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:
- Each electron has an effective cross section of interaction (each electron has some "size"). An average cross section of interaction may be defined.
- Electrons are distributed in some manner on the specimen. An average density of electrons per unit area may be defined.
- After an interaction with photon, each electron has some characteristic time during which a second interaction is possible (this time is very hard to estimate; in fact, I don't know if there are analytical methods for performing this estimation). An average characteristic time may be defined.
- The number of photons per unit area per unit time depends on the intensity of the light. An average number of photons per unit area per unit time may be defined.
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 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.
Best Answer
It is simple. What is intensity? Number of photons falling on a surface per unit time per unit area. This means you are supplying more number of photons. Each photon gives one emission of one electron -- then more number of photons will give emission of more number of electron