# Applications of Exponentials ### Advanced Level #### Anna Smith #### 10/01/2020 --- Question 5: Cesium-137 decays exponentially with a half-life of $30.17$ year. After a nuclear disaster, measurments in fish found amounts equaling $740,000$ becquerels/kilogram. The government safe limit is $100$ becquerels/kilogram. Given the Cesium-137 half-life and exponential decay, in how many years should we expect to reach that safe limit? --- So the half-life is $30.17$ years Half-life$=\frac{ln2}{k}$ $\rightarrow$ $(k)(30.17)=ln2$ $\rightarrow$ $k=\frac{ln2}{30.17}$ Now that we have found $k$ we can plug it into the equation $P(t)=P_0e^{-kt}$ --- $P(t)=P_0e^{-kt}$ $k$ is negative since it is exponentially decaying. Plug $k$ into the equation with $P(t)=100$ and $P_0=740,000$ $100=740000e^{-{\frac{ln2}{30.17}t}}$ --- Solve for $t$ (time)... $100=740000e^{-{\frac{ln2}{30.17}t}}$ $\frac{100}{740000}=e^{-{\frac{ln2}{30.17}t}}$ Multiply each side by $ln$ $(\frac{100}{740000}=e^{-{\frac{ln2}{30.17}t}})ln$ $\rightarrow$ $ln(\frac{100}{740000})=\frac{-ln2}{30.17}(t)$ $\rightarrow$ $t=(ln\frac{100}{740000})(\frac{30.17}{-ln2})$ $t= 387.8$ years.