Electron Probability distribution

[Arnav]

I was recently reading about atomic structure's journey, when i saw the electron probability distribution graphs for some orbitals.

Why is the electron probability density maximum "at the nucleus" for s subshell ? does it have any physical significance? the confusing part for me is that the probability of finding an electron would be the least at nucleus, so how come probability density is maximum?

P.S. I am a high school student, so it would be extremely good if someone could explain this to me in a simple way

Edited June 17, 2021 by Arnav

[swansont]

59 minutes ago, Arnav said:

Why is the electron probability density maximum "at the nucleus" for s subshell ?

It isn’t. You might have only been looking at the wave function itself. The probability multiplies this by r^2, making it go to zero at the origin.

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

[studiot]

Here is an explanation with simpler maths than swansont's link that is suitable for high school level.

Start reading at the star in the first attachment.

[exchemist]

2 hours ago, Arnav said:

I was recently reading about atomic structure's journey, when i saw the electron probability distribution graphs for some orbitals.

Why is the electron probability density maximum "at the nucleus" for s subshell ? does it have any physical significance? the confusing part for me is that the probability of finding an electron would be the least at nucleus, so how come probability density is maximum?

P.S. I am a high school student, so it would be extremely good if someone could explain this to me in a simple way

To build a bit on what @swansonthas said, yes it does have profound significance in chemistry.

Because the s orbital wave function has no node at the nucleus, it implies that the electron spends some of its time up close to the nucleus. This means that, in multi-electron atoms, the s electrons are exposed to the full nuclear charge more than electrons in p , d or f orbitals, which are more "shielded" from the full nuclear charge by the electrons in shells closer in. S-orbitals are said to "penetrate" the cloud of electrons surrounding the nucleus more than the others.

As one goes up* the Periodic Table, the increasing nuclear charge progressively pulls in the s orbitals and lowers their energy more than it does for the others. This is the reason why the Periodic Table has the shape it does. It results in s orbitals having lower energy than p, d or f orbitals of the same shell. This even happens to such a degree in the 4th row that at potassium, the 4s has lower energy than 3d. This is why the first row of the transition elements (d block) appears after K and Ca.  It is only then that the 3d has come down in energy enough to be filled, in preference to 4p.   

 

*This concept of progressively filling subshells with electrons as the nuclear charge increases from one element to the next is known as the Aufbauprinzip (= building up principle). 

Edited June 17, 2021 by exchemist

[Arnav]

Please note that I mentioned electron probability density in my question, not radial electron probability distribution. What I want to convey is, why is graph(b) maximum at the nucleus?

I understand how the actual probability is obtained by multiplying electron probability density at certain r by 4pir².dr

What I want too understand is why the graph of Ψ² vs r peaks at r = 0. Does it have any physical significance?

please bear with me if i missed something

[swansont]

1 hour ago, exchemist said:

Because the s orbital wave function has no node at the nucleus, it implies that the electron spends some of its time up close to the nucleus. This means that, in multi-electron atoms, the s electrons are exposed to the full nuclear charge more than electrons in p , d or f orbitals, which are more "shielded" from the full nuclear charge by the electrons in shells closer in. S-orbitals are said to "penetrate" the cloud of electrons surrounding the nucleus more than the others.

This also explains the relatively large hyperfine splitting of the S state.

52 minutes ago, Arnav said:

Please note that I mentioned electron probability density in my question, not radial electron probability distribution. What I want to convey is, why is graph(b) maximum at the nucleus?

That’s the solution to the Schrödinger equation.

This has to be multiplied by the volume element, which depends on r and is zero at r=0, to get the physically meaningful quantity.

[exchemist]

11 hours ago, Arnav said:

Please note that I mentioned electron probability density in my question, not radial electron probability distribution. What I want to convey is, why is graph(b) maximum at the nucleus?

I understand how the actual probability is obtained by multiplying electron probability density at certain r by 4pir².dr

What I want too understand is why the graph of Ψ² vs r peaks at r = 0. Does it have any physical significance?

please bear with me if i missed something

Ah, I may have misinterpreted what you were looking for in terms of physical significance. Let me try another angle. 

You may have come across the problem with the original Rutherford-Bohr model of the atom that it can't account for why a supposedly orbiting electron does not emit radiation, lose kinetic energy  and fall into the nucleus. In a sense you can view the s-orbitals as the QM version of exactly that scenario. Electrons in s orbitals have zero angular momentum, so they can't be said, in any sense, to be "orbiting " the nucleus. Instead, it is as if they continually fall towards it - even through it perhaps -  and come out the other side. Being QM entities (Uncertainty Principle and all that), one cannot say they follow any defined trajectory of course, but the overall sense is of being able to touch  the nucleus, rather as if they fall into it. Whereas p, d, f, etc orbitals have 1,2, 3 etc units of angular momentum and, lo and behold, all have a node at the nucleus, which is more consistent with some kind of "orbiting" motion, even though again, being QM entities, they have no defined trajectory. 

So I'be tempted to say the physical significance of non-zero ψ at the nucleus is a reflection of the absence of orbital angular momentum. 

[swansont]

2 hours ago, exchemist said:

So I'be tempted to say the physical significance of non-zero ψ at the nucleus is a reflection of the absence of orbital angular momentum. 

There’s a mundane mathematical aspect to it.

In a 1D system in cartesian coordinates, the square of the wave function gives us a probability. But going to 3D this is a density, so the probability is not just the square of the wave function. And we’ve moved to a spherical coordinate system. The volume element integral includes r^2dr, instead of the dx we had before.

If you solved for a 1/x potential in cartesian coordinates, you’re going to get a wave function that looks something like xe^-x/a (for x>0), i.e. something that goes to zero at the origin

 

[exchemist]

59 minutes ago, swansont said:

There’s a mundane mathematical aspect to it.

In a 1D system in cartesian coordinates, the square of the wave function gives us a probability. But going to 3D this is a density, so the probability is not just the square of the wave function. And we’ve moved to a spherical coordinate system. The volume element integral includes r^2dr, instead of the dx we had before.

If you solved for a 1/x potential in cartesian coordinates, you’re going to get a wave function that looks something like xe^-x/a (for x>0), i.e. something that goes to zero at the origin

 

Isn't that just tantamount to saying that a sphere of zero radius contains zero volume, so the chance of the electron being there is zero, the non-zero probability density function notwithstanding? 

[swansont]

20 minutes ago, exchemist said:

Isn't that just tantamount to saying that a sphere of zero radius contains zero volume, so the chance of the electron being there is zero, the non-zero probability density function notwithstanding? 

To the extent that “a sphere with zero volume” is meaningful, yes. But also that probability density lacks physical meaning if you ignore the volume you’re looking at.

[exchemist]

1 hour ago, swansont said:

To the extent that “a sphere with zero volume” is meaningful, yes. But also that probability density lacks physical meaning if you ignore the volume you’re looking at.

I'm not sure I agree. It seems to me that the concentration of a dissolved chemical substance, or the strength of a magnetic field - or indeed the density of a material -  has a physical meaning, regardless of what volume one considers.  

Edited June 18, 2021 by exchemist

[studiot]

19 hours ago, Arnav said:

I understand how the actual probability is obtained by multiplying electron probability density at certain r by 4pir².dr

Are you sure you fully understand it ?

You book does not say electron probability density or actual probability density on those graphs, which are basically the same as I posted.

 

2 hours ago, exchemist said:

I'm not sure I agree. It seems to me that the concentration of a dissolved chemical substance, or the strength of a magnetic field - or indeed the density of a material -  has a physical meaning, regardless of what volume one considers.  

I basically agree with you, +1 ,  but there is yet more to it than this, though swansont's terse replies are not actually wrong, just a bit short on explanation which is implied in the physics and maths.

Edited June 18, 2021 by studiot

[swansont]

2 hours ago, exchemist said:

I'm not sure I agree. It seems to me that the concentration of a dissolved chemical substance, or the strength of a magnetic field - or indeed the density of a material -  has a physical meaning, regardless of what volume one considers.  

But if it’s zero, are you considering a volume? How much mass do you have in zero volume, regardless of the density?

[exchemist]

4 minutes ago, swansont said:

But if it’s zero, are you considering a volume? How much mass do you have in zero volume, regardless of the density?

Well yes, sure, if it is literally zero, but equally one can say density has no physical meaning if one has a zero-sized lump of material. What I suppose I mean is it that has physical meaning if instead one considers, let us say, an arbitrarily small volume of space close to the nucleus.

What I'm rather more interested in, though, as I don't have the physics to know the answer to this, and I hope you might, is whether one can legitimately speak of an s-electron passing through the nucleus. I am not sure whether any of the interactions operating in the nucleus would prohibit this. Do you know? 

[studiot]

1 hour ago, swansont said:

But if it’s zero, are you considering a volume? How much mass do you have in zero volume, regardless of the density?

It's a limit as dv tends to zero.

 

[swansont]

1 hour ago, exchemist said:

Well yes, sure, if it is literally zero, but equally one can say density has no physical meaning if one has a zero-sized lump of material. What I suppose I mean is it that has physical meaning if instead one considers, let us say, an arbitrarily small volume of space close to the nucleus.

Of course density has a physical meaning, since you would typically have a volume of the material.

The question here was the probability at the origin as well as nearby. The volume element depends on the surface area of the shell, so it’s smaller close to the origin, and zero at the origin, and this artifact of the math has to be included when discussing the significance of the wave function’s value.

1 hour ago, exchemist said:

What I'm rather more interested in, though, as I don't have the physics to know the answer to this, and I hope you might, is whether one can legitimately speak of an s-electron passing through the nucleus. I am not sure whether any of the interactions operating in the nucleus would prohibit this. Do you know? 

The nucleus has a physical size, so sure. The electron interacts electromagnetically and the fact the the electron can be found there accounts for the large hyperfine splitting of the s state as opposed to the p states, where the wave function has a node, and so doesn’t have nearly as strong an interaction.

In alkali atoms the hyperfine splitting of the ground state is hundreds of MHz to >1 GHz, while the excited p-state is significantly smaller

[Arnav]

2 hours ago, studiot said:

Are you sure you fully understand it ?

You book does not say electron probability density or actual probability density on those graphs, which are basically the same as I posted.

Well I am a grade 11 student, studying these concepts for the first time,so I only have an abstract idea of it. What I posted was just a picture from the internet about the graphs

 

What I wanted to know was how come the electron probability density is maximum at the nucleus!?

How come as I go closer to the nucleus, the probability of finding an electron at a given point(or small cubic volume dV) gets bigger and bigger? And is greatest at the nucleus!? Is is just mathematical? How come the electron probability density is defined at the nucleus? Where there is no volume itself!?

I might have said some extremely wrong things in this last paragraph, so excuse me😅, analogies would work best for me as an explanation

[studiot]

9 minutes ago, Arnav said:

Well I am a grade 11 student, studying these concepts for the first time,so I only have an abstract idea of it. What I posted was just a picture from the internet about the graphs

 

What I wanted to know was how come the electron probability density is maximum at the nucleus!?

How come as I go closer to the nucleus, the probability of finding an electron at a given point(or small cubic volume dV) gets bigger and bigger? And is greatest at the nucleus!? Is is just mathematical? How come the electron probability density is defined at the nucleus? Where there is no volume itself!?

I might have said some extremely wrong things in this last paragraph, so excuse me😅, analogies would work best for me as an explanation

 graph (b) that you posted is a graph of 'probability density' against distance from the centre of the nucleus.

This is not the probability of finding an electron at a given point on the distance axis.

Clearly this would always be 1, if you waited long enough.

That is to say the electron will eventually be passing through this point (it will never be stationary there) so if you made your time frame long enough you would always momentarily spot the electron at some time or other.

In order to explain exactly what probability density means it would be helpful if you would  answer a couple of questions.

Have you done enough calculus to know what a derivative (or differential) and an integral are ?

Have you learned about mass, length and time as dimensions or units (called MLT)  ?

 

 

[Kartazion]

1 hour ago, Arnav said:

What I wanted to know was how come the electron probability density is maximum at the nucleus!?

Hello.

The electron can only be at the closer point of the nucleus only on the electronic layer 1 (n1). Its probability distribution being on the layer itself. No?

Or do you want to talk about the probability of distribution of the electron on the different electronic layers which would be very important on the layer 1 (at the most near the nuclei) rather than on a higher electronic layer further away from the nucleus?

[swansont]

10 minutes ago, Kartazion said:

The electron can only be at the closer point of the nucleus only on the electronic layer 1 (n1). Its probability distribution being on the layer itself. No?

Or do you want to talk about the probability of distribution of the electron on the different electronic layers which would be very important on the layer 1 (at the most near the nuclei) rather than on a higher electronic layer further away from the nucleus?

What are these layers you are referring to?

[studiot]

9 minutes ago, Kartazion said:

1 hour ago, Arnav said:

What I wanted to know was how come the electron probability density is maximum at the nucleus!?

Hello.

The electron can only be at the closer point of the nucleus only on the electronic layer 1 (n1). Its probability distribution being on the layer itself. No?

Or do you want to talk about the probability of distribution of the electron on the different electronic layers which would be very important on the layer 1 (at the most near the nuclei) rather than on a higher electronic layer further away from the nucleus?

 

Perhaps you should read what swansont posted before posting confusing material like this.

 

2 hours ago, swansont said:

Quote

What I'm rather more interested in, though, as I don't have the physics to know the answer to this, and I hope you might, is whether one can legitimately speak of an s-electron passing through the nucleus. I am not sure whether any of the interactions operating in the nucleus would prohibit this. Do you know? 

The nucleus has a physical size, so sure. The electron interacts electromagnetically and the fact the the electron can be found there accounts for the large hyperfine splitting of the s state as opposed to the p states, where the wave function has a node, and so doesn’t have nearly as strong an interaction.

In alkali atoms the hyperfine splitting of the ground state is hundreds of MHz to >1 GHz, while the excited p-state is significantly smaller

 

 

[Kartazion]

Just now, swansont said:

What are these layers you are referring to?

Oups ... It's shell not layer. Electron shell - Wikipedia

1 minute ago, studiot said:

Perhaps you should read what swansont posted before posting confusing material like this

Ah OK.

Therefore:

Quote

What I wanted to know was how come the electron probability density is maximum at the nucleus!?

Because 

Quote

The nucleus has a physical size, so sure. The electron interacts electromagnetically and the fact the the electron can be found there accounts for the large hyperfine splitting of the s state as opposed to the p states, where the wave function has a node, and so doesn’t have nearly as strong an interaction.

In alkali atoms the hyperfine splitting of the ground state is hundreds of MHz to >1 GHz, while the excited p-state is significantly smaller

 

[Arnav]

10 hours ago, studiot said:

Have you done enough calculus to know what a derivative (or differential) and an integral are ?

Have you learned about mass, length and time as dimensions or units (called MLT)  ?

Yes I have

11 hours ago, swansont said:

The nucleus has a physical size, so sure. The electron interacts electromagnetically and the fact the the electron can be found there accounts for the large hyperfine splitting of the s state as opposed to the p states, where the wave function has a node, and so doesn’t have nearly as strong an interaction.

In alkali atoms the hyperfine splitting of the ground state is hundreds of MHz to >1 GHz, while the excited p-state is significantly smaller

Could you please explain hyperfine splitting?

[swansont]

1 hour ago, Arnav said:

Could you please explain hyperfine splitting?

Some quantum particles, like electrons and protons, behave like little magnets, owing to having a charge and intrinsic angular momentum (spin). So a “spin up” electron will have a different energy than “spin down” in the presence of a magnetic field, which is provided by the nucleus. This energy splitting is the hyperfine splitting.

[Kartazion]

Is it fair to say that?

Quote

In atomic physics, fine structure describes the splitting of spectral lines of a particle. This fits the spectrum unless you take a high resolution look at fine structure where the electron spin and orbital quantum numbers are involved. There is a tiny dependence upon the orbital quantum number in the Lamb shift.
Hyperfine structure contrasts with fine structure, which results from the interaction between the magnetic moments associated with electron spin and the electrons orbital angular momentum. Hyperfine structure, with energy shifts typically orders of magnitudes smaller than those of a fine-structure shift, results from the interactions of the nucleus (or nuclei, in molecules) with internally generated electric and magnetic fields. The spin-up and spin-down states have different energies for the same "shell" (quantum numbers).