A patient is given 0.050 mg of technetium-99m, a radioactive isotope with a half-life of about 6.0 hours. How long until the radioactive isotope decays to 6.3 * 10^-3 mg?

Dec 5, 2015

$\text{44 h}$

Explanation:

As you know, a radioactive isotope's nuclear half-life tells you exactly how much time must pass in order for an initial sample of this isotope to be halved.

The equation that establishes a relationship between the initial mass of a radioactive isotope, the mass that remains undecayed after a given period of time, and the isotope's half-life looks like this

$\textcolor{b l u e}{A = {A}_{0} \cdot \frac{1}{2} ^ n} \text{ }$, where

$A$ - the initial mass of the sample
${A}_{0}$ - the mass remaining after a given time
$n$ - the number of half-lives that pass in the given period of time

In your case, the initial amount of technetium-99 is said to be equal to $\text{0.050 mg}$. You are asked to find the amount of time needed to reduce this initial sample to $6.3 \cdot {10}^{- 3} \text{mg}$.

Use the above equation to find the value of $n$

$6.3 \cdot {10}^{- 3} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{mg"))) = 0.050 * 10^(-3)color(red)(cancel(color(black)("mg}}}} \cdot \frac{1}{2} ^ n$

This is equivalent to

$\frac{6.3 \cdot {10}^{- 3}}{0.050} = \frac{1}{2} ^ n$

Take the natural log of both sides of the equation to get

$\ln \left(\frac{6.3 \cdot {10}^{- 3}}{0.050}\right) = \ln \left({\left(\frac{1}{2}\right)}^{n}\right)$

$\ln \left(\frac{6.3 \cdot {10}^{- 3}}{0.050}\right) = n \cdot \ln \left(\frac{1}{2}\right)$

$n = \ln \frac{\frac{1}{2}}{\ln} \left(\frac{6.3 \cdot {10}^{- 3}}{0.050}\right) = 7.31$

Since $n$ represens the number of half-lives that pass in a given period of time, you can say that

$n = \frac{t}{t} _ \text{1/2" implies t = n * t_"1/2}$

In your case, ${t}_{\text{1/2" = "6.0 h}}$, which means that

$t = 7.31 \cdot \text{6.0 h" = "43.86 h}$

Rounded to two sig figs, the answer will be

$t = \textcolor{g r e e n}{\text{44 h}}$