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.
![enter image description here](https://i.stack.imgur.com/6N5bY.jpg)
The important part of this experiment should be that there's a relatively sharp cutoff. "Classical" theory would say that the probility of electron ejection is proportional to the energy density, so a high-power long wavelength source would produce the same energy as a low-power, short wavelength source. In fact, photons below a certain energy level cannot eject electrons at all. This was one of the breakthrough experiments which showed that photons, or quantized packets of light energy, exist.
Best Answer
Intensity is the total amount of energy falling (or going through) per unit area per unit time i.e, $\frac{J}{m^2.s}$.
For monochromatic radiation,
$Total\space energy = Number\space of\space photons\space \times Energy\space of\space one\space \space photon$
and $E_{photon}=h\nu$
$Intensity = \frac{Number\space of\space photons\space \times Energy\space of\space one\space \space photon}{At}$
$I=\frac{nh\nu}{At}$,
For constant area and time,
$I \propto n.\nu$
This is a very important result. You can increase the intensity of the radiation by either increasing the number of photons in it or increasing energy of each photon, or both.
How the current can be increased ? :
A single photon with energy equal or greater to the work function $(\phi)$ of the cathode plate will knock out a single photo-electron (Considering no energy is lost by the electron in collisions with other electrons and atoms of the plate).
The statement that increasing the intensity increases the number of electrons emitted by the cathode plate is indeed correct, but incomplete.
$h\nu=\phi + KE_{max}$
As stated before, intensity can also be increased by increasing the frequency of the incident radiation while the number of photons in it is kept constant and this won't change the number of photoelectrons emitted. Although, this would increase the maximum kinetic energy of the photoelectrons.
We have to be a little bit more specific.
Specifically, At constant frequency, if the intensity is increased then the number of photoelectrons will increase.
In other words, if we set the frequency (or wavelength) of the radiation to the threshold frequency (or threshold wavelength) and then increase the intensity, the number of photo-electrons will surely increase.