The given information is:
- The function of decay is:
[tex]A(t)=A_0e^{-0.0244}[/tex]Where A0 is the initial amount of strontium-90, A is the amount present at time t (in years).
- The initial amount is 800 grams.
a. What is the decay rate of strontium-90?
The given formula is written in the form:
[tex]A(t)=A_0e^{rt}[/tex]Where r is the decay rate in decimal form, so:
[tex]r=-0.0244*100\%=-2.44\%[/tex]The decay rate is -2.44%.
b. How much strontium-90 is left after 30 years?
Replace t=30 and solve:
[tex]\begin{gathered} A(30)=800g*e^{-0.0244*30} \\ A(30)=800g*e^{-0.732} \\ A(30)=800g*0.481 \\ A(30)=384.8g \end{gathered}[/tex]There is 384.8 grams after 30 years.
c. When will only 200 g of strontium-90 be left?
A(t)=200g, then replace it and solve for t:
[tex]\begin{gathered} 200g=800g*e^{-0.0244t} \\ \frac{200g}{800g}=e^{-0.0244t} \\ \ln(\frac{200g}{800g})=\ln e^{-0.0244t} \\ -1.386=-0.0244t \\ t=\frac{-1.386}{-0.0244} \\ t=56.8 \end{gathered}[/tex]There will be 200 g left after 56.8 years.
d. What is the half-life of strontium-90?
The half-life is when A(t)=A0/2, then if we replace this into the decay formula we obtain:
[tex]\begin{gathered} \frac{A_0}{2}=A_0*e^{-0.0244t} \\ Simplify\text{ A0 on both sides} \\ \frac{1}{2}=e^{-0.0244t} \\ \ln(0.5)=\ln e^{-0.0244t} \\ -0.693=-0.0244t \\ t=\frac{-0.693}{-0.0244} \\ t=28.4 \end{gathered}[/tex]The half-life of strontium-90 is 28.4 years.
