Does a single photon's energy depend on frequency? (photoelectric effect) Essentially this question boils down to "why does the energy of light depend on frequency?".  
The analogy my textbook (pg. 272 principles of chemistry A molecular Approach 3rd edition by Nivaldo J. Tro) gave was that photons with higher frequency are like baseballs, whereas photons with lower frequency are like ping pong balls. Meaning that each photon has higher energy.
This StackExchange page (In the famous Einstein's Photoelectric effect, why does the intensity of light not raise the kinetic energy of the emitted electrons?) seems to say the opposite, that the energy of each photon is the same and what changes is the amount of photons per area.
 A: The energy $E$ of a photon is directly related to its frequency $f$ via $E = h f$.
This relation is a fact of Nature and is described by the Schrodinger equation for the time evolution of a quantum thing $|\Psi\rangle$:
$$H |\Psi(t)\rangle = i \hbar \frac{d}{dt}|\Psi(t)\rangle \, .$$
If $|\Psi\rangle$ happens to have a definite energy, then $H|\Psi(t)\rangle = E|\Psi(t)\rangle$ and we get
$$ E |\Psi(t)\rangle = i \hbar \frac{d}{dt} |\Psi(t)\rangle \, .$$
This ordinary differential equation has the solution
$$ |\Psi(t)\rangle = e^{-i E t / \hbar} |\Psi(0)\rangle$$
which shows that the quantum phase of the thing oscillates with angular frequency $\omega = E/\hbar$, or in other words $f = E / h$.$^{[a]}$
$[a]$: Remember that $\hbar \equiv h / (2\pi)$).
A: 
Essentially this question boils down to "why does the energy of light depend on frequency?

The answer is:This  is what has been observed, measured, for photons. Light is composed by zillions of photons and the energy of the light wave is dependent on the amplitude of the classical electromagnetic wave. This means many more photons are needed for low frequency light to carry the same energy, than for high frequency, which needs fewer. The energy of the photons  is additive.
The photoelectric effect you refer to:

seems to say the opposite, that the energy of each photon is the same and what changes is the amount of photons per area.

Is talking of a specific metal surface, in that case the energy level for releasing an electron is fixed. The more photons one has over that energy the more electrons. In the case of a source of  single frequency photons used in the experiment, this translates to electrons emitted proportional to the area illuminated .
A: If you are sending some light to a surface there are two ways to send more energy.
First, send more photons, each with the same energy. This increases the intensity, but keeps the frequency (and thus the energy per photon) fixed.
Second, increase the energy per photon. This requires increasing the frequency.
You could do neither (and thus not increase the energy) or either (you deliver more energy) or even both (that would be like sending more envelops and sending more money in each envelope).
For the photoelectric effect, a single electron needs to get at least a certain amount of energy to escape the metal. So when you tune the frequency down you can still send energy to the metal but not eject electrons out of it.
If the intensity was absolutely huge you might be able to get two photons to interact with the same electron and get it out, but that would be a very small number. And even if the intensity was quite low you could still get each photon to knock an electron out if the energy of each photon was high enough (which requires the frequency be high enough).
So just think of frequency as telling you the energy per photon, and intensity telling you the total energy being delivered.
For the photoelectric effect, the energy per photon needs to be high enough to eject the electrons and the total intensity will determine the total number of photons, which can scale up the number of electrons emitted (if any are).
