Question

a medical technician is working with the four samples of radionuclides listed in the table below. initially, each sample contains 7.00 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 1/4 of the initial amount. table with columns: sample, radionuclide (symbol, half - life), initial radioactivity, time for amount of radionuclide to decrease to 1/4 of initial amount; rows: a (51 24 cr, 28. days, select, days), b (59 26 fe, 45. days, select, days), c (55 27 co, 18. hours, select, hours), d (75 34 se, 120. days, select, days)

Explanation:

Step1: Determine initial radioactivity order

The initial radioactivity of a radionuclide is inversely proportional to its half - life. A shorter half - life means a higher initial radioactivity.
Sample C has the shortest half - life (18 hours), so it has the highest initial radioactivity. Then comes sample A (28 days), then sample B (45 days), and sample D (120 days) has the lowest initial radioactivity. So the order is \(C > A > B > D\).

Step2: Calculate the time for the amount to decrease to \(1/4\) of the initial amount

The formula for radioactive decay is \(N = N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}\), where \(N\) is the final amount, \(N_0\) is the initial amount, \(t\) is the time elapsed, and \(T_{1/2}\) is the half - life.
We want \(N=\frac{1}{4}N_0\), so \(\frac{1}{4}N_0=N_0(\frac{1}{2})^{\frac{t}{T_{1/2}}}\).
Dividing both sides by \(N_0\) gives \(\frac{1}{4}=(\frac{1}{2})^{\frac{t}{T_{1/2}}}\).
Since \(\frac{1}{4}=(\frac{1}{2})^2\), we have \(\frac{t}{T_{1/2}} = 2\), so \(t = 2T_{1/2}\).

For sample A:

\(T_{1/2}=28\) days. Using \(t = 2T_{1/2}\), we get \(t=2\times28 = 56\) days.

For sample B:

\(T_{1/2}=45\) days. Using \(t = 2T_{1/2}\), we get \(t=2\times45=90\) days.

For sample C:

\(T_{1/2}=18\) hours. Using \(t = 2T_{1/2}\), we get \(t=2\times18 = 36\) hours.

For sample D:

\(T_{1/2}=120\) days. Using \(t = 2T_{1/2}\), we get \(t=2\times120=240\) days.

Answer:

sample	initial radioactivity (order)	time for amount of radionuclide to decrease to \(1/4\) of initial amount
B	Third highest	\(90\) days
C	Highest	\(36\) hours
D	Lowest	\(240\) days