# Ionization energy of the dihydrogen

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## Homework Statement

Calculate the energy of ionization of the dihydrogen $H_2$, knowing that the binding energy of $H_2$ and $H_2^+$ are worth -4.48 eV and -2.65 eV respectively and that the ionization energy of the hydrogen atom is 13.6 eV.

2. Relevant thoughts
Energy of ionization of $H_2$ is the energy required to remove 1 electron from the system "2 protons+1 electron".
Binding energy of $H_2$ is the energy that the system :2 electrons+2 protons have. Since it's negative it means it's stable (when totally dissociated the molecule has 0 energy).

## The Attempt at a Solution

I don't really know how to use the given data.
A try: The binding energy increases by 1.83 eV when one removes an electron from $H_2$. The ion molecule is thus less stable (but still is). So I'm guessing that it's easier by an amount of 1.83 eV to remove an electron in the $H_2$ molecule compared to the hydrogen atom. This would make the answer to the problem 11.77 eV.
However in wikipedia I find 15.603 eV. I have absolutely no idea how to reach such a number (or 15.6 for that matter).
Any tip is welcome.

I like Serena
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What if you think up a process in 2 steps that matches your problem statement?

Gold Member
I'm really lost.
What 2 steps exactly?
Now that I think... The binding energy of the dihydrogen molecule increases by an amount of 1.83 eV when it gets ionized, i.e. when we remove an electron. So someone had to spend that energy to remove it? This would make the answer to the problem 1.83 eV but I didn't use the fact that removing an electron from an hydrogen atom costs 13.6 eV and the answer doesn't match wikipedia's article either.

I like Serena
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If you first break the binding energy of H2, you should be left with 2 H atoms.
Afterward, you can ionize them.

cepheid
Staff Emeritus
Gold Member

## Homework Statement

Calculate the energy of ionization of the dihydrogen $H_2$, knowing that the binding energy of $H_2$ and $H_2^+$ are worth -4.48 eV and -2.65 eV respectively and that the ionization energy of the hydrogen atom is 13.6 eV.

2. Relevant thoughts
Energy of ionization of $H_2$ is the energy required to remove 1 electron from the system "2 protons+1 electron".

Hmmm....shouldn't that be 2 protons + 2 electrons? Doesn't H2 consist of two full H atoms, with the two electrons shared between the two atoms in a covalent bond?

cepheid
Staff Emeritus
Gold Member
So here's what I'm thinking: to ionize a single hydrogen atom, the electron must be imparted with enough energy to escape a potential well that is 13.6 eV deep.

The question is, how much deeper is the potential well it must escape from in the case where it is escaping from an H2 molecule rather than an H atom?

Gold Member
Thanks guys!
If you first break the binding energy of H2, you should be left with 2 H atoms.
Afterward, you can ionize them.
Hmm but why would I need to unbind them and then ionize both of them? Wouldn't this require 4.48 eV +2x13.6 eV? I must be not understand you. Hmmm....shouldn't that be 2 protons + 2 electrons? Doesn't H2 consist of two full H atoms, with the two electrons shared between the two atoms in a covalent bond?
Yes, indeed.

I like Serena
Homework Helper
Thanks guys!

Hmm but why would I need to unbind them and then ionize both of them? Wouldn't this require 4.48 eV +2x13.6 eV? I must be not understand you. Yes, indeed.

Yes, you did understand me.

I interpret your problem statement to mean that H2 is broken into its 4 particles.
It does not matter how you do that.
Energy conservation guarantees that the end result will always be the same.

Btw, if H2 is fully ionized, the resulting 2 protons won't stay together. :)

Gold Member
Hmm is the following right?
First, I take under my hands a H2 molecule. I kick a proton out (unbind the molecule), I must pay 4.48 eV for that. With my remaining H atom under my fingers, I decide to remove an electron, this costs an additional 13.6 eV (ionization of H atom). Then I phone call the hydrogen atom I had kicked at first and he gives me back 2.65 eV (binding of H2 molecule).
In total I must pay (4.48+13.6-2.65)eV=15.43 eV, which would be slightly cheaper than the wikipedia's 15.6 eV. I'd therefore buy that argument.

I like Serena
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Where did you find 15.630 eV on wikipedia?
I don't see it.

In my interpretation the ionization energy of H2 should be when you split it into its protons and electrons for 4.48 eV + 2x13.6 eV = 31.68 eV.
If you want, that is 31.68/2=15.84 eV per hydrogen atom.

Or perhaps your explanation is the right one, and ionization of H2 means to ionize it into H2+.

Gold Member
Where did you find 15.630 eV on wikipedia?
I don't see it.

In my interpretation the ionization energy of H2 should be when you split it into its protons and electrons for 4.48 eV + 2x13.6 eV = 31.68 eV.
If you want, that is 31.68/2=15.84 eV per hydrogen atom.

Or perhaps your explanation is the right one, and ionization of H2 means to ionize it into H2+.

In the bottom of the page of http://en.wikipedia.org/wiki/Dihydrogen_cation.
Hmm I am not sure then what the exercise asks for.

I like Serena
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In the bottom of the page of http://en.wikipedia.org/wiki/Dihydrogen_cation.
Hmm I am not sure then what the exercise asks for.

That specific section of the article talks about the ionization of H2 into H2+.
I that context the term "ionization energy" will apply to that reaction.
And that's what you calculated.
Odd though that the number is slightly off.
Perhaps you got a discount rate from using the phone.

Gold Member
That specific section of the article talks about the ionization of H2 into H2+.
I that context the term "ionization energy" will apply to that reaction.
And that's what you calculated.
Odd though that the number is slightly off.
Perhaps you got a discount rate from using the phone. Indeed. In fact I got several discounts probably, one on the phone as you say (probably the binding of H2+ is lesser than the one I'm given) and according to hyperphysics the binding energy of H2 is -4.52 eV. So I should probably obtain a result closer to the one of wikipedia if I use more accurate values.