Question

a medical technician is working with the four samples of radionuclides listed in the table below. initially, each sample contains 2.00 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 1/16 of the initial amount.

sample	radionuclide		initial radioactivity	time for amount of radionuclide to decrease to 1/16 of initial amount
a	$\ce{^{62}_{30}zn}$	9.0 hours	select	hours
b	$\ce{^{62}_{29}cu}$	5.0 minutes	select	minutes
c	$\ce{^{67}_{31}ga}$	78.0 hours	select	hours
d	$\ce{^{88}_{39}y}$	107. days	select	days

Explanation:

Step1: Determine the relationship between half - life and initial radioactivity

The shorter the half - life of a radionuclide, the higher its initial radioactivity. Because a shorter half - life means the nuclei decay more quickly, and initially (with the same amount of substance), the decay rate (related to radioactivity) is higher.

Step2: Order the samples by decreasing initial radioactivity

Sample B (\(^{62}_{29}Cu\)) has a half - life of \(5.0\) minutes (the shortest among the four), sample A (\(^{62}_{30}Zn\)) has a half - life of \(9.0\) hours, sample C (\(^{67}_{31}Ga\)) has a half - life of \(78.0\) hours, and sample D (\(^{88}_{39}Y\)) has a half - life of \(107\) days.
So the order by decreasing initial radioactivity is \(B > A > C > D\)

Step3: Calculate the number of half - lives (\(n\)) for the amount to decrease to \(\frac{1}{16}\) of the initial amount

We use the formula \(N = N_0\times(\frac{1}{2})^n\), where \(N=\frac{1}{16}N_0\).
Substituting \(N=\frac{1}{16}N_0\) into \(N = N_0\times(\frac{1}{2})^n\), we get \(\frac{1}{16}=(\frac{1}{2})^n\).
Since \(\frac{1}{16}=\frac{1}{2^4}\), then \(n = 4\)

Step4: Calculate the time for each sample

• For sample A: \(t=n\times t_{1/2}\), \(n = 4\) and \(t_{1/2}=9.0\) hours. So \(t=4\times9.0 = 36\) hours
• For sample B: \(n = 4\) and \(t_{1/2}=5.0\) minutes. So \(t=4\times5.0=20\) minutes
• For sample C: \(n = 4\) and \(t_{1/2}=78.0\) hours. So \(t=4\times78.0 = 312\) hours
• For sample D: \(n = 4\) and \(t_{1/2}=107\) days. So \(t=4\times107 = 428\) days

Answer:

• Order of samples by decreasing initial radioactivity: \(B > A > C > D\)
• Time for sample A: \(36\) hours
• Time for sample B: \(20\) minutes
• Time for sample C: \(312\) hours
• Time for sample D: \(428\) days