We have a mass that follows the given exponential model:
[tex]A(t)=500e^{-0.032t}[/tex]
a) We have to calculate how much mass is present after 4 years. This correspond to the value of A when t = 4, as t is expressed in years.
We can replace t with 4 and calculate A(4) as:
[tex]\begin{gathered} A(4)=500e^{-0.032(4)} \\ A(4)=500e^{-0.128} \\ A(4)\approx500\cdot0.8799 \\ A(4)\approx440 \end{gathered}[/tex]
b) We have to calculate the remaining mass after 6 years. This is similar to the previous point, but with t = 6 years.
We can calculate A(6) as:
[tex]\begin{gathered} A(6)=500e^{-0.032(6)} \\ A(6)=500e^{-0.192} \\ A(6)\approx500\cdot0.8253 \\ A(6)\approx413 \end{gathered}[/tex]
c) We have to calaculate the mass after 15 years (t = 15):
[tex]\begin{gathered} A(15)=500\cdot e^{-0.032(15)} \\ A(15)=500\cdot e^{-0.48} \\ A(15)\approx500\cdot0.6188 \\ A(15)\approx309 \end{gathered}[/tex]
d) We have to calculate the half-life for this element.
This represents the interval of time t for which the mass is halved. This value is constant for an exponential decay model and we can find this value t as:
[tex]\begin{gathered} \frac{A(x+t)}{A(x)}=0.5 \\ \\ \frac{500\cdot e^{-0.032(x+t)}}{500\cdot e^{-0.032x}}=0.5 \\ \\ e^{-0.032t}=0.5 \\ \\ \ln(e^{-0.032t})=\ln(0.5) \\ \\ -0.032t=\ln(0.5) \\ \\ t=-\frac{\ln(0.5)}{0.032} \\ \\ t\approx21.66 \end{gathered}[/tex]
Answer:
a) 440 g
b) 413 g
c) 309 g
d) 21.66 years
