# Why does helium have a greater first ionization energy than hydrogen?

 P: 15 I have had this question in the back of my mind for a while. Hydrogen has 1 proton and 1 electron, and the atom is electrically neutral. So that means that the electrical charge from the electron and the proton cancel each other out as they have equal/opposite charge. Then when you add another proton to the nucleus, it seems like it would make sense that the overall charge of the atom would only be that of the one proton and thus the energy required to remove an electron from helium would only be the energy required to separate the electron from that one proton. So it doesn't make sense to me that the first ionization energies for hydrogen and helium aren't the same. Do you think you could clarify this for me? Thanks a lot, mikfig
 P: 4 I think it's because its is a noble gas and a the S² configuration is more stable than an s1 configuration.
 P: 836 Even though the atom in electrically neutral to an ouside particle, a particle inside the atom would not agree. Each electron in the Helium atom experiences a nuclear charge that is grater that the electron in the hydrogen atom experiences (1.67 times greater if my memory servers me right). This alone should account for the fact that helium has a higher ionization energy than hydrogen.
P: 15
Why does helium have a greater first ionization energy than hydrogen?

 Quote by espen180 Even though the atom in electrically neutral to an ouside particle, a particle inside the atom would not agree. Each electron in the Helium atom experiences a nuclear charge that is grater that the electron in the hydrogen atom experiences (1.67 times greater if my memory servers me right). This alone should account for the fact that helium has a higher ionization energy than hydrogen.
Ok, so essentially the charges of the proton and the electron don't cancel each other out when it comes to those distances from the nucleus that are considered "in the atom"? Just out of curiosity, do you know what equations I could plot in mathematica of the electrical field inside a helium atom if it was completely isolated?

Thanks again :D
 P: 836 The wavefunctions for the electrons in the helium atom are not analytically solvable, so it depends on the approximation you use. Maybe the wiki article will be useful. http://en.wikipedia.org/wiki/Helium_atom My suggestion is then to model the charge density of the electron as $$\rho=-e|\psi|^2$$, where $$\rho$$ is the charge density of the electron and $$\psi$$ is the wavefunction of the electron. $$|\psi|^2$$ is then the modulus of the wavefunction squared. Note that you need the normalized wavefunction. (Normalized to 2, i guess...). Then you can use Gauss' law to find the electric field at a given distance from the nucleus.
 P: 41 also, the helium atom is volumetricly smaller than the hydrogen atom. (Hydrogen atom is the LARGEST atom of all , volumetriclly .. its electrons are only dealing with the 'tug' of a single proton). In a 'graphical sense' we'd therefore expect more energy to be needed in the case of the helium atom.
 P: 8 Why do you think that hydrogen atom is the largest of all, volumetrically? I would say atoms of other elements, say chlorine, would disagree with you. Back to your point, I think you are a little bit confused between the neutrality of an atom and the first ionization energy. When you consider an atom as a system of the nucleus and the electrons, it is neutral because, as you suggested, the charge of the electrons cancels out the charge of the nucleus. However, when you are removing an electron from the atom, it's a completely different thing. In this process, you are dealing with the electrostatic forces of attraction between that one electron with the nucleus, as well as the repulsion between the other electrons with the electron you are removing. Helium has higher first ionization energy because the electrostatic forces of attraction between the nucleus (+2e) with the electron (-1e) are apparently higher than those between the H nucleus (+1e) and the electron (-1e) (by Coulomb's law), so is the first ionization energy. Even though there is repulsion between the remaining electron with the one being removed, it is not large enough to reduce the attraction. Hope that's helpful.
P: 137
 Quote by dorebase2006 Why do you think that hydrogen atom is the largest of all, volumetrically? I would say atoms of other elements, say chlorine, would disagree with you. Back to your point, I think you are a little bit confused between the neutrality of an atom and the first ionization energy. When you consider an atom as a system of the nucleus and the electrons, it is neutral because, as you suggested, the charge of the electrons cancels out the charge of the nucleus. However, when you are removing an electron from the atom, it's a completely different thing. In this process, you are dealing with the electrostatic forces of attraction between that one electron with the nucleus, as well as the repulsion between the other electrons with the electron you are removing. Helium has higher first ionization energy because the electrostatic forces of attraction between the nucleus (+2e) with the electron (-1e) are apparently higher than those between the H nucleus (+1e) and the electron (-1e) (by Coulomb's law), so is the first ionization energy. Even though there is repulsion between the remaining electron with the one being removed, it is not large enough to reduce the attraction. Hope that's helpful.
He means that for a given energy level the orbitals of hydrogen are the largest. Chlorine is only bigger because it has more electrons in higher shells, 2s for example. However, if you could excite the electrons in the hydrogen atom, promoting them to 2s and higher, then hydrogen's 2s orbital would be larger than chlorine's. (and also note that as someone pointed out earlier, analytically solving the wavefunctions of the higher atoms becomes a big n body problem and can't be done with our current math.)
P: 15
 Quote by espen180 The wavefunctions for the electrons in the helium atom are not analytically solvable, so it depends on the approximation you use. Maybe the wiki article will be useful. http://en.wikipedia.org/wiki/Helium_atom My suggestion is then to model the charge density of the electron as $$\rho=-e|\psi|^2$$, where $$\rho$$ is the charge density of the electron and $$\psi$$ is the wavefunction of the electron. $$|\psi|^2$$ is then the modulus of the wavefunction squared. Note that you need the normalized wavefunction. (Normalized to 2, i guess...). Then you can use Gauss' law to find the electric field at a given distance from the nucleus.
Ehh...I wish that I understood the math/physics behind that. But I'm working towards teaching it to myself right now :D. What I think I do understand though is that the wave function is the the Schrödinger wave equation and I understand what divergence means. But I don't see how you would go from the divergence back to the vector-valued function itself. And then there's the fact that I only understand some of the ideas behind quantum mechanics but not the actual math behind it.
But thanks to all you guys for a speedy reply :D.

-Mikfig
P: 15
 Quote by dorebase2006 Why do you think that hydrogen atom is the largest of all, volumetrically? I would say atoms of other elements, say chlorine, would disagree with you. Back to your point, I think you are a little bit confused between the neutrality of an atom and the first ionization energy. When you consider an atom as a system of the nucleus and the electrons, it is neutral because, as you suggested, the charge of the electrons cancels out the charge of the nucleus. However, when you are removing an electron from the atom, it's a completely different thing. In this process, you are dealing with the electrostatic forces of attraction between that one electron with the nucleus, as well as the repulsion between the other electrons with the electron you are removing. Helium has higher first ionization energy because the electrostatic forces of attraction between the nucleus (+2e) with the electron (-1e) are apparently higher than those between the H nucleus (+1e) and the electron (-1e) (by Coulomb's law), so is the first ionization energy. Even though there is repulsion between the remaining electron with the one being removed, it is not large enough to reduce the attraction. Hope that's helpful.
Ok, I think I'm starting to get it. I can't really explain it yet, but my assumption was that the charge being "felt" by the 2nd electron in helium's orbitals (the one being removed) from the nucleus was equal to +1e. And then I also didn't think about the repulsion between the two electrons, but once again, I was under the assumption that that charge was canceled by the protons in the nucleus. I think this may just be a complete misunderstanding on my part of how the electromagnetic force/field works. I don't really know how to visualize the fields that are being experienced by the different subatomic particles.
P: 15
 Quote by Galap He means that for a given energy level the orbitals of hydrogen are the largest. Chlorine is only bigger because it has more electrons in higher shells, 2s for example. However, if you could excite the electrons in the hydrogen atom, promoting them to 2s and higher, then hydrogen's 2s orbital would be larger than chlorine's. (and also note that as someone pointed out earlier, analytically solving the wavefunctions of the higher atoms becomes a big n body problem and can't be done with our current math.)
Your comment that our current math can't solve the wavefunctions of higher atoms really piqued my interest. What is the current bleeding edge of mathematical physics/math right now? The only recent news I've seen had to do with someone that solved one of the million dollar problems and turned down the prize. Oh, and if we did solve these equations, would we have to change our current system of electron orbitals? Like each atom has its own "kind" of s, p, d, f, g orbitals? What other kind of major improvements would solving these equations bring?

-Mikfig
P: 1,866
 Quote by mikfig So that means that the electrical charge from the electron and the proton cancel each other out as they have equal/opposite charge. Then when you add another proton to the nucleus, it seems like it would make sense that the overall charge of the atom would only be that of the one proton and thus the energy required to remove an electron from helium would only be the energy required to separate the electron from that one proton. So it doesn't make sense to me that the first ionization energies for hydrogen and helium aren't the same.
It's not a bad idea, but you're taking into account something that turns out to be quite important in the context, which is that the repulsion between electrons is very different from their attraction to the nucleus. For two reasons:
1) The electrons occupy a much larger volume of space. The size of an atom is about 5 orders of magnitude larger than that of the nucleus. The nucleus is practically a point-charge in the middle of a large electron density cloud.
2) The electrons are in motion (but the nucleus is so heavy and slow it's practically stationary relative the electrons). They can adapt their motion to avoid each other, lowering their energy.
(If you know a little quantum mechanics, you'll actually see that that's just one reason, really, since the degree to which a particle is spread out in space is related to its mass and momentum)

So, if an atom only has one electron, then its energy is -Z2/2. Hydrogen's electron has an energy of -0.5 atomic units. (Hartrees). He+ would have -2, which is indeed the second ionization energy. Now, if you add an electron to He+, and follow your assumption that the first electron cancels half the charge of the nucleus, then the total energy should be -2.5.

It's actually -2.9 - the existing electron doesn't cancel out the charge fully. You can calculate the effective nuclear charge that the second electron 'sees', and it's about 1.4 rather than 1. If you use an effective nuclear charge instead, you get -2.85.. almost right. But it still isn't completely correct because it only accounts for the average repulsion, which doesn't include the dynamical parts of (2).
P: 1,866
 Quote by mikfig Your comment that our current math can't solve the wavefunctions of higher atoms really piqued my interest. What is the current bleeding edge of mathematical physics/math right now?
Well, there's a difference between not being able to solve something analytically, which basically means having a mathematically exact expression (with no consideration of whether or not that expression can be evaluated easily or not), and being able to solve something numerically, i.e. an approximate solution, although one that can usually be made arbitrarily exact.

In practice it's not necessarily very important, which is well-illustrated by Sundman's solution to the classical three-body problem (which is related but a bit different from the quantum-mechanical three-body problem, as in Helium, which is still unsolved) The thing is, although Sundman's solution was analytic, it took the form of an infinite sum. Which means that any real calculation would (in this case) be approximate. Worse, this sum converges very slowly, which means you have to calculate very many terms to get a certain level of accuracy. Solving the same problem numerically is very simple, however. So from the standpoint of calculating something in practice, the analytical solution isn't useful.

We've been solving the equations involved numerically since the start of quantum mechanics. The ground-state energy of Helium was calculated to three decimal places by Hylleraas (using a hand-cranked mechanical desk calculator) in 1927, and to six places in 1929. And the development of better ways to do this has continued since; the Nobel prize in 1998 was given to Kohn and Pople for having developed such methods. Today we can calculate Helium's energy to within experimental error. Small molecules with light atoms can be calculated to within "chemical accuracy" as well. The largest systems we can solve to useful level of accuracy are about 100-200 (first and second-row) atoms.

In short, that's what quantum chemistry is about: Solving the equations, and developing better ways of doing it. So it's at the intersection of applied math, theoretical physics, computer science and chemistry.
 Oh, and if we did solve these equations, would we have to change our current system of electron orbitals? Like each atom has its own "kind" of s, p, d, f, g orbitals?
Nope. Orbitals come from making an approximation that makes the equations soluble (although you still have to do it numerially). In fact, I already mentioned it - by assuming that electrons only see the average repulsion from the other electrons. As you can see from the helium example it's a good approximation (2% error in energy). It's not good enough to be able to calculate specifics of chemistry, but it's good enough to get a qualitative understanding of what's going on, which is what your average chemist uses orbitals for. It's not exact in the strictest sense, but that's not a problem.
 What other kind of major improvements would solving these equations bring?
Being able to solve these equations quickly and exactly would means we'd be able to know every single chemical property of anything we'd want to calculate. Chemists wouldn't need to do experiments anymore, we'd just be able to calculate everything. That's not about to happen though, although quantum chem is becoming more important and useful by the day.

From the pure-math standpoint, the many-body problem isn't really a problem; it's a source of problems!
P: 15
 Quote by alxm Well, there's a difference between not being able to solve something analytically, which basically means having a mathematically exact expression (with no consideration of whether or not that expression can be evaluated easily or not), and being able to solve something numerically, i.e. an approximate solution, although one that can usually be made arbitrarily exact. In practice it's not necessarily very important, which is well-illustrated by Sundman's solution to the classical three-body problem (which is related but a bit different from the quantum-mechanical three-body problem, as in Helium, which is still unsolved) The thing is, although Sundman's solution was analytic, it took the form of an infinite sum. Which means that any real calculation would (in this case) be approximate. Worse, this sum converges very slowly, which means you have to calculate very many terms to get a certain level of accuracy. Solving the same problem numerically is very simple, however. So from the standpoint of calculating something in practice, the analytical solution isn't useful. We've been solving the equations involved numerically since the start of quantum mechanics. The ground-state energy of Helium was calculated to three decimal places by Hylleraas (using a hand-cranked mechanical desk calculator) in 1927, and to six places in 1929. And the development of better ways to do this has continued since; the Nobel prize in 1998 was given to Kohn and Pople for having developed such methods. Today we can calculate Helium's energy to within experimental error. Small molecules with light atoms can be calculated to within "chemical accuracy" as well. The largest systems we can solve to useful level of accuracy are about 100-200 (first and second-row) atoms. In short, that's what quantum chemistry is about: Solving the equations, and developing better ways of doing it. So it's at the intersection of applied math, theoretical physics, computer science and chemistry. Nope. Orbitals come from making an approximation that makes the equations soluble (although you still have to do it numerially). In fact, I already mentioned it - by assuming that electrons only see the average repulsion from the other electrons. As you can see from the helium example it's a good approximation (2% error in energy). It's not good enough to be able to calculate specifics of chemistry, but it's good enough to get a qualitative understanding of what's going on, which is what your average chemist uses orbitals for. It's not exact in the strictest sense, but that's not a problem. Being able to solve these equations quickly and exactly would means we'd be able to know every single chemical property of anything we'd want to calculate. Chemists wouldn't need to do experiments anymore, we'd just be able to calculate everything. That's not about to happen though, although quantum chem is becoming more important and useful by the day. From the pure-math standpoint, the many-body problem isn't really a problem; it's a source of problems!
That's very interesting :D. But just curious, if we did have analytical solutions of things then couldn't we use these functions/equations/whatever together in different forms, get the numerical solutions of those, and find the answers to some rather interesting problems? I mean numerical solutions only go so far right?

 Quote by alxm It's not a bad idea, but you're taking into account something that turns out to be quite important in the context, which is that the repulsion between electrons is very different from their attraction to the nucleus. For two reasons: 1) The electrons occupy a much larger volume of space. The size of an atom is about 5 orders of magnitude larger than that of the nucleus. The nucleus is practically a point-charge in the middle of a large electron density cloud. 2) The electrons are in motion (but the nucleus is so heavy and slow it's practically stationary relative the electrons). They can adapt their motion to avoid each other, lowering their energy. (If you know a little quantum mechanics, you'll actually see that that's just one reason, really, since the degree to which a particle is spread out in space is related to its mass and momentum) So, if an atom only has one electron, then its energy is -Z2/2. Hydrogen's electron has an energy of -0.5 atomic units. (Hartrees). He+ would have -2, which is indeed the second ionization energy. Now, if you add an electron to He+, and follow your assumption that the first electron cancels half the charge of the nucleus, then the total energy should be -2.5. It's actually -2.9 - the existing electron doesn't cancel out the charge fully. You can calculate the effective nuclear charge that the second electron 'sees', and it's about 1.4 rather than 1. If you use an effective nuclear charge instead, you get -2.85.. almost right. But it still isn't completely correct because it only accounts for the average repulsion, which doesn't include the dynamical parts of (2).
Ok I see, I guess I made a lot of assumptions and looked at the occurrences happening within the atom from a simple Newtonian physics perspective and also didn't think about fields and what not.

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