Problem 1

The element carbon consists of the stable isotopes

\ce{^{12}C}

(98.90 % of atoms) and

\ce{^{13}C}

(1.10 % 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}

and mixes with the stable isotopes via the natural

\ce{CO2}

cycle. The decay rate of

\ce{^{14}C}

is described by

-\mathrm{d}N/\mathrm{d}t = \lambda N

(N = number of

\ce{^{14}C}

atoms), whose integration gives

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

. 1.1 What is the mathematical relationship between

\lambda

and

t_{1/2}

? 1.2 The decay rate of carbon taking part in the natural

\ce{CO2}

cycle is 13.6 disintegrations per minute per gram of carbon. When a plant dies it leaves the

\ce{CO2}

cycle and its decay rate decreases. In 1983 a decay rate of 12.0 disintegrations per minute per gram was measured for a piece of wood from a Viking ship. In which year was the tree cut? 1.3 The error of the measured rate is 0.2 disintegrations per minute per gram. What is the corresponding error in the age? 1.4 What is the

\ce{^{12}C}/\ce{^{14}C}

isotope ratio of carbon taking part in the natural

\ce{CO2}

cycle (1 year = 365 days)?

Step 4 of 5: 14C atoms per gram

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N(\ce{^{14}C}) = \dfrac{A}{\lambda} = \dfrac{13.6 \times 365 \times 24 \times 60 \times 5730}{0.693} = 5.91\times 10^{10}\ \text{atoms g}^{-1}

Analysis

Steady-state living carbon has $A = \lambda N$ , so $N(\ce{^{14}C}) = A/\lambda$ with $A$ converted to disintegrations per year: $13.6\ \text{min}^{-1} \times 525600\ \text{min yr}^{-1} = 7.148\times 10^{6}$ yr $^{-1}$ , divided by $\lambda = 1.21\times 10^{-4}$ yr $^{-1}$ gives $5.91\times 10^{10}$ atoms per gram of carbon.