Nuclei: Real-World Scenarios (5)

Question

A medical PET scan uses Fluorine-18 ($^{18}$F), which decays via positron emission with a half-life of $110$ minutes. A hospital in Hyderabad receives a sample with initial activity $10$ mCi at 9 AM. What is the activity at 1 PM (4 hours later)? How long until activity falls below $1$ mCi? This is the kind of half-life calculation NEET tests reliably.

Solution — Step by Step

Why This Works

Radioactive decay is governed by $N(t) = N_0 e^{-\lambda t}$, with $\lambda = \ln 2 / T_{1/2}$. Activity is proportional to $N$, so it follows the same exponential decay. After every half-life, exactly half of the radioactive nuclei remain — independent of how many you started with.

This is why PET scans must be administered within a tight window after the isotope is delivered: by the time you account for transport, prep, and uptake, the dose is already decaying away. F-18 is one of the longer-lived PET tracers — others like O-15 (half-life 2 minutes) must be used at the cyclotron itself.

Alternative Method

Use the decay constant directly. $\lambda = \ln 2 / 110 \approx 0.0063$ /min. Then $A(240) = 10 e^{-0.0063 \times 240} = 10 e^{-1.51} \approx 10 \times 0.221 = 2.21$ mCi. Same answer.