Question

problem 14. technetium-99m is a radioactive element used in nuclear medicine. it decays continuously exponentially with a half life of 6 hours. suppose a medical researcher starts with 100mg of technetium-99m.
(a) find a function that models the amount of technetium-99m in terms of time in hours.
(b) in about how many hours will the scienctist have 22mg of technetium-99m.

Explanation:

Part (a)

Step1: Recall exponential decay formula

The formula for continuous exponential decay is \( A(t) = A_0 e^{kt} \), where \( A_0 \) is the initial amount, \( k \) is the decay constant, and \( t \) is time. For half - life, when \( t = T_{1/2} \) (half - life), \( A(T_{1/2})=\frac{A_0}{2} \).
Given \( A_0 = 100 \) mg and \( T_{1/2}=6 \) hours. Substitute into the formula: \( \frac{A_0}{2}=A_0e^{kT_{1/2}} \).
Since \( A_0
eq0 \), we can divide both sides by \( A_0 \): \( \frac{1}{2}=e^{6k} \).

Step2: Solve for k

Take the natural logarithm of both sides: \( \ln(\frac{1}{2})=\ln(e^{6k}) \).
Using the property \( \ln(e^x)=x \), we get \( \ln(\frac{1}{2}) = 6k \).
Since \( \ln(\frac{1}{2})=-\ln(2) \), then \( k=\frac{-\ln(2)}{6}\approx - 0.1155 \).

Step3: Write the model function

Substitute \( A_0 = 100 \) and \( k=\frac{-\ln(2)}{6} \) into \( A(t)=A_0e^{kt} \).
We get \( A(t)=100e^{-\frac{\ln(2)}{6}t} \). We can also rewrite this using the property \( e^{a\ln(b)} = b^a \). Since \( e^{-\frac{\ln(2)}{6}t}=(e^{\ln(2)})^{-\frac{t}{6}} = 2^{-\frac{t}{6}} \), so the function can also be written as \( A(t)=100\cdot2^{-\frac{t}{6}} \).

Step1: Set up the equation

We know that \( A(t) = 22 \) mg, \( A_0 = 100 \) mg, and the model \( A(t)=100e^{-\frac{\ln(2)}{6}t} \). So we set up the equation: \( 22=100e^{-\frac{\ln(2)}{6}t} \).

Step2: Solve for t

First, divide both sides by 100: \( \frac{22}{100}=e^{-\frac{\ln(2)}{6}t} \), which simplifies to \( 0.22 = e^{-\frac{\ln(2)}{6}t} \).
Take the natural logarithm of both sides: \( \ln(0.22)=\ln(e^{-\frac{\ln(2)}{6}t}) \).
Using the property \( \ln(e^x)=x \), we have \( \ln(0.22)=-\frac{\ln(2)}{6}t \).
Now, solve for \( t \): \( t=\frac{\ln(0.22)}{-\frac{\ln(2)}{6}}=\frac{6\ln(0.22)}{-\ln(2)}=\frac{6\ln(\frac{1}{0.22})}{\ln(2)}=\frac{6\ln(\frac{50}{11})}{\ln(2)} \).
Calculate the values: \( \ln(50/11)\approx\ln(4.545)\approx1.513 \), \( \ln(2)\approx0.693 \).
Then \( t=\frac{6\times1.513}{0.693}=\frac{9.078}{0.693}\approx13.1 \) hours.

Answer:

The function is \( A(t)=100e^{-\frac{\ln(2)}{6}t} \) (or \( A(t) = 100\cdot2^{-\frac{t}{6}} \))

Part (b)