Question

a medical technician is working with the four samples of radionuclides listed in the table below. initially, each sample contains 4.00 μmol of the radionuclide. first, order the samples by decreasing initial radioactivity. then calculate how long it will take for the amount of radionuclide in each sample to decrease to 1/8 of the initial amount.

sample	radionuclide
a	(_{7}^{13}\text{n})	10. minutes	select	(square) minutes
b	(_{15}^{33}\text{p})	25. days	select	(square) days
c	(_{39}^{90}\text{y})	64.0 hours	select	(square) hours
d	(_{53}^{131}\text{i})	8.0 days	select	(square) days

Explanation:

Step1: Use the half - life formula

The formula for the amount of a radioactive substance remaining after \(n\) half - lives is \(N = N_0\times(\frac{1}{2})^n\). We want \(N=\frac{1}{8}N_0\), so \(\frac{1}{8}=(\frac{1}{2})^n\). Solving for \(n\), we know that \(n = 3\) (since \((\frac{1}{2})^3=\frac{1}{8}\)).

Step2: Calculate the time for each sample

• Sample A (\(N\)):

Half - life \(t_{1/2}=10\) minutes. Time \(t=n\times t_{1/2}\). Substituting \(n = 3\) and \(t_{1/2}=10\) minutes, we get \(t=3\times10 = 30\) minutes.

• Sample B (\(P\)):

Half - life \(t_{1/2}=25\) days. Time \(t=n\times t_{1/2}\). Substituting \(n = 3\) and \(t_{1/2}=25\) days, we get \(t=3\times25=75\) days.

• Sample C (\(Y\)):

Half - life \(t_{1/2}=64.0\) hours. Time \(t=n\times t_{1/2}\). Substituting \(n = 3\) and \(t_{1/2}=64.0\) hours, we get \(t = 3\times64.0=192\) hours.

• Sample D (\(I\)):

Half - life \(t_{1/2}=8.0\) days. Time \(t=n\times t_{1/2}\). Substituting \(n = 3\) and \(t_{1/2}=8.0\) days, we get \(t=3\times8.0 = 24\) days.

Answer:

• Sample A: \(30\) minutes
• Sample B: \(75\) days
• Sample C: \(192\) hours
• Sample D: \(24\) days