Minimum voltage to ionize Xenon As we know, the ionization energy for Xenon is $12.13 \,\rm{eV}$. Is this  value obtained via experiment? Or can it be calculated? In the latter case, can you please share the formula? I tried to use
$$\Delta E =  R_H \left(\frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$$
taking $n_f = \infty$ and $n_i = 5$th orbit, as  valence electrons are  in the $5 p$ orbitals. 
Could you please correct me here? Thank you.
 A: The calculation you describe is the energy required to remove an electron from the $n=5$ orbital of a hydrogenic atom i.e. an atom that contains only a single electron. Clearly this isn't applicable to a Xenon atom because it contains $54$ electrons not just one.
For a hydrogenic atom the wavefunction can be calculated exactly because it is a two body problem, just the electron and the nucleus, and the equation describing this is straightforward. As soon as we introduce extra electrons life becomes far more complicated. Now the $5p$ electron feels not only the attraction due to the positive nucleus but the repulsion to the other $53$ electrons.
Calculating the wavefunction for a polyelectronic atom cannot be done analytically, but it is straightforward to calculate using numerical techniques. Typically we would start by using a Hartree-Fock approximation then refine it using a technique like configuration interaction. These allow us to calculate quantities like the ionisation energy to high precision.
So the $12.13$ eV ionisation energy can both be calculated and of course measured experimentally. However the calculation has to be done numerically and there is no simple equation like the one you quote.
