photoelectric-effectquantum mechanics
Provided it is above the threshold frequency of the metal, when electromagnetic radiation is shone onto a metals surface photo-electrons are emitted. This occurs because 1 photon is absorbed by 1 electron giving it enough energy to be ejected.
We know that the energy of the incident photons are all equal from the equation E = hf. If this is so why does the kinetic energy of the emitted photons vary? Why is there a maximum kinetic energy, is it not the same amount every time?
In general you're right - an electron being subject to interactions with more than a single photon may have a higher kinetic energy. However, in the vast majority of photoelectric setups you will observe that kinetic energy is independent of light's intensity.
The appropriate framework for this discussion is this of probability theory:
Now, you should ask the following question: "given the effective cross-section of interaction of electrons, the average number of electrons per unit area, the average characteristic time and the average number of photons per unit area per unit time, what is the probability for an electron to interact with more than one photon?".
The usual answer to the above question is "negligible". This happens, but so rarely that the current due to these electrons is below your measurement error.
However, in high intensity experiments (where the number of photons per unit area per unit time is enormous), multi-interaction-electrons were observed. See this for example.
Analogy:
The best analogy I can think of is this of rain. You may think about individual photons as drops of rain, about individual electrons as people in the crowd (each of whom has an effective cross-section of interaction which depends on how fat the person is :)), and about the characteristic time as of time it takes to open an umbrella over the head.
Now, if the rain is weak (usually when it just starts), each person in the crowd is hit by a single first drop. He takes his umbrella out of his bag and opens it above his head. If he does this sufficiently fast (short characteristic time), he will not be hit by more drops.
However, there are cases when the rain has no "few drops per minute" phase - it almost instantly starts and is very intensive. In this case, no matter how fast the people open their umbrellas, they will be hit by many drops.
There is a publication that calculates the photoelectric effect using Feynman diagrams.( Here is a pdf )
Here are two screen captures.
this one reduced so as to capture the Feynman diagram.
Anybody interested should go to the link above.
Best Answer
There are a couple of reasons for this. First and foremost, the electrons are ejected from the surface of the metal in random directions. When you measure things like the "stopping potential" you're only sensitive to motion in directions that would carry the photo-electron to the anode. Because you're only sensitive to motion in particular directions, you only see the part of the kinetic energy along that particular direction. It is this kinematic messiness that makes measuring the reverse potential needed to stop all current the preferred method of measuring the photoelectric effect - the last electrons stopped will be the ones where the largest possible fraction of the photon's energy went in to propelling the electron toward the anode.
Secondarily, the valence electrons in metals have energies in what are known as the "conduction band", which means the electron's energy can exist in a continuum. That fact, combined with the random messiness inherent in thermodynamics, means that no two electrons will have the same kinetic energy before the photons hit them. Thus, each electron will have a slightly different kinetic energy after it gets ejected from the metal.
Of approximately equal weight is the spread in frequencies for the incident light. See, even if your light is produced by a nice sharp atomic line, like in a low pressure mercury lamp, the mercury atoms in the gas will be undergoing thermal motion, leading to Doppler broadening of the line.