Question

A sample has a C614 activity of 0.0015 Bq per gram of carbon.
(a) Find the age of the sample, assuming that the activity per gram of carbon in a living organism has been constant at a value of 0.23 Bq.
(b) Evidence suggests that the value of 0.23 Bq might have been as much as 45% larger. Repeat part (a), taking into account this 45% increase.

Answer 1

Answer:

$\large \boxed{\text{(a) 42 000 yr;  (b) 45 000 yr}}$  

Explanation:

Two important equations in radioactive decay are

$\ln \dfrac{N_{0} }{N_{t}} = kt\\\\t_{\frac{1}{2}} = \dfrac{\ln2}{k }$

We use them for carbon dating.

(a) Initial activity = 0.23 Bq

(i) Calculate the rate constant

The half-life of ¹⁴C is 5730 yr.

$\begin{array}{rcl}t_{\frac{1}{2}}& = &\dfrac{\ln2}{k }\\\\k& = &\dfrac{\ln2}{t_{\frac{1}{2}}}\\\\ & = & \dfrac{\ln2}{\text{5730 yr}}\\\\ & = & 1.210 \times 10^{-4}\text{ yr}^{-1}\\\end{array}$

(ii) Calculate the age of the sample

$\begin{array}{rcl}\ln \dfrac{N_{0} }{N_{t}} & = & kt\\\\\ln \dfrac{0.23 }{0.0015} & = & k\times 1.210 \times 10^{-4}\text{ yr}^{-1}\\\\\ln 153 & = &  1.210 \times 10^{-4}k \text{ yr}^{-1}\\5.03 & = & 1.210 \times 10^{-4}k \text{ yr}^{-1}\\k & = & \dfrac{5.03}{1.210 \times 10^{-4} \text{ yr}^{-1}}\\\\ & = & \textbf{42 000 yr}\\\end{array}\\\text{The age of the sample is \large \boxed{\textbf{42 000 yr}} }$

(b) Initial activity =  45 % larger

N₀ = 1.45 × 0.230 Bq = 0.334 Bq

$\begin{array}{rcl}\ln \dfrac{N_{0} }{N_{t}} & = & kt\\\\\ln \dfrac{0.334 }{0.0015} & = & k\times 1.210 \times 10^{-4}\text{ yr}^{-1}\\\\\ln 222 & = &  1.210 \times 10^{-4}k \text{ yr}^{-1}\\5.40 & = & 1.210 \times 10^{-4}k \text{ yr}^{-1}\\k & = & \dfrac{5.40}{1.210 \times 10^{-4} \text{ yr}^{-1}}\\\\ & = & \textbf{45 000 yr}\\\end{array}\\\text{The age of the sample is \large \boxed{\textbf{45 000 yr}} }$