why does valence electrons not produce electric field in the space?


by ankities
Tags: electric, electrons, field, produce, space, valence
ankities
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#1
Dec16-12, 03:52 AM
P: 9
let X be any hypothetical isolated neutral atom and X- be the anion formed when X gain 1 electron

this extra electron is in the valence energy orbit(band) and X- being charged produces electric field

check me if i m wrng above


now X(neutral) also consists of electrons in the valence band but it does not produce electric field in the space ?

can u pl explain the mechanisms involved .....
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K^2
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#2
Dec16-12, 05:04 AM
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They do. That electric field cancels out the electric field created by positively charged nucleus. If you remove an electron instead, creating a cation, the cancellation is incomplete, and the overall charge is positive.
HomogenousCow
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#3
Dec16-12, 09:05 AM
P: 315
Do note that chemists use really wonky electrodynamics, we can't really define an electric field for a quantum object because it's motion is uncertain

K^2
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#4
Dec16-12, 10:13 AM
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why does valence electrons not produce electric field in the space?


Quote Quote by HomogenousCow View Post
Do note that chemists use really wonky electrodynamics, we can't really define an electric field for a quantum object because it's motion is uncertain
That is entirely false.
HomogenousCow
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#5
Dec16-12, 10:35 AM
P: 315
Well can you calculate the electric field of an electron in a hydrogen atom, or at least write a differential equation for it?
Nugatory
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#6
Dec16-12, 11:04 AM
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Quote Quote by ankities View Post
can u pl explain the mechanisms involved .....
The classical answer (the QM answer would be a lot more work but doesn't change anything qualitatively) is that there is an electromagnetic field if you get close enough to even a neutral atom, and indeed that's why some forms of atomic bonding happen.

However, for a neutral atom this field disappears if you move even a few tens of atomic radii away from the atom. Take the simplest case, hydrogen, with one proton and one neutron. At a point far away from the atom, the electrical fields of the proton are near as no never mind equal in strength (because the distance to the proton is near as no never mind equal to the distance to the electron) and pointing in near as no never mind in opposite directions (opposite charges, but very close to each other) so they pretty much cancel.

But if the atom has gained or lost an electron so that's it ionized, then the magnitudes of the two contributions no longer cancel - there's a net positive or negative charge. An ion with N protons in the nucleus and N-1 electrons around it looks more and more like a point charge of +1 the farther away from it you get.
K^2
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#7
Dec16-12, 09:55 PM
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Quote Quote by HomogenousCow View Post
Well can you calculate the electric field of an electron in a hydrogen atom, or at least write a differential equation for it?
Maxwell's Equations with ##\small \rho = q_e\psi^{*}\psi##. Here, ##\small q_e## is the electron charge. So long as there is no transition, magnetic field remains constant and we have ##\small \nabla\cdot E=\rho/\epsilon_0## and ##\small \nabla\times E=0##. The reason all of this works is that E is a linear field, so an electric field from superposition of sources is a superposition of fields from these sources, and that's just a sum of these fields. So probability density of electron is the charge density of electron. If you have entanglement, things can get interesting, but otherwise, it's extremely straight forward. So lets look at the hydrogen atom in ground state.

[tex]\psi_{100}(r) = \frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}[/tex]

Here, ##\small a_0## is the Bohr Radius, of course. This gives us a perfectly spherically symmetric distribution, so applying Gauss law, the equation for electric field follows.

[tex]E(r) = \frac{1}{4\pi \epsilon_0}\frac{q_e\hat{r}}{r^2}\int_0^r \psi_{100}^{*}(r)\psi_{100}(r) dr = \frac{1}{4\pi \epsilon_0}\frac{e\hat{r}}{r^2}\left( 1-\frac{e^{-2r/a_0}(a_0^2+2ra_0+2r^2)}{a^2} \right)[/tex]

Of course, that's just the electron charge. If we say that nucleus has charge ##\small q_p = -q_e##, and we assume it to be a point source, which it really close to being, then we have the net electric field in the neighborhood of a hydrogen atom.

[tex]E(r) = \frac{q_p\hat{r}}{4\pi \epsilon_0}\frac{a_0^2+2ra_0+2r^2}{a_0^2r^2}e^{-2r/a_0}[/tex]

Note that because of the exponent, this is basically zero for ##\small r>>a_0##.
GarageDweller
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#8
Dec16-12, 10:00 PM
P: 104
But then what happens when I go and perform a position measurment on the electron?
HomogenousCow
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#9
Dec16-12, 10:11 PM
P: 315
Quote Quote by K^2 View Post
Maxwell's Equations with ##\small \rho = q_e\psi^{*}\psi##. Here, ##\small q_e## is the electron charge. So long as there is no transition, magnetic field remains constant and we have ##\small \nabla\cdot E=\rho/\epsilon_0## and ##\small \nabla\times E=0##. The reason all of this works is that E is a linear field, so an electric field from superposition of sources is a superposition of fields from these sources, and that's just a sum of these fields. So probability density of electron is the charge density of electron. If you have entanglement, things can get interesting, but otherwise, it's extremely straight forward. So lets look at the hydrogen atom in ground state.

[tex]\psi_{100}(r) = \frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}[/tex]

Here, ##\small a_0## is the Bohr Radius, of course. This gives us a perfectly spherically symmetric distribution, so applying Gauss law, the equation for electric field follows.

[tex]E(r) = \frac{1}{4\pi \epsilon_0}\frac{q_e\hat{r}}{r^2}\int_0^r \psi_{100}^{*}(r)\psi_{100}(r) dr = \frac{1}{4\pi \epsilon_0}\frac{e\hat{r}}{r^2}\left( 1-\frac{e^{-2r/a_0}(a_0^2+2ra_0+2r^2)}{a^2} \right)[/tex]

Of course, that's just the electron charge. If we say that nucleus has charge ##\small q_p = -q_e##, and we assume it to be a point source, which it really close to being, then we have the net electric field in the neighborhood of a hydrogen atom.

[tex]E(r) = \frac{q_p\hat{r}}{4\pi \epsilon_0}\frac{a_0^2+2ra_0+2r^2}{a_0^2r^2}e^{-2r/a_0}[/tex]

Note that because of the exponent, this is basically zero for ##\small r>>a_0##.
I'm skeptical about the charge density simply being the probability density times the electron charge, that would imply that the electron be spread out throughout space. The electron does have a position upon measurment and we should get the usual 1/r^2 field if we measure the electric field right after.
Besides I really doubt we can apply classical EM to the electron, for example if one makes a position measurment that still leaves the momentum uncertain, hence we cannot know if there is a magnetic field.
K^2
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#10
Dec17-12, 12:16 AM
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What happens in a measurement is a separate question. It will depend on exactly what it is that you are measuring and how. Once you have the interaction associated with the measurement you can find its eigen functions and figure out what will happen to electron's wave function after the measurement.

That, up there, is how you compute an electric field due to an electron with a given wave function.
HomogenousCow
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#11
Dec17-12, 12:28 AM
P: 315
Could you elaborate on why ρ=qeψ∗ψ, I am aware that EM is linear and hence you can superposition sources, but I cannot see why the electron's charge is spread out throughout space. As I understand it, the electron is still a point particle in QM, just that it's position does not exist prior to observation, the state vector projected in the position basis only gives the probability density for finding it in some given volume, and does not represent some kind of "spread".
K^2
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#12
Dec17-12, 01:04 AM
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Quote Quote by HomogenousCow View Post
Could you elaborate on why ρ=qeψ∗ψ, I am aware that EM is linear and hence you can superposition sources, but I cannot see why the electron's charge is spread out throughout space. As I understand it, the electron is still a point particle in QM, just that it's position does not exist prior to observation, the state vector projected in the position basis only gives the probability density for finding it in some given volume, and does not represent some kind of "spread".
It's simply a mis-understanding of significance of wave function on your part, then. Electron is a point particle, and as any point particle, it has a specific location. Of course, the total state is a superposition of such locations. ψ(x) is the coefficient of the state located precisely at x. That component of the state produces electric field as if electron was precisely at x. Then you need to sum over components to get the total state. In the process, you accumulate an electric field equivalent to that produced by charge distribution ψ*ψ. Until you get entanglement into the picture or start looking at interactions, it really doesn't matter if you have a point particle with given probability distribution, or an actual distributed object. Charge density is still proportional to probability density.
HomogenousCow
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#13
Dec17-12, 06:14 AM
P: 315
could you link to some website that agrees with you, because this is all news to me
mfb
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#14
Dec17-12, 09:35 AM
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Most books for an introduction to QM should have this in some way.

<- not a website, but agrees with K^2, too.


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