Question 30.4: Radiocarbon dating In this example we will apply the radioac...

SET UP We know that the half-life of \mathrm{^{14}C  is  T_{1/2}=5730  y} We convert years to seconds and then find the decay constant l from Equation 30.7, \mathrm{T_{1/2}=0.693/\lambda.} The number of decays per second in the atmosphere is 0.255, and we can find the total number of \mathrm{^{14}C} atoms from Equation 30.5:

\mathrm{T_{1/2}=\frac{\ln}{\lambda } =\frac{0.693}{\lambda } }               (30.7)

\mathrm{\frac{\Delta N}{\Delta t} =-\lambda N.}

Finally, we compare this N with the total number of all carbon atoms, using the fact that 1 gram of the common isotope \mathrm{^{12}C  is  \frac{1}{12}  mol} . For the second part of the problem we know from the notes in Equation 30.5 that the activity is equal to Nλ. Therefore, if we multiply both sides of Equation 30.6 by l, then we obtain an equation that gives us the activity as a function of time, \mathrm{N\lambda =(N_0\lambda )e^{-\lambda t}} .

SOLVE First, the half-life is

\mathrm{T_{1/2}=(5730  y)(3.156 \times 10^7  s/y) = 1.81 \times 10^{11}  s.}

The decay constant is \mathrm{\lambda = 0.693/T_{1/2} = 3.83 \times 10^{-12}  s^{-1}} From Equation 30.5 we note that the activity of the atmospheric carbon is simply λN. Therefore, we know that Nλ = 0.255 Bq, and solving for N gives

\mathrm{N=\frac{0.255}{\lambda }=6.66 \times 10^{10}. }

The total number of C atoms in \mathrm{1 g (= 1/12  mol)  is  (1/12  mol) × (6.023 × 10^{23}  atoms/mol),  or  5.02 × 10^{22}.} The ratio of ^{14}C atoms to all carbon atoms is

\mathrm{\frac{6.66\times 10^{10}}{5.02\times 10^{22}}=1.33 \times 10^{-12}. }

The activity of the archeological specimen, per unit mass, is one-fourth the activity of ^{14}C in the present atmosphere. Thus, the ^{14}C in the specimen has been decaying for a time equal to twice the half-life, or about 11,400 y.
Of course, we can use the fact that the activity decays exponentially to calculate the age of the specimen directly. We begin with

\mathrm{N\lambda =(N_0\lambda )e^{-\lambda t}},

where Nλ = 0.0637 Bq is the final activity and N_0λ = 0.255 Bq is the intial activity associated with the carbon that was ingested by the specimen when it was alive. Since we know λ, it is an easy matter to calculate t. First we isolate the exponential function,

\mathrm{\frac{0.0637}{0.255 }=e^{-\lambda t} ,}

and then take the natural log of both sides:

\mathrm{\ln\left(\frac{0.0637}{0.255} \right) =-\lambda t,}

\mathrm{t=\frac{\ln(0.250)}{-3.83 \times 10^{-12}  s^{-1}} = 3.62 \times 10^{11}  s =11,470  y.}

REFLECT Only about one carbon atom in a trillion in the earth’s atmosphere is \mathrm{^{14}C} , yet such small concentrations can be used to date archeological specimens with surprising precision.

Practice Problem: An archeological specimen with carbon mass 0.25 g has an activity of 8.0 \times 10^{-3}  Bq  due  to  ^{14}C decays. Find the approximate age of the specimen. Answer: 17,000 y.