Let's say we have a metallic plate and a light source so that each atom of the metal will receive $15'6\space \frac{meV}{s}$ (mili electron volts per second). Assuming the work function of the metal is $\Phi=2\space eV$, we can calculate how much time we would need for the photoelectric effect to happen according to classic theory: $$2\space eV=\int_{0}^t15'6·10^{-3}dt<=>t=128\space seconds$$ Now, we actually know there is no delay in the photoelectric effect, so a photon of at least $E=2\space eV$ (assuming our light's wave length is $\lambda=\frac{hc}{2\space eV}=620'8\space nm $) should hit an atom once the light is lit up. My question is, how is it possible for an atom to receive $2\space eV$ at an instant if we said that it receives $15'6\space \frac{meV}{s}$?
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according to classic theory ... how is it possible for an atom to receive 2 eV at an instant
It's not. That's one the reasons the photoelectric effect was considered to be evidence against classical mechanics.
In quantum theory, the 15.6 meV/s is the average across all atoms, not the amount that each atom gets; each atom either gets a whole photon's worth of energy, or none at all.
