I The atomic Coulomb potential extends to infinity?

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The discussion centers on the interpretation of the statement that the Coulomb potential and the resulting electronic charge distribution extend to infinity. It is clarified that the "charge distribution" refers to the electron's wave function, which theoretically has a non-zero probability of being found at any distance from the nucleus. However, in practical terms, this probability decreases rapidly beyond a few times the Bohr radius. Concerns are raised about the clarity of the text from "Introduction to Nuclear Physics" by Kenneth S. Krane, particularly regarding its interpretation of wave functions. The conversation highlights the distinction between theoretical concepts and practical observations in atomic physics.
aaronll
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I'm studying nuclear physics in a text, but at one point that is said: "Both the Coulomb potential that binds the atom and the resulting electronic charge distribution extends to infinity" , I don't understand what is that "resulting electronic charge distribution extends to infinity" what they mean? ( maybe I misunderstand the phrase but i don't know)
thanks
 
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aaronll said:
I'm studying nuclear physics in a text, but at one point that is said: "Both the Coulomb potential that binds the atom and the resulting electronic charge distribution extends to infinity" , I don't understand what is that "resulting electronic charge distribution extends to infinity" what they mean? ( maybe I misunderstand the phrase but i don't know)
thanks

I assume by "charge distribution" it means the wave-function for the electron. And, theoretically the spatial wave-function for the electron is non-zero everywhere. I.e. no matter how far the distance from the nucleus, there is still a non-zero probability of detecting the electron there.

In practical terms, of course, the electron probability distribution drops off to nearly zero very quickly - of the order of magnitude of a few times the Bohr radius.
 
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PeroK said:
I assume by "charge distribution" it means the wave-function for the electron. And, theoretically the spatial wave-function for the electron is non-zero everywhere. I.e. no matter how far the distance from the nucleus, there is still a non-zero probability of detecting the electron there.

In practical terms, of course, the electron probability distribution drops off to nearly zero very quickly - of the order of magnitude of a few times the Bohr radius.
Thank you
 
Which text is this? I guess, I'll like to avoid its use ;-)).
 
vanhees71 said:
Which text is this? I guess, I'll like to avoid its use ;-)).
Is the "Introduction to Nuclear Physics" written by Kennet S. Krane, I believe is a good book
 
Yes, I like it too, but are there really such statements as that the long-ranged nature of the Coulomb potential of the nucleus in an atom were "resulting electronic charge distribution extends to infinity"? That doesn't make sense or at least hints at an pretty unusual interpretation of the (energy-eigen) wave functions of the electron(s) in atoms. It sounds something like Schrödinger's very first interpretation of his ##\psi(t,\vec{x})## before Born's probabilistic interpretation. Even then it's strange since as the hydrogen wave functions show, the bound states all fall exponentially for ##r \rightarrow \infty##, and from that the typical atomic length scales are determined by the Born radius of about ##0.5 \mathring{\text{A}}##.
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...