# Chemistry - In helium-2, would each electron experience a single positive charge from the nucleus?

## Solution 1:

To answer this question, I am going to assume that a $\ce{^2He}$ nucleus actually exists long enough for two electrons to surround it — this is probably not a valid assumption since Webelements.com does not list $\ce{^2He}$ while it does list the isotopes $\ce{^6He}$ and $\ce{^8He}$ (although they admittedly have long half-lifes measured in tenths of seconds). But for our case, let us assume so.

The charge radius of a proton — our closest experimental measurement of its diameter, if ‘diameter’ is even a valid term at subatomic scales — is given on Wikipedia as $\pu{8.75e-16 m}$ or $\pu{0.875fm}$. At this point, take a step back to remember that the Bohr radius, approximately the most probable distance between the electron and the proton in a hydrogen atom, is $\pu{5.29e-11 m}$ or $\pu{52.9pm}$. This is a difference in scale by a factor of $10^5$.

Therefore, a reasonable assumption for the $\ce{^2He}$ nucleus — and indeed for practically any nucleus — is that it can be approximated by a dimensionless point of $+2$ charge and not any sphere, ovaloid, double-sphere or other three-dimensional object. Therefore, if we want to use quantum mechanics to calculate the wave function of $\ce{^2He+}$, we assume a spheric potential caused by a positive charge in the origin of a coordinate system; the value of the positive charge is $2e$ with $e$ being the elementary charge.

A two-electron system cannot, to the best of my knowledge, accurately be calculated. However, we may approximate it as just taking the wave function solutions of $\ce{^2He+}$ and adding another spin-opposite electron. (Then, we slightly adjust the energy values to account for electron-electron repulsion.) This way of approaching the problem shows that the helium atom should really better be considered a central positively charged dot with the outer space occupied by the negatively charged electrons, each of which see a $+2$ charge in the centre and an additional negative charge ‘buzzing around them’ (Note: physically incorrect macroscopic picture).

Replacing $\ce{^2He}$ (as is present in your question) with either of the stable isotopes $\ce{^3He}$ or $\ce{^4He}$ would not change the picture; the size of a neutron (see my note in the second paragraph on subatomic scales) is roughtly that of a proton. If anything, adding one or two neutrons to the two protons will make the nucleus smaller since neutrons increase the intranuclear attractive forces.

In many introductory chemistry books, the core and shell model of an atom and the relevant size differences are exemplified by macroscopic comparisons. My chemistry book offered: if you enlargen an atom so that it would fit the Munich Olympic Tower inside it, the nucleus would inflate to the size of a cherry stone, match head or something of that size. This again shows how good the assumption of poiut-sized charges in the nucleus is.

## Solution 2:

Electrostatic interaction is like gravitation: A spherical charge (or body) gives the same attraction as an equal point charge (or mass) at the centre of the sphere. If you are in the sphere, gravity is the same as for a mass with the radius of your current distance to the centre. All the mass or charge closer to the surface than yourself has zero net interaction with you.

The electrons in helium are in the spherical s-orbital. In every instant, one electron feels the full charge of two, while the other one, which is a bit further out, only feels a charge of one, because the other electron shields the core.

This is a bit oversimplistic (hello quantum mechanics), but each electron feels a core charge roughly around 1.5 in average.

## Solution 3:

First of all, helium-2 is an extremely short-lived species - if you wanted to speak about its existence at all. So I assume that you had a more stable isotope in mind, e.g. helium-3 or helium-4.

... is it true that each electron in helium experiences the same force ...

Both electrons in a helium atom are indistinguishable from each other. If you wanted to apply the concept of a force exerted on an electron at microscopic scale, then both electrons would experience the same force.

... would each electron experience a single positive charge from the nucleus?

No. The charge experienced is the charge of the two protons and the electrons that shield each other.

The second ionization energy of helium is four times the ionization energy of hydrogen, as predicted by quantum mechanics. The first ionization energy of helium is greater than that of hydrogen and smaller than its second ionization energy. This is in accordance with the statement above and suggests an "effective charge" roughly about $1.3$ to $1.4$.

• $\pu{1312.0 kJ/mol}$ Hydrogen 1. ionization energy
• $\pu{2372.3 kJ/mol}$ Helium 1. ionization energy
• $\pu{5250.5 kJ/mol}$ Helium 2. ionization energy

## Solution 4:

No, the protons are not "shared" and the attractive force from the nucleus on each electron is about 6 times higher for helium than for hydrogen. This is due to the fact that there are now 2 protons pulling on each electron and they are considerably closer, so each one pulls about 3 times as hard. There is a repelling force between both electrons, but there is no simple analytical expression for it.

In a classical description of this system, the force for each particle-pair is given by Coulomb's law. A single electron around a Hydrogen nucleus feels an attractive force that is proportional to $$\frac{1}{r^2}$$ and each electron around the Helium-nucleus will feel this force for each proton. The smaller radius of the $$\ce{He}$$ atom (roughly $$\frac{1}{\sqrt{3}}$$ times the Bohr radius*) will lead to a ~3 times higher force per proton, so a ~6 times higher force between the entire nucleus and each electron.

If you wanted to relate this to ionization energy, you can effectively assume that the nuclear attraction force on the departing electron is shielded by the remaining electron. As an approximation, this energy is given by the shielded nuclear charge divided by the radius in Rydberg atomic units : $$\frac{Z-1}{r} \approx\sqrt{3}$$, or $$\pu{24 eV}$$. The remaining electron of what is now $$\ce{He+}$$ will have moved closer to the nucleus, at a distance of $$\frac{1}{\sqrt{4}}$$, i.e. $$\frac{1}{2}$$. Its ionization potential is $$\frac{Z-0}{r} = 4$$, or $$\pu{54 eV}$$. For completeness sake, in these units the radius of $$\ce{H}$$ is $$1$$ and its ionization potential is $$\frac{Z-0}{r} = 1$$, or $$\pu{13.6 eV}$$.

* the $$\frac{1}{\sqrt{3}}$$ expression for the radius of $$\ce{He}$$ is just a shortcut for ease of calculation. It is within 2% of theoretical/experimental values.