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Atomic orbitals


Butch

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Consider the following:

The probability of an electron being at any single point within an orbital domain approaches zero.

The probability of an electron being in a defined area of an orbital domain is directly proportional to the dimensions of the defined area and inversely proportional to velocity.

The area of an orbital intersection with the plane of the nucleus is a single point, thus the probability of an electron being in the orbital plane approaches zero.

Comments?

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4 minutes ago, Butch said:

The probability of an electron being at any single point within an orbital domain approaches zero.

Really? Where do you get that from?

11 minutes ago, Butch said:

Comments?

It seems to be based on a complete misunderstanding.

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Hey Butch it's that calculus fish. It bit ya again.

 

You are probably (ha ha) thinking of the following

 

The probability of finding the electron between x and (x+dx) is f(x) as we let dx tend to zero (ie take a limit).

That is the way this sort of probability is calculated.

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10 minutes ago, Strange said:

Really? Where do you get that from?

The electron has velocity.

 

11 minutes ago, Strange said:

It seems to be based on a complete misunderstanding.

Please elaborate, it will be much appreciated.

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21 minutes ago, Butch said:

Consider the following:

The probability of an electron being at any single point within an orbital domain approaches zero.

The probability of an electron being in a defined area of an orbital domain is directly proportional to the dimensions of the defined area and inversely proportional to velocity.

Velocity is not a well-defined attribute of an electron in an orbital.

21 minutes ago, Butch said:

The area of an orbital intersection with the plane of the nucleus is a single point, thus the probability of an electron being in the orbital plane approaches zero.

Comments?

The nucleus is not a point object.

4 minutes ago, Butch said:

Please elaborate, it will be much appreciated.

You need to learn about quantum mechanics.

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22 minutes ago, swansont said:

Velocity is not a well-defined attribute of an electron in an orbital.

The nucleus is not a point object.

But the intersection of the orbital with the plane of the nucleus is, as I understand it the probability of an electron being in this plane is zero, would it not be correct to say rather that the probability approaches zero?

Although velocity is not well defined, still an electron has a wave function and particle manifestation, so it has velocity.

25 minutes ago, studiot said:

Hey Butch it's that calculus fish. It bit ya again.

 

You are probably (ha ha) thinking of the following

 

The probability of finding the electron between x and (x+dx) is f(x) as we let dx tend to zero (ie take a limit).

That is the way this sort of probability is calculated.

Lol, yes indeed! But that limit approaches zero, which as you know the calculus fish says is much different than zero.

48 minutes ago, Butch said:

Consider the following:

The probability of an electron being at any single point within an orbital domain approaches zero.

The probability of an electron being in a defined area of an orbital domain is directly proportional to the dimensions of the defined area and inversely proportional to velocity.

The area of an orbital intersection with the plane of the nucleus is a single point, thus the probability of an electron being in the orbital plane approaches zero.

Comments?

Let me be a little clearer the intersection of the orbital with the plane of the nucleus is perhaps an infinite number of points, however the set of points in the orbital set is infinitely greater. Thus the intersection approaches zero. 

That calculus fish again.

Edited by Butch
Elboration
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10 minutes ago, Butch said:

But the intersection of the orbital with the plane of the nucleus is, as I understand it the probability of an electron being in this plane is zero, would it not be correct to say rather that the probability approaches zero?

What is the plane of the nucleus? The nucleus is an extended, 3D object. It doesn't have a plane.

An electron has an orbital, it doesn't have a plane, either.

10 minutes ago, Butch said:

Although velocity is not well defined, still an electron has a wave function and particle manifestation, so it has velocity.

If you can calculate the velocity given the wave function, let's see it. Try hydrogen, in the ground state (n=1), for simplicity

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.html

 

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3 minutes ago, swansont said:

What is the plane of the nucleus? The nucleus is an extended, 3D object. It doesn't have a plane.

An electron has an orbital, it doesn't have a plane, either.

If you can calculate the velocity given the wave function, let's see it. Try hydrogen, in the ground state (n=1), for simplicity

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydwf.html

 

I am not trying to calculate velocity... Yet.

I am trying to demonstrate that an electron can pass through the orbital plane of the nucleus, and still have a probability of being in the plane that approaches zero.

The nucleus has a center of mass a plane of the nucleus would intersect the center of mass perpendicular to the orbital axis.

Edited by Butch
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1 minute ago, Butch said:

I am not trying to calculate velocity... Yet.

I am trying to demonstrate that an electron can pass through the orbital plane of the nucleus, and still have a probability of being in the plane that approaches zero.

As I said, there is no "orbital plane of the nucleus". e.g. the orbital in a hydrogen atom is spherically symmetric. 

There is no trajectory one can assign to the electron. 

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29 minutes ago, swansont said:

As I said, there is no "orbital plane of the nucleus". e.g. the orbital in a hydrogen atom is spherically symmetric. 

There is no trajectory one can assign to the electron. 

Perhaps we cross posted...

I am not speaking in terms of trajectory or velocity, only probability... It is constantly stated that the probability of an electron in the nucleus is zero, that is quite different than approaches zero... As Strange will agree "Words have meaning".

1 hour ago, swansont said:

You need to learn about quantum mechanics.

I am, thanks very much to the members of this forum!

29 minutes ago, swansont said:

As I said, there is no "orbital plane of the nucleus". e.g. the orbital in a hydrogen atom is spherically symmetric.

I do apologise... The "plane of the nucleus", rather than the "orbital plane of the nucleus".

The 1s orbital is unique, what of 2s and up?

Edited by Butch
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7 hours ago, Butch said:

Perhaps we cross posted...

I am not speaking in terms of trajectory or velocity, only probability... It is constantly stated that the probability of an electron in the nucleus is zero, that is quite different than approaches zero... As Strange will agree "Words have meaning".

The probability is not zero. I don't know who is constantly saying that it is, but they're wrong.

7 hours ago, Butch said:

I do apologise... The "plane of the nucleus", rather than the "orbital plane of the nucleus".

Stll makes no sense to me. What is the plane of a sphere? (With no other references)

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17 hours ago, Strange said:

Do you mean a plane (any plane) through the nucleus? (Not that it makes any more sense.)

The plane perpendicular to the axis of the orbital, this plane is probably not static. 

17 hours ago, swansont said:

The probability is not zero. I don't know who is constantly saying that it is, but they're wrong.

Stll makes no sense to me. What is the plane of a sphere? (With no other references)

In the case of an s orbital of 1h the plane is likely not in play as the nucleus is a single proton. It is my feeling at this time that the electron and proton do not merge... That could however be incorrect.

To elaborate I would be treading in a closed thread.

You have answered my question, thank you! 

Can we say then that an electron can be located in the nucleus?

On 1/13/2018 at 11:34 AM, swansont said:

As I said, there is no "orbital plane of the nucleus". e.g. the orbital in a hydrogen atom is spherically symmetric. 

There is no trajectory one can assign to the electron. 

There is no trajectory, however there is a vector, correct?

∆p ∆t...

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50 minutes ago, Butch said:

The plane perpendicular to the axis of the orbital, this plane is probably not static. 

There is no axis of the orbital. S orbitals are spherically symmetric.

 

50 minutes ago, Butch said:

In the case of an s orbital of 1h the plane is likely not in play as the nucleus is a single proton. It is my feeling at this time that the electron and proton do not merge... That could however be incorrect.

They rarely merge, but the electron can be found inside the nucleus a nonzero fraction of the time 

50 minutes ago, Butch said:

Can we say then that an electron can be located in the nucleus?

Yes

50 minutes ago, Butch said:

There is no trajectory, however there is a vector, correct?

∆p ∆t...

No vector. The velocity is not defined.

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28 minutes ago, swansont said:

There is no axis of the orbital. S orbitals are spherically symmetric.

You are pushing my thinking, thank you!  There is discussion for this, however I will reserve it for a later time.

28 minutes ago, swansont said:

No vector. The velocity is not defined.

If we can know the position at t0 and the position at t1 we can construct a vector, we cannot know the actual path. Correct? 

My question has been answered, there will be more, but I will confine them to new topics.

Again, thank you very much... Y'ALL are awesome!

28 minutes ago, swansont said:

 

Edited by Butch
Oops
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