Homework 16
Score: $2 / 8 \quad 2 / 4$ answered
Question 2
To check for a blockage in a patient's veins, $2.25 \mathrm{ml}$ of a radioactive liquid with a half-life of 3.1 hours is injected into the patient's blood stream.
If the liquid is administered at 7:30 am, at what time will the amount of the liquid decrease to $0.46 \mathrm{ml}$ ?
Type your answer as a time: $5: 14 \mathrm{pm}$ or $11: 03 \mathrm{am}$.
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Step 1 ：Given that the half-life of the radioactive liquid is 3.1 hours, the initial amount of the liquid is 2.25 ml and we want to find out when it will decrease to 0.46 ml.

Step 2 ：We can use the formula for exponential decay to solve this problem: \(N = N0 * (1/2)^{t/T}\) where \(N\) is the final amount of the substance, \(N0\) is the initial amount of the substance, \(t\) is the time, and \(T\) is the half-life of the substance.

Step 3 ：Substitute the given values into the formula: \(0.46 = 2.25 * (1/2)^{t/3.1}\)

Step 4 ：Divide both sides by 2.25: \(0.46 / 2.25 = (1/2)^{t/3.1}\)

Step 5 ：Take the natural logarithm of both sides: \(\ln(0.46 / 2.25) = \ln((1/2)^{t/3.1})\)

Step 6 ：Using the property of logarithms that \(\ln(a^b) = b * \ln(a)\), we can simplify the right side: \(\ln(0.46 / 2.25) = (t/3.1) * \ln(1/2)\)

Step 7 ：Solve for \(t\): \(t = 3.1 * \ln(0.46 / 2.25) / \ln(1/2)\)

Step 8 ：Calculating the above expression gives us \(t \approx 6.8\) hours.

Step 9 ：The liquid was administered at 7:30 am, so 6.8 hours later would be approximately 2:18 pm. Therefore, the amount of the liquid will decrease to 0.46 ml at around 2:18 pm.

Step 10 ：\(\boxed{2:18 \text{ pm}}\) is the time when the amount of the liquid will decrease to 0.46 ml.