Say a photon of frequency $\nu_1$ strikes the metal, and we have certain kinetic energy which is stopped by $V_{stop1}$. Now for frequency $\nu_2$ for same metal it will have $V_{stop2}$. Now $V_{stop}=\frac{1}{2}mu^2$. So $V_{stop}$ determines the $u$. Now current is defined as rate of flow of charge. So the flow of charge per unit time depends upon the number of electrons (here intensity) and how fast do they move (here kinetic energy or $u$). So for same intensity and different frequency, my logic says that current will differ and vice versa when we change intensity keeping frequency same. This seems to contradict photoelectric effect observation. Can you tell me where I have gone wrong?
Thanks in advance
[Physics] Photoelectric effect and variation of current
photoelectric-effectphotons
Related Solutions
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.
In order to answer your question you must think through the entire measurement. In this case you are measuring the photoelectric effect by placing a photodiode into a circuit. What kind of circuit element is the photodiode? It is a source. And as you have pointed out above, it is a constant voltage source (with voltage determined by the energy of the photons and the work function of the metal). It is also current limited (with no load it still cannot draw more current than there are incoming photons). I think you may be slightly confusing yourself because current is charge/time so it depends on how fast the electrons are moving (i.e. the voltage). Always recall that it is a constant voltage source.
To proceed with understanding your lab you need to understand your circuit. There is a schematic diagram of the photodiode, op-amp, and associated circuitry in your pdf. I am sure you are familiar with the basic idea: the circuit acts as a capacitor which builds up voltage until the current is 0 (or approaches 0). The 'current-limited' feature of the diode power supply effects the capacitor charging only at the start of a measurement. The trick to this equipment is understanding the op-amp. There is plenty of literature out there on op-amps. Wiki is always a good place to start.
If you have done your lab by now you will have noticed that there IS some intensity dependence. How can you explain this? (as an aside... ideally you have recorded voltage vs time for each measurement and fit it to an exponential--this assumes it is capacitor charging at constant voltage) I usually tell my students to chalk it up to the finite input impedance of the amp causing the current at the input to the amp to be non-zero and changing the feedback mechanism. I got this explanation from a professor a long time ago, and I am not sure I entirely buy it (Pasco claims there is an enormous input impedance). I never came up with a better answer while I was teaching the lab though.
Related Question
- [Physics] Photoelectric Effect – Dependence of current on frequency
- [Physics] Why do electrons move towards anode in the Photoelectric effect experiment
- [Physics] How is photo-current related to stopping voltage in the Photoelectric Effect
- Why photocurrent doesnt increase after saturation limit in Photoelectric effect on increasing Voltage
Best Answer
A photon that strikes the metal either has enough energy to release a valence electron from its bond with the atoms in the metal, from a certain frequency and up, or its energy is just not enough to eject the electron from the metal. The surplus of energy over the energy needed to eject the electron is used as kinetic energy of the electron and determines the speed with which it escapes the metal.
The current that starts running, however, is determined by the number of electrons that is ejected from the metal, as this creates a potential difference within the metal. The speed at which they move away from the metal does not matter in general, as they move in any random direction away from the metal.
The number of ejected electrons is determined by the number of photons striking the metal and therefore by the intensity of the light, not by its frequency.
I hope this helps you.