In photoelectric emission,
"The number of photoelectrons emitted per unit area per unit time is directly proportional to the intensity of light used."
Is this true if the frequency is not held constant? If yes, please explain how.
If a light of frequency f and intensity I emits n photoelectrons from a metal surface(of threshold frequency f/4) per unit area per unit time, how many photoelectrons will be emitted from the same metal surface per unit area per unit time when a lights of frequency and intensity
1)f/2, I
2)2f, 2I
3)f/2, 2I
are separately used.
[Physics] Number of photoelectrons emitted and the intensity of light used
photoelectric-effect
Related Solutions
In the photoelectric effect one photon displaces one electron, so the way to understand it is to consider the properties of the photons.
If you take light of a fixed frequency, $\nu$, then the energy of the photons is fixed at $h\nu$. That means the intensity of light is proportional to the number of photons, and because one photon = one displaced electron, the intensity of light is proportional to the number of photoelectrons.
When a photon ejects a photoelectron the energy of the electron is equal to the energy of the photon, $h\nu$, minus the work function, $\phi$. This explains why the energy of the electrons, $E$, increases with frequency of the light, $\nu$.
$$E = h\nu - \phi$$
In the last part of your question you say:
means that the further up the EM spectrum the radiation is, the higher the voltage (electric potential) produced?
The photoelectric effect ejects electrons, so whatever you are illuminating will get a positive charge and therefore a potential difference relative to ground. However the potential difference is just related to the total charge, i.e. the number of electrons ejected, so it isn't affected by the energy of the photoelectrons.
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.
Best Answer
Yes, the number of photoelectrons is proportional to the light intensity. At a fixed frequency.
The rest of this question is impossible to answer. The workfunction threshold is given by the OP as $f/4$, so $f$ should be in the far UV. It is then likely that the photoelectric yield is higher at $f/2$. And that it would be lower at $2f$ (likely above the plasmon cutoff).
But photoemission intensities depend on details of the electronic structure of the material: its density of occupied states and also on the final states reached by vertical transitions in $k$-space. And at $2f$ (eight times the work function) there might be emission from a shallow core level.