Question 21.5: Indium-111 (t1/2 = 2.805 d) and gallium-67 (t1/2 = 3.26 d) a...
Indium-111 (t_{1/2} = 2.805 d) and gallium-67 (t_{1/2} = 3.26 d) are both used in radioimaging to diagnose chronic infections. Which isotope decays faster? If we start with 10.0 mg of each isotope, how much of each remains after 24 h?
Collect and Organize We are given the half-lives of two radionuclides and asked to predict which one will decay faster, and to calculate how much of each isotope remains after 24 h of decay.
Analyze An isotope with a shorter half-life decays faster. Radioactive decay follows first-order kinetics. Quantitatively, the relationship between half-life and the amount of material remaining is described by the following equation from Chapter 19:
\ln \frac{N_{t}}{N_{0}}= – 0.693 \frac{t}{t_{1/2}}
Here N_{0} and N_{t} refer to the amount of material present initially and the amount at time t, respectively. If our prediction for the relative decay rates of the two isotopes is correct, then more of the isotope with the longer half-life should remain after 24 h. We need to complete two calculations to determine the amount of each sample present after 24 h.
Solve Indium-111 has the shorter half-life, so it should decay faster.
For the amount of indium-111 remaining after 24 h, we have
\ln \frac{N_{t}}{10.0 mg}=\frac{(-0.693)(24 \sout{\text{h}})}{(24 \sout{\text{h}}/\sout{\text{d}}) (2.805 \sout{\text{d}})}=-0.247
Taking the antilog of both sides, we get
\frac{N_{t}}{10.0 mg}=0.781
N_{t}=(0.781)(10.0 mg) = 7.81 mg
For gallium-67:
\ln \frac{N_{t}}{10.0 mg}= \frac{(-0.693) (24 \sout{\text{h}})}{(24 \sout{\text{h}}/\sout{\text{d}}) (3.26 \sout{\text{d}})}=-0.213
\frac{N_{t}}{10.0 mg}=0.808
N_{t}=(0.808)(10.0 mg)= 8.08 mg
Think About It We predicted that indium-111 would decay faster, which means that after 24 h the quantity of this isotope should be less than the quantity of gallium-67, and it is.