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L=0, L=1ħ, 2ħ, 3ħ Suborbital Configuration

by euclideanspace
Tags: , , configuration, l1ħ, suborbital
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euclideanspace
#1
Jul16-14, 05:45 PM
P: 22
L=0 Suborbital Configuration


These questions are directed at resident theoretical quantum physicists in PF


Regarding suborbitals L 0 IL 0 VII corresponding to Periods 1 - 7


A total of 14 electron positions exist within Azimuthal Quantum Energy State 0 in Periods 1 through 7

Additionally, Azimuthal Quantum State 0 represents a orbital plane angle along an x-y axis,
and a limit of 1 electron pair perL 0 suborbital;

This suggests 7 discrete coaxial suborbital shells for L 0 without violating the Pauli exclusion principle.

Figure 1 illustrates 7 coaxial L 0 suborbital shells and 14 electron positions,
numbered L 0 IL 0 VII, with 7 electron pairs aligned along the y axis, each electron representing 1 standing wave nodal loop,
and the crest of each node being 1 negative point charge.

Question 1 : Are the 7 suborbital shells illustrated in Figure 1 in agreement with the current standard model?


Regarding Azimuthal Quantum Energy States : L+1ħ, +2ħ, +3ħ → ]L -1ħ, -2ħ, -3ħ

with the point charges at each node counted as discrete electrons
which would result in :

3 suborbital shells (x 2) for L +1ħ and -1ħ respectively, with 6 electrons (nodal point charges) per suborbital shell per suborbital shell
a total of 36 electron positions for Periods 2 → 7

L +1ħ and -1ħ illustrated in Figure 2 and Figure 3

2 suborbital shells (x 2) for L +2ħ and -2ħ respectively, with 10 electrons (nodal point charges) per suborbital shell,
a total of 40 electron positions for Periods 4 → 7

L +2ħ and -2ħ illustrated in Figure 4 and Figure 5

and

1 orbital shell (x 2) for L +3ħ and -3ħ respectively, with 14 electrons (nodal point charges) per orbital shell
a total of 28 electron positions for Periods 6 → 7

L +3ħ and -3ħ illustrated in Figure 6 and Figure 7


Question 2 : Are the suborbital shells illustrated in Figures 2 → 7 in agreement with the current standard model?


Thanks in advance

~ E:S
Attached Thumbnails
Fig1.png  
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euclideanspace
#2
Jul16-14, 05:45 PM
P: 22
L 0 refers to Quantum Number 0 typically known as s shell (ground state)

I use Quantum Number rather than the outdated and misconceived sharp, diffuse, principle and fundamental convention.

There are 14 electron positions in Periods 1 → 7

Figure 1 presumes these 14 positions require 7 L 0 sub-shells


following the Aufbau principle :

The 14 electron positions are more clearly explained through the shell structure of element 118 Ununoctium

02→02 16 → 02 16 210 → 02 16 210 314 → 02 16 210 314 → 02 16 210 → 02 16

The arrows (→) indicate Aufbau per Period

For Question 2 :

Please See Figures 2 → 7
Attached Thumbnails
Fig2.png   Fig3.png   Fig4.png   Fig5.png   Fig6.png  

Fig7.png  
ChrisVer
#3
Jul16-14, 06:09 PM
P: 919
Although the thing is already answered, I don't find the reason to repost. However I will make this as clear as possible.
The planetary motion model for the electrons in atoms is not correct. So your questions are wrong from their basis. The Standard Model describes the particle interactions and so it's irrelevant. Your question is about quantum mechanics. So back to the point, there are no point balls going around the big ball - these images in books are just illustrative to see the energy differences of orbitals and not to see the actual atom.. The model was lacking a lot of information actually observed. It's latest model being the Bohr's one failed to:
http://en.wikipedia.org/wiki/Bohr_model#Shortcomings
Of course as a semiclassical model (but we know it today), it was supposed to give some hints.
To these, came QM to give a better description and explain them.
For that reason you don't treat electrons as balls moving in trajectories and so being alligned or not, but you are trying to study their wavefunction... The [itex]L_{z}=0[/itex] doesn't really mean a planar motion, as already given to you, it represents a spherical symmetry ( https://en.wikipedia.org/wiki/Atomic_orbital ) . [itex]L_{z}[/itex] is actually the generator of rotations, it being zero means nothing else than that there is non angular dependence in your wavefunction, and that's why it's described as a sphere (of course the sphere corresponds to the angular coords).
So what's the point ? A failed model is not necessarily able to be in agreement with QM...

euclideanspace
#4
Jul16-14, 06:30 PM
P: 22
L=0, L=1ħ, 2ħ, 3ħ Suborbital Configuration

Although the thing is already answered, I don't find the reason to repost. However I will make this as clear as possible.
The planetary motion model for the electrons in atoms is not correct. So your questions are wrong from their basis. The Standard Model describes the particle interactions and so it's irrelevant. Your question is about quantum mechanics. So back to the point, there are no point balls going around the big ball - these images in books are just illustrative

Sorry, I didn't get a chance to read the replies in the old thread.

The planetary model is long dead, however the illustrative images are still useful tools
for understanding structure. I need a model to illustrate the point charge locations,
Quantum Physics is incomprehensible without visual tools.

Fortunately I was able to back button my browser to the deleted thread :

mfb

There are no "paths" or "positions" for electrons in an atom (or for any other quantum-mechanical object), and no "electron pairs aligned along [something]".

Those 14 electrons correspond to the principal quantum numbers n=1, n=2 and so on, with 2 electrons of opposite spin in each orbital. They have different radial distributions of their orbitals.
I know the principal quantum numbers n=1, n=2 n=3 relate to spherical harmonics
(See Attached Image)

the question is mainly regarding the ground state electrons
Attached Thumbnails
Spherical_Harmonics.png  
ChrisVer
#5
Jul16-14, 07:02 PM
P: 919
The illustrative images are not good tools to understand the atom... They are good tools to see the energy differences let's say... Otherwise they can be totally misleading .
You can't put locations to something that can be anywhere in some volume with the same probability anywhere around a sphere...The charge is distributed around the nucleus..
The ground state electrons means that you have no excited electron, nothing more nothing less....The answer of mfb didn't have anything to do with that distinction. The radial distribution is all about the 3rd coordinate to complete the volume- the radius... In that case the different n numbers correspond to different radial distributions - each more extended outwards with no vanishing point (for s orbitals).
Your image doesn't say anything about n's either.... it's the spherical harmonics.... for example the top one is the s ... there exists s for n=1, n=2, .... the n number has more to do with energy, or BETTER put with the radial distribution... because it characterizes the radial part of the wavefunction (the energy can have L dependencies ).
euclideanspace
#6
Jul16-14, 09:45 PM
P: 22
Are you saying then that the 14 s-shell electrons are a spherical shell of distributed radial charges,
rather than point charges within concentric s-shells?

My understanding of the Pauli exclusion principle, is that a finite number of electron locations
exist at each quantum energy level : 0 = 2, 1 = 6, 2 = 10 and 3 = 14
and that lower shells must be filled before filling higher energy states, according to the Aufbau principle.

Are the Pauli exclusion princple and Aufbau are still valid (?)

If we look at the shell structure of element 118, we find 14 distinct electron locations within energy state L=0 (aka. s-shell)

02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 210 314 → 02 16 210 → 02 16

you may substitute s, d, p and f for 0, 1, 2, 3 if you wish - I see s, d, p and f as obsolete terms.

Aufbau fills by order 0, 3, 2, 1 which describes the Periodic Table arrangement.

There cannot be 118 electrons in element 118, without 14 electrons in shell 0,
36 electrons in shell 1, 40 electrons in shell 2, and 28 electrons in shell 3.

14 + 36 + 40 + 28 = 118

There cannot be a total of 118 electrons in element 118 any other way.

each Period in the table adds 2 electron locations to shell 0
A total of 7 Periods = 14 electron locations

This is what is represented in Figure 1.

So I need to know how and why this concept is not correct?
(by concept I do not mean actual geometry - only fill)

I visualise electrons as point charges at nodal peaks, with counter-directional ½ spin
analogous to (+) and (-) nodes on a sine wave, is this interpretation incorrect?
PeterDonis
#7
Jul16-14, 10:35 PM
Physics
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P: 6,172
Quote Quote by euclideanspace View Post
Are you saying then that the 14 s-shell electrons are a spherical shell of distributed radial charges, rather than point charges within concentric s-shells?
They're neither. The "shells" refer to particular electron wave functions. Wave functions describe probability amplitudes, not electron positions or charge distributions. See further comments below.

Quote Quote by euclideanspace View Post
My understanding of the Pauli exclusion principle, is that a finite number of electron locations
exist at each quantum energy level : 0 = 2, 1 = 6, 2 = 10 and 3 = 14
and that lower shells must be filled before filling higher energy states, according to the Aufbau principle.
You have the principle right, but you appear to be mistaken about the ordering of the energy levels. I'll use the s, p, d, f nomenclature since that's how I originally learned this stuff. The ordering of energy levels in that nomenclature looks like this (from lower to higher energy): 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, ...

So the L=0 states are not all the lowest energy states: there are L=1 states that are lower in energy than some L=0 states, etc. But you have the number of electrons that can occupy each set of states correct: each set of s states (1s, 2s, etc.) can be occupied by 2 electrons (of opposite spins), each set of p states (2p, 3p, etc.) can be occupied by 6 electrons (3 pairs each with opposite spins), and so on.

If you then fill the states in the order I gave above, from lowest to highest energy, one electron at a time, you will generate the periodic table of the elements: H is 1s1, He is 1s2, Li is 1s2-2s1, Be is 1s2-2s2, B is 1s2-2s2-2p1, etc. So the Aufbau principle works, but you have to get the ordering of the states by energy correct to apply it. For element 118, the configuration would be 1s2-2s2-2p6-3s2-3p6-4s2-3d10-4p6-5s2-4d10-5p6-6s2-4f14-5d10-6p6-7s2-5f14-6d10-7p6. I think this matches what you have, but if, for example, we looked at some other element in period 7, you might get an incorrect answer because of the order in which you appear to think the shells get filled.

Quote Quote by euclideanspace View Post
I visualise electrons as point charges at nodal peaks, with counter-directional ½ spin
analogous to (+) and (-) nodes on a sine wave, is this interpretation incorrect?
Yes. One easy way to see that it must be incorrect is to note that an L=0 state must be spherically symmetric, but a spherically symmetric wave cannot have any nodes as you have drawn them: so if electrons are point charges at nodal peaks, where are they in an L=0 state?

The fact is that there is *no* easy classical visualization that corresponds to what the wave function actually means in quantum mechanics. The best you can do is to think of the wave function, as I said above, as describing a distribution of probability amplitude--roughly speaking, the electron is more likely to be detected (if we were to try to detect it) at locations where the wave function's amplitude is larger. The various L=0 (s) states represent spherically symmetric distributions with amplitudes that take on their maximum value at some particular radius from the nucleus: the smaller the radius, the lower the energy of the state. The L=1 (p) and higher states represent distributions with different symmetry properties.
euclideanspace
#8
Jul17-14, 07:22 AM
P: 22
You have the principle right, but you appear to be mistaken about the ordering of the energy levels
I actually wrote out the shell structure for 118 as an illustration of the 14 L=0 locations,
as an attempt to clarify my location question.
admittedly it was overly simplified, and not according to Aufbau.

my Aufbau Abschluss illustration is attached below, which shows the fill order for all but 20 anomalous elements :
Cr, Mn, Cu, Zn, Pd, Ag, Cd, La, Lu, Eu, Gd, Yb, Ac, Th, Au, Hg, Am, Cm, Rg, Cn

They're neither. The "shells" refer to particular electron wave functions. Wave functions describe probability amplitudes, not electron positions or charge distributions.
Are you refering to the Schrödinger wave function or Dirac wave function?

If probability amplitudes means that the 14 electron positions are actually 7 amplitudes,
and magnitudes of elementary charge, then I understand perfectly.

I have a clear understanding of the principles of quantum mechanics, it's the visualisation problem
of the quantum conundrum I have some difficulty wrapping my head around.
As an engineer, I can resolve problems mathematically as long as I have a clear understanding of the process.
however, as a visual artist, I have a need to visualise things in order to understand them.

I'm getting gradually closer to making sense of the location problem - Thanks Peter!
Attached Thumbnails
Aufbau Abschluss.jpg  
PeterDonis
#9
Jul17-14, 04:21 PM
Physics
Sci Advisor
PF Gold
P: 6,172
Quote Quote by euclideanspace View Post
my Aufbau Abschluss illustration is attached below, which shows the fill order for all but 20 anomalous elements :
Cr, Mn, Cu, Zn, Pd, Ag, Cd, La, Lu, Eu, Gd, Yb, Ac, Th, Au, Hg, Am, Cm, Rg, Cn
I agree with the order shown in your illustration. I'm not sure about the exact list of "anomalous" elements; do you have a specific source for that?

Quote Quote by euclideanspace View Post
Are you refering to the Schrödinger wave function or Dirac wave function?
Schrodinger. The wave functions are solutions of the Schrodinger equation for the case of electrons in the static Coulomb field of the nucleus.

Quote Quote by euclideanspace View Post
If probability amplitudes means that the 14 electron positions are actually 7 amplitudes,
and magnitudes of elementary charge
That's not what it means. A wave function is a function of position with respect to the nucleus; the value of the function at a particular position is the probability amplitude for the electron to be at that position. "Probability amplitude" here is a technical term that was adopted in quantum mechanics because, in order to get the actual probability of detecting an electron at a particular position, you have to take the square of the amplitude (more precisely, the squared modulus, since in general the amplitude can be a complex number).

So the diagrams and drawings that you often see of electron orbitals (an "orbital" is a particular electron wave function, i.e., a particular solution of the Schrodinger equation with the appropriate constraints) are diagrams of probability amplitude as a function of position. But the position of an electron in an atom never actually gets measured, so the electrons don't actually have positions; the states they are in are not position eigenstates. They're energy eigenstates. (More precisely, they are eigenstates of energy and orbital angular momentum.)
euclideanspace
#10
Jul17-14, 04:55 PM
P: 22
I agree with the order shown in your illustration. I'm not sure about the exact list of "anomalous" elements; do you have a specific source for that?
coming..
euclideanspace
#11
Jul18-14, 04:52 AM
P: 22
I agree with the order shown in your illustration. I'm not sure about the exact list of "anomalous" elements; do you have a specific source for that?
We can let the elements speak for themselves

The obvious anomalous elements have been highlighted in blue

For example Palladium has 10 Valence electrons
Lanthanum is an f-block element with no f-block electrons!

if the periodic table were arranged strictly according to block and fill electrons
the following elements : Cr, Cu, Nb, Mo, Tc, Ru, Rh, Pd, Ag, La, Gd, Ac, Th, Au, Pu, Am, Ds, Rg are out of place.




K.......... 02 → 02 16 → 02 16 20 → 01
Ca........ 02 → 02 16 → 02 16 20 → 02
Sc......... 02 → 02 16 → 02 16 21 → 02
Ti.......... 02 → 02 16 → 02 16 22 → 02
V........... 02 → 02 16 → 02 16 23 → 02
Cr......... 02 → 02 16 → 02 16 25 → 01
Mn........ 02 → 02 16 → 02 16 25 → 02
Fe......... 02 → 02 16 → 02 16 26 → 02
Co........ 02 → 02 16 → 02 16 27 → 02
Ni......... 02 → 02 16 → 02 16 28 → 02
Cu........ 02 → 02 16 → 02 16 210 → 01
Zn........ 02 → 02 16 → 02 16 210 → 02


Y........... 02 → 02 16 → 02 16 210 → 02 16 21 30 → 02
Zr......... 02 → 02 16 → 02 16 210 → 02 16 22 30 → 02
Nb........ 02 → 02 16 → 02 16 210 → 02 16 24 30 → 01
Mo........ 02 → 02 16 → 02 16 210 → 02 16 25 30 → 01
Tc......... 02 → 02 16 → 02 16 210 → 02 16 26 30 → 01
Ru........ 02 → 02 16 → 02 16 210 → 02 16 27 30 → 01
Rh........ 02 → 02 16 → 02 16 210 → 02 16 28 30 → 01
Pd........ 02 → 02 16 → 02 16 210 → 02 16 210 30 → 00
Ag........ 02 → 02 16 → 02 16 210 → 02 16 210 30 → 01

Cd........ 02 → 02 16 → 02 16 210 → 02 16 210 30 → 02

Cs........ 02 → 02 16 → 02 16 210 → 02 16 210 30 → 02 16 20 30 → 01
Ba........ 02 → 02 16 → 02 16 210 → 02 16 210 30 → 02 16 20 30 → 02
La........ 02 → 02 16 → 02 16 210 → 02 16 210 30 → 02 16 21 30 → 02
Ce........ 02 → 02 16 → 02 16 210 → 02 16 210 32 → 02 16 20 30 → 02


Pm........ 02 → 02 16 → 02 16 210 → 02 16 210 35 → 02 16 20 30 → 02
Sm........ 02 → 02 16 → 02 16 210 → 02 16 210 36 → 02 16 20 30 → 02
Eu........ 02 → 02 16 → 02 16 210 → 02 16 210 37 → 02 16 20 30 → 02
Gd........ 02 → 02 16 → 02 16 210 → 02 16 210 37 → 02 16 21 30 → 02
Tb......... 02 → 02 16 → 02 16 210 → 02 16 210 39 → 02 16 20 30 → 02
Dy......... 02 → 02 16 → 02 16 210 → 02 16 210 310 → 02 16 20 30 → 02
Ho........ 02 → 02 16 → 02 16 210 → 02 16 210 311 → 02 16 20 30 → 02
Er......... 02 → 02 16 → 02 16 210 → 02 16 210 312 → 02 16 20 30 → 02
Tm........ 02 → 02 16 → 02 16 210 → 02 16 210 313 → 02 16 20 30 → 02
Yb......... 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 20 30 → 02

Lu......... 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 21 30 → 02
Hf.......... 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 22 30 → 02
Ta......... 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 23 30 → 02
W.......... 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 24 30 → 02
Re......... 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 25 30 → 02
Os......... 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 26 30 → 02
Ir........... 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 27 30 → 02
Pt.......... 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 29 30 → 01
Au......... 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 210 30 → 01

Hg......... 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 210 30 → 02

Ra........ 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 210 30 → 02 16 20 30 → 02
Ac........ 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 210 30 → 02 16 21 30 → 02
Th........ 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 210 30 → 02 16 22 30 → 02
Pa........ 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 210 32 → 02 16 21 30 → 02


U........... 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 210 33 → 02 16 21 30 → 02
Np......... 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 210 34 → 02 16 21 30 → 02
Pu......... 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 210 36 → 02 16 20 30 → 02
Am........ 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 210 37 → 02 16 20 30 → 02
Cm........ 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 210 37 → 02 16 21 30 → 02


Ds........ 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 210 314 → 02 16 29 30 → 01
Rg........ 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 210 314 → 02 16 210 30 → 01
Cn........ 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 210 314 → 02 16 210 30 → 02
Fl......... 02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 210 314 → 02 16 210 30 → 02 12
euclideanspace
#12
Jul18-14, 05:05 AM
P: 22
the electrons don't actually have positions; the states they are in are not position eigenstates. They're energy eigenstates. (More precisely, they are eigenstates of energy and orbital angular momentum.)
so the 14 L=0 electrons in element 118 are simply scattered at the L=0 energy level??
ChrisVer
#13
Jul18-14, 05:11 AM
P: 919
it's not "Scattered"... it's you don't know.... they may be within some volume. the energy level doesn't define any position for you to say "somewhere in that energy level" and that's also why I told you before that those sketches in the book although they look like radii don't correspond to radii but to energy levels.

Of course the radial wavefunction depends on the energy levels, but this doesn't give much insight.... a higher L0 energy level has its radial mean value at some greater radius than the smaller... However the electrons are not just on the mean value radius...Otherwise the wavefunction would have to be a delta function : [itex]\delta(r-r_{mean})[/itex], which doesn't work for having a definite energy.
http://hyperphysics.phy-astr.gsu.edu...gmod2/hy2s.gif
The diagram is for a Hydrogen atom... you can see that the 1s, 2s, 3s (and in general so on) can share the same radii
euclideanspace
#14
Jul18-14, 05:21 AM
P: 22
If one cannot 'know' the position of specific electrons, how is it possible to specify the shell structure?

in element 118 for example - we just assume the electrons are distributed at these energy levels, just by calculation alone?

02 → 02 16 → 02 16 210 → 02 16 210 314 → 02 16 210 314 → 02 16 210 → 02 16


It would be more intuitive if the electron quantitative number were simply a charge level amplitude (magnitude in units of elementary charge), then it would make perfect sense.
ChrisVer
#15
Jul18-14, 05:28 AM
P: 919
because the shell structure tells you almost nothing about the radius of the electron in the Quantum Mechanical framework, in order to write these things (02 → 02 16 → 02...... ) you just need to know the energies and not the positions/radii. To build them you just say "I want at the ground state to have the least energy availabe" and so you make it up. It can tell you about the radii meanvalues alone, because the mean values behave classically and the classical treatment of the Hydrogen atom tells you that an electron further away from the nucleus will have more energy.
Your numbers correspond to [itex]n[/itex] and [itex]l[/itex] quantum number, which is the number that defines your energy and not your position. The energy of a hydrogen-like atom for example is like [itex]E \propto \frac{1}{n^{2}}[/itex], you can't say that for the position in general (because it's not definite). You may also have [itex]l[/itex] dependence since you can use an effective potential to account the centrifugal force, and that's why at some points you have the [itex]l[/itex] quantum numbers not following exactly the scheme of 1/n^2 (the [itex](n-1),d[/itex] for example can be filled before the next : [itex]n,s[/itex]. So nevertheless, the energy of the Hydrogen atom electron cannot be measured simultaneously with its position. Or the Hamiltonian doesn't commute with the position operator. The only way you can relate the [itex]n[/itex] with the radii, is by relating it to the mean values alone.
PeterDonis
#16
Jul18-14, 09:46 AM
Physics
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PF Gold
P: 6,172
Quote Quote by euclideanspace View Post
so the 14 L=0 electrons in element 118 are simply scattered at the L=0 energy level??
L=0 is not an energy level; it's an angular momentum eigenstate. Electrons with L=0 can be in different energy levels: 1s, 2s, 3s, etc. Each of those states is a different wave function, i.e., a different distribution of probability amplitude in space: the only thing they have in common is that they are all spherically symmetric (since they all have L=0).
euclideanspace
#17
Jul18-14, 10:46 AM
P: 22
L=0 is not an energy level; it's an angular momentum eigenstate. Electrons with L=0 can be in different energy levels: 1s, 2s, 3s, etc. Each of those states is a different wave function, i.e., a different distribution of probability amplitude in space: the only thing they have in common is that they are all spherically symmetric (since they all have L=0).
I was previously trying to model the shell according to concentric rms amplitudes by energy level, with a nodal concept in mind.

Now I understand the principle perfectly.

so, is Richard Feynman's description of an "electron cloud" a reasonable analogy?
PeterDonis
#18
Jul18-14, 11:34 AM
Physics
Sci Advisor
PF Gold
P: 6,172
Quote Quote by euclideanspace View Post
is Richard Feynman's description of an "electron cloud" a reasonable analogy?
I would say it's ok as a very rough, heuristic description. I don't know that it really helps much in making actual predictions.


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