Question

The number of grams A of a certain radioactive substance present at time, in years from the present, this given by the formula A = 45e^-0.0045tb. What is half-life of this substance?c. How much will be around in 2500 years?

Answer 1

b.

To determine the half life we plug half the original amount in the expression given and solve for t:

`\begin{gathered} (45)/(2)=45e^(-0.0045t) \\ e^(-0.0045t)=(45)/(2\cdot45) \\ e^(-0.0045t)=(1)/(2) \\ \ln e^(-0.0045t)=\ln ((1)/(2)) \\ -0.0045t=\ln ((1)/(2)) \\ t=(1)/(-0.0045)\ln ((1)/(2)) \\ t=154.03 \end{gathered}`

Therefore the half-life of the substance is approximately 154 years.

c.

To determine how much of the substance will be after 2500 years we just plug this value in the expression:

`\begin{gathered} A=45e^(-0.0045\cdot2500) \\ A=5.85*10^(-4) \end{gathered}`

Therefore after 2500 years there will be:

`5.85*10^(-4)\text{ gr}`

of the substance.