**Core Concept**
The half-life of a radioactive substance is the time required for its radioactivity to decrease by half due to nuclear decay. In the context of Fluorine-18 (18F), a positron-emitting radionuclide used in nuclear medicine, understanding its half-life is crucial for imaging and dosimetry calculations.

**Why the Correct Answer is Right**
The half-life of Fluorine-18 is approximately 110 minutes. This is due to the radioactive decay of 18F, which undergoes positron emission to transform into Oxygen-18. The half-life of 110 minutes is a critical parameter in the production and use of 18F-labeled compounds for positron emission tomography (PET) imaging. It determines the optimal time frame for imaging and the required dose of the radioactive tracer.

**Why Each Wrong Option is Incorrect**
**Option A:** This option is incorrect because it does not accurately represent the half-life of Fluorine-18.
**Option B:** This option is incorrect because it is significantly shorter than the actual half-life of Fluorine-18.
**Option C:** This option is incorrect because it is longer than the actual half-life of Fluorine-18.

**Clinical Pearl / High-Yield Fact**
The half-life of Fluorine-18 is an essential consideration in the production and use of 18F-labeled compounds for PET imaging. It affects the timing of imaging studies and the required dose of the radioactive tracer.

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