#set document(title: "31.5 Half-Life and Activity", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 31.5#h(0.6em)Half-Life and Activity === Learning Objectives By the end of this section, you will be able to: - Define half-life. - Define dating. - Calculate age of old objects by radioactive dating. Unstable nuclei decay. However, some nuclides decay faster than others. For example, radium and polonium, discovered by the Curies, decay faster than uranium. This means they have shorter lifetimes, producing a greater rate of decay. In this section we explore half-life and activity, the quantitative terms for lifetime and rate of decay. === Half-Life Why use a term like half-life rather than lifetime? The answer can be found by examining , which shows how the number of radioactive nuclei in a sample decreases with time. The #emph[time in which half of the original number of nuclei decay] is defined as the #strong[half-life], #math.equation(block: false, alt: "t sub 1 / 2")[$t_(1 / 2)$]. Half of the remaining nuclei decay in the next half-life. Further, half of that amount decays in the following half-life. Therefore, the number of radioactive nuclei decreases from #math.equation(block: false, alt: "N")[$N$] to #math.equation(block: false, alt: "N / 2")[$N / 2$] in one half-life, then to #math.equation(block: false, alt: "N / 4")[$N / 4$] in the next, and to #math.equation(block: false, alt: "N / 8")[$N / 8$] in the next, and so on. If #math.equation(block: false, alt: "N")[$N$] is a large number, then #emph[many] half-lives (not just two) pass before all of the nuclei decay. Nuclear decay is an example of a purely statistical process. A more precise definition of half-life is that #emph[each nucleus has a 50% chance of living for a time equal to one half-life #math.equation(block: false, alt: "t sub 1 / 2")[$t_(1 / 2)$]]. Thus, if #math.equation(block: false, alt: "N")[$N$] is reasonably large, half of the original nuclei decay in a time of one half-life. If an individual nucleus makes it through that time, it still has a 50% chance of surviving through another half-life. Even if it happens to make it through hundreds of half-lives, it still has a 50% chance of surviving through one more. The probability of decay is the same no matter when you start counting. This is like random coin flipping. The chance of heads is 50%, no matter what has happened before. #figure(figph[The figure shows a radioactive decay graph of number of nuclides in thousands versus time in multiples of half-life. The number of radioactive nuclei decreases exponentially and finally approaches zero after about ten half-lives.], alt: "The figure shows a radioactive decay graph of number of nuclides in thousands versus time in multiples of half-life. The number of radioactive nuclei decreases exponentially and finally approaches zero after about ten half-lives.", caption: [Radioactive decay reduces the number of radioactive nuclei over time. In one half-life #math.equation(block: false, alt: "t sub 1 / 2")[$t_(1 / 2)$], the number decreases to half of its original value. Half of what remains decay in the next half-life, and half of those in the next, and so on. This is an exponential decay, as seen in the graph of the number of nuclei present as a function of time.]) There is a tremendous range in the half-lives of various nuclides, from as short as #math.equation(block: false, alt: "10 to the power minus 23")[$"10"^(− "23")$] s for the most unstable, to more than #math.equation(block: false, alt: "10 to the power 16")[$"10"^("16")$] y for the least unstable, or about 46 orders of magnitude. Nuclides with the shortest half-lives are those for which the nuclear forces are least attractive, an indication of the extent to which the nuclear force can depend on the particular combination of neutrons and protons. The concept of half-life is applicable to other subatomic particles, as will be discussed in Particle Physics. It is also applicable to the decay of excited states in atoms and nuclei. The following equation gives the quantitative relationship between the original number of nuclei present at time zero (#math.equation(block: false, alt: "N sub 0")[$N_(0)$]) and the number (#math.equation(block: false, alt: "N")[$N$]) at a later time #math.equation(block: false, alt: "t")[$t$]: #math.equation(block: true, alt: "N equals N sub 0 e to the power minus λt ,")[$N = N_(0) e^(− "λt") ","$] where #math.equation(block: false, alt: "e equals 2.71828 ...")[$e = "2.71828" "..."$] is the base of the natural logarithm, and #math.equation(block: false, alt: "λ")[$λ$] is the #strong[decay constant] for the nuclide. The shorter the half-life, the larger is the value of #math.equation(block: false, alt: "λ")[$λ$], and the faster the exponential #math.equation(block: false, alt: "e to the power minus λt")[$e^(− "λt")$] decreases with time. The relationship between the decay constant #math.equation(block: false, alt: "λ")[$λ$] and the half-life #math.equation(block: false, alt: "t sub 1 / 2")[$t_(1 / 2)$] is #math.equation(block: true, alt: "λ equals the fraction ln(2) over t sub 1/2 approximately equals the fraction 0 . 693 over t sub 1 / 2 .")[$λ = frac("ln"(2), t_(1/2)) ≈ frac(0 "." "693", t_(1 / 2)) "."$] To see how the number of nuclei declines to half its original value in one half-life, let #math.equation(block: false, alt: "t equals t sub 1 / 2")[$t = t_(1 / 2)$] in the exponential in the equation #math.equation(block: false, alt: "N equals N sub 0 e to the power minus λt")[$N = N_(0) e^(− "λt")$]. This gives #math.equation(block: false, alt: "N equals N sub 0 e to the power minus λt equals N sub 0 e to the power −0.693 equals 0.500 N sub 0")[$N = N_(0) e^(− "λt") = N_(0) e^(−0.693) = 0.500 N_(0)$]. For integral numbers of half-lives, you can just divide the original number by 2 over and over again, rather than using the exponential relationship. For example, if ten half-lives have passed, we divide #math.equation(block: false, alt: "N")[$N$] by 2 ten times. This reduces it to #math.equation(block: false, alt: "N / 1024")[$N / "1024"$]. For an arbitrary time, not just a multiple of the half-life, the exponential relationship must be used. #strong[Radioactive dating] is a clever use of naturally occurring radioactivity. Its most famous application is #strong[carbon-14 dating]. Carbon-14 has a half-life of 5730 years and is produced in a nuclear reaction induced when solar neutrinos strike #math.equation(block: false, alt: "to the power 14 N")[$"14" N$] in the atmosphere. Radioactive carbon has the same chemistry as stable carbon, and so it mixes into the ecosphere, where it is consumed and becomes part of every living organism. Carbon-14 has an abundance of 1.3 parts per trillion of normal carbon. Thus, if you know the number of carbon nuclei in an object (perhaps determined by mass and Avogadro’s number), you multiply that number by #math.equation(block: false, alt: "1 . 3 × 10 to the power minus 12")[$1 "." 3 × "10"^(− "12")$] to find the number of #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] nuclei in the object. When an organism dies, carbon exchange with the environment ceases, and #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] is not replenished as it decays. By comparing the abundance of #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] in an artifact, such as mummy wrappings, with the normal abundance in living tissue, it is possible to determine the artifact’s age (or time since death). Carbon-14 dating can be used for biological tissues as old as 50 or 60 thousand years, but is most accurate for younger samples, since the abundance of #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] nuclei in them is greater. Very old biological materials contain no #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] at all. There are instances in which the date of an artifact can be determined by other means, such as historical knowledge or tree-ring counting. These cross-references have confirmed the validity of carbon-14 dating and permitted us to calibrate the technique as well. Carbon-14 dating revolutionized parts of archaeology and is of such importance that it earned the 1960 Nobel Prize in chemistry for its developer, the American chemist Willard Libby (1908–1980). One of the most famous cases of carbon-14 dating involves the Shroud of Turin, a long piece of fabric purported to be the burial shroud of Jesus. This relic was first displayed in Turin in 1354 and was denounced as a fraud at that time by a French bishop. Its remarkable negative imprint of an apparently crucified body resembles the then-accepted image of Jesus, and so the shroud was never disregarded completely and remained controversial over the centuries. Carbon-14 dating was not performed on the shroud until 1988, when the process had been refined to the point where only a small amount of material needed to be destroyed. Samples were tested at three independent laboratories, each being given four pieces of cloth, with only one unidentified piece from the shroud, to avoid prejudice. All three laboratories found samples of the shroud contain 92% of the #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] found in living tissues, allowing the shroud to be dated. #figure(figph[The figure shows two images of Jesus. Left image is very faint and hardly visible but the right image shows a much clearer picture.], alt: "The figure shows two images of Jesus. Left image is very faint and hardly visible but the right image shows a much clearer picture.", caption: [Part of the Shroud of Turin, which shows a remarkable negative imprint likeness of Jesus complete with evidence of crucifixion wounds. The shroud first surfaced in the 14th century and was only recently carbon-14 dated. It has not been determined how the image was placed on the material.]) #examplebox("Example 1")[How Old Is the Shroud of Turin?][ Calculate the age of the Shroud of Turin given that the amount of #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] found in it is 92% of that in living tissue. Strategy Knowing that 92% of the #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] remains means that #math.equation(block: false, alt: "N / N sub 0 equals 0 . 92")[$N / N_(0) = 0 "." "92"$]. Therefore, the equation #math.equation(block: false, alt: "N equals N sub 0 e to the power minus λt")[$N = N_(0) e^(− "λt")$] can be used to find #math.equation(block: false, alt: "λt")[$"λt"$]. We also know that the half-life of #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] is 5730 y, and so once #math.equation(block: false, alt: "λt")[$"λt"$] is known, we can use the equation #math.equation(block: false, alt: "λ equals the fraction 0 . 693 over t sub 1 / 2")[$λ = frac(0 "." "693", t_(1 / 2))$] to find #math.equation(block: false, alt: "λ")[$λ$] and then find #math.equation(block: false, alt: "t")[$t$] as requested. Here, we postulate that the decrease in #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] is solely due to nuclear decay. Solution Solving the equation #math.equation(block: false, alt: "N equals N sub 0 e to the power minus λt")[$N = N_(0) e^(− "λt")$] for #math.equation(block: false, alt: "N / N sub 0")[$N / N_(0)$] gives #math.equation(block: true, alt: "the fraction N over N sub 0 equals e to the power minus λt .")[$frac(N, N_(0)) = e^(− "λt") "."$] Thus, #math.equation(block: true, alt: "0 . 92 equals e to the power minus λt .")[$0 "." "92" = e^(− "λt") "."$] Taking the natural logarithm of both sides of the equation yields #math.equation(block: true, alt: "ln 0 . 92 equals –λt")[$"ln" #h(0.25em) 0 "." "92" = "–λt"$] so that #math.equation(block: true, alt: "minus 0 . 0834 equals minus λt .")[$− 0 "." "0834" = − "λt" "."$] Rearranging to isolate #math.equation(block: false, alt: "t")[$t$] gives #math.equation(block: true, alt: "t equals the fraction 0 . 0834 over λ .")[$t = frac(0 "." "0834", λ) "."$] Now, the equation #math.equation(block: false, alt: "λ equals the fraction 0 . 693 over t sub 1 / 2")[$λ = frac(0 "." "693", t_(1 / 2))$] can be used to find #math.equation(block: false, alt: "λ")[$λ$] for #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$]. Solving for #math.equation(block: false, alt: "λ")[$λ$] and substituting the known half-life gives #math.equation(block: true, alt: "λ equals the fraction 0 . 693 over t sub 1 / 2 equals the fraction 0 . 693 over 5730 y .")[$λ = frac(0 "." "693", t_(1 / 2)) = frac(0 "." "693", "5730 y") "."$] We enter this value into the previous equation to find #math.equation(block: false, alt: "t")[$t$]: #math.equation(block: true, alt: "t equals the fraction 0 . 0834 over the fraction 0 . 693 over 5730 y equals 690 y.")[$t = frac(0 "." "0834", frac(0 "." "693", "5730 y")) = "690 y."$] Discussion This dates the material in the shroud to 1988–690 = a.d. 1300. Our calculation is only accurate to two digits, so that the year is rounded to 1300. The values obtained at the three independent laboratories gave a weighted average date of a.d. #math.equation(block: false, alt: "1320 plus or minus 60")[$"1320" ± "60"$]. The uncertainty is typical of carbon-14 dating and is due to the small amount of #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] in living tissues, the amount of material available, and experimental uncertainties (reduced by having three independent measurements). It is meaningful that the date of the shroud is consistent with the first record of its existence and inconsistent with the period in which Jesus lived. ] There are other forms of radioactive dating. Rocks, for example, can sometimes be dated based on the decay of #math.equation(block: false, alt: "to the power 238 U")[$"238" "U"$]. The decay series for #math.equation(block: false, alt: "to the power 238 U")[$"238" "U"$] ends with #math.equation(block: false, alt: "to the power 206 Pb")[$"206" "Pb"$], so that the ratio of these nuclides in a rock is an indication of how long it has been since the rock solidified. The original composition of the rock, such as the absence of lead, must be known with some confidence. However, as with carbon-14 dating, the technique can be verified by a consistent body of knowledge. Since #math.equation(block: false, alt: "to the power 238 U")[$"238" "U"$] has a half-life of #math.equation(block: false, alt: "4 . 5 times 10 to the power 9")[$4 "." 5 × "10"^(9)$] y, it is useful for dating only very old materials, showing, for example, that the oldest rocks on Earth solidified about #math.equation(block: false, alt: "3 . 5 times 10 to the power 9")[$3 "." 5 × "10"^(9)$] years ago. === Activity, the Rate of Decay What do we mean when we say a source is highly radioactive? Generally, this means the number of decays per unit time is very high. We define #strong[activity] #math.equation(block: false, alt: "R")[$R$] to be the #strong[rate of decay] expressed in decays per unit time. In equation form, this is #math.equation(block: true, alt: "R equals the fraction Δ N over Δ t")[$R = frac(Δ N, Δ t)$] where #math.equation(block: false, alt: "Δ N")[$"Δ" N$] is the number of decays that occur in time #math.equation(block: false, alt: "Δ t")[$"Δ" t$]. The SI unit for activity is one decay per second and is given the name #strong[becquerel] (Bq) in honor of the discoverer of radioactivity. That is, #math.equation(block: true, alt: "1 Bq equals 1 decay/s.")[$1 #h(0.25em) "Bq" = "1 decay/s."$] Activity #math.equation(block: false, alt: "R")[$R$] is often expressed in other units, such as decays per minute or decays per year. One of the most common units for activity is the #strong[curie] (Ci), defined to be the activity of 1 g of #math.equation(block: false, alt: "to the power 226 Ra")[$"226" "Ra"$], in honor of Marie Curie’s work with radium. The definition of curie is #math.equation(block: true, alt: "1 Ci equals 3 . 70 times 10 to the power 10 Bq,")[$" 1 Ci" = 3 "." "70" × "10"^("10") #h(0.25em) " Bq,"$] or #math.equation(block: false, alt: "3 . 70 times 10 to the power 10")[$3 "." "70" × "10"^("10")$] decays per second. A curie is a large unit of activity, while a becquerel is a relatively small unit. #math.equation(block: false, alt: "1 MBq equals 100 microcuries open parenthesis μ Ci close parenthesis")[$"1 MBq" = "100 microcuries " #h(0.25em) ( μ "Ci" )$]. In countries like Australia and New Zealand that adhere more to SI units, most radioactive sources, such as those used in medical diagnostics or in physics laboratories, are labeled in Bq or megabecquerel (MBq). Intuitively, you would expect the activity of a source to depend on two things: the amount of the radioactive substance present, and its half-life. The greater the number of radioactive nuclei present in the sample, the more will decay per unit of time. The shorter the half-life, the more decays per unit time, for a given number of nuclei. So activity #math.equation(block: false, alt: "R")[$R$] should be proportional to the number of radioactive nuclei, #math.equation(block: false, alt: "N")[$N$], and inversely proportional to their half-life, #math.equation(block: false, alt: "t sub 1 / 2")[$t_(1 / 2)$]. In fact, your intuition is correct. It can be shown that the activity of a source is #math.equation(block: true, alt: "R equals the fraction 0 . 693 N over t sub 1 / 2")[$R = frac(0 "." "693" N, t_(1 / 2))$] where #math.equation(block: false, alt: "N")[$N$] is the number of radioactive nuclei present, having half-life #math.equation(block: false, alt: "t sub 1 / 2")[$t_(1 / 2)$]. This relationship is useful in a variety of calculations, as the next two examples illustrate. #examplebox("Example 2")[How Great Is the #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] Activity in Living Tissue?][ Calculate the activity due to #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] in 1.00 kg of carbon found in a living organism. Express the activity in units of Bq and Ci. Strategy To find the activity #math.equation(block: false, alt: "R")[$R$] using the equation #math.equation(block: false, alt: "R equals the fraction 0 . 693 N over t sub 1 / 2")[$R = frac(0 "." "693" N, t_(1 / 2))$], we must know #math.equation(block: false, alt: "N")[$N$] and #math.equation(block: false, alt: "t sub 1 / 2")[$t_(1 / 2)$]. The half-life of #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] can be found in Appendix B, and was stated above as 5730 y. To find #math.equation(block: false, alt: "N")[$N$], we first find the number of #math.equation(block: false, alt: "to the power 12 C")[$"12" "C"$] nuclei in 1.00 kg of carbon using the concept of a mole. As indicated, we then multiply by #math.equation(block: false, alt: "1 . 3 × 10 to the power minus 12")[$1 "." 3 × "10"^(− "12")$] (the abundance of #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] in a carbon sample from a living organism) to get the number of #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] nuclei in a living organism. Solution One mole of carbon has a mass of 12.0 g, since it is nearly pure #math.equation(block: false, alt: "to the power 12 C")[$"12" "C"$]. (A mole has a mass in grams equal in magnitude to #math.equation(block: false, alt: "A")[$A$] found in the periodic table.) Thus the number of carbon nuclei in a kilogram is #math.equation(block: true, alt: "N open parenthesis to the power 12 C close parenthesis equals the fraction 6.02 times 10 to the power 23 mol to the power –1 over 12.0 g/mol times (1000 g) equals 5.02 times 10 to the power 25 .")[$N ( 12 "C" ) = frac(6.02 × "10"^("23") #h(0.25em) "mol"^(–1), 12.0 g/"mol") × "(1000 g)" = "5.02" × "10"^("25") "."$] So the number of #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] nuclei in 1 kg of carbon is #math.equation(block: true, alt: "N open parenthesis to the power 14 C close parenthesis equals open parenthesis 5.02 times 10 to the power 25 close parenthesis open parenthesis 1.3 times 10 to the power −12 close parenthesis equals 6.52 times 10 to the power 13 .")[$N ( "14" "C" ) = ( 5.02 × "10"^("25") ) ( 1.3 × "10"^("−12") ) = 6.52 × "10"^("13") "."$] Now the activity #math.equation(block: false, alt: "R")[$R$] is found using the equation #math.equation(block: false, alt: "R equals the fraction 0 . 693 N over t sub 1 / 2")[$R = frac(0 "." "693" N, t_(1 / 2))$]. Entering known values gives #math.equation(block: true, alt: "R equals the fraction 0 . 693 open parenthesis 6 . 52 × 10 to the power 13 close parenthesis over 5730 y equals 7 . 89 × 10 to the power 9 y to the power –1 ,")[$R = frac(0 "." "693" ( 6 "." "52" × "10"^("13") ), "5730 y") = 7 "." "89" × "10"^(9) #h(0.25em) "y"^(–1) ,$] or #math.equation(block: false, alt: "7 . 89 times 10 to the power 9")[$7 "." "89" × "10"^(9)$] decays per year. To convert this to the unit Bq, we simply convert years to seconds. Thus, #math.equation(block: true, alt: "R equals open parenthesis 7.89 × 10 to the power 9 y to the power –1 close parenthesis the fraction 1.00 y over 3 . 16 × 10 to the power 7 s equals 250 Bq,")[$R = ( "7.89" × "10"^(9) #h(0.25em) "y"^(–1) ) frac(1.00 y, 3 "." "16" × "10"^(7) #h(0.25em) " s") = "250 Bq,"$] or 250 decays per second. To express #math.equation(block: false, alt: "R")[$R$] in curies, we use the definition of a curie, #math.equation(block: true, alt: "R equals the fraction 250 Bq over 3.7 × 10 to the power 10 Bq/Ci equals 6.76 × 10 to the power minus 9 Ci.")[$R = frac("250 Bq", 3.7 × "10"^("10") #h(0.25em) " Bq/Ci") = 6.76 × "10"^(− 9) #h(0.25em) " Ci."$] Thus, #math.equation(block: true, alt: "R equals 6.76 nCi.")[$R = 6.76 #h(0.25em) "nCi."$] Discussion Our own bodies contain kilograms of carbon, and it is intriguing to think there are hundreds of #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] decays per second taking place in us. Carbon-14 and other naturally occurring radioactive substances in our bodies contribute to the background radiation we receive. The small number of decays per second found for a kilogram of carbon in this example gives you some idea of how difficult it is to detect #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] in a small sample of material. If there are 250 decays per second in a kilogram, then there are 0.25 decays per second in a gram of carbon in living tissue. To observe this, you must be able to distinguish decays from other forms of radiation, in order to reduce background noise. This becomes more difficult with an old tissue sample, since it contains less #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$], and for samples more than 50 thousand years old, it is impossible. ] Human-made (or artificial) radioactivity has been produced for decades and has many uses. Some of these include medical therapy for cancer, medical imaging and diagnostics, and food preservation by irradiation. Many applications as well as the biological effects of radiation are explored in Medical Applications of Nuclear Physics, but it is clear that radiation is hazardous. A number of tragic examples of this exist, one of the most disastrous being the meltdown and fire at the Chernobyl reactor complex in the Ukraine. Several radioactive isotopes were released in huge quantities, contaminating many thousands of square kilometers and directly affecting hundreds of thousands of people. The most significant releases were of #math.equation(block: false, alt: "to the power 131 I")[$"131" "I"$], #math.equation(block: false, alt: "to the power 90 Sr")[$"90" "Sr"$], #math.equation(block: false, alt: "to the power 137 Cs")[$"137" "Cs"$], #math.equation(block: false, alt: "to the power 239 Pu")[$"239" "Pu"$], #math.equation(block: false, alt: "to the power 238 U")[$"238" "U"$], and #math.equation(block: false, alt: "to the power 235 U")[$"235" "U"$]. Estimates are that the total amount of radiation released was about 100 million curies. === Human and Medical Applications #figure(figph[A person holding a hand held radiation detector near the Chernobyl reactor.], alt: "A person holding a hand held radiation detector near the Chernobyl reactor.", caption: [The Chernobyl reactor. More than 100 people died soon after its meltdown, and there will be thousands of deaths from radiation-induced cancer in the future. While the accident was due to a series of human errors, the cleanup efforts were heroic. Most of the immediate fatalities were firefighters and reactor personnel.]) #examplebox("Example 3")[What Mass of #math.equation(block: false, alt: "to the power 137 Cs")[$"137" "Cs"$] Escaped Chernobyl?][ It is estimated that the Chernobyl disaster released 6.0 MCi of #math.equation(block: false, alt: "to the power 137 Cs")[$"137" "Cs"$] into the environment. Calculate the mass of #math.equation(block: false, alt: "to the power 137 Cs")[$"137" "Cs"$] released. Strategy We can calculate the mass released using Avogadro’s number and the concept of a mole if we can first find the number of nuclei #math.equation(block: false, alt: "N")[$N$] released. Since the activity #math.equation(block: false, alt: "R")[$R$] is given, and the half-life of #math.equation(block: false, alt: "to the power 137 Cs")[$"137" "Cs"$] is found in Appendix B to be 30.2 y, we can use the equation #math.equation(block: false, alt: "R equals the fraction 0 . 693 N over t sub 1 / 2")[$R = frac(0 "." "693" N, t_(1 / 2))$] to find #math.equation(block: false, alt: "N")[$N$]. Solution Solving the equation #math.equation(block: false, alt: "R equals the fraction 0 . 693 N over t sub 1 / 2")[$R = frac(0 "." "693" N, t_(1 / 2))$] for #math.equation(block: false, alt: "N")[$N$] gives #math.equation(block: true, alt: "N equals the fraction Rt sub 1/2 over 0.693 .")[$N = frac("Rt"_(1/2), "0.693") "."$] Entering the given values yields #math.equation(block: true, alt: "N equals the fraction open parenthesis 6.0 MCi close parenthesis open parenthesis 30 . 2 y close parenthesis over 0 . 693 .")[$N = frac(( 6.0 "MCi" ) ( "30" "." 2 y ), 0 "." "693") "."$] Converting curies to becquerels and years to seconds, we get #math.equation(block: true, alt: "N, equals, the fraction open parenthesis 6 . 0 × 10 to the power 6 Ci close parenthesis open parenthesis 3 . 7 × 10 to the power 10 Bq/Ci close parenthesis open parenthesis 30.2 y close parenthesis open parenthesis 3 . 16 × 10 to the power 7 s/y close parenthesis over 0.693; equals, 3 . 1 times 10 to the power 26 .")[$N & = & frac(( 6 "." 0 × "10"^(6) #h(0.25em) " Ci" ) ( 3 "." 7 × "10"^("10") #h(0.25em) " Bq/Ci" ) ( "30.2 y" ) ( 3 "." "16" × "10"^(7) #h(0.25em) " s/y" ), "0.693") \ & = & 3 "." 1 × "10"^("26") "."$] One mole of a nuclide #math.equation(block: false, alt: "to the power A X")[$A X$] has a mass of #math.equation(block: false, alt: "A")[$A$] grams, so that one mole of #math.equation(block: false, alt: "to the power 137 Cs")[$"137" "Cs"$] has a mass of 137 g. A mole has #math.equation(block: false, alt: "6 . 02 times 10 to the power 23")[$6 "." "02 " × "10"^("23")$] nuclei. Thus the mass of #math.equation(block: false, alt: "to the power 137 Cs")[$"137" "Cs"$] released was #math.equation(block: true, alt: "m, equals, the fraction 137 g over 6.02 times 10 to the power 23 open parenthesis 3 . 1 times 10 to the power 26 close parenthesis equals 70 times 10 cubed g; equals, 70 kg.")[$m & = & frac("137 g", "6.02 " × "10"^("23")) ( 3 "." 1 × "10"^("26") ) = "70" × "10"^(3) #h(0.25em) " g" \ & = & "70 kg."$] Discussion While 70 kg of material may not be a very large mass compared to the amount of fuel in a power plant, it is extremely radioactive, since it only has a 30-year half-life. Six megacuries (6.0 MCi) is an extraordinary amount of activity but is only a fraction of what is produced in nuclear reactors. Similar amounts of the other isotopes were also released at Chernobyl. Although the chances of such a disaster may have seemed small, the consequences were extremely severe, requiring greater caution than was used. More will be said about safe reactor design in the next chapter, but it should be noted that more recent reactors have a fundamentally safer design. ] Activity #math.equation(block: false, alt: "R")[$R$] decreases in time, going to half its original value in one half-life, then to one-fourth its original value in the next half-life, and so on. Since #math.equation(block: false, alt: "R equals the fraction 0 . 693 N over t sub 1 / 2")[$R = frac(0 "." "693" N, t_(1 / 2))$], the activity decreases as the number of radioactive nuclei decreases. The equation for #math.equation(block: false, alt: "R")[$R$] as a function of time is found by combining the equations #math.equation(block: false, alt: "N equals N sub 0 e to the power minus λt")[$N = N_(0) e^(− "λt")$] and #math.equation(block: false, alt: "R equals the fraction 0 . 693 N over t sub 1 / 2")[$R = frac(0 "." "693" N, t_(1 / 2))$], yielding #math.equation(block: true, alt: "R equals R sub 0 e to the power minus λt ,")[$R = R_(0) e^(− "λt") ","$] where #math.equation(block: false, alt: "R sub 0")[$R_(0)$] is the activity at #math.equation(block: false, alt: "t equals 0")[$t = 0$]. This equation shows exponential decay of radioactive nuclei. For example, if a source originally has a 1.00-mCi activity, it declines to 0.500 mCi in one half-life, to 0.250 mCi in two half-lives, to 0.125 mCi in three half-lives, and so on. For times other than whole half-lives, the equation #math.equation(block: false, alt: "R equals R sub 0 e to the power minus λt")[$R = R_(0) e^(− "λt")$] must be used to find #math.equation(block: false, alt: "R")[$R$]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Alpha Decay] Watch alpha particles escape from a polonium nucleus, causing radioactive alpha decay. See how random decay times relate to the half life. #link("https://openstax.org/l/02alphadecay")[Click to view content]. ] === Test Prep for AP Courses A radioactive sample has N atoms initially. After 3 half-lives have elapsed, how many atoms remain? + N/3 + N/6 + N/8 + N/27 When #math.equation(block: false, alt: "P 84 215 o")[$attach("P", tl: 215, bl: 84) "o"$] decays, the product is #math.equation(block: false, alt: "P 82 211 b.")[$attach("P", tl: 211, bl: 82) "b."$] The half-life of this decay process is 1.78 ms. If the initial sample contains 3.4 x 10#super[17] parent nuclei, how many are remaining after 35 ms have elapsed? What kind of decay process is this (alpha, beta, or gamma)? #solutionbox[ This must be alpha decay since 4 nucleons (2 positive charges) are lost from the parent nucleus. The number remaining is found from: #math.equation(block: true, alt: "N open parenthesis t close parenthesis equals N sub 0 e open parenthesis the fraction minus 0.693 t over the fraction t sub 1 over 2 close parenthesis equals 3.4 times 10 to the power 17 e open parenthesis the fraction minus open parenthesis 0.693 close parenthesis open parenthesis 0.035 close parenthesis over 0.00173 close parenthesis")[$"N" ( t ) = "N"_(0) e ( frac(− 0.693 t, frac(t_(1), 2)) ) = 3.4 × 10^(17) e ( frac(− ( 0.693 ) ( 0.035 ), 0.00173) )$] #math.equation(block: false, alt: "N open parenthesis t close parenthesis equals 4.1 times 10 to the power 11")[$"N" ( t ) = 4.1 × 10^(11)$] nuclei ] === Section Summary - Half-life #math.equation(block: false, alt: "t sub 1 / 2")[$t_(1 / 2)$] is the time in which there is a 50% chance that a nucleus will decay. The number of nuclei #math.equation(block: false, alt: "N")[$N$] as a function of time is#math.equation(block: true, alt: "N equals N sub 0 e to the power minus λt ,")[$N = N_(0) e^(− "λt") ,$] where #math.equation(block: false, alt: "N sub 0")[$N_(0)$] is the number present at #math.equation(block: false, alt: "t equals 0")[$t = 0$], and #math.equation(block: false, alt: "λ")[$λ$] is the decay constant, related to the half-life by #math.equation(block: true, alt: "λ equals the fraction 0 . 693 over t sub 1 / 2 .")[$λ = frac(0 "." "693", t_(1 / 2)) .$] - One of the applications of radioactive decay is radioactive dating, in which the age of a material is determined by the amount of radioactive decay that occurs. The rate of decay is called the activity #math.equation(block: false, alt: "R")[$R$]:#math.equation(block: true, alt: "R equals the fraction Δ N over Δ t .")[$R = frac("Δ" N, "Δ" t) .$] - The SI unit for #math.equation(block: false, alt: "R")[$R$] is the becquerel (Bq), defined by#math.equation(block: true, alt: "1 Bq equals 1 decay/s.")[$" 1 Bq" = "1 decay/s."$] - #math.equation(block: false, alt: "R")[$R$] is also expressed in terms of curies (Ci), where #math.equation(block: true, alt: "1 Ci equals 3 . 70 times 10 to the power 10 Bq.")[$1 #h(0.25em) "Ci" = 3 "." "70" × "10"^("10") #h(0.25em) " Bq."$] - The activity #math.equation(block: false, alt: "R")[$R$] of a source is related to #math.equation(block: false, alt: "N")[$N$] and #math.equation(block: false, alt: "t sub 1 / 2")[$t_(1 / 2)$] by#math.equation(block: true, alt: "R equals the fraction 0 . 693 N over t sub 1 / 2 .")[$R = frac(0 "." "693" N, t_(1 / 2)) .$] - Since #math.equation(block: false, alt: "N")[$N$] has an exponential behavior as in the equation #math.equation(block: false, alt: "N equals N sub 0 e to the power minus λt")[$N = N_(0) e^(− "λt")$], the activity also has an exponential behavior, given by#math.equation(block: true, alt: "R equals R sub 0 e to the power minus λt ,")[$R = R_(0) e^(− "λt") ,$] where #math.equation(block: false, alt: "R sub 0")[$R_(0)$] is the activity at #math.equation(block: false, alt: "t equals 0")[$t = 0$]. === Conceptual Questions In a #math.equation(block: false, alt: "3 times 10 to the power 9")[$3 × "10"^(9)$]-year-old rock that originally contained some #math.equation(block: false, alt: "to the power 238 U")[$"238" "U"$], which has a half-life of #math.equation(block: false, alt: "4.5 times 10 to the power 9")[$4.5 × "10"^(9)$] years, we expect to find some #math.equation(block: false, alt: "to the power 238 U")[$"238" "U"$] remaining in it. Why are #math.equation(block: false, alt: "to the power 226 Ra")[$"226" "Ra"$], #math.equation(block: false, alt: "to the power 222 Rn")[$"222" "Rn"$], and #math.equation(block: false, alt: "to the power 210 Po")[$"210" "Po"$] also found in such a rock, even though they have much shorter half-lives (1600 years, 3.8 days, and 138 days, respectively)? Does the number of radioactive nuclei in a sample decrease to #emph[exactly] half its original value in one half-life? Explain in terms of the statistical nature of radioactive decay. Radioactivity depends on the nucleus and not the atom or its chemical state. Why, then, is one kilogram of uranium more radioactive than one kilogram of uranium hexafluoride? Explain how a bound system can have less mass than its components. Why is this not observed classically, say for a building made of bricks? Spontaneous radioactive decay occurs only when the decay products have less mass than the parent, and it tends to produce a daughter that is more stable than the parent. Explain how this is related to the fact that more tightly bound nuclei are more stable. (Consider the binding energy per nucleon.) To obtain the most precise value of BE from the equation #math.equation(block: false, alt: "BE= ZM to the power 1 H plus Nm sub n c squared minus m to the power A X c squared")[$"BE=" "ZM" 1 "H" + "Nm"_(n) c^(2) − m A X c^(2)$], we should take into account the binding energy of the electrons in the neutral atoms. Will doing this produce a larger or smaller value for BE? Why is this effect usually negligible? How does the finite range of the nuclear force relate to the fact that #math.equation(block: false, alt: "BE / A")[$"BE" / A$] is greatest for #math.equation(block: false, alt: "A")[$A$] near 60? === Problems & Exercises Data from the appendices and the periodic table may be needed for these problems. An old campfire is uncovered during an archaeological dig. Its charcoal is found to contain less than 1/1000 the normal amount of #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$]. Estimate the minimum age of the charcoal, noting that #math.equation(block: false, alt: "2 to the power 10 equals 1024")[$2^("10") = "1024"$]. #solutionbox[ 57,300 y ] A #math.equation(block: false, alt: "to the power 60 Co")[$"60" "Co"$] source is labeled 4.00 mCi, but its present activity is found to be #math.equation(block: false, alt: "1 . 85 times 10 to the power 7")[$1 "." "85" × "10"^(7)$] Bq. (a) What is the present activity in mCi? (b) How long ago did it actually have a 4.00-mCi activity? (a) Calculate the activity #math.equation(block: false, alt: "R")[$R$] in curies of 1.00 g of #math.equation(block: false, alt: "to the power 226 Ra")[$"226" "Ra"$]. (b) Discuss why your answer is not exactly 1.00 Ci, given that the curie was originally supposed to be exactly the activity of a gram of radium. #solutionbox[ (a) 0.988 Ci (b) The half-life of #math.equation(block: false, alt: "to the power 226 Ra")[$"226" "Ra"$] is now better known. ] Show that the activity of the #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] in 1.00 g of #math.equation(block: false, alt: "to the power 12 C")[$"12" "C"$] found in living tissue is 0.250 Bq. Mantles for gas lanterns contain thorium, because it forms an oxide that can survive being heated to incandescence for long periods of time. Natural thorium is almost 100% #math.equation(block: false, alt: "to the power 232 Th")[$"232" "Th"$], with a half-life of #math.equation(block: false, alt: "1 . 405 times 10 to the power 10 y")[$1 "." "405" × "10"^("10") #h(0.25em) "y"$]. If an average lantern mantle contains 300 mg of thorium, what is its activity? #solutionbox[ #math.equation(block: true, alt: "1.22 times 10 cubed Bq")[$1.22 × "10"^(3) #h(0.25em) "Bq"$] ] Cow’s milk produced near nuclear reactors can be tested for as little as 1.00 pCi of #math.equation(block: false, alt: "to the power 131 I")[$"131" "I"$] per liter, to check for possible reactor leakage. What mass of #math.equation(block: false, alt: "to the power 131 I")[$"131" "I"$] has this activity? (a) Natural potassium contains #math.equation(block: false, alt: "to the power 40 K")[$"40" "K"$], which has a half-life of #math.equation(block: false, alt: "1 . 277 times 10 to the power 9")[$1 "." "277" × "10"^(9)$] y. What mass of #math.equation(block: false, alt: "to the power 40 K")[$"40" "K"$] in a person would have a decay rate of 4140 Bq? (b) What is the fraction of #math.equation(block: false, alt: "to the power 40 K")[$"40" "K"$] in natural potassium, given that the person has 140 g in his body? (These numbers are typical for a 70-kg adult.) #solutionbox[ (a) 16.0 mg (b) 0.0114% ] There is more than one isotope of natural uranium. If a researcher isolates 1.00 mg of the relatively scarce #math.equation(block: false, alt: "to the power 235 U")[$"235" "U"$] and finds this mass to have an activity of 80.0 Bq, what is its half-life in years? #math.equation(block: false, alt: "to the power 50 V")[$"50" "V"$] has one of the longest known radioactive half-lives. In a difficult experiment, a researcher found that the activity of 1.00 kg of #math.equation(block: false, alt: "to the power 50 V")[$"50" "V"$] is 1.75 Bq. What is the half-life in years? #solutionbox[ #math.equation(block: true, alt: "1.48 times 10 to the power 17 y")[$1.48 × "10"^("17") #h(0.25em) "y"$] ] You can sometimes find deep red crystal vases in antique stores, called uranium glass because their color was produced by doping the glass with uranium. Look up the natural isotopes of uranium and their half-lives, and calculate the activity of such a vase assuming it has 2.00 g of uranium in it. Neglect the activity of any daughter nuclides. A tree falls in a forest. How many years must pass before the #math.equation(block: false, alt: "to the power 14 C")[$"14" "C"$] activity in 1.00 g of the tree’s carbon drops to 1.00 decay per hour? #solutionbox[ #math.equation(block: true, alt: "5.6 times 10 to the power 4 y")[$5.6 × "10"^(4) #h(0.25em) y$] ] What fraction of the #math.equation(block: false, alt: "to the power 40 K")[$"40" "K"$] that was on Earth when it formed #math.equation(block: false, alt: "4 . 5 times 10 to the power 9")[$4 "." 5 × "10"^(9)$] years ago is left today? A 5000-Ci #math.equation(block: false, alt: "to the power 60 Co")[$"60" "Co"$] source used for cancer therapy is considered too weak to be useful when its activity falls to 3500 Ci. How long after its manufacture does this happen? #solutionbox[ 2.71 y ] Natural uranium is 0.7200% #math.equation(block: false, alt: "to the power 235 U")[$"235" "U"$] and 99.27% #math.equation(block: false, alt: "to the power 238 U")[$"238" "U"$]. What were the percentages of #math.equation(block: false, alt: "to the power 235 U")[$"235" "U"$] and #math.equation(block: false, alt: "to the power 238 U")[$"238" "U"$] in natural uranium when Earth formed #math.equation(block: false, alt: "4 . 5 times 10 to the power 9")[$4 "." 5 × "10"^(9)$] years ago? The #math.equation(block: false, alt: "β to the power minus")[$β^(−)$] particles emitted in the decay of #math.equation(block: false, alt: "cubed H")[$3 "H"$] (tritium) interact with matter to create light in a glow-in-the-dark exit sign. At the time of manufacture, such a sign contains 15.0 Ci of #math.equation(block: false, alt: "cubed H")[$3 "H"$]. (a) What is the mass of the tritium? (b) What is its activity 5.00 y after manufacture? #solutionbox[ (a) 1.56 mg (b) 11.3 Ci ] World War II aircraft had instruments with glowing radium-painted dials. The activity of one such instrument was #math.equation(block: false, alt: "1.0 times 10 to the power 5")[$1.0 × "10"^(5)$] Bq when new. (a) What mass of #math.equation(block: false, alt: "to the power 226 Ra")[$"226" "Ra"$] was present? (b) After some years, the phosphors on the dials deteriorated chemically, but the radium did not escape. What is the activity of this instrument 57.0 years after it was made? (a) The #math.equation(block: false, alt: "to the power 210 Po")[$"210" "Po"$] source used in a physics laboratory is labeled as having an activity of #math.equation(block: false, alt: "1.0 μ Ci")[$1.0 #h(0.25em) μ "Ci"$] on the date it was prepared. A student measures the radioactivity of this source with a Geiger counter and observes 1500 counts per minute. She notices that the source was prepared 120 days before her lab. What fraction of the decays is she observing with her apparatus? (b) Identify some of the reasons that only a fraction of the #math.equation(block: false, alt: "α")[$α$] s emitted are observed by the detector. #solutionbox[ (a) #math.equation(block: false, alt: "1.23 times 10 to the power minus 3")[$"1.23" × "10"^(− 3)$] (b) Only part of the emitted radiation goes in the direction of the detector. Only a fraction of that causes a response in the detector. Some of the emitted radiation (mostly #math.equation(block: false, alt: "α")[$α$] particles) is observed within the source. Some is absorbed within the source, some is absorbed by the detector, and some does not penetrate the detector. ] Armor-piercing shells with depleted uranium cores are fired by aircraft at tanks. (The high density of the uranium makes them effective.) The uranium is called depleted because it has had its #math.equation(block: false, alt: "to the power 235 U")[$"235" "U"$] removed for reactor use and is nearly pure #math.equation(block: false, alt: "to the power 238 U")[$"238" "U"$]. Depleted uranium has been erroneously called non-radioactive. To demonstrate that this is wrong: (a) Calculate the activity of 60.0 g of pure #math.equation(block: false, alt: "to the power 238 U")[$"238" "U"$]. (b) Calculate the activity of 60.0 g of natural uranium, neglecting the #math.equation(block: false, alt: "to the power 234 U")[$"234" "U"$] and all daughter nuclides. The ceramic glaze on a red-orange Fiestaware plate is #math.equation(block: false, alt: "U sub 2 O sub 3")[$"U"_(2) "O"_(3)$] and contains 50.0 grams of #math.equation(block: false, alt: "to the power 238 U")[$238 "U"$] , but very little #math.equation(block: false, alt: "to the power 235 U")[$235 "U"$]. (a) What is the activity of the plate? (b) Calculate the total energy that will be released by the #math.equation(block: false, alt: "to the power 238 U")[$238 "U"$] decay. (c) If energy is worth 12.0 cents per #math.equation(block: false, alt: "kW times h")[$"kW" ⋅ "h"$], what is the monetary value of the energy emitted? (These plates went out of production some 30 years ago, but are still available as collectibles.) #solutionbox[ (a) #math.equation(block: false, alt: "1.68 times 10 to the power minus 5 Ci")[$1.68 × 10^(− 5) #h(0.25em) "Ci"$] (b) #math.equation(block: false, alt: "8.65 times 10 to the power 10 J")[$8.65 × 10^(10) #h(0.25em) "J"$] (c) #math.equation(block: false, alt: "$ 2.9 times 10 cubed")[$\$ 2.9 × 10^(3)$] ] Large amounts of depleted uranium (#math.equation(block: false, alt: "to the power 238 U")[$238 "U"$]) are available as a by-product of uranium processing for reactor fuel and weapons. Uranium is very dense and makes good counter weights for aircraft. Suppose you have a 4000-kg block of #math.equation(block: false, alt: "to the power 238 U")[$238 "U"$]. (a) Find its activity. (b) How many calories per day are generated by thermalization of the decay energy? (c) Do you think you could detect this as heat? Explain. The #emph[Galileo] space probe was launched on its long journey past several planets in 1989, with an ultimate goal of Jupiter. Its power source is 11.0 kg of #math.equation(block: false, alt: "to the power 238 Pu")[$238 "Pu"$], a by-product of nuclear weapons plutonium production. Electrical energy is generated thermoelectrically from the heat produced when the 5.59-MeV #math.equation(block: false, alt: "α")[$"α"$] particles emitted in each decay crash to a halt inside the plutonium and its shielding. The half-life of #math.equation(block: false, alt: "to the power 238 Pu")[$238 "Pu"$] is 87.7 years. (a) What was the original activity of the #math.equation(block: false, alt: "to the power 238 Pu")[$238 "Pu"$] in becquerel? (b) What power was emitted in kilowatts? (c) What power was emitted 12.0 y after launch? You may neglect any extra energy from daughter nuclides and any losses from escaping #math.equation(block: false, alt: "γ")[$"γ"$] rays. #solutionbox[ (a) #math.equation(block: false, alt: "6.97 times 10 to the power 15 Bq")[$6.97 × 10^(15) #h(0.25em) "Bq"$] (b) 6.24 kW (c) 5.67 kW ] Construct Your Own Problem Consider the generation of electricity by a radioactive isotope in a space probe, such as described in . Construct a problem in which you calculate the mass of a radioactive isotope you need in order to supply power for a long space flight. Among the things to consider are the isotope chosen, its half-life and decay energy, the power needs of the probe and the length of the flight. Unreasonable Results A nuclear physicist finds #math.equation(block: false, alt: "1.0 μ g")[$1.0 #h(0.25em) μ "g"$] of #math.equation(block: false, alt: "to the power 236 U")[$236 "U"$] in a piece of uranium ore and assumes it is primordial since its half-life is #math.equation(block: false, alt: "2.3 times 10 to the power 7 y")[$2.3 × 10^(7) #h(0.25em) "y"$]. (a) Calculate the amount of #math.equation(block: false, alt: "to the power 236 U")[$236 "U"$]that would had to have been on Earth when it formed #math.equation(block: false, alt: "4.5 times 10 to the power 9 y")[$4.5 × 10^(9) #h(0.25em) "y"$] ago for #math.equation(block: false, alt: "1.0 μ g")[$1.0 #h(0.25em) μ "g"$] to be left today. (b) What is unreasonable about this result? (c) What assumption is responsible? Unreasonable Results (a) Repeat but include the 0.0055% natural abundance of #math.equation(block: false, alt: "to the power 234 U")[$234 "U"$] with its #math.equation(block: false, alt: "2.45 times 10 to the power 5 y")[$2.45 × 10^(5) #h(0.25em) "y"$] half-life. (b) What is unreasonable about this result? (c) What assumption is responsible? (d) Where does the #math.equation(block: false, alt: "to the power 234 U")[$234 "U"$] come from if it is not primordial? Unreasonable Results The manufacturer of a smoke alarm decides that the smallest current of #math.equation(block: false, alt: "α")[$"α"$] radiation he can detect is #math.equation(block: false, alt: "1.00 μ A")[$1.00 #h(0.25em) μ "A"$]. (a) Find the activity in curies of an #math.equation(block: false, alt: "α")[$"α "$] emitter that produces a #math.equation(block: false, alt: "1.00 μ A")[$1.00 #h(0.25em) μ "A"$] current of #math.equation(block: false, alt: "α")[$"α"$] particles. (b) What is unreasonable about this result? (c) What assumption is responsible? #solutionbox[ (a) 84.5 Ci (b) An extremely large activity, many orders of magnitude greater than permitted for home use. (c) The assumption of #math.equation(block: false, alt: "1.00 μA")[$1.00 #h(0.25em) "μA"$] is unreasonably large. Other methods can detect much smaller decay rates. ]