Step 1: Understanding the Question:
The problem presents four lanthanoid elements and requires the identification of the single member that possesses an unstable nucleus and exhibits radioactivity.
Step 2: Key Formula or Approach:
The lanthanoids (or inner transition elements belonging to the 4f-series) extend from Lanthanum ($Z=57$) or Cerium ($Z=58$) to Lutetium ($Z=71$). Nearly all naturally occurring lanthanoids are highly stable with stable isotopes, with exactly one notable exception.
Step 3: Detailed Explanation:
Let's look at the given options:
Pr (Praseodymium, $Z=59$), Nd (Neodymium, $Z=60$), and Sm (Samarium, $Z=62$) are stable elements with non-radioactive isotopes found in relative abundance in the Earth's crust.
Pm (Promethium, $Z=61$) is unique among the lanthanoids. It has no stable isotopes and is entirely man-made/synthetic or formed temporarily via radioactive decay chains. Because its longest-lived isotope has a half-life of only 17.7 years, it is highly radioactive.
Step 4: Final Answer:
The radioactive lanthanoid element is Promethium (Pm), matching option (C).