Localization of photon Is the photon localized at the instant of its Emission (release)?


*

*by an accelarating charge 

*by Atom´s emission

 A: Yes and no. It takes a finite time for a given amount of energy to emerge, and the emission process is continuous in time. While the emitter and photon are still interacting, you can't really imagine the photon as an independent thing. But your question also touches on the wave-particle duality. If you have a device which can measure where the photon is with sufficient precision, then at early times in the emission process such a device would only register the presence of a photon at positions close to the emitter.
In the case of spontaneous emission from an atom, if the atomic state natural lifetime is $\tau$ then the emitted light comes out as a pulse whose amplitude decays exponentially with characteristic time $\tau$ (e.g. a few nanoseconds for some cases, or seconds, days or even years for others). However a light detector can itself disturb this process, and that is what happens when $\tau$ is long. That is also why people speak of 'quantum jumps'. There is a wavepacket of light of some finite duration, but the detection process may itself impose another timescale, so that it looks as if first there is no photon, and then suddenly there is one. Really what happens here is that there is a non-zero value for the photon wavefunction over a duration $\tau$, giving a finite probability of a photon being detected at any time, and the detector  suddenly announces a 'click' at some time within this interval. It is similar in radioactive decay.
In the case of the photoelectric effect the process of ejecting a photoelectron from a metal surface is fast but not instantaneous.
See also: What does a photon emitted by an atom "look" like?
