#set document(title: "10.3 Radioactive Decay", 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")) == 10.3#h(0.6em)Radioactive Decay In 1896, Antoine Henri #strong[Becquerel] discovered that a uranium-rich rock emits invisible rays that can darken a photographic plate in an enclosed container. Scientists offer three arguments for the nuclear origin of these rays. First, the effects of the radiation do not vary with chemical state; that is, whether the emitting material is in the form of an element or compound. Second, the radiation does not vary with changes in temperature or pressure—both factors that in sufficient degree can affect electrons in an atom. Third, the very large energy of the invisible rays (up to hundreds of eV) is not consistent with atomic electron transitions (only a few eV). Today, this radiation is explained by the conversion of mass into energy deep within the nucleus of an atom. The spontaneous emission of radiation from nuclei is called nuclear #strong[radioactivity] . #figure(figph[A yellow triangle with a black outline, enclosing a fan shaped structure is shown. The “fan” is black and has three blades.], alt: "A yellow triangle with a black outline, enclosing a fan shaped structure is shown. The “fan” is black and has three blades.", caption: [The international ionizing radiation symbol is universally recognized as the warning symbol for nuclear radiation.]) === Radioactive Decay Law When an individual nucleus transforms into another with the emission of radiation, the nucleus is said to #strong[decay]. Radioactive decay occurs for all nuclei with #math.equation(block: false, alt: "Z greater than 82 ,")[$Z > 82 ,$] and also for some unstable isotopes with #math.equation(block: false, alt: "Z less than 83 .")[$Z < 83 .$] The decay rate is proportional to the number of original (undecayed) nuclei #emph[N] in a substance. The number of nuclei lost to decay, #math.equation(block: false, alt: "− d N")[$"−" d N$] in time interval #emph[dt], is written #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "minus the fraction d N over d t equals λ N")[$− frac(d N, d t) = λ N$] ] where #math.equation(block: false, alt: "λ")[$λ$] is called the #strong[decay constant]. (The minus sign indicates the number of original nuclei decreases over time.) In other words, the more nuclei available to decay, the more that do decay (in time #emph[dt]). This equation can be rewritten as #math.equation(block: true, alt: "the fraction d N over N equals − λ d t .")[$frac(d N, N) = "−" λ d t .$] Integrating both sides of the equation, and defining #math.equation(block: false, alt: "N sub 0")[$N_(0)$] to be the number of nuclei at #math.equation(block: false, alt: "t equals 0")[$t = 0$], we obtain #math.equation(block: true, alt: "∫ N sub 0 N the fraction d N prime over N equals − ∫ 0 t λ d t prime .")[$display(∫_(N_(0))^(N) frac(d N^(′), N)) = "−" display(∫_(0)^(t) λ d t^(′)) .$] This gives us #math.equation(block: true, alt: "ln the fraction N over N sub 0 equals − λ t .")[$"ln" #h(0.2em) frac(N, N_(0)) = "−" λ t .$] Taking the left and right sides of the equation as a power of #emph[e], we have the #strong[radioactive decay law]. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Radioactive Decay Law] The total number #emph[N] of radioactive nuclei remaining after time #emph[t] is #math.equation(block: true, alt: "N equals N sub 0 e to the power − λ t")[$N = N_(0) e^("−" λ t)$] where #math.equation(block: false, alt: "λ")[$λ$] is the decay constant for the particular nucleus. ] The total number of nuclei drops very rapidly at first, and then more slowly . #figure(figph[A graph of N versus t is shown. It is labeled N equal to N subscript 0 e to the power minus lambda t. The value of N is maximum, N subscript 0, at t =0 and it reduces with time till it reaches 0. At t = T subscript half, N = N subscript 0 by 2 and at t = 2T subscript half, N = N subscript 0 by 4.], alt: "A graph of N versus t is shown. It is labeled N equal to N subscript 0 e to the power minus lambda t. The value of N is maximum, N subscript 0, at t =0 and it reduces with time till it reaches 0. At t = T subscript half, N = N subscript 0 by 2 and at t = 2T subscript half, N = N subscript 0 by 4.", caption: [A plot of the radioactive decay law demonstrates that the number of nuclei remaining in a decay sample drops dramatically during the first moments of decay.]) The #strong[half-life] #math.equation(block: false, alt: "open parenthesis T sub 1 / 2 close parenthesis")[$( T_(1 "/" 2) )$] of a radioactive substance is defined as the time for half of the original nuclei to decay (or the time at which half of the original nuclei remain). The half-lives of unstable isotopes are shown in the chart of nuclides in . The number of radioactive nuclei remaining after an integer (#emph[n]) number of half-lives is therefore The decay law N = N₀e^(−λt) with the decay constant written the way this section derives it, λ = ln 2/T½, so the curve is 100·2^(−t/T½) percent of the original nuclei still undecayed. The rising dashed curve is the percentage that has decayed; the two always add to 100. The slider is the half-life, defaulting to the 28.8 y of strontium-90 from this section's example, and the vertical marker sits at exactly one half-life — where the falling curve crosses 50%, which is the definition rather than a coincidence. Two half-lives leave a quarter and three leave an eighth, so 86.4 years of ⁹⁰Sr leaves 12.5%. Drag the half-life and notice that the shape never changes, only the horizontal scale: one number fixes the whole decay, which is what makes ¹⁴C at 5730 y a usable clock for objects thousands of years old and ⁹⁰Sr a hazard for a human lifetime.{"functions":\[{"color":"\#1f4e79","domain":\[0,170\],"expression":"100\*exp(-ln(2)\*t/Thalf)","expression2":"","id":"remaining","inequality":"lt","lineDash":\[\],"lineWidth":2.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"t","visible":true},{"color":"\#2e6b3f","domain":\[0,170\],"expression":"100 - 100\*exp(-ln(2)\*t/Thalf)","expression2":"","id":"decayed","inequality":"lt","lineDash":\[6,4\],"lineWidth":2,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"t","visible":true},{"color":"\#8a8a8a","domain":null,"expression":"50","expression2":"","id":"half-level","inequality":"lt","lineDash":\[3,3\],"lineWidth":1.5,"mode":"cartesian","tMax":10,"tMin":-10,"variable":"t","visible":true},{"color":"\#b8442a","domain":null,"expression":"Thalf","expression2":"t","id":"half-life-marker","inequality":"lt","lineDash":\[6,4\],"lineWidth":2,"mode":"parametric","tMax":50,"tMin":0,"variable":"t","visible":true}\],"grid":{"axisColor":"\#333333","gridColor":"\#cccccc","majorSpacing":10,"minorGridColor":"\#eeeeee","showAxes":true,"showLabels":true},"parameters":\[{"id":"up3-10-4-Thalf","label":"Half-life T\\u00bd (strontium-90 = 28.8 y)","max":60,"min":5,"name":"Thalf","step":0.1,"unit":"y","value":28.8}\],"title":"Radioactive decay and the half-life","version":1,"viewport":{"centerX":76,"centerY":48,"scale":4.1228}}#math.equation(block: true, alt: "N equals the fraction N sub 0 over 2 to the power n")[$N = frac(N_(0), 2^(n))$] If the decay constant (#math.equation(block: false, alt: "λ")[$λ$]) is large, the half-life is small, and vice versa. To determine the relationship between these quantities, note that when #math.equation(block: false, alt: "t equals T sub 1 / 2")[$t = T_(1 "/" 2)$], then #math.equation(block: false, alt: "N equals N sub 0 / 2")[$N = N_(0) "/" 2$]. Thus, can be rewritten as #math.equation(block: true, alt: "the fraction N sub 0 over 2 equals N sub 0 e to the power − λ T sub 1 / 2 .")[$frac(N_(0), 2) = N_(0) e^("−" λ T_(1 "/" 2)) .$] Dividing both sides by #math.equation(block: false, alt: "N sub 0")[$N_(0)$] and taking the natural logarithm yields #math.equation(block: true, alt: "ln the fraction 1 over 2 equals ln e to the power − λ T sub 1 / 2")[$"ln" #h(0.2em) frac(1, 2) = "ln" #h(0.2em) e^("−" λ T_(1 "/" 2))$] which reduces to #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "λ equals the fraction 0.693 over T sub 1 / 2 .")[$λ = frac(0.693, T_(1 "/" 2)) .$] ] Thus, if we know the half-life #emph[T]#sub[1/2] of a radioactive substance, we can find its decay constant. The #strong[lifetime] #math.equation(block: false, alt: "T minus")[$accent(T, −)$] of a radioactive substance is defined as the average amount of time that a nucleus exists before decaying. The lifetime of a substance is just the reciprocal of the decay constant, written as #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "T minus equals the fraction 1 over λ .")[$accent(T, −) = frac(1, "λ") .$] ] The #strong[activity] #emph[A] is defined as the magnitude of the decay rate, or #math.equation(block: true, alt: "A equals minus the fraction d N over d t equals λ N equals λ N sub 0 e to the power − λ t .")[$A = − frac(d N, d t) = λ N = λ N_(0) e^("−" λ t) .$] The infinitesimal change #emph[dN] in the time interval #emph[dt] is negative because the number of parent (undecayed) particles is decreasing, so the activity (#emph[A]) is positive. Defining the initial activity as #math.equation(block: false, alt: "A sub 0 equals λ N sub 0")[$A_(0) = λ N_(0)$], we have #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "A equals A sub 0 e to the power − λ t .")[$A = A_(0) e^("−" λ t) .$] ] Thus, the activity #emph[A] of a radioactive substance decreases exponentially with time . #figure(figph[Figure a shows a graph of A versus t. It starts at point A subscript 0 and reduces with time. The rate of reduction decreases slowly till A is very close to 0, making a curved plot on the graph. The plot is labeled A = A subscript 0 e to the power minus lambda t. Figure b shows a graph of ln A versus t. It starts at ln A subscript 0 and slopes downwards in a straight line. The slope of the line is labeled minus lambda t.], alt: "Figure a shows a graph of A versus t. It starts at point A subscript 0 and reduces with time. The rate of reduction decreases slowly till A is very close to 0, making a curved plot on the graph. The plot is labeled A = A subscript 0 e to the power minus lambda t. Figure b shows a graph of ln A versus t. It starts at ln A subscript 0 and slopes downwards in a straight line. The slope of the line is labeled minus lambda t.", caption: [(a) A plot of the activity as a function of time (b) If we measure the activity at different times, we can plot ln #emph[A] versus #emph[t], and obtain a straight line.]) #examplebox("Example 1")[Decay Constant and Activity of Strontium-90][ The half-life of strontium-90, #math.equation(block: false, alt: "38 90 Sr")[$38 90 "Sr"$], is 28.8 y. Find (a) its decay constant and (b) the initial activity of 1.00 g of the material. Strategy We can find the decay constant directly from . To determine the activity, we first need to find the number of nuclei present. Solution + The decay constant is found to be #math.equation(block: true, alt: "λ equals the fraction 0.693 over T sub 1 / 2 equals open parenthesis the fraction 0.693 over T sub 1 / 2 close parenthesis open parenthesis the fraction 1 yr over 3.16 times 10 to the power 7 s close parenthesis equals 7.61 times 10 to the power −10 s to the power − 1 .")[$λ = frac(0.693, T_(1 "/" 2)) = ( frac(0.693, T_(1 "/" 2)) ) ( frac(1 #h(0.2em) "yr", 3.16 #h(0.2em) × #h(0.2em) 10^(7) #h(0.2em) "s") ) = 7.61 #h(0.2em) × #h(0.2em) 10^(−10) #h(0.2em) "s"^("−" 1) .$] + The atomic mass of #math.equation(block: false, alt: "38 90 S r")[$38 90 "S" "r"$] is 89.91 g. Using Avogadro’s number #math.equation(block: false, alt: "N sub A equals 6.022 times 10 to the power 23")[$N_(A) = 6.022 #h(0.2em) × #h(0.2em) 10^(23)$] atoms/mol, we find the initial number of nuclei in 1.00 g of the material: #math.equation(block: true, alt: "N sub 0 equals the fraction 1.00 g over 89.91 g open parenthesis N sub A close parenthesis equals 6.70 times 10 to the power 21 nuclei .")[$N_(0) = frac(1.00 #h(0.2em) "g", 89.91 #h(0.2em) "g") ( N_(A) ) = 6.70 #h(0.2em) × #h(0.2em) 10^(21) #h(0.2em) "nuclei" .$] From this, we find that the activity #math.equation(block: false, alt: "A sub 0")[$A_(0)$] at #math.equation(block: false, alt: "t equals 0")[$t = 0$] for 1.00 g of strontium-90 is #math.equation(block: true, alt: "A sub 0, equals λ N sub 0; equals open parenthesis 7.61 times 10 to the power −10 s to the power −1 close parenthesis open parenthesis 6.70 times 10 to the power 21 nuclei close parenthesis; equals 5.10 times 10 to the power 12 decays/s .")[$A_(0) & = λ N_(0) \ & = ( 7.61 #h(0.2em) × #h(0.2em) 10^(−10) #h(0.2em) "s"^(−1) ) ( 6.70 #h(0.2em) × #h(0.2em) 10^(21) "nuclei" ) \ & = 5.10 #h(0.2em) × #h(0.2em) 10^(12) #h(0.2em) "decays/s" .$] ] Expressing #math.equation(block: false, alt: "λ")[$λ$] in terms of the half-life of the substance, we get #math.equation(block: true, alt: "A equals A sub 0 e to the power − open parenthesis 0.693 / T sub 1 / 2 close parenthesis T sub 1 / 2 equals A sub 0 e to the power −0.693 equals A sub 0 / 2 .")[$A = A_(0) e^("−" ( 0.693 "/" T_(1 "/" 2) ) T_(1 "/" 2)) = A_(0) e^(−0.693) = A_(0) "/" 2 .$] Therefore, the activity is halved after one half-life. We can determine the decay constant #math.equation(block: false, alt: "λ")[$λ$] by measuring the activity as a function of time. Taking the natural logarithm of the left and right sides of , we get #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "ln A equals − λ t plus ln A sub 0 .")[$"ln" #h(0.2em) A = "−" λ t + "ln" #h(0.2em) A_(0) .$] ] This equation follows the linear form #math.equation(block: false, alt: "y equals m x plus b")[$y = m x + b$]. If we plot ln #emph[A] versus #emph[t], we expect a straight line with slope #math.equation(block: false, alt: "− λ")[$"−" λ$] and #emph[y]-intercept #math.equation(block: false, alt: "ln A sub 0")[$"ln" #h(0.2em) A_(0)$] ((b)). Activity #emph[A] is expressed in units of #strong[becquerels] (Bq), where one #math.equation(block: false, alt: "1 Bq equals 1 decay per second")[$1 #h(0.2em) "Bq" = 1 #h(0.2em) "decay per second"$]. This quantity can also be expressed in 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")[$226 "Ra"$]. The relationship between the Bq and Ci is #math.equation(block: true, alt: "1 Ci equals 3.70 times 10 to the power 10 Bq .")[$1 #h(0.2em) "Ci" = 3.70 #h(0.2em) × #h(0.2em) 10^(10) "Bq" .$] #examplebox("Example 2")[What is #math.equation(block: false, alt: "to the power 14 C")[$14 "C"$] Activity in Living Tissue?][ Approximately #math.equation(block: false, alt: "20 %")[$20 "%"$] of the human body by mass is carbon. 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 The activity of #math.equation(block: false, alt: "to the power 14 C")[$14 "C"$] is determined using the equation #math.equation(block: false, alt: "A sub 0 equals λ N sub 0")[$A_(0) = λ N_(0)$], where #emph[λ] is the decay constant and #math.equation(block: false, alt: "N sub 0")[$N_(0)$] is the number of radioactive nuclei. The number of #math.equation(block: false, alt: "to the power 14 C")[$14 "C"$] nuclei in a 1.00-kg sample is determined in two steps. First, we determine the number of #math.equation(block: false, alt: "to the power 12 C")[$12 "C"$] nuclei using the concept of a mole. Second, we multiply this value by #math.equation(block: false, alt: "1.3 times 10 to the power −12")[$1.3 #h(0.2em) × #h(0.2em) 10^(−12)$] (the known abundance of #math.equation(block: false, alt: "to the power 14 C")[$14 "C"$] in a carbon sample from a living organism) to determine the number of #math.equation(block: false, alt: "to the power 14 C")[$14 "C"$] nuclei in a living organism. The decay constant is determined from the known half-life of #math.equation(block: false, alt: "to the power 14 C")[$14 "C"$] (available from ). 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")[$12 "C"$]. 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 open parenthesis 1000 g close parenthesis equals 5.02 times 10 to the power 25 .")[$N ( 12 "C" ) = frac(6.02 #h(0.2em) × #h(0.2em) 10^(23) #h(0.2em) "mol"^(−1), 12.0 #h(0.2em) "g/mol") #h(0.2em) × #h(0.2em) ( 1000 #h(0.2em) "g" ) = 5.02 #h(0.2em) × #h(0.2em) 10^(25) .$] The number of #math.equation(block: false, alt: "to the power 14 C")[$14 "C"$] nuclei in 1 kg of carbon is therefore #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 .")[$N ( 14 "C" ) = ( 5.02 #h(0.2em) × #h(0.2em) 10^(25) ) ( 1.3 #h(0.2em) × #h(0.2em) 10^(−12) ) = 6.52 #h(0.2em) × #h(0.2em) 10^(13) .$] Now we can find the activity #emph[A] by using the equation #math.equation(block: false, alt: "A equals the fraction 0.693 N over t sub 1 / 2 .")[$A = frac(0.693 #h(0.2em) N, t_(1 "/" 2)) .$] Entering known values gives us #math.equation(block: true, alt: "A equals the fraction 0.693 open parenthesis 6.52 times 10 to the power 13 close parenthesis over 5730 y equals 7.89 times 10 to the power 9 y to the power − 1")[$A = frac(0.693 ( 6.52 #h(0.2em) × #h(0.2em) 10^(13) ), 5730 #h(0.2em) "y") = 7.89 #h(0.2em) × #h(0.2em) 10^(9) #h(0.2em) "y"^("−" 1)$] or #math.equation(block: false, alt: "7.89 times 10 to the power 9")[$7.89 #h(0.2em) × #h(0.2em) 10^(9)$] decays per year. To convert this to the unit Bq, we simply convert years to seconds. Thus, #math.equation(block: true, alt: "A equals open parenthesis 7.89 times 10 to the power 9 y to the power − 1 close parenthesis the fraction 1.00 y over 3.16 times 10 to the power 7 s equals 250 Bq ,")[$A = ( 7.89 #h(0.2em) × #h(0.2em) 10^(9) #h(0.2em) "y"^("−" 1) ) frac(1.00 #h(0.2em) "y", 3.16 #h(0.2em) × #h(0.2em) 10^(7) #h(0.2em) "s") = 250 #h(0.2em) "Bq" ,$] or 250 decays per second. To express #emph[A] in curies, we use the definition of a curie, #math.equation(block: true, alt: "A equals the fraction 250 Bq over 3.7 times 10 to the power 10 Bq/Ci equals 6.76 times 10 to the power −9 Ci .")[$A = frac(250 #h(0.2em) "Bq", 3.7 #h(0.2em) × #h(0.2em) 10^(10) #h(0.2em) "Bq/Ci") #h(0.2em) = 6.76 #h(0.2em) × #h(0.2em) 10^(−9) #h(0.2em) "Ci" .$] Thus, #math.equation(block: true, alt: "A equals 6.76 nCi .")[$A = 6.76 #h(0.2em) "nCi" .$] Significance Approximately #math.equation(block: false, alt: "20 %")[$20 "%"$] of the human body by weight is carbon. Hundreds of #math.equation(block: false, alt: "to the power 14 C")[$14 "C"$] decays take place in the human body every second. Carbon-14 and other naturally occurring radioactive substances in the body compose a person’s background exposure to nuclear radiation. As we will see later in this chapter, this activity level is well below the maximum recommended dosages. ] === Radioactive Dating #strong[Radioactive dating] is a technique that uses naturally occurring radioactivity to determine the age of a material, such as a rock or an ancient artifact. The basic approach is to estimate the original number of nuclei in a material and the present number of nuclei in the material (after decay), and then use the known value of the decay constant #math.equation(block: false, alt: "λ")[$λ$] and to calculate the total time of the decay, #emph[t]. An important method of #strong[radioactive dating] is #strong[carbon-14 dating]. Carbon-14 nuclei are produced when high-energy solar radiation strikes #math.equation(block: false, alt: "to the power 14 N")[$14 "N"$] nuclei in the upper atmosphere and subsequently decay with a half-life of 5730 years. Radioactive carbon has the same chemistry as stable carbon, so it combines with the ecosphere and eventually becomes part of every living organism. Carbon-14 has an abundance of 1.3 parts per trillion of normal carbon. Therefore, if you know the number of carbon nuclei in an object, you multiply that number by #math.equation(block: false, alt: "1.3 times 10 to the power −12")[$1.3 #h(0.2em) × #h(0.2em) 10^(−12)$] to find the number of #math.equation(block: false, alt: "to the power 14 C")[$14 "C"$] nuclei in that 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 mummy’s age (or the time since the person’s death). Carbon-14 dating can be used for biological tissues as old as 50,000 years, but is generally 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. The validity of carbon dating can be checked by other means, such as by historical knowledge or by tree-ring counting. #examplebox("Example 3")[An Ancient Burial Cave][ In an ancient burial cave, your team of archaeologists discovers ancient wood furniture. Only #math.equation(block: false, alt: "80 %")[$80 "%"$] of the original #math.equation(block: false, alt: "to the power 14 C")[$14 "C"$] remains in the wood. How old is the furniture? Strategy The problem statement implies that #math.equation(block: false, alt: "N / N sub 0 equals 0.80 .")[$N "/" N_(0) = 0.80 .$] Therefore, the equation #math.equation(block: false, alt: "N equals N sub 0 e to the power − λ t")[$N = N_(0) e^("−" λ t)$] can be used to find the product, #math.equation(block: false, alt: "λ t")[$λ t$]. We know the half-life of #math.equation(block: false, alt: "to the power 14 C")[$14 "C"$] is 5730 y, so we also know the decay constant, and therefore the total decay time #emph[t]. Solution Solving the equation #math.equation(block: false, alt: "N equals N sub 0 e to the power − λ t")[$N = N_(0) e^("−" λ t)$] for #math.equation(block: false, alt: "N / N sub 0")[$N "/" N_(0)$] gives us #math.equation(block: true, alt: "the fraction N over N sub 0 equals e to the power − λ t .")[$frac(N, N_(0)) = e^("−" λ t) .$] Thus, #math.equation(block: true, alt: "0.80 equals e to the power − λ t .")[$0.80 = e^("−" λ t) .$] Taking the natural logarithm of both sides of the equation yields #math.equation(block: true, alt: "ln 0.80 equals minus λ t ,")[$"ln" #h(0.2em) 0.80 = − λ t ,$] so that #math.equation(block: true, alt: "−0.223 equals − λ t .")[$−0.223 = "−" λ t .$] Rearranging the equation to isolate #emph[t] gives us #math.equation(block: true, alt: "t equals the fraction 0.223 over λ ,")[$t = frac(0.223, λ) ,$] where #math.equation(block: true, alt: "λ equals the fraction 0.693 over t sub 1 / 2 equals the fraction 0.693 over 5730 y .")[$λ = frac(0.693, t_(1 "/" 2)) = frac(0.693, 5730 #h(0.2em) "y") .$] Combining this information yields #math.equation(block: true, alt: "t equals the fraction 0.223 over open parenthesis the fraction 0.693 over 5730 y close parenthesis equals 1844 y .")[$t = frac(0.223, ( frac(0.693, 5730 #h(0.2em) "y") )) = 1844 #h(0.2em) "y" .$] Significance The furniture is almost 2000 years old—an impressive discovery. The typical uncertainty on carbon-14 dating is about #math.equation(block: false, alt: "5 %")[$5 "%"$], so the furniture is anywhere between 1750 and 1950 years old. This date range must be confirmed by other evidence, such as historical records. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ A radioactive nuclide has a high decay rate. What does this mean for its half-life and activity? #solutionbox[ Half-life is inversely related to decay rate, so the half-life is short. Activity depends on both the number of decaying particles and the decay rate, so the activity can be great or small. ] ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Visit the #link("https://openstax.org/l/21raddatgame")[Radioactive Dating Game] to learn about the types of radiometric dating and try your hand at dating some ancient objects. ] === Summary - In the decay of a radioactive substance, if the decay constant (#math.equation(block: false, alt: "λ")[$λ$]) is large, the half-life is small, and vice versa. - The radioactive decay law, #math.equation(block: false, alt: "N equals N sub 0 e to the power − λ t ,")[$N = N_(0) e^("−" λ t) ,$] uses the properties of radioactive substances to estimate the age of a substance. - Radioactive carbon has the same chemistry as stable carbon, so it mixes into the ecosphere and eventually becomes part of every living organism. By comparing the abundance of #math.equation(block: false, alt: "to the power 14 C")[$14 "C"$] in an artifact with the normal abundance in living tissue, it is possible to determine the artifact’s age. === Conceptual Questions How is the initial activity rate of a radioactive substance related to its half-life? For the carbon dating described in this chapter, what important assumption is made about the time variation in the intensity of cosmic rays? #solutionbox[ That it is constant. ] === Problems A sample of radioactive material is obtained from a very old rock. A plot ln#emph[A] verses #emph[t] yields a slope value of #math.equation(block: false, alt: "− 10 to the power −9 s to the power −1")[$"−" 10^(−9) "s"^(−1)$] (see (b)). What is the half-life of this material? #solutionbox[ The decay constant is equal to the negative value of the slope or #math.equation(block: false, alt: "10 to the power −9 s to the power −1 .")[$10^(−9) "s"^(−1) .$] The half-life of the nuclei, and thus the material, is #math.equation(block: false, alt: "T sub 1 / 2 equals 693 million years .")[$T_(1 "/" 2) = 693 #h(0.2em) "million years" .$] ] Show that: #math.equation(block: false, alt: "T minus equals the fraction 1 over λ")[$accent(T, −) = frac(1, "λ")$]. The half-life of strontium-91, #math.equation(block: false, alt: "38 91 S r")[$38 91 "S" "r"$] is 9.70 h. Find (a) its decay constant and (b) for an initial 1.00-g sample, the activity after 15 hours. #solutionbox[ a. The decay constant is #math.equation(block: false, alt: "λ equals 1.99 times 10 to the power −5 s to the power − 1")[$λ = 1.99 #h(0.2em) × #h(0.2em) 10^(−5) #h(0.2em) "s"^("−" 1)$]. b. Since strontium-91 has an atomic mass of 90.90 g, the number of nuclei in a 1.00-g sample is initially #linebreak() #math.equation(block: false, alt: "N sub 0 equals 6.63 times 10 to the power 21 nuclei .")[$N_(0) = 6.63 #h(0.2em) × #h(0.2em) 10^(21) #h(0.2em) "nuclei" .$] #linebreak() The initial activity for strontium-91 is #linebreak() #math.equation(block: false, alt: "A sub 0, equals λ N sub 0; equals 1.32 times 10 to the power 17 decays/s")[$A_(0) & = λ N_(0) \ & = 1.32 #h(0.2em) × #h(0.2em) 10^(17) #h(0.2em) "decays/s"$] #linebreak() The activity at #math.equation(block: false, alt: "t equals 15.0 h equals 5.40 times 10 to the power 4 s")[$t = 15.0 #h(0.2em) "h" = 5.40 #h(0.2em) × #h(0.2em) 10^(4) #h(0.2em) "s"$] is #linebreak() #math.equation(block: false, alt: "A equals 4.51 times 10 to the power 16 decays/s .")[$A = 4.51 #h(0.2em) × #h(0.2em) 10^(16) #h(0.2em) "decays/s" .$] ] A sample of pure carbon-14 #math.equation(block: false, alt: "open parenthesis T sub 1 / 2 equals 5730 y close parenthesis")[$( T_(1 "/" 2) = 5730 #h(0.2em) "y" )$] has an activity of #math.equation(block: false, alt: "1.0 μ Ci .")[$1.0 #h(0.2em) μ #h(0.2em) "Ci" .$] What is the mass of the sample? A radioactive sample initially contains #math.equation(block: false, alt: "2.40 times 10 to the power − 2")[$2.40 #h(0.2em) × #h(0.2em) 10^("−" 2)$] mol of a radioactive material whose half-life is 6.00 h. How many moles of the radioactive material remain after 6.00 h? After 12.0 h? After 36.0 h? #solutionbox[ #math.equation(block: true, alt: "1.20 times 10 to the power − 2 mol")[$1.20 #h(0.2em) × #h(0.2em) 10^("−" 2) #h(0.2em) "mol"$]; #math.equation(block: true, alt: "6.00 times 10 to the power − 3 mol")[$6.00 #h(0.2em) × #h(0.2em) 10^("−" 3) "mol"$]; #math.equation(block: true, alt: "3.75 times 10 to the power − 4 mol")[$3.75 #h(0.2em) × #h(0.2em) 10^("−" 4) "mol"$] ] 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 .")[$2^(10) = 1024 .$] 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 R a .")[$226 "R" "a" .$] (b) Explain 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 R a")[$226 "R" "a"$] is more precisely known than it was when the Ci unit was established. ] Natural uranium consists of #math.equation(block: false, alt: "to the power 235 U")[$235 "U"$] #math.equation(block: false, alt: "open parenthesis percent abundance equals 0.7200 %")[$\( "percent abundance" = 0.7200 "%"$], #math.equation(block: false, alt: "λ equals 3.12 times 10 to the power −17 / s close parenthesis")[$λ = 3.12 #h(0.2em) × #h(0.2em) 10^(−17) "/" "s" \)$] and #math.equation(block: false, alt: "to the power 238 U")[$238 "U"$] #math.equation(block: false, alt: "open parenthesis percent abundance equals 99.27 %")[$\( "percent abundance" = 99.27 "%"$], #math.equation(block: false, alt: "λ equals 4.92 times 10 to the power −18 / s close parenthesis .")[$λ = 4.92 #h(0.2em) × #h(0.2em) 10^(−18) "/" "s" \) .$] What were the values for percent abundance of #math.equation(block: false, alt: "to the power 235 U")[$235 "U"$] and #math.equation(block: false, alt: "to the power 238 U")[$238 "U"$] when Earth formed #math.equation(block: false, alt: "4.5 times 10 to the power 9")[$4.5 #h(0.2em) × #h(0.2em) 10^(9)$] years ago? 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")[$1.0 #h(0.2em) × #h(0.2em) 10^(5)$] Bq when new. (a) What mass of #math.equation(block: false, alt: "to the power 226 R a")[$226 "R" "a"$] 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? #solutionbox[ a. #math.equation(block: false, alt: "2.73 μ g")[$2.73 μ "g"$]; b. #math.equation(block: false, alt: "9.76 times 10 to the power 4 Bq")[$9.76 #h(0.2em) × #h(0.2em) 10^(4) #h(0.2em) "Bq"$] ] The #math.equation(block: false, alt: "to the power 210 P o")[$210 "P" "o"$] source used in a physics laboratory is labeled as having an activity of #math.equation(block: false, alt: "1.0 μCi")[$1.0 #h(0.2em) "μCi"$] 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? 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 nonradioactive. 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. #solutionbox[ a. #math.equation(block: false, alt: "7.46 times 10 to the power 5 Bq")[$7.46 #h(0.2em) × #h(0.2em) 10^(5) #h(0.2em) "Bq"$]; b. #math.equation(block: false, alt: "7.75 times 10 to the power 5 Bq")[$7.75 #h(0.2em) × #h(0.2em) 10^(5) #h(0.2em) "Bq"$] ]