What happens to the $3d$ and $4s$ levels during the ionisation of a transition metal? I'm just wondering about the transition state of a transitional metal before (ex: $[\mathrm{Ar}]\, 3d^3 \,4s^2$) and after ionisation (which means $[\mathrm{Ar}]\, 3d^3 \,4s^1$) because before ionisation the $4s$ level is closer to the radius (the $4s$ orbital is filled before the $3d$ orbital). However, during ionisation, a $4s$ electron is knocked off which means that $4s$ has to be the outermost shell of the atom. 


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*Scenario 1: We blast the atom with photon of different energy and record the energy of the photon that causes the atom to be ionised. We found out that the energy of this photon is equal to that of a $4s$ orbital. (The atom somehow can tell that we are going to ionised it and therefore changes its electron configuration)? 

*Scenario 2: The energy of the specific photon is equal to that of a $3d$ orbital but then due to increase of effective nuclear charge and the decrease of repulsive term, $$E_\mathrm{rep}(3d,3d) - E_\mathrm{rep}(3d,4s),$$ in total energy of an ionised atom the electron configuration is modified by having a $4s$ electron drops to $3d$ level. 
(See, in particular, this chemguide page regarding why the assumption that $4s$ electron levels are lower than of $3d$ is wrong.)
 A: This depends on the energy of the photon.


*

*If the photon has less energy than the difference $E_{g, \rm V^+} - E_{g, \rm V}$ between the ground state $[\mathrm{Ar}]\, 3d^3 \,4s^2$ (assuming we're doing vanadium) of the neutral atom and the ground state of the singly-ionized cation $\rm V^+$, then no ionization will take place. (Instead, the photon will only be able to excite the atom to higher eigenstates of the neutral, assuming that it is resonant with any relevant transitions.)

*If the photon has enough energy $\hbar\omega$ to go from the neutral's ground state $E_{g, \rm V}$ to the ground state $E_{g, \rm V^+}$ of the cation, then ionization will take place, with any excess energy $E_{\mathrm{kin},g} = \hbar \omega - (E_{g, \rm V^+} - E_{g, \rm V})$ going into the kinetic energy of the electron.

*If the photon has enough energy $\hbar\omega$ to go from the ground state to the neutral to an excited state of the cation, $E_{e,\rm V^+}$, then you will also have ionization onto that channel, where the ion is left in that excited state and the photoelectron carries away the remaining energy,
$$E_{\mathrm{kin},e} = \hbar \omega - (E_{e, \rm V^+} - E_{g, \rm V}),$$
as kinetic energy. This will be less than the kinetic energy carried away on the ground-state channel $E_{\mathrm{kin},g}$ by $E_{e, \rm V^+} - E_{g, \rm V^+}$, so if you look at the photoelectron energy spectrum, you will see two distinct peaks separated by that energy.


Here it is important to note that it is the energies of the eigenstates of the ion that matter, and those are equally subject to the vicissitudes of transition-metal electron-configuration rules as the neutrals are.
In particular, for vanadium, while it would indeed be reasonable to think that the ground-state configuration of the $\rm V^+$ ion would be $[\mathrm{Ar}]\, 3d^3 \,4s^1$, it appears (as explained here) that the added bonus of getting closer to a half-filled $3d$ shell makes it more beneficial to go to a $[\mathrm{Ar}]\, 3d^4$ configuration as the ground state.
I'm unsure what the electron configurations would be for the first few excited states - for an atom that's this complex, there's no substitute for actually going and doing the experiment, and my go-to resource for this (the NIST Atomic Spectra Database) is currently closed.
Ultimately, you're not that wrong in your second scenario.

The energy of the specific photon is equal to that of a $3d$ orbital 

This is wrong - this single-electron picture doesn't work. It's the global energies of the entire system that matter, and trying to split them up into single-electron contributions is only going to lead you up all sorts of wrong alleys.

but then due to increase of effective nuclear charge and the decrease of repulsive term ($E_\mathrm{rep}(3d,3d) - E_\mathrm{rep}(3d,4s)$) in total energy of an ionised atom the electron configuration is modified by having a $4s$ electron drops to $3d$ level.

This isn't that wrong. Within the Hartree-Fock picture, it is indeed correct to note that the transition in the ionization to the $\rm V^+$ ground state,
$$
[\mathrm{Ar}]\, 3d^3 \,4s^1 \longrightarrow [\mathrm{Ar}]\, 3d^4 + e^-,
$$
is a two-particle transition, and those are in principle forbidden in the initial naive theory. Your interpretation ("$4s$ electron drops to $3d$ level") is ultimately wrong, though it does represent the kind of hand-waving language that's used at professional levels to explain what happens in the process. 
However, the fact that the process is forbidden in the naive theory (Hartree-Fock with single-photon single-particle processes) means that a full understanding of this ionization pathway is going to be tricky to get, and it won't fit within any theoretical framework that you're likely to be taught before graduate school. There are plenty of post-Hartree-Fock theory frameworks that are able to handle this (generally by allowing states to contain superpositions of multiple configurations) but they just don't fit into simple hand-wavy language - they're just a bunch of technical quantum chemistry, and there's ultimately no way around that.
