# Can an electron in the 1s orbital be indefinitely far from the nucleus?

• I
Wrong. By "moving the electron" (what you are really describing is changing the energy level the electron is in, which is not the same as "moving" it, but we don't need to delve into all those complications right now) you added energy to the system. So the system's total energy is now -3.4 eV (potential energy -6.8 eV, kinetic energy +3.4 eV) instead of -13.6 eV.

Sorry, I didn't use the right words. The situation I was describing was a bound electron in the position x=R going to the position x=4R by itself, while still bounded to the atom and in the 1s orbital with that characteristic energy of -13,6eV, no external work being made. This is possible by my understanding right? At least it seems to be possible by Schrodinger equations if I'm not misunderstanding it.

PeroK
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Sorry, I didn't use the right words. The situation I was describing was a bound electron in the position x=R going to the position x=4R by itself, while still bounded to the atom and in the 1s orbital with that characteristic energy of -13,6eV, no external work being made. This is possible by my understanding right? At least it seems to be possible by Schrodinger equations if I'm not misunderstanding it.

This does not make sense in QM. If the electron is in a bound state it has no well-defined position. It cannot "go from A to B"; it cannot "be at A" and it cannot "move to B".

None of these classical descriptions make any sense in terms of the SDE, which describes the wavefunction.

• vanhees71
ZapperZ
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This does not make sense in QM. If the electron is in a bound state it has no well-defined position. It cannot "go from A to B"; it cannot "be at A" and it cannot "move to B".

None of these classical descriptions make any sense in terms of the SDE, which describes the wavefunction.

I've been thinking of the same thing and have been scratching my head at all this. Look at the radial part of the wavefunction, even for the 1s orbital, r=4R is INCLUDED in that "spread"! The electron distribution includes that distance from the nucleus.

So this whole scenario makes no sense.

Zz.

PeterDonis
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The situation I was describing was a bound electron in the position x=R going to the position x=4R by itself

This is meaningless if you are not measuring the position. If the electron is in the ground state energy level, the only thing you can say about it is that its energy is -13.6eV. You cannot say where it is, and you cannot talk about it "moving". You can only meaningfully talk about its energy.

This is possible by my understanding right?

No. See above.

it seems to be possible by Schrodinger equations

Why?

I don't think I messed up with the frames. I stick with my frame of reference, in which I consider the potential energy to be zero at infinity. Considering the same frame of reference, without changing it, I know that at Bohr radius the potential energy would be -27,2eV and the kinetic energy +13,6 eV. The total energy would be -13,6 eV, and that's no problem with that being negative, I know it's based on the frame of reference, the kinetic energy is the one that cannot be negative because it doesn't depend on any inertial reference frame. When I move the electron, keeping it in the 1s orbital, to 4R, the potential energy, considering the same frame of reference, becames -6,8eV. By conservation of energy the kinetic energy must be -6,8eV for the sum to be -13,6eV.

If you consider the potential energy at infinity to any other value, say +100eV, then the potential energy at Bohr radius would be 72,8eV, the kinetic energy would remain 13,6eV and the total energy would be 86,4 eV. At 4R, the potential energy would be 93,2eV and the kinetic energy should be -6,8eV for the sum to be 86,4eV. Kinetic energy doesn't depend of the frame of reference chosen (as it shouldn't), and would be negative in both scenarios in the problem.

My question actually is why is this paradox happening. But I think, as @Nugatory said, the measurement is changing the previous state of the atom, and hat is giving the electron some amount of energy

Use a different convention, let the potential energy at infinity be equal to X. We can write energy equations for the particles at infinity and at the ground state.

Infinity P.E. = X , K.E = 0
Bohr radius P.E. = X -27.2 ,K.E = +13.6, Energy lost = 13.6

Whatever convention is used energy is conserved. I think a problem is that when some people look at the convention of setting X equal to zero they think there is no potential energy at all. In this example the maximum at an infinite separation.

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PeterDonis
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The situation I was describing was a bound electron in the position x=R going to the position x=4R by itself, while still bounded to the atom and in the 1s orbital with that characteristic energy of -13,6eV, no external work being made.

Even if we put aside all the valid issues that have been raised with this in terms of QM, and just consider a classical electron orbiting a classical proton with the given energy, and rule out the electron radiating any energy away (which classically it would, it would not remain in a stable orbit but spiral into the nucleus while emitting radiation), we still need to be clear about how classical orbits work.

If the electron's total orbital energy is -13.6eV, that's its total energy at all positions in its orbit. If its orbit is elliptical, so that its position is not always x=R, its total energy is still -13.6eV at all positions in its orbit. That means that it is impossible for the classical orbit to reach any position where the potential energy would be greater than or equal to -13.6eV, no matter how elliptical it gets, since, as you correctly believe, it is impossible for its orbital kinetic energy to be zero or negative in the classical model. That means that even the most elongated possible elliptical orbit would have to have its greatest distance from the proton less than the distance at which the potential energy is -13.6eV. In particular, it would be impossible for a classical elliptical orbit to reach x=4R, since the potential energy there is -6.8eV, which is greater than -13.6eV.

So another way of illustrating the issue you are having is that you are trying to combine two inconsistent views of the electron. You are trying to think of it like a classical particle with a position that is moving in some kind of orbit; but you are also trying to make its orbit have properties that a classical orbit simply cannot have. The "negative kinetic energy" at x=4R is simply a manifestation of this underlying problem. The way to resolve it is to stop trying to combine two inconsistent views of the electron: i.e., to understand that the QM model of the electron is simply incompatible with any way of thinking of the electron in a hydrogen atom as a classical particle "orbiting" the proton. That simply doesn't work.

Stephen Tashi
If we detected an electron on the moon, how would we know which atom it "belonged to"? Does answering the question in OP require setting up a situation where we somehow know we are detecting an electron that belongs to the the atom in question and not to an atom that's part of the moon?

• PeroK
PeterDonis
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Use a different convention, let the potential energy at infinity be equal to X

The potential energy zero point, as has already been said, has nothing to do with any of the issues being discussed in this thread. Only differences in potential energy matter for this discussion.

No, the zero point of potential energy makes no difference at all to the calculation, since the difference in potential energy between the unbound state and the bound state is the only thing that appears in the calculation.
Of course the choice of convention makes no difference to the calculations. I never said it did. But it can be confusing whatever convention is used. Some people may look at the chosen values and think they are absolute values.

PeterDonis
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Some people may look at the chosen values and think they are absolute values.

I don't think the OP is doing that since he has already agreed, in the post that you quoted, that changing the zero point of potential energy does not change anything in his calculations.

I don't think the OP is doing that since he has already agreed, in the post that you quoted, that changing the zero point of potential energy does not change anything in his calculations.
Yes I had another look at his post and I see what you mean. It seems that he forgot to take into account the fact that during a transition to the ground state the ionisation energy is radiated to the surroundings.

If we detected an electron on the moon, how would we know which atom it "belonged to"? Does answering the question in OP require setting up a situation where we somehow know we are detecting an electron that belongs to the the atom in question and not to an atom that's part of the moon?
I think a good reply to your question has been given by ZapperZ in post 17.

PeterDonis
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It seems that he forgot to take into account the fact that during a transition to the ground state the ionisation energy is radiated to the surroundings.

And that, in the reverse process, you have to add that energy back, yes.

Stephen Tashi
I think a good reply to your question has been given by ZapperZ in post 17.

Post #17 contrasts emprical facts with theoretical probabilities. How does this answer the question of determining which atom an electron "belongs to". Are we saying that, empirically, a detected electron can be assigned to one particular atom?

PeterDonis
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Are we saying that, empirically, a detected electron can be assigned to one particular atom?

I don't think this is the case in general. In some particular situations it might be, for example, if we have a single atom confined in a trap. But nobody is going to build a single-atom trap the size of the Earth-Moon distance.

bhobba
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Does that mean if you have measured where it is, it's meaningful to ask where it is?

That is to some extent interpretation dependent. From the formalism Peter is of course correct. But let's take say Consistent History's (CH) interpretation. It says if you measure it then it has that position. But after the measurement its wavefunction will then start to spread. Again the formalism is silent on it's position, but the CH interpretation says it has a position somewhere between the areas of the wavefunction that are not zero. Over time my view of QM has changed quite a bit and my view now is like Peters - but interpretations have various takes on the issue. It's part of the reason Wienberg doesn't really like any interpretation. Maybe we just have to accept reality is just a mathematical description, and that's all you can really say, especially in areas beyond our direct experience from which our everyday view of the world is formed. As quantum effects leak into that world because of advancing technology expect to see some really weird things eg a well known one is the behaviour of liquid Helium.

Thanks
Bill

PeterDonis
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That is to some extent interpretation dependent.

While this is true, if we are going to go down that rabbit hole, it should be in a separate thread in the interpretations forum. For this thread and this forum, "the formalism" is what we should use.

• bhobba
Nugatory
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The situation I was describing was a bound electron in the position x=R going to the position x=4R by itself, while still bounded to the atom and in the 1s orbital with that characteristic energy of -13,6eV, no external work being made. This is possible by my understanding right? At least it seems to be possible by Schrodinger equations if I'm not misunderstanding it.
You are misunderstanding it. The Schrodinger equation gives us the probability of finding the electron at a given position, but that doesn't imply that that it "goes to" that position from somewhere else - not even if we had previously measured the position and gotten a different result.

Let's say that we measure the position at time ##T_1## and find the electron at ##x=R##. Terrific..... we can say that electron was at that position at the time of the measurement. But what about a moment later? We don't have a position measurement from a moment later so we cannot say that the electron is still at that position a moment later - no measurement, no position. Instead we use the Schrodinger equation to calculate the wave function of an electron at time ##t=T_1+\Delta{t}## given that it was at ##x=R## at time ##T_1##; we'll get a probability distribution that says it's likely to be fairly close to ##R## but continues to spread out over time.

So we measure it again, at time ##T_2##, and this time we find it at ##x=4R##. That doesn't mean that it moved from ##x=R## to ##x=4R## during the time interval between ##T_1## and ##T_2##; that interpretation would imply that it had a definite position between measurements, and it doesn't. Instead, the first measurement took the electron from the state "no definite position" to the state "at position ##R## at time ##T_1##"; and the second measurement took it from the state "no definite position" to the state "at position ##4R## at time ##T_2##". Both measurements necessarily changed the energy of the electron (although the total energy of the system including the measuring device is conserved).

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