# Energy of hydrogen atoms

1. Dec 13, 2009

### 6021023

Why is it that when two hydrogen atoms combine together, their combined energy is less than the energy of one hydrogen atom by itself? Why wouldn't the combined energy be twice as much?

2. Dec 13, 2009

### Staff: Mentor

To what combined energy is one referring?

3. Dec 13, 2009

### 6021023

The combined energy of the two hydrogen atoms when they come together to form H2.

Last edited: Dec 13, 2009
4. Dec 14, 2009

### alxm

I.e. the binding energy. Okay, well how detailed an answer do you want?

A simple "conceptual" hand-waving answer would be something like:
1) Because electrons are in motion, they don't fully screen the nuclear charge. So even though a hydrogen atom is neutral, the electron on the other hydrogen atom will 'see' a charge from the other nucleus. Therefore, the total electronic energy will be lower if the two nuclei are closer together, up to the point where the nuclear-nuclear repulsion becomes greater than this effect.
2) The electron-electron repulsion, on the other hand, is offset by the correlation of electron motion, in other words, the electrons 'choose' patterns of motion such that they avoid each other.

In-between you've got explanations in terms of Lewis structures (binding gives a full shell), MO theory (the two 1s orbitals combine to form a bonding $$\sigma$$ orbital), Valence-Bond theory, Hartree-Fock, etc.

Last edited by a moderator: Apr 24, 2017
5. Dec 14, 2009

### Staff: Mentor

I'm guessing one means binding energy as alxm mentions.

physics.nist.gov/cgi-bin/Ionization/table.pl?ionization=H2

This might be of use - http://cnx.org/content/m14777/latest/

6. Dec 14, 2009

### bcrowell

Staff Emeritus
There is a change in the classical electrostatic energy associated with the rearranged charge distribution, which others have described above. But there is also a purely quantum-mechanical effect, which is basically just a particle-in-a-box effect. The electrons' wavefunctions spread out to encompass both atoms, increasing their wavelengths and reducing their kinetic energies. Roughly, you're doubling the wavelength along one axis. Suppose the original kinetic energy corresponding to the wavelength of each electron along each of the three axes is K. Then the separated atoms have total kinetic energy 6K. If you treat the atom as a particle in a box, with a length of 0.1 nm, then you get K=38 eV. When the two atoms form a molecule, the wavelengths parallel to the bond axis are both roughly doubled. This reduces two of the K terms to K/4, giving a total kinetic energy of about 4.5K. This makes the binding energy 1.5K=57 eV. The actual binding energy is 15.43 eV. So the particle-in-a-box argument is crude, but it does give a result on the right order of magnitude to be an important contribution to the total binding energy.

7. Dec 14, 2009

### RedX

So if the total energy of two atoms apart is 6*38 eV=228 eV, and a particle-in-a-box says that close together, the total energy is 57 eV, but the real answer is 15.43 eV, would that mean the majority of the energy reduction is the particle-in-a-box effect? The other stuff would then only account for the remaining drop of 57 eV-15.43 eV, while the particle-in-a-box accounts for a drop of 228 eV-57 eV. In fact, the latter drop is 4 times as much as the former drop.

8. Dec 14, 2009

### alxm

bcrowell's approach doesn't quite work as a rationale (except in showing that it's significant)
I'm not sure where that number came from, but the electronic energy of two hydrogen atoms is 1.0 Hartree = 27.2 eV. The binding energy of H2 is 4.52 eV.

The kinetic energy of the electrons is an order of magnitude greater than the binding energy and so is significant, but so is the nuclear attraction and electronic repulsion. The exchange/correlation energy is about on the same order.

So you can't really neglect anything in the electronic Hamiltonian and still get a decent picture of chemical bonding; (or quantum chemistry would be a lot easier) Or to put it another way: Chemistry is a very subtle effect.

9. Dec 14, 2009

### bcrowell

Staff Emeritus
That's all I was claiming -- that it was an argument to show it was significant.

You have a particle in a box of length L=0.1 nm. $\lambda=2L, p=h/\lambda, K=p^2/2m$

10. Dec 14, 2009

### bcrowell

Staff Emeritus
No, the whole thing is just an order-of-magnitude estimate that shows that the change in kinetic energy is of the right order of magnitude to be relevant to the binding energy. The effect is also in the right direction to help explain the binding. (It makes the bound system more bound.)

11. Dec 14, 2009

### RedX

So it's sort of like measuring the strength of nuclear forces by knowing the size of a nucleus and modeling it as a particle-in-a-box (or just using the uncertainty principle) and ignoring all the nuclear physics that goes on? Everything is just based on size?

12. Dec 14, 2009

### bcrowell

Staff Emeritus
Right. It's actually a better approximation in nuclear physics, because the nuclear potential has a flat bottom, not a singularity like the one an atomic potential has in the middle.