Question

Radioactive carbon-14 has a half life of 5730 years. Suppose a peice of wood has a decay rate of 15 disintegrations per minute. How many years would it take for the rate to decrease to 4 disintegrations per minute?

Answer 1

Answer:

$\large \boxed{\text{2920 yr}}$

Explanation:

Two important formulas in radioactive decay are

$(1) \qquad t_{\frac{1}{2}} = \dfrac{\ln 2}{k}\\\\(2) \qquad \ln \left(\dfrac{N_{0}}{N}\right) = kt$

1. Calculate the decay constant k

$\begin{array}{rcl}t_{\frac{1}{2}} &=& \dfrac{\ln 2}{k}\\\\\text{1530 yr} &= &\dfrac{\ln 2}{k}\\\\k & = & \dfrac{\ln 2}{\text{1530 yr}}\\\\& = & 4.530 \times 10^{-4} \text{ yr}^{-1}\\\end{array}$

2. Calculate the time

$\begin{array}{rcl}\ln \left(\dfrac{N_{0}}{N}\right) &= &kt \\\\\ln \left(\dfrac{15}{4}\right) &= &4.530 \times 10^{-4} \text{ yr}^{-1}\times t \\\\\ln 3.75 &= &4.530 \times 10^{-4} \text{ yr}^{-1}\times t \\t &= &\dfrac{\ln 3.75}{4.530 \times 10^{-4} \text{ yr}^{-1}}\\\\& = & \textbf{2920 yr}\\\end{array}\\\text{It would take \large \boxed{\textbf{2920 yr}} for the rate to decrease to 4 dis/min.}$