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

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

Step-by-step explanation:

Two important formulas in radioactive decay are

`(1) \qquad t_{(1)/(2)} = (\ln 2)/(k)\\\\(2) \qquad \ln \left((N_(0))/(N)\right) = kt`

1. Calculate the decay constant k

`\begin{array}{rcl}t_{(1)/(2)} &=& (\ln 2)/(k)\\\\\text{1530 yr} &= &(\ln 2)/(k)\\\\k & = & \frac{\ln 2}{\text{1530 yr}}\\\\& = & 4.530 * 10^(-4) \text{ yr}^(-1)\\\end{array}`

2. Calculate the time

`\begin{array}{rcl}\ln \left((N_(0))/(N)\right) &= &kt \\\\\ln \left((15)/(4)\right) &= &4.530 * 10^(-4) \text{ yr}^(-1)* t \\\\\ln 3.75 &= &4.530 * 10^(-4) \text{ yr}^(-1)* t \\t &= &\frac{\ln 3.75}{4.530 * 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.}`