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 2 of 5: Age of the wood

← Previous Next →

Intuition

Activity is proportional to N, so the activity ratio directly gives the elapsed decay time.

t = \dfrac{t_{1/2}}{\ln 2}\ln\dfrac{A_0}{A} = \dfrac{5730}{0.693}\ln\dfrac{13.6}{12.0} = 1035\ \text{yr}\;\Rightarrow\;\text{cut in } 1983 - 1035 \approx 948

Analysis

Since $A = \lambda N$ , the ratio $A_0/A = 13.6/12.0$ equals $N_0/N = e^{\lambda t}$ ; solving gives $t = (t_{1/2}/\ln 2)\ln(13.6/12.0) = 1035$ years, so the tree was felled around $1983 - 1035 = 948$ AD.