Photons are massless meaning they exist only with speed c. When the electron is exited the photon doesn't exist. In order to be born a photon from the electron there must be an external field. Such fields exists even in vacuum. So the energy of that field is enhanced at some instant by hv e.g, you can say

that a photon is born. There is acceleration of the photon from 0 to c.

There is NO acceleration of the photon from 0 to c.

But considering the facts, the photons have rest mass zero. But photons don't exist in rest. What i see is a contradiction of facts.

I would appreciate if you can brief me a little about the field that you have mentioned in your answer.

There is no such thing as rest mass of photon. It is always moving even if you move with it at c. Rest mass have the electron and the proton. If they move their mass increases.

The fields are in fact EM fields comprised from photons. They exist even in empty space.

There is no "orbiting around the nucleus".

There are no "Photons (as) particles", meaned as small little things.

In 1926 in the Physical Review, Erwin Schrödinger had already answered your questions. Alas, he had (temporarily) given up the relativistic frame, so he had no more intrinsic and definite energy-per-state, so no intrinsic frequencies either. He did not corrected this thoughtless mistake before 1930, but when he did, only P.A.M. Dirac took care. Bad luck for everybody, for ever after.

Quantum world is not completely accessible to our intuition built on everyday experience.  Yet in case of photoelectric effect there is such a chance.  It works nearly as a familar billiards:  incoming photon strikes an electron and ejects it from the material but at a cost of changing its own energy and momentum.  The peculiarity of photon's energy is that it is not related to its velocity (always equal to c) but rather to frequency.  Scattered photon simply runs away.  That is all -- when the ejected electron was residing in the outer shell of an ion.  When the electron was burried deeper, then its place will be soon filled by other electron, coming from the shell with higher energy.  The excess energy is radiated as still another photon, with conservation of system's momentum.  Its again like a billiards ball entering pot.  It makes noise, but noise itself is not a part of the billiards ball.

What Marek describes is inelastic photon scattering such as in the Compton effect. This is actually a nice piece of physics to consider, since the kinematics, as Marek says, is correctly described by a "particle billard" like, quasi-mechanic description, while the change of energy of the photon is actually detected as a change of wavelength evidenced by a monochromator (\lambda *E / \hbar = c). [In an adequate theoretical framework, this would actually rather be described as the destruction of the original photon and the creation of a new one with different energy and momentum, rather than a change of the properties of some specific photon.]

As was said before, in the photoelectric effect, the elementary process is photon absorption, i.e. the photon is being destroyed in the process, and its energy (and momentum) is being taken up by "the system" in such a way that the end product comprises an electron escaping from the absorbing atom (or solid).

As I understand the original question, the inquiry is about the details in time (evolution) of this process. One usual attempt to do this is to use so-called time dependent perturbation theory, where the perturbation is assumed to be an electromagnetic field, for example in the shape of a plane wave of given frequency. This is a so-called semi-classical approach to the problem, since the electomagnetic field is modeled as a classical field sufficiently weak to not perturb the quantum states of "the system". Carrying out this calculation leads indeed to a time evolution of the quantum system, resulting in an oscillation between the two quantum levels being coupled by the radiation field (Rabi-oscillations). However, one inherently assumes in this approximation that the e/m field itself is unchanged, so that the detailed energy conservation issues are not properly treated. Nevertheless, the approach is useful for many computations (transition matrix elements are correctly computed in this approximation, the absorption probabilities derived from them "survive" the transition to the fully quantum description).

The time evolution thus computed derives from the explicit time dependence of the classical electromagnetic field. The quantum nature of radiation is not included here (so there are no photons in this description). Photons, in turn, are to be understood as quantized modes of the electromagnetic field. It is an inherent property of single photon states that we cannot attribute classical field properties like E(r,t) to these states. The expectation values of the classical field quantities E and H actually vanish at all r and t (in other words: the phase of the e/m field in a single photon state is not determined. Messiah's books on QM would probably be a suitable place to look this up). As I understand it, the time evolution of the absorption (annihilation) of a single photon therefore concerns a question that cannot be answered in the full quantum description (the framework for this would be quantum electrodynamics). This is not dissimilar to the case of radioactive decay, where we know the probability for something to happen, but not the moment in time when a particular individual decay will occur.

QM states of the e/m field with known/defined phase (i.e. coherent radiation) can nevertheless be constructed, they are referred to as Glauber states. The price to pay for this information is that the number of photons is actually unknown (this is yet another version of the Heisenberg uncertainty principle), because to obtain such states one needs to superimpose states with 1,2,3,..,n,... photons. This means that the energy of the e/m field in such a state is indetermined...  Lasers produce an approximation to such coherent e/m fields. Therefore, using such coherent light sources one can nevertheless observe Rabi-oscillations (in sufficiently simple systems).

This answer has neither been short, nor particularly simple, sorry. Hope it nevertheless helps as an orientation. (And of course, it only reflects my understanding of the subject...)

Note added: To come back to the original formulation of the question and to reinforce a statement made earlier in this thread, the formulation "removal of photons from electrons" is not adequate. Rather, e.g. in considering the process of photon absorption by a hydrogen atom, we would have to view the entire system of matter (proton & electron) and the electromagnetic field as one system. In modeling it, we first describe them separately (e.g. by computing all the stationary solutions of the Hydrogen atom problem on the one hand, and the quantized modes of the e/m radiation field on the other, described as continua of e/m harmonic oscillators). The basis states in this description would just be product states of atomic excitations and e/m excitations (photons), and these states would be stationary (i.e. 'live' forever). Next we introduce the coupling between the two (it must be finite, otherwise there could be no interaction/absorption). In principle, we would now have to solve the coupled system (as a result the original basis states are no more stationary). If the coupling is weak enough it is permissible and a very good approximation indeed to continue using the product states as the basis of the language in which we describe the processes. There remains the problem of photon creation out of an excited state, which was also mentioned in the original question. The initial state would consist of "excited atomic state" * "no photon". Now with (apparently) no electromagnetic field, how can a photon be created? The fascinating answer is that the state "no photon" is actually not equivalent with "no e/m field. The rates/probabilities of this "sponataneous" emission are adequately described by considering the zero point fluctuations of the electromagnetic field" (related to the fact that QM harmonic oscillators in their lowest energy state have nonzero energy).