Question 25.6: Medical Imaging with ^18F The isotope ^18F is widely used in...
Medical Imaging with { }^{18} F
The isotope { }^{18} F is widely used in medical imaging, because its 110-min half-life means it doesn’t stay radioactive in the patient s body for too long. (a) If a patient is administered 10.0 μg of what’s the initial activity? (b) How much { }^{18} F remains in the body after 24 h?
ORGANIZE AND PLAN Equation 25.4 gives the activity R=\lambda N; while Equation 25.6 gives the decay constant in terms of half-life: \lambda=\ln 2 / t_{1 / 2}. You can find the initial number N_{0} of radioactive nuclei using the dose and { }^{18} F‘s molar mass, 18 g/ mol. Then Equation 25.5 gives the number of nuclei remaining at any time: N=N_{0} e^{-\lambda t}.
R=\left|\frac{\Delta N}{\Delta t}\right|=\lambda N (Activity; SI unit: Bq) (25.4).
N=N_{0} e^{-\lambda t} (Radioactive decay) (25.5).
t_{1 / 2}=\frac{\ln 2}{\lambda} (Half-life and decay constant; SI unit: s) (25.6)
\text { Known: } t_{1 / 2}=110 min =6600 s ; \text { mass } m=10.0 \mu g ; \text { molar mass }=18.0 g / mol.
(a) The initial number of nuclei is
N_{0}=1.0 \times 10^{-5} g \times \frac{1 mol }{18.0 g } \times \frac{6.022 \times 10^{23} \text { nuclei }}{1 mol }=3.35 \times 10^{17} \text { nuclei }.
Then the initial activity is
R=\lambda N_{0}=\frac{\ln 2}{t_{1 / 2}} N_{0}=\frac{\ln 2}{6600 s }\left(3.35 \times 10^{17}\right)=3.52 \times 10^{13} s ^{-1}=3.52 \times 10^{13} Bq.
\text { where } 1 Bq =1 s ^{-1} .
(b) After 24 h 86,400 s , the fraction of the original sample remaining is
\frac{N}{N_{0}}=e^{-\lambda t}=e^{-\ln (2) t / t_{1 / 2}}=e^{-\ln (2)(86,400 s ) / 6600 s }=1.15 \times 10^{-4}.
Only about one ten-thousandth of the original sample remains, about a nanogram.
REFLECT The initial activity is on the order of 10^{13} Bq high enough for diagnostic imaging. The 635-keV positrons from { }^{18} \text { F's } decay travel only about 2 mm in tissue before they annihilate, which enables high-resolution imaging. Our calculation at 24 h may be an overestimate, because biological processes help flush the isotope out of the body; the so-called biological half life depends on the specific chemical compound into which the { }^{18} Fis incorporated.