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Why does valence electrons not produce electric field in the space? 
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#1
Dec1612, 03:52 AM

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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 ..... 


#2
Dec1612, 05:04 AM

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P: 2,470

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.



#3
Dec1612, 09:05 AM

P: 366

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



#4
Dec1612, 10:13 AM

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



#5
Dec1612, 10:35 AM

P: 366

Well can you calculate the electric field of an electron in a hydrogen atom, or at least write a differential equation for it?



#6
Dec1612, 11:04 AM

Mentor
P: 3,969

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 N1 electrons around it looks more and more like a point charge of +1 the farther away from it you get. 


#7
Dec1612, 09:55 PM

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[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##. 


#8
Dec1612, 10:00 PM

P: 104

But then what happens when I go and perform a position measurment on the electron?



#9
Dec1612, 10:11 PM

P: 366

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. 


#10
Dec1712, 12:16 AM

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P: 2,470

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. 


#11
Dec1712, 12:28 AM

P: 366

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".



#12
Dec1712, 01:04 AM

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#13
Dec1712, 06:14 AM

P: 366

could you link to some website that agrees with you, because this is all news to me



#14
Dec1712, 09:35 AM

Mentor
P: 12,113

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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