## **Core Concept**
The half-life of a radioactive isotope is the time required for half of the initial amount of the isotope to decay. This concept is crucial in nuclear medicine for determining the duration of diagnostic or therapeutic procedures. Fluorine-18 (F-18) is a positron-emitting isotope used in PET scans.

## **Why the Correct Answer is Right**
F-18 has a half-life of approximately 109.77 minutes, which is roughly 110 minutes. This half-life is ideal for PET imaging because it allows enough time for the radiopharmaceutical to be administered, imaged, and for the isotope to decay to safe levels, but not so long that it poses a prolonged radiation risk. The half-life of F-18 makes it suitable for studying various physiological and pathological processes.

## **Why Each Wrong Option is Incorrect**
- **Option A:** 10 minutes - This is much too short for F-18 and does not accurately reflect its half-life.
- **Option B:** 6 hours - This is longer than the actual half-life of F-18 and would imply a significantly slower decay rate.
- **Option D:** 24 hours - This is much too long for the half-life of F-18 and is more characteristic of longer-lived radioactive isotopes.

## **Clinical Pearl / High-Yield Fact**
A key point to remember is that the half-life of F-18 is about 110 minutes. This makes F-18 labeled compounds, like FDG (fluorodeoxyglucose), very useful for PET scans because they can provide high-quality images of metabolic activity within this timeframe.

## **Correct Answer:** C. 110 minutes.