Question

question 13
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you go to the doctor and he gives you 16 milligrams of radioactive dye. after 12 minutes, 5.5 milligrams of dye remain in your system. to leave the doctor’s office, you must pass through a radiation detector without sounding the alarm. if the detector will sound the alarm if more than 2 milligrams of the dye are in your system, how long will your visit to the doctor take, assuming you were given the dye as soon as you arrived? give your answer to the nearest minute.
you will spend minutes at the doctor’s office.

Explanation:

Step1: Define the decay model

Radioactive decay follows the exponential model \( N(t) = N_0 e^{-kt} \), where \( N_0 = 16 \) (initial amount), \( N(t) \) is the amount at time \( t \), and \( k \) is the decay constant.

Step2: Find the decay constant \( k \)

We know at \( t = 12 \), \( N(12)=5.5 \). Substitute into the model: \( 5.5 = 16 e^{-12k} \). Divide both sides by 16: \( \frac{5.5}{16}=e^{-12k} \). Take natural log: \( \ln(\frac{5.5}{16})=-12k \). Solve for \( k \): \( k = -\frac{\ln(\frac{5.5}{16})}{12}\approx0.087 \).

Step3: Find time when \( N(t)=2 \)

Substitute \( N(t)=2 \), \( N_0 = 16 \), \( k\approx0.087 \) into \( N(t)=N_0 e^{-kt} \): \( 2 = 16 e^{-0.087t} \). Divide by 16: \( \frac{2}{16}=e^{-0.087t} \). Take natural log: \( \ln(\frac{1}{8})=-0.087t \). Solve for \( t \): \( t = -\frac{\ln(\frac{1}{8})}{0.087}\approx24 \).

Answer:

24