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If an electron falls through the nucleus of an atom...


Butch

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6 hours ago, Butch said:

In my atom the greatest probability for the position of an electron is at the apogee, note that the probability of its position in the nucleus is not 0. Note also that if an electron is at apogee another is at perigee in the nucleus. This does not hold true for all my atoms (without naming the atom) in an atom composed of 3 pairs when an electron is at apogee, 2 are near the nucleus but one is post perigee another is pre perigee.

Trying to understand wave function, in this atom:

IMG_20171216_143123.jpg

How would I describe the relationship of e1 and e2 as wave functions?

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16 minutes ago, Butch said:

Trying to understand wave function, in this atom:

 

How would I describe the relationship of e1 and e2 as wave functions?

 

You wouldn't directly.

The wave function of anything describes its interaction with the whole of its environment.

 

The protons constitute the dominant part of the environment of these electrons.

This accounts for main quantum numbers.

Only the Pauli exclusion number describes the fact they (must) have opposite spins.

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15 minutes ago, studiot said:

Only the Pauli exclusion number describes the fact they (must) have opposite spins.

So... The wave function would be the same for both but the spin would be opposite and if the wave function for e1 was 2, the wave function for e2 would have to be 2 for this system to maintain balance. 

I take it that even though my electrons "fall" through the nucleus,  the wave function really is about cyclical time, not altitude of orbit and even though my electrons accelerate and decelerate unless their average energies change their wave function remains the same?

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The solution of the Schrodinger equation results in a system of four quantum numbers for electrons in atoms.

 

The principle quantum number   symbol n    values n    1, 2 ,3, 4.........

The subsidiary quantum number   symbol l   values  0, 1, 2, ....(n-1)

The magnetic quantum number   symbol m  values  -l,  (-1+1), ....-1, 0, +1, ...(l-1), l

The spin quantum number    symbol s   values -1/2,  +1/2

Energy increases with increasing n or l

Each value of n refers to a separate  quantum shell. The shells correspond to the largest jumps in energy.
Within each shell the subsidiary quantum number divides the energy levels.
So n=1, l=0 corresponds to the 1s shell.

Strictly the solution is for hydrogen, and the quantum numbers refer to the one available electron.

Although all the states represented by all the quantum numbers exist, they are just empty except for one.

Luckily the quantum solutions are sufficiently similar for the atoms of all the other elements that we can simply populate them with the appropriate number of electrons.

 

 

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22 hours ago, swansont said:

What part of "stick to hydrogen" is causing confusion?

Ok, so no matter the path of the electron, it must cycle true to it's wave function. Any excess energies are attributed to spin... Correct?

A hydrogen1 atom at ground state would have a wave function of 1?

On 12/15/2017 at 3:22 PM, Butch said:
On 12/15/2017 at 1:52 PM, swansont said:

At what speed? Not anywhere close to the speed of an electron in your model.

 

Correct, but it would wobble and in more complex atoms that would have to be accounted for because of the periodically close proximity of the electrons to the nucleus.

Ok, thanks to qm, the wobble does not matter and thanks to wave function the math is simpler. 

Thank you... Again!

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2 hours ago, Butch said:

Ok, so no matter the path of the electron, it must cycle true to it's wave function. Any excess energies are attributed to spin... Correct?

It has no path in QM. The energy difference due to spin shows up in the hyperfine structure, and only because the proton has a magnetic moment. The overall energy levels are due to the electrostatic interaction.

2 hours ago, Butch said:

A hydrogen1 atom at ground state would have a wave function of 1?

I'm not sure what that is supposed to mean.

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He is reading the earlier chapters of volume 1 of the Feymann lectures which is a bit confusing until you get into the later chapters. In the earlier chapters Feyman is stepping the reader into how QM would describe the atom. 

Albeit it will cause some confusion. He is currently on the first 5 chapters of volume 1

http://www.feynmanlectures.caltech.edu

The way Feymann put these chapters together in its attempt to simplify is a little sketchy on details. He clarifies and refines the more complete and accurate physics involved later on

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15 hours ago, swansont said:
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A hydrogen1 atom at ground state would have a wave function of 1?

I'm not sure what that is supposed to mean.

The ground state electron is most likely to be found at a distance from the nucleus equal to the Bohr radius, 1s atomic orbital (sorry for using that word again, I am transitioning to qm) If the single electron in the hydrogen atom is at ground state would it be incorrect to refer to the atom as being at ground state?

Regardless, if the electron is at ground state would I be correct to say it has a wave function of 1?

Would wave function relate to angular momenta akin to the relation of a circle and a sine wave? That is to say, angular momenta indicates a changing vector if the electron is to remain with the atom that vector must transit until it returns to its original state relative to the nucleus.

 

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54 minutes ago, Butch said:

The ground state electron is most likely to be found at a distance from the nucleus equal to the Bohr radius, 1s atomic orbital (sorry for using that word again, I am transitioning to qm) If the single electron in the hydrogen atom is at ground state would it be incorrect to refer to the atom as being at ground state?

That would be correct terminology

54 minutes ago, Butch said:

Regardless, if the electron is at ground state would I be correct to say it has a wave function of 1?

A wave function is just that: a function. So labeling it as 1 is meaningless; it sounds like you are assigning it a value.

You could say the ground state wave function, or the n = 1 wave function (since n = 1 is the lowest energy level, i.e. the ground state)

54 minutes ago, Butch said:

Would wave function relate to angular momenta akin to the relation of a circle and a sine wave? That is to say, angular momenta indicates a changing vector if the electron is to remain with the atom that vector must transit until it returns to its original state relative to the nucleus.

The wave function includes the angular momentum, but there is no trajectory, so there is no vector involved here. "Returning to a state" makes sense in terms of energy levels, since there is a ground state in which a system preferentially resides.

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1 hour ago, swansont said:

The wave function includes the angular momentum, but there is no trajectory, so there is no vector involved here. "Returning to a state" makes sense in terms of energy levels, since there is a ground state in which a system preferentially resides.

Because the angular momentum is an average or perhaps a constant in qm, could we refer to it as a pseudo vector? That is as a vector set that cycles through 360 degrees and may or may not vary in amplitude?

1 hour ago, swansont said:

A wave function is just that: a function. So labeling it as 1 is meaningless; it sounds like you are assigning it a value.

You could say the ground state wave function, or the n = 1 wave function (since n = 1 is the lowest energy level, i.e. the ground state)

I see your point, I was still thinking paths in that the path length(in terms of time) would equal wavelength, sorry hard keeping my head in qm rather than classical.

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44 minutes ago, Butch said:

Because the angular momentum is an average or perhaps a constant in qm, could we refer to it as a pseudo vector?

Without messing about like that, angular momentum is a vector.

45 minutes ago, Butch said:

That is as a vector set that cycles through 360 degrees and may or may not vary in amplitude?

No.

It's a conserved quantity so it doesn't usually vary.

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34 minutes ago, John Cuthber said:

Without messing about like that, angular momentum is a vector.

No.

It's a conserved quantity so it doesn't usually vary.

As far as qm is concerned, no it does not vary... However an electron can have many states, qm derives the highest probability state... So would it not be possible that the amplitude of the angular momentum vector could vary?

 

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2 hours ago, swansont said:

A wave function is just that: a function. So labeling it as 1 is meaningless; it sounds like you are assigning it a value.

You could say the ground state wave function, or the n = 1 wave function (since n = 1 is the lowest energy level, i.e. the ground state)

 

@Butch

 

This analogy might help get your head round quantum numbers.

Take standard screw sizes.

 

Standard Screw sizes are numbered thus

1,2,3,4 ...12...14

More detail is provided by subsidiary characteristics

Slotted
Crosspoint

Brass
Stainless
Plated
 

Length 1/2", 1", 2"   ....

If you want to know the more exact details such as pitch, thread size, thread length, etc you look up a standard table or drawing.

 

This is almost exactly the same for quantum numbers.
They give you the entry points on the look up tables or the probability distribution plots, which replace the thread drawings.

 

Does this help?

 

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3 minutes ago, studiot said:

 

@Butch

 

This analogy might help get your head round quantum numbers.

Take standard screw sizes.

 

Standard Screw sizes are numbered thus

1,2,3,4 ...12...14

More detail is provided by subsidiary characteristics

Slotted
Crosspoint

Brass
Stainless
Plated
 

Length 1/2", 1", 2"   ....

If you want to know the more exact details such as pitch, thread size, thread length, etc you look up a standard table or drawing.

 

This is almost exactly the same for quantum numbers.
They give you the entry points on the look up tables or the probability distribution plots, which replace the thread drawings.

 

Does this help?

 

Yes, I restated my previous post, is it correct?

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23 minutes ago, Butch said:

As far as qm is concerned, no it does not vary... However an electron can have many states, qm drives the highest probability state... So would it not be possible that the amplitude of the angular momentum vector could vary?

 

It's strictly quantised in units of h/2pi

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31 minutes ago, John Cuthber said:

It's strictly quantised in units of h/2pi

And the component vectors of this quantity could vary?

In the atomic model I will be attempting to describe in terms of qm "sister" electrons would have synchrony in that at any instant the sum of the angular momentum of the electrons would equal the energy state of the atom, while the angular momentum of the individual electrons would vary the sum would not.

In hydrogen1 the electrons energy would not vary, however hydrogen2 is a different story.

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3 hours ago, John Cuthber said:

Without messing about like that, angular momentum is a vector.

And classically that's fine. But in QM there's is no defined trajectory, which limits what you can say about the angular momentum.

You have information about one component of the angular momentum. That's the limit of vector information.

3 hours ago, Butch said:

As far as qm is concerned, no it does not vary... However an electron can have many states, qm derives the highest probability state... So would it not be possible that the amplitude of the angular momentum vector could vary?

If one could assign a vector (if that had meaning), it must vary. So we would have limited information about it. All we can say is e.g. we have h/2pi , or - h/2pi of angular momentum with respect to the z axis.

2 hours ago, Butch said:

 In hydrogen1 the electrons energy would not vary, however hydrogen2 is a different story.

Which is not in accordance with experiment, so it is wrong. There is a shift because of the different mass of the nucleus, but the values do not vary.

2 hours ago, Butch said:

In the atomic model I will be attempting to describe in terms of qm "sister" electrons would have synchrony in that at any instant the sum of the angular momentum of the electrons would equal the energy state of the atom, while the angular momentum of the individual electrons would vary the sum would not.

Half of the elements have an odd number of electrons, so there an electron which has no sister.

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1 hour ago, swansont said:
4 hours ago, Butch said:

 In hydrogen1 the electrons energy would not vary, however hydrogen2 is a different story.

Which is not in accordance with experiment, so it is wrong. There is a shift because of the different mass of the nucleus, but the values do not vary.

I stand corrected, the energy would have to vary... However the sum of the energies would have to equal the energy state of the atom.

I have some of the math for my model, indeed a good chunk of it and the rest is not difficult thanks to qm. The shift is somewhat more intriguing than you might think. 

Checking my thinking, is the diameter of deuterium different than that of hydrogen1? I suppose there are other explanations, the space between protons could vary or (unlikely) the density of the proton could change along with it's diameter. Any of these ring true?

1 hour ago, swansont said:

Half of the elements have an odd number of electrons, so there an electron which has no sister.

I did not say their could only be 2 sisters, tritium has 3 sisters. 

1 hour ago, swansont said:

If one could assign a vector (if that had meaning), it must vary. So we would have limited information about it. All we can say is e.g. we have h/2pi , or - h/2pi of angular momentum with respect to the z axis.

I think that indirectly qm solves this for us, take all those vectors together and we have h/2pi or -h/2pi, with qm it is just so much simpler.

Checking my understanding, when an electron shifts to a higher energy level it moves to a lower "shell" or a higher shell(still trying to pull my head out of my clASSical).

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15 hours ago, Butch said:

Checking my thinking, is the diameter of deuterium different than that of hydrogen1? I suppose there are other explanations, the space between protons could vary or (unlikely) the density of the proton could change along with it's diameter. Any of these ring true?

Ok, scratch all that... I should be thinking of particles as magnetic dipoles and centers of mass, correct? Still, the radius of the atom... 

The entity that I have referred to as the nucleus area would be the charge radius of the nucleus, this fits with my atom.

IMG_20171220_084950.jpg

Certainly not to scale. e1 and e2 are sister electrons when e1 is at its highest probability position e2 is near the center of the charge radius of the nucleus. e1 is the valent electron e2 is contributing to the charge and mass of the nucleus, the charge of the nucleus is +1 and the mass of the nucleus is equal to the rest mass of the 2 protons + the rest mass of e2 + the apparent mass due to the velocity of e2. The apparent mass(and thus the velocity) of e2 should be the difference in rest mass of a proton and a neutron. As e1 begins to fall towards the nucleus e2 which has greater velocity moves away from the nucleus, thus the charge radius of the nucleus increases. When e2 "passes" e1 the charge radius of the nucleus begins to decrease.

In this atom:

Valence electrons = 1

Charge of the nucleus = +1

Mass of the nucleus = 3.3475517 e-27 kilograms.

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15 hours ago, Butch said:

 I did not say their could only be 2 sisters, tritium has 3 sisters. 

Tritium is an isotope of hydrogen and has only one electron.

If you can't solve hydrogen, there's no point in trying to solve more complex systems. You don't have a viable model.

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11 minutes ago, swansont said:

Is there any experimental evidence of this?

Tritium is unstable. What happens when the neutron decays?

The instability of a free neutron (Chadwick).

In the tritium atom, the interaction of 2 electrons with the nucleus causes a greater expansion (or perhaps I should say oscillation) of the nucleus charge radius, causing greater instability of the atom, indeed deuterium is less stable than hydrogen1. Hydrogen one is ultimately stable.

The thing that really draws me to this model is that if we add another proton electron pair to the tritium atom we either get a very unstable isotope of hydrogen or a helium4 atom!

Note that in my model the neutron is actually an electron inside the charge radius of the nucleus, hence if the neutron decayed it would simply mean that a proton electron pair (or pairs) would be ejected from the nucleus. The pair would be tightly bound however the electron would be high energy, hence they would immediately dissociate, observed as the decay of a free neutron.

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