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What is the relationship between Intensity (amplitude) and energy (frequency/wavelength) in a quantum system?

Physics Asked on March 12, 2021

The photoelectric effect showed us that it is only the energy (frequency x wavelength) of a photon that will eject an electron, NOT the intensity (wave amplitude). But in everyday language it seems that wave amplitude would also be important. And we know that wave amplitude decreases with the square of the distance.
How is wave amplitude related to the actual energy put into the creation of the quantum wave? Why do we use the terms energy and intensity, and how are they related to each other?

One Answer

In more detail to the questions in the comments:

The intensity is the number of photons, while the the energy $E=hf$ is the energy of one photon.

To our question, if one also can eject an electron by using photons whose energies as their owns are enough to eject the electron but their total energy (sum of all) would be enough: Yes this is possible, but more unlikely to happen and only possible for certain energies-configurations of the photons depending on the atom you are looking at. Say you consider the H-Atom and say the electron is in the ground state and we neclegt special relativity and so on...

Then the energy of the electron for all bounded states (labelled by n) is $E_n=-frac{13,6}{n^2}ev$. If the electron is in the ground state (n=1), the energy is -13.6ev. So you can hit the atom with an photon of having this energy or more to eject the electron (ejecting means the electron can get a positive energy). That's clear. But you could also hit it with an photon with energy $E_2-E_1$, which is the energy difference between the 1st excited state $(n=2)$ and the ground state $(n=2)$. After having excited it on the first state you could hit the atom now with a photon of energy $E_1-E_2$ to eject the electron from it's excited state i.e the sum of the absorbed energies would be $E_1$, so the electron is unbounded. But you could also instead excited the electron to the 2nd excited state and so on... Note to eject the electron the sum of all photons energies must sum to $E_1=13.6ev$. So there are a lot of possibilities here to eject the electron. However, if you calculate the quantum mechanically probabilities of the events involving more than one photon it turns out that these are more unlikely. Moreover one assumed in these more-photons-processes that the 1st photons hits the atom, excites the electron immediately and the excited electron is immediately hitted by the next and so on...here one neglegts the fact, that the electron might fall back in the lower states and emits by that a corresponding photon...so in reality this fact makes whose processes more unlikely...

So to summarize: It is possible, but it is unlikely.

Correct answer by CVJM on March 12, 2021

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