Question

a sample of a radioactive substance has an initial mass of 339.1 mg. this substance follows a continuous exponential decay model and has a half - life of 16 minutes. (a) let t be the time (in minutes) since the start of the experiment, and let y be the amount of the substance at time t. write a formula relating y to t. use exact expressions to fill in the missing parts of the formula. do not use approximations. y = e^( t) (b) how much will be present in 13 minutes? do not round any intermediate computations, and round your answer to the nearest tenth. mg

Explanation:

Step1: Recall the continuous - exponential decay formula

The general form of continuous - exponential decay is $y = y_0e^{kt}$, where $y_0$ is the initial amount, $k$ is the decay constant, and $t$ is the time. We know that when $t =$ half - life ($T$), $y=\frac{y_0}{2}$. Given $y_0 = 339.1$ mg and $T = 16$ minutes. Substitute into $y = y_0e^{kt}$: $\frac{y_0}{2}=y_0e^{kT}$.

Step2: Solve for the decay constant $k$

Divide both sides of $\frac{y_0}{2}=y_0e^{kT}$ by $y_0$ (since $y_0
eq0$), we get $\frac{1}{2}=e^{kT}$. Take the natural logarithm of both sides: $\ln(\frac{1}{2})=\ln(e^{kT})$. Using the property $\ln(e^{x}) = x$, we have $\ln(\frac{1}{2})=kT$. Since $\ln(\frac{1}{2})=-\ln(2)$ and $T = 16$, then $k=-\frac{\ln(2)}{16}$. So the formula is $y = 339.1e^{-\frac{\ln(2)}{16}t}$.

Step3: Find the amount at $t = 13$ minutes

Substitute $t = 13$ into the formula $y = 339.1e^{-\frac{\ln(2)}{16}t}$. First, calculate $-\frac{\ln(2)}{16}\times13=-\frac{13\ln(2)}{16}$. Then $y = 339.1e^{-\frac{13\ln(2)}{16}}$. Since $a\ln(b)=\ln(b^{a})$, $-\frac{13\ln(2)}{16}=\ln(2^{-\frac{13}{16}})$. So $y = 339.1\times2^{-\frac{13}{16}}$.
$2^{-\frac{13}{16}}=\frac{1}{2^{\frac{13}{16}}}$, and $2^{\frac{13}{16}}=\sqrt[16]{2^{13}}$. Calculate $y = 339.1\times\frac{1}{2^{\frac{13}{16}}}\approx339.1\times0.547\approx185.4$.

Answer:

(a) $y = 339.1e^{-\frac{\ln(2)}{16}t}$
(b) $185.4$ mg