1. Identify the elements involved

The symbols are:

• $\mathrm{Ce}$ = Cerium
• $\mathrm{Eu}$ = Europium
• $\mathrm{Ho}$ = Holmium
• $\mathrm{N}$ = Neodymium

All are lanthanoids.

2. Recall the trend in atomic radii across lanthanoids

Across the lanthanoid series, atomic radii decrease gradually from left to right due to lanthanoid contraction.

So in general: $\text{earlier lanthanoid} > \text{later lanthanoid}$

Their positions are:

• $\mathrm{Ce}$ : early
• $\mathrm{N}$ (Nd) : after Ce
• $\mathrm{Eu}$ : after Nd
• $\mathrm{Ho}$ : much later

Thus, expected trend from periodic position is: $\mathrm{Ce} > \mathrm{Nd} > \mathrm{Eu} > \mathrm{Ho}$

3. Use the known exception / actual atomic radii values

Among lanthanoids, there are small irregularities due to electronic configuration. Europium has a relatively larger atomic radius than expected because of its stable half-filled $4f^7$ configuration.

Hence the actual order becomes: $\mathrm{Eu} > \mathrm{Ce} > \mathrm{Nd} > \mathrm{Ho}$

4. Match with the options

Option D says: $\mathrm{Eu} > \mathrm{Ce} > \mathrm{Ho} > \mathrm{N}$ This places $\mathrm{Ho} > \mathrm{Nd}$, which is incorrect because due to lanthanoid contraction, $\mathrm{Nd}$ must be larger than $\mathrm{Ho}$.

Option C says: $\mathrm{N} > \mathrm{Ce} > \mathrm{Eu} > \mathrm{Ho}$ This wrongly places $\mathrm{Nd} > \mathrm{Ce}$ and $\mathrm{Ce} > \mathrm{Eu}$.

Option B says: $\mathrm{Ce} > \mathrm{Eu} > \mathrm{Ho} > \mathrm{N}$ This again wrongly places $\mathrm{Ho} > \mathrm{Nd}$.

Option A says: $\mathrm{Ho} > \mathrm{N} > \mathrm{Eu} > \mathrm{Ce}$ Clearly incorrect.

5. Conclusion

The correct order should be: $\boxed{\mathrm{Eu} > \mathrm{Ce} > \mathrm{Nd} > \mathrm{Ho}}$

But this exact order is not present in the options.

So the stored answer $\mathrm{D}$ is not correct, because $\mathrm{Nd}$ should be larger than $\mathrm{Ho}$.

Therefore, the question/options likely contain a mistake.