Why doesn't the electron fall into the nucleus?

[Achilles]

If there is an electrostatic force between the electron and proton, why doesn't the electron go all the way into the nucleus, what is stopping the electron from doing that.

[Sensei]

Sometimes electron is captured by proton-rich isotope.

https://en.wikipedia.org/wiki/Electron_capture

In this process one of protons is changed to neutron and there is released electron neutrino with fixed energy.

[math]p^+ + e^- \rightarrow n^0 + v_e[/math]

If you will calculate energy, this process cannot happen spontaneously, because free neutron has larger rest-mass (939.565 MeV/c^2) than sum of rest-masses of proton (938.272 MeV/c^2) and electron (0.511 MeV/c^2).

 

Beryllium-7 is the first one isotope (going by proton and neutron quantity) which undergo this decay mode.

[math]_4^7Be + e^- \rightarrow _3^7Li + v_e + 0.861893 MeV[/math]

 

 

Edited August 14, 2018 by Sensei

[Markus Hanke]

2 hours ago, Achilles said:

If there is an electrostatic force between the electron and proton, why doesn't the electron go all the way into the nucleus, what is stopping the electron from doing that.

Because the electron is not a classical particle (“little ball of mass and charge”), but a quantum object. As a first approximation, you can picture an electron as a 3D standing wave around the nucleus - you can only get standing waves of a given wavelength in specific places, which is why orbitals come in discrete levels. Crucially, there is a lowest energy level, which corresponds to the minimum distance an electron can be with respect to the nucleus (let’s assume here there is only one electron) - and that lowest energy level is not zero. Therefore the electron cannot fall all the way to the nucleus, it can only fall into its lowest energy level, which corresponds to an orbital that is still some distance outside the nucleus. This is a direct consequence of the laws of quantum mechanics, and coincidentally one of the questions that motivated the development of quantum mechanics in the first place.

Edited August 14, 2018 by Markus Hanke

[Strange]

Just to add a detail to Markus's (excellent) description, no two electrons can share the same "quantum state". What this means is that all the electrons in an atom cannot be in the lowest energy level. You can have up to two electrons at each energy level (because they can have opposite spin and hence a different quantum sate) and so once each energy level is occupied the other electrons have to be at progressively higher energy levels.

This, ultimately, explains all of chemistry, the mechanical properties of materials and why computers work!

[Carrock]

Just to clarify, while electrons in a stable atom can't fall into the nucleus, s orbital electrons can pass through it. The nucleus is not inherently a forbidden space for electrons.

From https://en.wikipedia.org/wiki/Atomic_orbital#Qualitative_understanding_of_shapes

(scroll down for some good graphics)

Quote

This [s orbital] antinode means the electron is most likely to be at the physical position of the nucleus (which it passes straight through without scattering or striking it), since it is moving (on average) most rapidly at that point, giving it maximal momentum.

 

Edited August 14, 2018 by Carrock
grammar and graphics

[swansont]

1 hour ago, Carrock said:

Just to clarify, while electrons in a stable atom can't fall into the nucleus, s orbital electrons can pass through it. The nucleus is not inherently a forbidden space for electrons.

From https://en.wikipedia.org/wiki/Atomic_orbital#Qualitative_understanding_of_shapes

(scroll down for some good graphics)

 

"This [s orbital] antinode means the electron is most likely to be at the physical position of the nucleus"

That's not right. The probability of finding the electron is very small at the origin. It's not just the wave function (squared), it's the integral over a shell. The most probable radius of finding the electron is the Bohr radius.

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydr.html#c1

The probability of the electron being at r=0 is zero, but the nucleus is not a point, so there is a small probability of finding the electron there.

[Carrock]

7 minutes ago, swansont said:

"This [s orbital] antinode means the electron is most likely to be at the physical position of the nucleus"

That's not right. The probability of finding the electron is very small at the origin. It's not just the wave function (squared), it's the integral over a shell. The most probable radius of finding the electron is the Bohr radius.

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydr.html#c1

The probability of the electron being at r=0 is zero, but the nucleus is not a point, so there is a small probability of finding the electron there.

I agree the most probable radius of finding the electron is the Bohr radius.

From http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydr.html#c1

Quote

The radial probability density for the hydrogen ground state is obtained by multiplying the square of the wavefunction by a spherical shell volume element.

So there is a factor of r^2 corresponding to the surface area at any radial distance. The surface area corresponding to the nuclear radius is tiny compared to e.g. the Bohr radius so the ratio of these surfaces is tiny. If the electron density was uniform throughout space, the probability of finding the electron at a given radius would be proportional to r^2.

 

The probability per unit volume (for s orbitals) increases as the nucleus is approached and is higher inside the nucleus than anywhere else.

Probability per unit volume is not used much since it doesn't provide much of use like e.g. the Bohr radius and the calculation of probability very near the antinode would be difficult and probably pointless.

 

Probability per unit volume might be useful here....

11 hours ago, Sensei said:

Sometimes electron is captured by proton-rich isotope.

https://en.wikipedia.org/wiki/Electron_capture

Quote

Electrons in s orbitals (regardless of shell or primary quantum number), have a probability antinode at the nucleus, and are thus far more subject to electron capture than p or d electrons, which have a probability node at the nucleus.

 

The problem with radial probability, especially for s orbitals, is that it is often conflated with probability per unit volume.

 

[swansont]

9 minutes ago, Carrock said:

Probability per unit volume might be useful here....

Quite true. Not being specific about what is being claimed could be a problem, and lead to contradictory answers. 

[beecee]

2 hours ago, Brett Nortj said:

The electron is pushing against the other electrons like a 'fullerene.'

So pray tell, what is the electron in an atom of hydrogen pushing against?

Leaving that pseudo claim........

14 hours ago, Markus Hanke said:

Because the electron is not a classical particle (“little ball of mass and charge”), but a quantum object. As a first approximation, you can picture an electron as a 3D standing wave around the nucleus - you can only get standing waves of a given wavelength in specific places, which is why orbitals come in discrete levels. Crucially, there is a lowest energy level, which corresponds to the minimum distance an electron can be with respect to the nucleus (let’s assume here there is only one electron) - and that lowest energy level is not zero. Therefore the electron cannot fall all the way to the nucleus, it can only fall into its lowest energy level, which corresponds to an orbital that is still some distance outside the nucleus. This is a direct consequence of the laws of quantum mechanics, and coincidentally one of the questions that motivated the development of quantum mechanics in the first place.

Great detailed answer Marcus! This is obviously Pauli's exclusion principle in operation, correct?

[Markus Hanke]

10 hours ago, beecee said:

Great detailed answer Marcus! This is obviously Pauli's exclusion principle in operation, correct?

No, not really. The Pauli principle states that no two fermions can share the same quantum state, but I assumed in my answer that there is only one electron anyway. I guess my answer comes down to the fact that the ground state (i.e. lowest possible excitation) of a single electron around a nucleus is non-trivial, meaning it is not just a point centered at the origin where the nucleus is; instead, it’s a spatially distributed probability cloud with corresponding non-vanishing energy.

Note that we are talking bound states here - it is of course still possible to fire an electron at the nucleus, and hit it in the process, but that is not a stable bound state, and won’t happen spontaneously.

[beecee]

2 hours ago, Markus Hanke said:

No, not really. The Pauli principle states that no two fermions can share the same quantum state, but I assumed in my answer that there is only one electron anyway. I guess my answer comes down to the fact that the ground state (i.e. lowest possible excitation) of a single electron around a nucleus is non-trivial, meaning it is not just a point centered at the origin where the nucleus is; instead, it’s a spatially distributed probability cloud with corresponding non-vanishing energy.

Note that we are talking bound states here - it is of course still possible to fire an electron at the nucleus, and hit it in the process, but that is not a stable bound state, and won’t happen spontaneously.

OK, thanks. got it.

[Carrock]

 

3 hours ago, Markus Hanke said:

.... it is of course still possible to fire an electron at the nucleus, and hit it in the process, but that is not a stable bound state, and won’t happen spontaneously.

It is possible for an electron in any s state (lowest energy or any higher bound state) in an atom to pass through the nucleus spontaneously, as I've said several times in this thread.

Radial probability is not probability per unit volume.

18 hours ago, swansont said:

...The probability of the electron being at r=0 is zero, but the nucleus is not a point, so there is a small probability of finding the electron there.

The radial probability density at r=0 is 0, wherever you choose the origin of the coordinates to be. Mathematically, r=0 is a point at the origin, here inside the nucleus, and the probability of finding an electron at r=0 is 0.

You could choose the origin to be at any point in the nucleus or elsewhere, and the radial probability of finding the electron at r=0 is always 0.

As calculus is involved, it would be somewhat more precise to say

"As r approaches 0, the electron radial probability approaches 0, for any location of the centre of the radial coordinate system."

As electron capture in metastable atoms shows, the concept of s electrons passing through the nucleus (and having a higher probability per unit volume of being in the nucleus than anywhere else) is not just an abstraction without physical meaning.

Quote

Electrons in s orbitals (regardless of shell or primary quantum number), have a probability antinode at the nucleus, and are thus far more subject to electron capture than p or d electrons, which have a probability node at the nucleus.

If you scroll down to 's-type drum modes and wave functions' , https://en.wikipedia.org/wiki/Atomic_orbital#Qualitative_understanding_of_shapes

really does have some good graphics of orbitals and its easy to see the maximum electron probability per unit volume for s orbitals is at the nucleus.

 

Edited August 15, 2018 by Carrock

[swansont]

1 minute ago, Carrock said:

 

Radial probability is not probability per unit volume.

The radial probability density at r=0 is 0, wherever you choose the origin of the coordinates to be. Mathematically, r=0 is a point at the origin, here inside the nucleus, and the probability of finding an electron at r=0 is 0.

You could choose the origin to be at any point in the nucleus or elsewhere, and the radial probability of finding the electron at r=0 is always 0.

You aren't going to get the same solution if you choose another point to be your origin. You will have (r-r₀) terms. The function is going to be radially symmetric about the center of the nucleus, and not symmetric in r.

 

[Carrock]

4 minutes ago, swansont said:

You aren't going to get the same solution if you choose another point to be your origin. You will have (r-r₀) terms. The function is going to be radially symmetric about the center of the nucleus, and not symmetric in r.

 

The only point in choosing a different origin is to show that with any arbitrary origin the radial probability of finding the electron at r=0 is always 0.

The solution is different of course but asymmetry does not cause the electron to have nonzero probability at r=0.

I was trying to fight the idea that electrons in stable atoms do not interact with nucleons because the radial probability of them being inside nucleons is almost 0 but it seems that meme is too powerful for me.

[swansont]

4 hours ago, Carrock said:

The only point in choosing a different origin is to show that with any arbitrary origin the radial probability of finding the electron at r=0 is always 0.

Then the relevant calculation is being found in a shell of thickness dr

Quote

The solution is different of course but asymmetry does not cause the electron to have nonzero probability at r=0.

I was trying to fight the idea that electrons in stable atoms do not interact with nucleons because the radial probability of them being inside nucleons is almost 0 but it seems that meme is too powerful for me.

And since dP varies as e^-r/a₀r²dr the probability is indeed very small near r=0

The probability of being between 0 and 0.001a₀ is 1.3 x 10^-9
The probability of being between a₀ and 1.001a₀ is 0.54 x 10^-3

Same radial range, very different results, for "the probability of finding an electron at a given distance from the nucleus" which is what the wikipedia page is saying they are looking at

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydrng.html#c1

 

[Carrock]

2 hours ago, swansont said:

Then the relevant calculation is being found in a shell of thickness dr

And since dP varies as e^-r/a₀r²dr the probability is indeed very small near r=0

The probability of being between 0 and 0.001a₀ is 1.3 x 10^-9
The probability of being between a₀ and 1.001a₀ is 0.54 x 10^-3

Same radial range, very different results, for "the probability of finding an electron at a given distance from the nucleus" which is what the wikipedia page is saying they are looking at

http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydrng.html#c1

 

I'm honestly not clear what point you're making.

If you don't agree with either of these I'll respond in detail to your last post.

 

 

1:

[ Origin is centre of nuceus.]

The probability of an electron in the hydrogen ground state being between 0 and 0.001a₀ is 1.3 x 10^-9
The probability of an electron in the hydrogen ground state being between a₀ and 1.001a₀ is 0.54 x 10^-3

 

2:

The probability of an s orbital electron in a hydrogen atom in any (very small) unit volume is a maximum at the antinode (i.e. centre of the nucleus) and is higher throughout the nucleus than anywhere else.

(The quantitative calculation of probability at or near an antinode is not simple but measurements e.g. of the half life of an atom where an s electron is captured by a proton-rich nucleus can give an experimental value.)

 

 

[Markus Hanke]

19 hours ago, Carrock said:

It is possible for an electron in any s state (lowest energy or any higher bound state) in an atom to pass through the nucleus spontaneously, as I've said several times in this thread.

This is true, but it is not what the OP has asked. The original question was why the electron does not fall into the nucleus, i.e. how is an atom different from a purely classical system of a charge in free fall towards another (opposite) charge, which of course is not a stable situation in the classical domain. So the OP wanted to know how this is possible, so I have attempted to answer the question.

The spontaneous tunnelling through the nucleus - or any other classically forbidden region - is not the same as the electron “falling in”.

[Sensei]

In classic physics there is no electron. There is no quantization of charge e=1.6021766*10^-19 C. Classic physics don't deal with particles.. etc. etc.

Edited August 16, 2018 by Sensei

[Strange]

1 hour ago, Sensei said:

In classic physics there is no electron. There is no quantization of charge e=1.6021766*10^-19 C. Classic physics don't deal with particles.. etc. etc.

I'm not sure I agree with that. The electron was discovered long before quantum theory was developed. 

[swansont]

1 hour ago, Strange said:

I'm not sure I agree with that. The electron was discovered long before quantum theory was developed. 

I definitely don't agree with it. To say that the electron doesn't exist in classical physics is wrongness of a staggering degree. 

[Carrock]

3 hours ago, Markus Hanke said:

23 hours ago, Carrock said:

It is possible for an electron in any s state (lowest energy or any higher bound state) in an atom to pass through the nucleus spontaneously, as I've said several times in this thread. 

This is true, but it is not what the OP has asked. The original question was why the electron does not fall into the nucleus, i.e. how is an atom different from a purely classical system of a charge in free fall towards another (opposite) charge, which of course is not a stable situation in the classical domain. So the OP wanted to know how this is possible, so I have attempted to answer the question.

The spontaneous tunnelling through the nucleus - or any other classically forbidden region - is not the same as the electron “falling in”.

I don't know the meaning of 'classically forbidden' in this context. The nucleus is not a forbidden region for an s orbital electron. The electron does not tunnel through the nucleus - that would require some sort of barrier which the electron had insufficient energy to cross classically.

In stable atoms, for quantum mechanical reasons there is simply no energetically favourable reaction the electron can have with the nucleus.

That is an incomplete answer to the OP but accurate as far as it goes.

On 14/08/2018 at 4:51 AM, Achilles said:

If there is an electrostatic force between the electron and proton, why doesn't the electron go all the way into the nucleus, what is stopping the electron from doing that.

I've answered that: nothing stops the electron going all the way into (and out of) the nucleus and it does exactly that.

A classically analogous question would be: If I jump from an aircraft, why can't I go all the way into a cloud?

The answer is that you can, but a cloud is not a solid object and there is no energetically favourable reaction which stops you leaving the cloud.

 

On 14/08/2018 at 6:59 AM, Markus Hanke said:

Crucially, there is a lowest energy level, which corresponds to the minimum distance an electron can be with respect to the nucleus (let’s assume here there is only one electron) - and that lowest energy level is not zero. Therefore the electron cannot fall all the way to the nucleus, it can only fall into its lowest energy level, which corresponds to an orbital that is still some distance outside the nucleus.

 

23 hours ago, Carrock said:

I was trying to fight the idea that electrons in stable atoms do not interact with nucleons because the radial probability of them being inside nucleons is almost 0 but it seems that meme is too powerful for me.

...and also the idea in your quote that "there is a lowest energy level, which corresponds to the minimum distance an electron can be with respect to the nucleus." Bound s electrons of any energy have a maximum probability, per unit volume, of being found inside the nucleus.

[studiot]

On 16/08/2018 at 6:47 AM, Markus Hanke said:

This is true, but it is not what the OP has asked. The original question was why the electron does not fall into the nucleus, i.e. how is an atom different from a purely classical system of a charge in free fall towards another (opposite) charge, which of course is not a stable situation in the classical domain. So the OP wanted to know how this is possible, so I have attempted to answer the question.

The spontaneous tunnelling through the nucleus - or any other classically forbidden region - is not the same as the electron “falling in”.

I can't see any valid reason for the negative vote so +1.

Further I agree that much of the discussion was not what the OP asked for.

I disagree that there is no purely classical explanation in terms of coulomb forces and classical mechanics however.

The Rutherford- Bohr atom did that.

The problem requiring resolution was due to classical electrodynamics, not electrostatics (coulomb forces).

 

Achilles, if you are still interested we can follow the development of the theory in the terms you asked for.

 

Carrock,

On 15/08/2018 at 7:41 PM, Carrock said:

If you don't agree with either of these I'll respond in detail to your last post.

I would be interested in the mathematical detail you offer since I wonder if you are misplacing the use of electron density as an observable?

Edited August 18, 2018 by studiot

[Carrock]

8 hours ago, studiot said:

Carrock,

On 15/08/2018 at 7:41 PM, Carrock said:

If you don't agree with either of these I'll respond in detail to your last post.

I would be interested in the mathematical detail you offer since I wonder if you are misplacing the use of electron density as an observable?

Like any quantum effect electron density is not directly observable but it can be  'observed' by firing high energy electrons (accurate position, inaccurate momentum) at hydrogen atoms with an s1 orbital electron and observing the scattering positions. Also fire electrons at proton to calculate nuclear scattering and allow for it in result. More practical ways are available....

 

If you construct a 3d rectangular grid around the atom you will find the most scattering per unit volume at and near the nucleus.

You will also find that the greatest number of scatterings is at the Bohr radius.

 

I'll do the Swansont type calculations for an imaginary atom which has uniform electron probability up to the Bohr radius a₀ and zero elsewhere. I'm not going to bother saying things which should be obvious but would lose marks in an exam.

For that atom, the equation in http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/hydrng.html#c1 simplifies to

 

P = k * (integral b to c of r^2 dr) where k is a constant.

1 = k * (integral 0 to a₀ of r^2 dr) - probability of finding electron in this range is unity.

1  = k * (a₀)³/3

i.e. k = 3/(a₀)³

(integral 0 to 0.001a₀ of r^2 dr) = 10-9  * (a₀)³/3  * k = 10^-9.

(integral 0.1 to 0.101a₀ of r^2 dr) = ( 0.001030301  * (a₀)³/3 )  - ( 0.001 * (a₀)³/3 )) * k = 3.0301 * 10^-5

So for uniform electron probability density the radial probability for a 0.001a₀ shell increases with distance from the nucleus.

In the Bohr model as the radius decreases the volume of shell decreases faster than the probability per unit volume increases.

Do you maintain that one of these is not true?

On 15/08/2018 at 7:41 PM, Carrock said:

The probability of an electron in the hydrogen ground state being between 0 and 0.001a₀ is 1.3 x 10^-9
The probability of an electron in the hydrogen ground state being between a₀ and 1.001a₀ is 0.54 x 10^-3

 

2:

The probability of an s orbital electron in a hydrogen atom in any (very small) unit volume is a maximum at the antinode (i.e. centre of the nucleus) and is higher throughout the nucleus than anywhere else.

(The quantitative calculation of probability at or near an antinode is not simple but measurements e.g. of the half life of an atom where an s electron is captured by a proton-rich nucleus can give an experimental value.)

 

 

 

 

[studiot]

14 minutes ago, Carrock said:

Do you maintain that one of these is not true?

Thank you for responding. Did you see my reply to your test in the sandbox?

I will think about your comments overnight, but two things to go away with.

1) The OP specifically asked for a classical discussion so anything else would be off topic (eg discussion of orbitals and probability.)

2) As I understand X ray diffraction of crystals electron density is directly observable, and calculable.
Electron density maps are very useful between nuclei (atoms) in molecules or crystals, but of very limited value within an individual atom.

What units are you measuring electron density in? Perhaps we mean something entirely different?

My definition refers to a number (of electrons or moles) per unit volume.
As such it has three important characteristics.
The number part is one of the basic 'dimensions' in dimensional analysis
Because it is discrete it is automatically 'quantized', without the need to refer to QM.
 

Edited August 18, 2018 by studiot

[Carrock]

2 hours ago, studiot said:

Thank you for responding. Did you see my reply to your test in the sandbox?

I will think about your comments overnight, but two things to go away with.

1) The OP specifically asked for a classical discussion so anything else would be off topic (eg discussion of orbitals and probability.)

2) As I understand X ray diffraction of crystals electron density is directly observable, and calculable.
Electron density maps are very useful between nuclei (atoms) in molecules or crystals, but of very limited value within an individual atom.

Thanks for very useful reply to sandbox test which I've just read. I actually have latex on my computer but every time I try it here I feel I'm starting from scratch. I'll give it another go and probably give up again when I spend two minutes on a reply and twenty minutes finding an 'obvious' latex error such as the one you pointed out.

 

1) The OP posted in quantum theory and did not ask for a classical discussion. When I started reading about forbidden regions, barriers which cannot be tunnelled through etc I didn't want the OP to end up totally confused.

2) I just used a simple example for clarity and hinted 'More practical ways are available....' that there were a lot of other ways.

2 hours ago, studiot said:

What units are you measuring electron density in? Perhaps we mean something entirely different?

My definition refers to a number (of electrons or moles) per unit volume.
As such it has three important characteristics.
The number part is one of the basic 'dimensions' in dimensional analysis
Because it is discrete it is automatically 'quantized', without the need to refer to QM.
 

I'm referring, probably imprecisely, to the square of the probability wave amplitude. On the small scale of an atomic orbital, you can only talk of probabilities unless you destroy its stability by localising it with a high energy probe.

Your definition is essentially macroscopic which isn't useful here. At sufficiently small unit volume, e.g. part of an atom, you can only say something like 'there is a 30% probability that if I measure, there will be an electron in this region.'

 

 

11 hours ago, studiot said:

I can't see any valid reason for the negative vote so +1.

 

Some reasons I gave Hanke a downvote...

 

On 14/08/2018 at 3:36 PM, Carrock said:

Just to clarify, while electrons in a stable atom can't fall into the nucleus, s orbital electrons can pass through it. The nucleus is not inherently a forbidden space for electrons......

 

On 16/08/2018 at 6:47 AM, Markus Hanke said:

On 15/08/2018 at 10:48 AM, Carrock said:

It is possible for an electron in any s state (lowest energy or any higher bound state) in an atom to pass through the nucleus spontaneously, as I've said several times in this thread....

[Part of this post not quoted by Hanke]

...As electron capture in metastable atoms shows, the concept of s electrons passing through the nucleus (and having a higher probability per unit volume of being in the nucleus than anywhere else) is not just an abstraction without physical meaning.

Quote

Electrons in s orbitals (regardless of shell or primary quantum number), have a probability antinode at the nucleus, and are thus far more subject to electron capture than p or d electrons, which have a probability node at the nucleus.

 

This is true, but it is not what the OP has asked. The original question was why the electron does not fall into the nucleus, i.e. how is an atom different from a purely classical system of a charge in free fall towards another (opposite) charge, which of course is not a stable situation in the classical domain. So the OP wanted to know how this is possible, so I have attempted to answer the question.

The spontaneous tunnelling through the nucleus - or any other classically forbidden region - is not the same as the electron “falling in”.

I get rather irritated with ...this is true ... except it's not.

It's perfectly reasonable to claim I've got it wrong, but to edit my post and then imply I've said meaningless rubbish like 'The spontaneous tunnelling through the nucleus - or any other classically forbidden region - is the same as the electron “falling in”' is not.

The point I make that the electron doesn't significantly interact with the nucleus but does fall through it has been ignored.

Claiming that the OP is correct in the idea that the electron is never in the nucleus requires justification by Heinke.

Edited August 19, 2018 by Carrock