Waveform of an electron

[petrushka.googol]

What does the waveform of an electron in orbit look like ?

 

And what if it is dislodged (made free) ?

 

Please advise.

[ajb]

What does the waveform of an electron in orbit look like ?

For hydrogen? Or more generally?

 

Anyway, you can simply google pictures of the orbitals for hydrogen. For example

 

[petrushka.googol]

Plot of motion vs time.

Is there a perigee and apogee ?

[swansont]

Plot of motion vs time.

Is there a perigee and apogee ?

 

There is no plot of motion vs time. Position is not a good quantum number. There are no classical trajectories. Perigee and apogee are not concepts that apply as they do in astronomy. Some orbits have the electron spending some fraction of its time inside the nucleus.

[Enthalpy]

What does the waveform of an electron in orbit look like ?

The orbital IS the waveform of the electron. And in a slightly integrist wording: the electron is an orbital, when stable around a nucleus. The electron is a wave, you know?

 

By the very definition of an orbital, it's a stationary solution. It does not evolve over time. Perfectly static, immobile - put all synonyms here.

[studiot]

 

the electron is an orbital

 

Not convinced.

 

1) An orbital can contain more than one electron,

 

2) What are you suggesting empty orbitals to be?

3) Please describe the 'orbital' of a free electron.

4) If an electron becomes free, ie leaves an orbital, what is left behind? Does the electron leave part of itself behind?

Edited April 7, 2016 by studiot

[DrP]

1 - so what...

2 - nothing... would you describe a hole as an actual thing or the lack of a thing?

3 - free - unbound.. it is no longer in 'orbit' around the nucleus.

4 - no - nothing left behind

 

I think you are arguing semantics or just not getting it. He made it clear that the wording was a bit funny. I think I get what he means - he said that the orbital was the waveform of the electron - is this not true?

[studiot]

 

DrP

1 - so what...

2 - nothing... would you describe a hole as an actual thing or the lack of a thing?

3 - free - unbound.. it is no longer in 'orbit' around the nucleus.

4 - no - nothing left behind

 

I think you are arguing semantics or just not getting it. He made it clear that the wording was a bit funny. I think I get what he means - he said that the orbital was the waveform of the electron - is this not true?

 

 

Wow, that is rather peremptory. Did I offend you?

 

To answer your comments in reverse order

 

4) So you are saying that if I remove 1 electron from the 1s orbital of Helium, the orbital I took it from collapses? So what happens to the other electron in the 1s orbital?

 

3) So is this a claim that a 'free, unbound' electron no longer obeys quantum mechanics in general and the wave equation in particular?

 

2) So how do you explain bonding orbitals and anti-bonding orbitals?

 

1) So what? well this is tied in with question 4, already answered.

Edited April 7, 2016 by studiot

[swansont]

I think you are arguing semantics or just not getting it. He made it clear that the wording was a bit funny. I think I get what he means - he said that the orbital was the waveform of the electron - is this not true?

The waveform of the electron and the deBroglie wavelength of the electron are not synonymous. Consider a free electron moving with some speed — the deBroglie wavelength is given by v, but the wave function shape is dictated by the uncertainty in the speed (it could be a delta function). So when one says the electron is a wave, which wave does one mean? Diffraction depends on the deBroglie wavelength, so that suggests that it's the wave one should associate with the electron itself.

 

In hydrogen you get the Bohr radius of the electron ground state orbital by assuming it has h-bar of angular momentum, so the deBroglie wavelength is the circumference, but the actual angular momentum of that state is zero. So again, saying that the wave function/waveform is the electron because the electron is a wave seems to be a dubious claim.

 

[studiot]

DrP, you should not imagine the comparison with your guitar string.

Orbitals as solutions to the Schrodinger equation only bears a passing resemblence to the solution to classical wave equation which your cuitar string obeys.

 

If you twang the string you can see a shape in space describing the vibrations.

If you are inhumanly very very quick you can catch the string at the top of its vibration between your fingers.

At this time ther is no possibility that the string will elude your grasp simply because it is at the opposite side of its vibration.

That is what you see is a time averaged blur comprising mostly empty space.

 

The orbital is not like that.

Wherever you 'grab' it you will have some interaction with any electron in that orbital, as any atom approaching to bond will experience.

The electron is not on one side of the vibration at one time and on the other side at another as the string is.

Edited April 7, 2016 by studiot

[swansont]

 

The orbital is not like that.

Wherever you 'grab' it you will have some interaction with any electron in that orbital, as any atom approaching to bond will experience.

The electron is not one one side of the vibration at one time and on the other side at another as the string is.

 

 

Further, if one is asserting that the string represents the probability (which is the wave function), you get the wrong answer. The probability of finding the ground state hydrogen electron in a differential volume is spherically symmetric, only having a radial dependence.

[DrP]

Thanks - I'll digest all that later and read around.. (it's been about 20 years since studying it)

 

I was just under the impression that an empty orbital was like an empty hole. i.e. - not actually there.

[studiot]

An empty orbital has more in common with an empty bookshelf.

 

Whilst the book is on the shelf it is on a ledge in a potential energy field.

If I remove the book does the shelf disappear?

 

But as swansont has pointed out there is more to quantum orbitals than just energy levels so they are more complicated than bookshelves.

[Enthalpy]

 

Not convinced.

 

1) An orbital can contain more than one electron,

 

2) What are you suggesting empty orbitals to be?

3) Please describe the 'orbital' of a free electron.

4) If an electron becomes free, ie leaves an orbital, what is left behind? Does the electron leave part of itself behind?

1) Yes. I nearly mentioned it but decided to keep simple. I wouldn't have formulated it "contain" then, rather like "a pair of electrons can be an orbital".

1b) It isn't the same orbital by the way. With two electrons, the orbital differs formally for being a function of two positions and time, and observationally for having a different shape.

2) A mathematical construction for instance. Nothing observable.

3) You wrote this nonsense. I didn't. I wrote: "the electron is an orbital, when stable around a nucleus."

4) I'd say it leaves nothing physical. Do you hope to observe an orbital that isn't an electron nor a pair?

[studiot]

 

enthalpy

3) You wrote this nonsense

 

Everyone deserves the right of reply.

 

But not to call the other persons statement nonsense.

 

Particularly when you specifically said "The orbital IS the waveform of the electron", emphasising the word 'is'.

 

Now some electrons are free.

So I asked you to describe the obital that is the electron, in this condition.

Edited April 7, 2016 by studiot

[Enthalpy]

You changed my meaningful sentence into a nonsense that you attributed to me. This is unfair and I underline it. Expect me to do it everytime.

 

I did not write that free electrons are orbitals. You did and attributed it to me.

[swansont]

 

2) A mathematical construction for instance. Nothing observable.

 

 

The AC Stark shift pushes levels around, and this includes unoccupied states. And there are second (and higher) order interactions that couple through unoccupied states. It's all a mathematical construct, so that's not a particularly revealing statement. But there are observable effects.

[studiot]

 

You changed my meaningful sentence into a nonsense that you attributed to me. This is unfair and I underline it. Expect me to do it everytime.

 

I did not write that free electrons are orbitals. You did and attributed it to me.

 

 

Perhaps you don't understand plain English.

 

You clearly misinterpreted what I said because at no point in my life have I ever claimed that a free electron is an orbital, or even connected to one, even when I was at school.

 

I actually asked a question because you claimed congruence between electrons and orbitals and I asked question 3 because I wanted you to demonstate this claimed congruence for a free electron.

 

You cannot have it both ways.

 

Either an electron is an orbital or it is not.

 

Which are you claiming?

 

You have been very fond of telling me that I'm wrong on this forum, yet I did not initially flatly contradict you. Rather I offered you the opportunity to think again by expressing doubts and requesting more details, indicating the areas of my doubts.

 

Your response was a direct personal attack, which has continued in that vein since.

Edited April 8, 2016 by studiot

[Enthalpy]

I wrote in #5: "the electron is an orbital, when stable around a nucleus.". I'm still pleased with that formulation.

 

In #6 you misquoted me as "the electron is an orbital", which is bad practice in English too, and misused it to drift to

"Please describe the 'orbital' of a free electron."

 

It still holds that:

• The "orbital of a free electron" is a nonsense
• I didn't write this nonsense
• You did write this nonsense and attributed it to me.
• I showed your bad practice. I have no regret about it.

[swansont]

I wrote in #5: "the electron is an orbital, when stable around a nucleus.". I'm still pleased with that formulation.

 

 

You've ignored the objection about the differences between an orbital and a deBroglie wave

[Enthalpy]

It wasn't a good objection because I had not wroten such a thing.

 

Sure, an orbital is the waveform around a nucleus and if stationary. If the (lone) electron is captive around a nucleus but not stationary, the wave is a linear combination of orbitals. If the electron is free, the wave can be written for instance as a linear combination of plane waves. Under other circumstances, different solutions appear.

[swansont]

It wasn't a good objection because I had not wroten such a thing.

 

It's you failure to address the issue that's my concern. Why is the electron the orbital rather than the deBroglie wave?

[studiot]

OK, let us examine this claim that the electron is the orbital and the orbital is the electron under any circumstances.

 

 

I have a pendulum clock.

 

The pendulum swings to and fro marking out a sector of a circle.

This sector does not have a particular name (that I am aware of) but I will call it the swingspace.

 

I hope you will not claim that the swingspace is the same as the pendulum itself, but it has the same, or very similar, characteristics as an orbital.

 

The properties involved in the equation of motion of the pendulum reduce to zero outside the swingspace and defined everywhere within it.

The pendulum can considered as a point mass in a potential field, for the purpose of obtaining solutions to this equation.

The boundary conditions are set by the potential field not the mass of the pendulum.

 

The equation of motion of the electron in a potential field is set by the boundary conditions, not the electron charge or mass.

In this equation the electron is considered a point mass and a point charge.

Solution of this equation yields a more complicated property than the pendulum solution that has a value of all space (ie every point in space).

Most of this value is concetrated into a particular region of space we call an orbital.

The usual graphic that we draw as s, p, d etc orbitals corresponds to a surface in 3D space that contains 90% of the probability density of this solution, given by the square of the wave function that is the solution to the electron's equation of motion.

So the orbital corresponds to the swingspace of my pendulum.

[altsci]

DeBroglie's wave length/frequency: f=mcv/h (1) is usually attributed to an electron as its second nature (particle-wave dualism). But according to the formula (1) DeBroglie's frequency proportional to the velocity of electron which always relative. That means that electron beam's target (crystal) also involved. Suppose that electron having linear momentum p=mv hits a crystal and produces photon that takes the linear momentum of the electron. Then we have the formula (1). DeBroglie's frequency is not the property of electron alone! In atom the second to electron object is nucleus. An electron is orbiting in atom and psy-function represents yet another physical field object that relevant to nucleus and to electron

 

[swansont]

DeBroglie's wave length/frequency: f=mcv/h (1) is usually attributed to an electron as its second nature (particle-wave dualism). But according to the formula (1) DeBroglie's frequency proportional to the velocity of electron which always relative. That means that electron beam's target (crystal) also involved. Suppose that electron having linear momentum p=mv hits a crystal and produces photon that takes the linear momentum of the electron. Then we have the formula (1). DeBroglie's frequency is not the property of electron alone! In atom the second to electron object is nucleus. An electron is orbiting in atom and psy-function represents yet another physical field object that relevant to nucleus and to electron

 

 

 

What crystal? What electron beam? You seem to have conjured a specific example, and even then, the deBroglie wave is not the property of the target. How would you get a collision where the photon takes the momentum of the electron, anyway? Not from a stationary target, I think. The momentum of a 10^-15 J electron is 4.27 x 10^23 kg-m/s. The energy of a photon with that momentum is 1.28 x 10^-14 J. A factor of 8 larger. So unless you don't believe in conservation of energy, this doesn't happen with a stationary target.