Actinium-227 is a radioactive substance that decays according to the following function, where $y_{0}$ is the initial amount present, and $y$ is the amount present at time $t$ (in years).
\[
y=y_{0} e^{-0.032 t}
\]
Find the half-life of Actinium-227. Do not round any intermediate computations, and round your answer to the nearest tenth.
years

Step 1 ：Given that Actinium-227 is a radioactive substance that decays according to the function \(y=y_{0} e^{-0.032 t}\), where \(y_{0}\) is the initial amount present, and \(y\) is the amount present at time \(t\) (in years).

Step 2 ：We are asked to find the half-life of Actinium-227. The half-life of a radioactive substance is the time it takes for half of the substance to decay. In other words, it's the time \(t\) at which \(y = y_0 / 2\).

Step 3 ：We can set up the equation \(y_0 / 2 = y_0 e^{-0.032 t}\) and solve for \(t\).

Step 4 ：Let's assume \(y_0 = 1\) and \(y = 0.5\).

Step 5 ：Solving the equation gives \(t = 21.66084939249829\).

Step 6 ：Rounding to the nearest tenth, we get \(t = 21.7\).

Step 7 ：Final Answer: The half-life of Actinium-227 is approximately \(\boxed{21.7}\) years.