IChO 1984 · Theory problem 1

Radioactive dating with 14C and 87Rb/87Sr

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A)

The element carbon consists of the stable isotopes \(\ce{^{12}C}\) (98.90 percent of atoms) and \(\ce{^{13}C}\) (1.10 percent of atoms). In addition, carbon contains a small fraction of the radioisotope \(\ce{^{14}C}\) (\(t_{1/2}=5730\) years), which is continuously formed in the atmosphere by cosmic rays as \(\ce{CO2}\). \(\ce{^{14}C}\) mixes with the isotopes \(\ce{^{12}C}\) and \(\ce{^{13}C}\) via the natural \(\ce{CO2}\) cycle.

The decay rate of \(\ce{^{14}C}\) is described by (\(N\) = number of \(\ce{^{14}C}\) atoms; \(t\) = time; \(\lambda\) = decay constant):

\[ \text{decay rate}=-\frac{dN}{dt}=\lambda N \tag{1} \]

Integration of (1) leads to the well-known rate law (2) for the radioactive decay:

\[ N=N_0\,\mathrm{e}^{-\lambda t} \tag{2} \]

\(N_0\) = number of \(\ce{^{14}C}\) atoms at \(t=0\)

1.1 What is the mathematical relationship between the parameters \(\alpha\) and \(t_{1/2}\) (= half l life)?

1.2 The decay rate of carbon, which is a part of the natural \(\ce{CO2}\) cycle, is found to be 13.6 disintegrations per minute and gram of carbon. When a plant (e. g. a tree) dies, it no longer takes part in the \(\ce{CO2}\) cycle. As a consequence, the decay rate of carbon decreases.

In 1983, a decay rate of 12.0 disintegrations per minute and gram of carbon was measured for a piece of wood which belongs to a ship of the Vikings. In which year was cut the tree from which this piece of wood originated?

1.3 Assume that the error of the decay rate of 12.0 disintegrations per minute and gram of carbon is 0.2 disintegrations per minute and gram of carbon. What is the corresponding error in the age of the wood in question b)?

1.4 What is the isotope \(\ce{^{12}C}/\ce{^{14}C}\) ratio of carbon, which takes part in the natural \(\ce{CO2}\) cycle (1 year = 365 days)?

B)

The elements strontium and rubidium have the following isotope composition:

Strontium: 0.56 % \(\ce{^{84}Sr}\) ; 9.86 % \(\ce{^{86}Sr}\) ; 7.00 % \(\ce{^{87}Sr}\) ; 82.58 % \(\ce{^{88}Sr}\) (these isotopes are all stable).

Rubidium: 72.17 % \(\ce{^{85}Rb}\) (stable) ; 27.83 % \(\ce{^{87}Rb}\) (radioactive; \(t_{1/2}=4.7\times10^{10}\) years).

The radioactive decay of \(\ce{^{87}Rb}\) leads to \(\ce{^{87}Sr}\).

In Greenland one finds a gneiss (= silicate mineral) containing both strontium and rubidium.

1.5 What is the equation rate law describing the formation of \(\ce{^{87}Sr}\) from \(\ce{^{87}Rb}\) as a function of time?

1.6 Assume that the isotope ratio \(\ce{^{87}Sr}/\ce{^{86}Sr}\) (as determined by mass spectrometry) and the isotope ratio \(\ce{^{87}Rb}:\ce{^{86}Sr}\) are known for the gneiss. What is the mathematical relationship with which one can calculate the age of the gneiss?

Official English text of the problem.

Answers

  • 1.1 Interpreting \(\alpha\) as the decay constant \(\lambda\), \[ \boxed{\alpha=\lambda=\frac{\ln 2}{t_{1/2}}.} \]

  • 1.2 The wood was approximately \(1.03\times10^3\) years old in 1983. The tree was cut around 948 CE.

  • 1.3 The age uncertainty is approximately \[ \boxed{\Delta t\approx\pm1.4\times10^2\ \mathrm{years}.} \]

  • 1.4 The isotope number ratio is \[ \boxed{\frac{N(\ce{^{12}C})}{N(\ce{^{14}C})}\approx8.39\times10^{11}.} \]

  • 1.5 Let \(P(t)\) and \(D(t)\) be the numbers of \(\ce{^{87}Rb}\) and \(\ce{^{87}Sr}\) atoms, respectively. Then \[ \boxed{\frac{dD}{dt}=\lambda_{\mathrm{Rb}}P_0\mathrm{e}^{-\lambda_{\mathrm{Rb}}t}, \qquad D(t)=D_0+P_0\left(1-\mathrm{e}^{-\lambda_{\mathrm{Rb}}t}\right),} \] where \(\lambda_{\mathrm{Rb}}=\ln2/(4.7\times10^{10}\ \mathrm{yr})\).

  • 1.6 Define the present isotope number ratios \[ R=\left(\frac{\ce{^{87}Sr}}{\ce{^{86}Sr}}\right)_t, \qquad Q=\left(\frac{\ce{^{87}Rb}}{\ce{^{86}Sr}}\right)_t. \] For a closed system with initial strontium ratio \(R_0\), \[ \boxed{t=\frac{4.7\times10^{10}\ \mathrm{yr}}{\ln2} \ln\!\left(1+\frac{R-R_0}{Q}\right).} \] Taking the listed strontium composition as the initial composition gives \(R_0=7.00/9.86\approx0.710\).

Solution

1.1

At one half-life, half the original radioactive atoms remain:

\[ \frac{N(t_{1/2})}{N_0}=\frac12 =\mathrm{e}^{-\lambda t_{1/2}}. \]

Taking natural logarithms gives

\[ -\ln2=-\lambda t_{1/2}, \]

and therefore

\[ \boxed{\lambda=\frac{\ln2}{t_{1/2}}.} \]

For carbon-14,

\[ \lambda_{\mathrm C} =\frac{\ln2}{5730\ \mathrm{yr}} =1.20968\times10^{-4}\ \mathrm{yr^{-1}}. \]

1.2

The activity is proportional to the number of carbon-14 atoms:

\[ A=\lambda_{\mathrm C}N. \]

Consequently, the specific activity of the wood follows the same exponential law:

\[ A(t)=A_0\mathrm{e}^{-\lambda_{\mathrm C}t}. \]

Taking \(A_0=13.6\ \mathrm{min^{-1}\,g^{-1}}\) at the time the tree was cut and \(A=12.0\ \mathrm{min^{-1}\,g^{-1}}\) in 1983,

\[ t=\frac{1}{\lambda_{\mathrm C}}\ln\frac{A_0}{A} =\frac{5730\ \mathrm{yr}}{\ln2} \ln\frac{13.6}{12.0} =1034.68\ \mathrm{yr}. \]

The corresponding calendar year is

\[ 1983-1034.68=948.32. \]

Thus, within the dating model specified in the problem, the tree was cut around

\[ \boxed{948\ \text{CE}.} \]

1.3

Treating the initial activity and half-life as fixed, differentiate the age formula with respect to the measured activity:

\[ t(A)=\frac{1}{\lambda_{\mathrm C}}\ln\frac{A_0}{A}, \qquad \frac{dt}{dA}=-\frac{1}{\lambda_{\mathrm C}A}. \]

The magnitude of the first-order age uncertainty is therefore

\[ \Delta t \approx\left|\frac{dt}{dA}\right|\Delta A =\frac{\Delta A}{\lambda_{\mathrm C}A}. \]

Substitution gives

\[ \Delta t =\frac{5730\ \mathrm{yr}}{\ln2} \frac{0.2\ \mathrm{min^{-1}\,g^{-1}}} {12.0\ \mathrm{min^{-1}\,g^{-1}}} =137.78\ \mathrm{yr}. \]

Hence

\[ \boxed{t\approx(1.03\pm0.14)\times10^3\ \mathrm{yr}.} \]

The exact endpoint calculation gives

\[ t(12.2)=898.04\ \mathrm{yr}, \qquad t(11.8)=1173.62\ \mathrm{yr}, \]

or \(1034.68^{+138.94}_{-136.64}\ \mathrm{yr}\), consistent with the rounded uncertainty of approximately \(\pm140\) years.

1.4

Consider \(1.000\ \mathrm g\) of carbon participating in the natural cycle. Convert the carbon-14 decay constant to inverse minutes:

\[ \lambda_{\mathrm C} =\frac{\ln2} {5730\times365\times24\times60\ \mathrm{min}} =2.30152\times10^{-10}\ \mathrm{min^{-1}}. \]

Since its activity is \(13.6\ \mathrm{min^{-1}}\), the number of carbon-14 atoms is

\[ N(\ce{^{14}C}) =\frac{A}{\lambda_{\mathrm C}} =\frac{13.6\ \mathrm{min^{-1}}} {2.30152\times10^{-10}\ \mathrm{min^{-1}}} =5.90913\times10^{10}. \]

Neglecting the minute carbon-14 contribution and using the isotope mass numbers to approximate their molar masses, the mean carbon molar mass is

\[ M_{\mathrm C} =\left(0.9890\times12+0.0110\times13\right) \mathrm{g\,mol^{-1}} =12.011\ \mathrm{g\,mol^{-1}}. \]

Using the standard Avogadro constant, \(N_{\mathrm A}\approx6.022\times10^{23}\ \mathrm{mol^{-1}}\),

\[ N(\ce{^{12}C}) =0.9890\, \frac{1.000\ \mathrm g}{12.011\ \mathrm{g\,mol^{-1}}} N_{\mathrm A} \approx4.959\times10^{22}. \]

Thus the required atom-number ratio is

\[ \boxed{ \frac{N(\ce{^{12}C})}{N(\ce{^{14}C})} =\frac{4.959\times10^{22}}{5.90913\times10^{10}} \approx8.39\times10^{11}. } \]

1.5

Rubidium-87 undergoes beta-minus decay to strontium-87:

\[ \ce{^{87}_{37}Rb -> ^{87}_{38}Sr + ^{0}_{-1}e} +\bar{\nu}_e, \]

where \(\bar{\nu}_e\) is an electron antineutrino. The mass number remains 87, and charge is conserved because \(37=38-1\).

Let \(P(t)\) denote the number of rubidium-87 atoms and \(D(t)\) the number of strontium-87 atoms. Each parent decay produces one stable daughter atom. In a closed system,

\[ \frac{dP}{dt}=-\lambda_{\mathrm{Rb}}P, \qquad \frac{dD}{dt}=\lambda_{\mathrm{Rb}}P, \qquad \lambda_{\mathrm{Rb}} =\frac{\ln2}{4.7\times10^{10}\ \mathrm{yr}}. \]

Since \(P(t)=P_0\mathrm{e}^{-\lambda_{\mathrm{Rb}}t}\),

\[ \frac{dD}{dt} =\lambda_{\mathrm{Rb}}P_0\mathrm{e}^{-\lambda_{\mathrm{Rb}}t}. \]

Integrating from time zero to time \(t\) gives

\[ D(t)-D_0 =\int_0^t \lambda_{\mathrm{Rb}}P_0\mathrm{e}^{-\lambda_{\mathrm{Rb}}\tau}\,d\tau =P_0\left(1-\mathrm{e}^{-\lambda_{\mathrm{Rb}}t}\right). \]

Therefore,

\[ \boxed{D(t)=D_0+P_0\left(1-\mathrm{e}^{-\lambda_{\mathrm{Rb}}t}\right).} \]

Here \(D_0\) includes the strontium-87 already present when the system formed.

1.6

The initial number of rubidium-87 atoms can be expressed using its present number:

\[ P_0=P(t)\mathrm{e}^{\lambda_{\mathrm{Rb}}t}. \]

Substituting this into the daughter-growth equation yields

\[ D(t)=D_0+P(t)\left(\mathrm{e}^{\lambda_{\mathrm{Rb}}t}-1\right). \]

Strontium-86 is stable and is neither produced nor consumed by this decay. Its number, \(S=N(\ce{^{86}Sr})\), therefore remains constant in a closed system. Dividing by \(S\) gives

\[ \frac{D(t)}{S} =\frac{D_0}{S} +\frac{P(t)}{S}\left(\mathrm{e}^{\lambda_{\mathrm{Rb}}t}-1\right). \]

In terms of isotope ratios,

\[ \boxed{R=R_0+Q\left(\mathrm{e}^{\lambda_{\mathrm{Rb}}t}-1\right).} \]

Rearranging,

\[ \mathrm{e}^{\lambda_{\mathrm{Rb}}t} =1+\frac{R-R_0}{Q}, \]

so

\[ \boxed{ t=\frac{1}{\lambda_{\mathrm{Rb}}} \ln\!\left(1+\frac{R-R_0}{Q}\right). } \]

Taking the supplied strontium isotope composition to represent the initial composition,

\[ R_0=\frac{7.00}{9.86}=0.709939\ldots, \]

and the requested relationship becomes

\[ \boxed{ t=\frac{4.7\times10^{10}\ \mathrm{yr}}{\ln2} \ln\!\left[ 1+ \frac{ \left(\ce{^{87}Sr}/\ce{^{86}Sr}\right)_t-7.00/9.86 }{ \left(\ce{^{87}Rb}/\ce{^{86}Sr}\right)_t } \right]. } \]

This age represents the elapsed time since the mineral system became closed to rubidium and strontium exchange.

The solution was checked against the official solution.