Does a photon instantaneously gain $c$ speed when emitted from an electron? An excited electron loses energy in the form of radiation. The radiation constitutes photons which move at a speed $c$. But is the process of conversion of the energy of the electron into the kinetic energy of the photon instantaneous? Is there a a simple way to visualize this process rather than math?
 A: You have to understand that this is Quantum Mechanics. A classical view is by definition just an approximation.
The Heisenberg uncertainty means that you cannot know both the exact time and energy of a given event. At macroscopic scales, this doesn't matter. But consider that the radius of an atom is a few attolightseconds. The energy involved is about 1eV. Planck's constant is 4 feV's. So, to the degree that you can say something meaningful, it's instant.
Note that this is just the photon's energy. Its speed is always c, the speed of the associated EM field. But that field isn't precisely localized in space (again, Heisenberg). 
Edit:
dj_mummy's answer reminds me of another point. The electron emission event is a quantum event. That means you also get the superposition of states. Like Schrödinger's cat, the atom will and will not have emitted the photon, at the same time. But once you see the photon, the wave function collapses and the no-photon state has zero probability. Also, all states in which the photon was emitted at a different time also have zero probability; only one photon can be emitted.
The important thing to understand is that this collapse happens only on observation. Until you see the photon, there are multiple states with non-zero probability, corresponding with different times at which the photon is emitted. So, you can't just talk about the time when it's emitted.
A: The best way to visualize this is to think of a 'before' and an 'after'. The 'before' is the state the electron is prepared in (with a definite momentum and spin orientation). After some time (can be arbitrarily small BUT NOT ZERO) one can observe the system again, what is found is the 'after' state. Now in both these states total momentum, energy, charge etc. are conserved. 
What happened in between these 2 states? It is irrelevant, since there was no observer in between. 
So the photon does not actually 'come out of the electron. It is just detected with an electron in the 'after' state. There is no continuously observable emission of the photon.
A: Imagine throwing a stone into water and asking the question how fast do the waves accelerate?
The answer is that the waves don't accelerate. Any given body has a range of possible waves that can propagate along its surface, and the total energy of the surface is related to the amplitudes of all those possible oscillatory modes. In principle at least there are always waves present due to thermal excitations, though in practice the amplitude of the waves will be vanishingly small. When you are throwing in the stone you are transferring energy to some of the already existent oscillatory modes. So you are not accelerating anything in the sense you accelerate an object – you are just transferring energy to waves that are already moving.
Now consider your atom generating light. As the electron falls to a lower energy level, it transfers energy into the quantum field that constitutes light, just as the stone transfers energy to the water. This energy goes into exciting modes that are already travelling at $c$, so no acceleration is necessary.
I would guess that much of the confusion arises from considering light as a photon. Light is a quantum field, and the photon is an approximation for its behaviour that works well in some cases but not in others. As a general rule, the particle approximation is useful when considering the transfer of energy but the wave approximation is better when considering the propagation of energy. In this case the photon is the transfer of energy from the electron to the photon field, but the subsequent motion of that energy through spacetime is best described as a wave. So the photon is never accelerated in the usual sense of the word.
A: Taking an atom as an example, the initial state must already be a state of the combined atom-field system. It cannot be an eigenstate, then the atom would not decay. This with the exception where the initial state involves the atomic ground state and the field vacuum state.
But how is such a state created? One possibility is a scattering situation with the atom in its ground state in the distant past and a field wavepacket moving towards it. Since c, the speed of light, is finite, there is initially, as $t{\rightarrow}-\infty$, no interaction. Finally, as the time $t{\rightarrow}+\infty$, we observe an atom in its ground state and a second field wavepacket. 
Note that the initial field wavepacket must be localized, a momentum eigenstate would extend through all space. This does not mean that momentum eigenstates play no role, they emerge if a Fourier decomposition (plane wave expansion) is made. Then the scattering event is described in terms of the scattering operator kernel S(k,k'). Energy conservation (the initial and final atomic states are both the ground state) tells us that k and k' have equal magnitude.
In this case all information is contained in the operator S. The scattering formalism does not seem to allow to pose the original question nor does it lead to a simple answer.
A: MC Physics would suggest that photons are formed from the joining of 2 opposite electro-static charged mono-charges that are emitted from particles or atoms due to excessive vibration. Those joined mono-charges are (almost) instantly accelerated to c by surrounding electric-magnetic forces then any excess force is applied to rotate the real particle to frequency.  This is further described at:
https://fs23.formsite.com/viXra/files/f-1-2-9037551_RhNE84zB_MC_Physics-_Model_of_A_Real_Photon_With_Structure_and_Mass.pdf
A: Quantum Mechanics tells us that electrons only lose or gain energy equal to the energy of an incoming or outgoing photon. And by default, all photons travel at speed c in vacuum. As I understand it, there is no "conversion" time for energy. Photons are energy and energy comes in photons. What we choose to call them is more a reflection of the state of the system than how much or what wavelength of photons are emitted. Hope this helps!
