One last question...even though no one seems to be willing to help me out without being rude.
The blood volume in a cancer patient was measured by injecting 5.0 mL of Na2SO4(aq) labeled with 35S (t1/2 = 87.4 d). The activity of the sample was 300 µCi. After 22 min, 12.9 mL of blood was withdrawn from the man and the activity of that sample was found to be 0.75 µCi. Report the blood volume of the patient.
I am unsure how the activity fits in here. I know it is equal to the rate of decay=k*N but I am getting really confused with these variables as in the ohter problems. I tried finding k by dividing ln2 by the half life 87.4d (converted to minutes). I then plugged into N'=Ne^-kt, solving for N. my answer just came up to be 12.9mL which clearly is not right....someone please help me with at least one o these problems....I am trying to learn it and cannot get a hold of my professor. I guess that's what happens when the student ratio is 240:1.
DrBob222 had good advice to you. He was not being rude. Go go bed and get a good night's sleep. These problems will look easier then.
Yes, a student/teacher ratio of 240:1 is an impossible situation -- both for the students and the instructor.
I wasn't talking about DrBob being rude...that would be bobpursely...DrBob has been nothing but helpful while I have used this site as a resource.
I also posted the above well before Dr. Bob had answered one of my other questions...
I understand that you're struggling with this question, and I'll do my best to explain the steps to solve it.
To find the blood volume of the patient, we need to use the concept of radioactive decay and apply the formula: N(t) = N₀ * e^(-kt), where N(t) is the amount of radioactive substance at time t, N₀ is the initial amount, k is the decay constant, and e is the base of the natural logarithm.
In this case, we have two samples with different volumes and activities. One sample is injected into the patient, while the other is withdrawn after 22 minutes. We can use the radioactive decay formula for each sample and set up a system of equations to solve for the blood volume.
Let's start with the first sample, where 5.0 mL of Na2SO4(aq) with an activity of 300 µCi is injected. We can assume that all the activity is due to the radioactive isotope of sulfur-35 (35S). We need to determine the decay constant (k) for this isotope.
Since the half-life (t1/2) of 35S is given as 87.4 days, we can convert it to minutes by multiplying it by 24 hours/day * 60 minutes/hour:
t1/2 = 87.4 days * 24 hours/day * 60 minutes/hour ≈ 125760 minutes
Now, we can find the decay constant (k) using the formula: k = ln(2) / t1/2:
k = ln(2) / 125760 ≈ 5.52 x 10^(-6) min^(-1)
Next, we can use the activity (A) of the sample to determine the initial amount (N₀) of the radioactive substance using the formula: A = k * N₀:
300 µCi = (5.52 x 10^(-6) min^(-1)) * N₀
N₀ = (300 µCi) / (5.52 x 10^(-6) min^(-1)) ≈ 54225
Now, for the second sample, 12.9 mL of blood is withdrawn after 22 minutes, and its activity is measured as 0.75 µCi. We can use the same decay constant (k) to find the amount of radioactive substance remaining in this sample (N(t)).
0.75 µCi = (5.52 x 10^(-6) min^(-1)) * N(t)
Now, since we know the amount of radioactive substance remaining (N(t)) in the second sample and the initial amount (N₀) in the first sample is equal to the blood volume (V) in mL, we can set up the following equation:
N(t) = N₀ * e^(-kt) = V
Plugging the values, we have:
0.75 µCi = (54225) * e^(-5.52 x 10^(-6) min^(-1) * 22 min) = V
Now you can solve this equation to find the blood volume (V) of the patient. Remember to convert the result to the desired units (mL or any other specified unit).
I hope this explanation helps you understand how to approach this problem. Feel free to ask if you have any further questions or need additional assistance.