We know that the ionization energy of hydrogen atom is 13.6 eV. What bothers is this energy corresponds to infinite seperation between the electron and proton. If we assume that this is true then in photoelectric effect we couldn't measure current as the electron would be far away (infinite distance) from the amperemeter. So what happens when the electron gains exactly 13.6 eV?
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asked Jan 2, 2020 at 14:03
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$\begingroup$ The statement: "What bothers is this energy corresponds to infinite seperation between the electron and proton. " is just wrong or very unclear. Please improve the formulation. $\endgroup$– ohneValCommented Jan 2, 2020 at 14:06
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$\begingroup$ Slightly more, and the photoelectron goes off with some kinetic energy. Slightly less, and one creates a Rydberg atom, maybe larger than a bacterium. (In atomic physics, there is no amperemeter connected.) $\endgroup$– user137289Commented Jan 2, 2020 at 14:20
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The Hydrogen atom model is a very basic one. It consists of a an electron in a central potential (which represents the proton), and assumes nothing else exists in the universe. This is of course a simplification, but a necessary one. In this model, the electron has bound states, which are states characterized by energy less than zero, and more importantly, by the property that as $r\to \infty$ the wave function of the electron decreases exponentially. This means that for these states the probability to find the electron at a very large distance from the proton is rapidly going to zero. Of these states, the one with the lowest energy has $E=-13.6\, \mathrm{eV}$.
There are also unbound states with energy greater than zero. These states describe the electron as a wave stretching throughout the entire space, and indeed, if we do not consider anything else in the model, the electron in this state can be found anywhere in the universe! Within this model, if we magically give the electron in the hydrogen model slightly more than $13.6\, \mathrm{eV}$ of energy, it will transition into one of the low-energy unbound states and we would be able to find it with some probability everywhere.
This is of course not a proper physical picture. For example, if there is nothing else in the universe, how are we to give the electron energy? Also - what is the meaning of these "unbound states" and how can we consider them if they cannot be normalized? There are ways to carefully treat these issues, by applying controlled approximations and adding terms that will allow us to carry out meaningful calculations.
For the purpose of the photoelectric effect with hydrogen atoms, when considering an experiment, we can think of the following scenario -- there are many hydrogen atoms, and we can think of each of them separately using the isolated single-electron-in-a-central-potential-model (when can we do that? not always! it depends on their density, for example). Then we "pump energy" into the electrons using light, where the light carries $13.6\, \mathrm{eV}$ of energy. Some of it does nothing, but some of it allows the electrons to transition into the unbound state. There are many of these electrons, and since they have a probability to be found anywhere inside our experimental setup, some of them will be measured at the ampermeter, giving rise to a reading of finite current.
There are many more delicate points we can consider but for the naïve picture of the experiment this should be enough to describe the results.
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$\begingroup$ Two typesetting comments. It is usual to set until off from their quantity with a thin-space (
\,in LateX/MathJax) and to use an upright typeface for units (either\mathrmor\text). The result is $-13.6\,\mathrm{eV}$ (-13.6\,\mathrm{eV}). $\endgroup$ Commented Jan 2, 2020 at 15:15 -
$\begingroup$ @dmckee thanks. fixed it (i hope) $\endgroup$– user245141Commented Jan 2, 2020 at 15:25