Is ionization energy enthalpy change?

In physics, we learn how to calculate the ionization energy of a hydrogen atom [using the Bohr model](https://en.wikipedia.org/wiki/Ionization_energy#Bohr_model_for_hydrogen_atom). The result is the following equation:


$$
E = - \frac{1}{n^2} \frac{Z^2e^2}{2a\_0} = - \frac{Z^2 R\_H}{n^2} = - \frac{Z^213.6\ \mathrm{eV}}{n^2}
$$


This equation shows that the ionization energy of a hydrogen atom is calculated as 13.6 eV.


On the other hand, in chemistry, ionization energy is treated as a type of enthalpy change, such as in the [Born-Haber cycle](https://en.wikipedia.org/wiki/Born%E2%80%93Haber_cycle).


When calculating ionization energy using the Bohr model, it appears that only the internal energy change ($\Delta U$) at 0K is considered and calculated. However, for the Born-Haber cycle, the enthalpy change ($\Delta H = \Delta U + p\Delta V$) at 25°C and 1 atm is required.


This raises the question: when defining "ionization energy," is the volume change of the gas due to the electron produced by the ionization of hydrogen--assuming such a volume change occurs--taken into account?


Or is "ionization energy" and "ionization enthalpy" distinct in concept but so close in value that the difference can be ignored? If so, why is this the case?


