The process of ionization for Sodium is: $Na \rightarrow Na^{+} + e^{-}$. The energy required for this is the ionization potential ($5.1\text{ eV}$). The reverse process is: $Na^{+} + e^{-} \rightarrow Na$. The energy released in this reverse process is the electron gain enthalpy of $Na^{+}$, which is equal in magnitude but opposite in sign to the ionization potential. Therefore, the value is $5.1\text{ eV}$.