Q:

Why emission of an electron does not depend on light's intensity?

There's something that confuses me about the photoelectric effect. In an article I read, it's stated that the emission of an electron does not depend on the light's intensity. I'm not sure on which parameters does the light intensity depend on, after an online search I found this connection:
$$
I = \frac{1}{2}E h
$$
where I is the intensity, $E$ is the energy and $h$ is planck's constant. when I see this formula, and consider the following one as well:
$$
 E = h\nu
$$
where $E$ is the energy of the photon, $h$ is planck's constant and $\nu$ is the light's frequency. I can write the light's frequency as a function of the intensity.
what I don't understand is, if the emission of the electron experimentally is independent of the light's intensity, and depend solely on the light's frequency, why is this connection: $I = \frac{1}{2}h^2\nu$ is wrong? or why can't I use it in order to calculate if electron will be emitted depending on the intensity?
I just want to emphasis: I'm not talking about the number of emitted electrons, I consider only the question whether a specific electron will break free from the atom or not.

A:

If the intensity of the light is changed, without changing the frequency, this merely reflects the number of photons present, and not their individual energies. i.e.
$$ I \propto N h\nu,$$
where $N$ is something like photons per second per unit area.
The formula you have written down is dimensionally incorrect. The units of energy are Joules and the units of the Planck constant are Joule-seconds. So your intensity has units of J$^2$s, which doesn't make sense.
Whether an electron is emitted just depends on whether an atom can absorb enough energy to release it from a metal. At anything other than ludicrous intensities one considers only single-photon interactions, so all that matters is whether each individual photon has the requisite amount of energy. 

