Question

the half-life of technetium-99m is 6.00 hours. a 105.-mg sample of technetium-99m is used for a diagnostic procedure.
part 1 of 2
how much tc-99m remains after 18.0 hours? be sure your answer has the correct number of significant figures.
□ mg of tc-99m
part 2 of 2
how much tc-99m remains after exactly 2 days? be sure your answer has the correct number of significant figures.
□ mg of tc-99m

Explanation:

Part 1 of 2:

Step1: Calculate the number of half - lives ($n$)

The half - life of Tc - 99m ($t_{1/2}=6.00$ hours). The time elapsed $t = 18.0$ hours.
The formula for the number of half - lives is $n=\frac{t}{t_{1/2}}$.
Substitute $t = 18.0$ hours and $t_{1/2}=6.00$ hours into the formula:
$n=\frac{18.0}{6.00}=3$.

Step2: Use the radioactive decay formula

The radioactive decay formula is $N = N_{0}\times(\frac{1}{2})^{n}$, where $N_{0}=105$ mg (initial amount) and $n = 3$.
$N=105\times(\frac{1}{2})^{3}$.
Since $(\frac{1}{2})^{3}=\frac{1}{8}$, then $N = 105\times\frac{1}{8}=13.125\approx13.1$ mg (three significant figures).

Part 2 of 2:

Step1: Calculate the number of half - lives ($n$)

The time elapsed $t = 2$ days. Since 1 day = 24 hours, $t=2\times24 = 48$ hours.
$n=\frac{t}{t_{1/2}}$, with $t_{1/2}=6.00$ hours.
$n=\frac{48}{6.00}=8$.

Step2: Use the radioactive decay formula

Using $N = N_{0}\times(\frac{1}{2})^{n}$, where $N_{0}=105$ mg and $n = 8$.
$N=105\times(\frac{1}{2})^{8}$.
Since $(\frac{1}{2})^{8}=\frac{1}{256}$, then $N=\frac{105}{256}\approx0.410$ mg (three significant figures).

Answer:

Part 1 of 2: $13.1$ mg of Tc - 99m.
Part 2 of 2: $0.410$ mg of Tc - 99m.