Instantaneous ejection of photoelectrons indicative of particle nature of light? My physics textbook under a section explaining about How wave nature of light does not explain Photoelectric effect, mentions the following:

No matter how small is the intensity, photoelectrons are ejected and that too without any appreciable time delay.

The italicized line above kind of points to the conclusion that near instantaneous emmission of photoelectrons is a behaviour of particles and not waves.
Also the author states a few lines ago that if energy would have been transfered in a wave like manner, then the ejection would not have been near instantenous.
I am not getting this point, that How near intantaneous ejection of electrons would point towards particle nature of light?

Also, in my Chemistry Textbook, the author mentions that time lag between incidence of light and ejection of photoelectron is less than $10^{-8}$ seconds, which he says is negligible, which generally is true, but in comparison to the time periods of the electromagnetic waves shined onto the photoelectric plates, is even millions of times larger!
So, the tranfer of energy(greater than the work function) to the electron by the wave(behaving as a wave only) would have been completed [in some cases] in less than a few hundred oscillations of the wave, the total time taken in which would still have been tens of thousands of times smaller than the mentioned $10^{-8}$ seconds , no where indicating the particle nature.

Where am I wrong?
 A: I'll give an example, How the calculation of time is done through the classical point of view(means wave point of view),
Photoelectron may be emitted from $\phi=2.36 eV$ even for light intensities as low as $10^{-8} W/m^2$. We will calculate here the time taken so the photoelectrons have $1.00 eV$ of kinetic energy.
We will assume that all the light is absorbed in the first layer of atoms in the surface. Then we calculate the number of sodium atoms per unit area in a layer one atom thick. We assume that each atom in a single atomic layer
absorbs equal energy, but a single electron in each of these
atoms receive all the energy. We then calculate how long it
takes these electrons to attain the energy ($2.36 eV +1.00 eV
= 3.36 eV$) needed for the electron to escape.
Now as an exercise, I leave to varify the calculation,
$$\mathrm{Number \  of \  Na \ atoms/ volume}=\frac{N_A}{\mathrm{Na \ gram \ molecular \ weight}}\times \mathrm{density}=2.54\times 10^{28} \frac{\mathrm{atoms}}{m^3}$$
To estimate the thickness of one layer of atoms, we will assume a cubic structure.
$$\frac{\mathrm{1 \ atom}}{d^3}=2.54\times 10^{28} \frac{\mathrm{atoms}}{m^3}$$
$$d=3.40\times 10^{-10}m$$
If all the light is absorbed in the first layer of atoms, the number of exposed atom per $m^2$ is
$$2.54\times \frac{\mathrm{atoms}}{m^3}\times 3.40\times 10^{-10}m=8.64\times 10^{18}\frac{\mathrm{atoms}}{m^2}$$
With the intensity of $10^{-8} \ W/m^2$, each atom will receive energy at the rate of
$$1.00\times 10^{-8}\frac{W}{m^2}\times \frac{1}{8.64\times 10^{18} \mathrm{atoms/m}^2}=7.25\times 10^{-9} \ eV/s$$
We can calculate the time $t$ needed to absorb $3.36$ eV:
$$t=\frac{3.36}{7.25\times 10^{-9}}=14.7 \ \mathrm{year}$$
Based on classical calculations , the time required to eject a photoelectron should be $15$ years!

That's what they mean by times lag!
