Question

Iron-59 has a half-life of 44.5 days. How much of a 3.00 mg sample will be left after 222.5 days?

Answer 1

Using exponential decay formula:

`N=N_o((1)/(2))^{(t)/(\tau)}`

Where:

`\begin{gathered} N_o=3mg \\ t=222.5_{\text{ }}days \\ \tau=44.5_{\text{ }}days \end{gathered}`

Therefore:

`\begin{gathered} N=3((1)/(2))^{(222.5)/(44.5)} \\ N\approx0.09mg \end{gathered}`

Answer:

About 0.09 miligrams