2 (25)

where c is the speed of light in vacuum, c = 299, 792, 458 m/s. In particular, when a nuclear reaction loses some mass ∆M, it releases energy E = ∆M × c2. The fusion reaction in question loses mass

∆M = 0.024858 amu = 0.024858 × 1.66053886 · 10−27 kg ≈ 4.12793 · 10−29 kg, (26)

so it releases energy

2 −29 8 2 −12 E3D = ∆M ×c = (4.12793·10 kg)×(2.99792458·10 m/s) ≈ 3.7100·10 J. (27)

At ﬁrst blush, this energy looks rather small. But please note that this is the energy released by fusion of a single trio of deuterons! In terms of energy-per-atom or energy-per-kilogram, the fusion reaction is millions times more energetic than any chemical reaction.

(c) The Ivy Mike explosion released 47 · 1015 J of energy — equivalent to 11 megatons of TNT or 759 Hiroshima bombs — and obliterated the Elugelab island where it happened. 23% of that energy — Efusion =0.23 × 47 · 1015 J ≈ 10.8 · 1015 J — came from the deuterium fusion reaction. To ﬁnd how many deuterons had fused to release this enormous energy, we simply divide it by the energy-per-deuteron. In light of part (b),

E 3.7100 · 10−12 J = = 1.2367 · 10−12 J (28) #D 3

9 hence the net number of deuterons that fused during the Ivy Mike test was

Efusion 10.8 · 10+15 J #D = = =≈ 8.74 · 1027. (29) E/#D 1.2367 · 10−12 J

A single deuterium atom has mass

2 −27 M(1H) = M(D) + M(e) ≈ 2.015 amu ≈ 3.345 · 10 kg, (30) so the net mass of deuterium fused in Ivy Mike explosion was

M(fused deuterium) = 8.74 · 1027 × 3.345 · 10−27 kg = 29.2 kg. (31)

Now you can see the power of nuclear fusion: Fusing just 65 pounds of deuterium produced enough energy to obliterate an island — and it could have vaporized a major city.

Of course, making those 65 pounds of deuterium fuse took a big lot of other things: the whole Mike device weighed about 82 tons, including 5 tons of uranium tamper. But that was an experiment, not a weapon. A bomb with a similar explosive power weighed about 5 tons.

BTW, nobody makes or keeps the multi-megaton nuclear weapons any more. The US and the FUSSR made them in 1950s and 1960s for delivery by bomber planes and single- warhead missiles of poor accuracy. The development of multi-warhead missiles and better targeting accuracy made those monster bombs obsolete, although a few were kept in reserve as bunker-busters. In 1990s, they were taken out of stockpiles and disassembled.