Question

a specific radioactive substance follows a continuous exponential decay model. it has a half - life of 11 minutes. at the start of the experiment, 179.8 g is present.
(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=square e^{square t})
(b) how much will be present in 17 minutes?
do not round any intermediate computations, and round your answer to the nearest tenth.

Explanation:

Step1: Recall the 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.

Step2: Find the decay constant $k$ using the half - life

We know that at half - life $t = T_{1/2}$, $y=\frac{y_0}{2}$. Given $T_{1/2}=11$ minutes and $y_0$ is the initial amount. Substituting into $y = y_0e^{-kt}$, we get $\frac{y_0}{2}=y_0e^{-k\times11}$. Divide both sides by $y_0$ (since $y_0
eq0$), we have $\frac{1}{2}=e^{-11k}$. Take the natural logarithm of both sides: $\ln(\frac{1}{2})=\ln(e^{-11k})$. Using the property $\ln(e^x)=x$, we get $\ln(\frac{1}{2})=- 11k$. Since $\ln(\frac{1}{2})=-\ln(2)$, then $k = \frac{\ln(2)}{11}$.

Step3: Write the formula relating $y$ to $t$

Substitute $y_0 = 179.8$ and $k=\frac{\ln(2)}{11}$ into $y = y_0e^{-kt}$, we get $y = 179.8e^{-\frac{\ln(2)}{11}t}$.

Step4: Find the amount at $t = 17$ minutes

Substitute $t = 17$ into $y = 179.8e^{-\frac{\ln(2)}{11}t}$. First, calculate the exponent: $-\frac{\ln(2)}{11}\times17=-\frac{17\ln(2)}{11}$. Then $y = 179.8e^{-\frac{17\ln(2)}{11}}$.
We know that $a\ln(b)=\ln(b^a)$, so $-\frac{17\ln(2)}{11}=\ln(2^{-\frac{17}{11}})$. Then $y = 179.8\times2^{-\frac{17}{11}}$.
$2^{-\frac{17}{11}}=\frac{1}{2^{\frac{17}{11}}}$, and $2^{\frac{17}{11}}=2^{1 + \frac{6}{11}}=2\times2^{\frac{6}{11}}$.
$y = 179.8\times\frac{1}{2^{\frac{17}{11}}}\approx179.8\times\frac{1}{3.047}\approx59.0$.

Answer:

(a) $y = 179.8e^{-\frac{\ln(2)}{11}t}$
(b) $59.0$ g