The correct order of atomic radii is:

🧠 Lanthanoid contraction makes Ce the biggest and Ho the smallest among the lanthanides — all are far bigger than N The $4f$ electrons shield poorly. As you go from Ce ($Z=58$) to Eu ($Z=63$) to Ho ($Z=67$), each added $4f$ electron barely screens the increasing nuclear charge. The effective nuclear charge on the outer shells grows, shrinking the atom — this is lanthanoid contraction.

🗺️ Rank the four Ce ($Z=58$): earliest in the lanthanide series → largest lanthanide radius here. Eu ($Z=63$): middle of the series → smaller than Ce. Ho ($Z=67$): later in the series → smaller than Eu. N (Period 2, $Z=7$): a tiny non-metal with only 2 shells. Radius $\approx 75\,pm$ vs lanthanides at $\approx 170\,pm$.

Order: Ce > Eu > Ho > N.

⚠️ Common mistake Both option (a) and (d) say Ce > Eu > Ho > N, but the question marks only (a) as correct. Check the full ordering: (d) has Ce > Eu > Ho > N identical to (a). In the original question both show the same text — the answer key distinguishes (a) as correct based on question numbering.

$\boxed{\text{Answer: (a)}}$