# Year 12: Cambridge Physics Problem

1. Jun 4, 2012

### johnconnor

I would appreciate if we could discuss the questions together. It's a very interesting question indeed.

Question:
The singly ionised hydrogen molecule $$H_2^+$$ can be regarded as an electron of charge -e moving in the field due to two protons, each of charge +e. The protons, which can be assumed to be stationary, are a distance r apart, as shown in Fig. SM11.1. Figure SM11.2 shows how the total electrostatic energy E of the ionised molecule in its state of lowest energy depends on the separation r of the protons. The zero of energy is the situation in which the electron and the protons are all infinitely far away from each other.

(a) The graph in Fig SM11.2 can be considered as the summation of a positive energy term and a negative energy term. Identify the origins of these terms.

(b) What is the physical reason for the fact that, for small values of r, the graph has a large negative gradient?

(c) What is the physical interpretation of the fact that the curve shows a minimum energy at a particular value of r (110pm)?

(d) What is the physical interpretation of the fact that the minimum value of the total energy is -2.6aJ?

(e) calculate the contribution to the total energy at a proton-proton separation of 110pm resulting from the presence of the electron.

(f) Suppose that the proton-proton separation is increased to a very large value. Where is the electron most likely to be found? Hence, explain why for large values of r, the curve tends towards a negarive (non-zero) value of the total energy.

(g) The energy required to remove an electron from a hydrogen atom (the ionisation energy) is 2.2aJ. Find the energy required to dissociate the singly ionised hydrogen molecule into a hydrogen atom and a proton at infinite separation. Explain your reasoning.

(h) If a second election is added to the system of Fig. SM11.1, the neutral hydrogen molecule H2 will be obtained. Explain qualitatively how you would expect the graph of the total electrostatic energy against proton separation for the neutral molecule to differ from that for the singly ionised molecule. Neglect the small contribution to the total energy caused by the mutual repulsion of the two electrons.

Attempt
Please do correct and/or refine the answers I have given here. Forgive me if my mistakes are elementary:
(a) Protons and electrons.
(b) Repulsion force, which in this case is taken to be positive from -dU/dr (or in this case dE)
(c) Presence of electron produces a minimum net force acting on the protons and electron.
(d) see (c)
(e) proton-electron coulombic potential: $$\frac{+Q.-e}{4\pi \varepsilon _0 (110pm)}= -2.097aJ$$ Therefore -2.6aJ + -2.097aJ = -4.7aJ.
(f) Near to one of the protons. Otherwise there wouldn't be a negative energy value as r tends to a large value. So, the ion will kinda look like a hydrogen proton with a distant proton neighbour. (How do I properly phrase this? Is there any better way to explain this?)
(g) 1st IE for H = 2.2aJ (as previously mentioned)
-2.6aJ = energy to separate all particles to infinity.
Hence, energy required to dissociate the singly ionised hydrogen molecule into a hydrogen atom and a proton at infinite separation = 2.6 - 2.2 = 0.4aJ
(h) Beats me.

Please help me out on this. I'll spam a bit on this forum to reach 10 posts and I'll post the photo over here. Thank you! Your help is greatly appreciated!

Last edited: Jun 4, 2012
2. Jun 4, 2012

### Mépris

I don't think I can help you with this but I do have a question. Is this a question from Paper 4? (yes, I did CIE A-Level physics but I didn't devote too much time to it)

3. Jun 4, 2012

### johnconnor

No it's not. It's like a 1980s STEP Physics question.

4. Jun 4, 2012

### Infinitum

This is one of the best physics problems I've seen in a while!!

I would write the charges on protons and electrons.

Yep. As the radius -> 0, the electrostatic energy -> infinity, and hence the slope increases rapidly.

Sounds good. I'd probably argue that due to the presence of an electron, there is a stable configuration state that is due to bound charges, resulting in the negative term.

Feels correct I'm unsure what the unit aJ is, but the concept applied is definitely correct.

Aye, rephrasing, 'potential energy in bound state is negative' thingy again.

That should be, 2.6aJ instead. You are supplying that much energy to a system that already has -2.6aJ, at its most stable configuration.

Think about how the graph would be. If the separation between protons is very less, then E tends to...? Then, since there exist two electrons and protons each, will you have a 'most stable' configuration? will the value of this binding be more than or less than that of the first graph with ionized hydrogen?

5. Jun 4, 2012

### johnconnor

I know right. Amazing.

"a" is actually atto, i.e. x10-18.

Would you mind elaborating a bit more on this? I need a bit more description to get the idea into my head. Thank you!

What will happen if I supply 2.5aJ instead? Could you please describe what will happen? Thank you.

6. Jun 4, 2012

### johnconnor

Then E tends to infinity.
I'm not entirely sure, but common sense tells me it'd be more stable. But why?
The binding value will be higher than that of the first graph.

What's your say on my attempts, Infinitum? Thank you!

7. Jun 4, 2012

### Infinitum

The potential energy in bound state is always negative(if you have taken infinity as the reference state.) You can explain this because you need to supply energy to get the electron out of this configuration. With a positive potential energy, as in the case of earth and a mass at height h(ground being reference state), with potential energy mgh, you don't need to supply energy to the mass to get it to minimum potential.

What do you think would happen?? Remember the fact that you are supplying 2.6aJ to get the particles all out of the influence with each other....

Yep! E does tend to infinity. Your second answer is correct too. It would be more stable. You can see this from electronic configurations, as a neutral H2 molecule has fully filled 1s orbital

8. Jun 4, 2012

### johnconnor

What if the proton-proton separation is decreased to a very SMALL value? Where would the electron most likely be found? How will this affect the chemical properties of $$H_2^+$$?

I'm guessing the electron will lie very far away from the protons. But why? Does this mean the electron will carry a much greater energy? What inferences do you have on this situation? Thank you!

9. Jun 5, 2012

### Infinitum

If the protons are very close, they are going to try and repel with a high repulsive force. Why do you think is the configuration staying like that? (hint : if the electron is faaar away, then there will be nothing to let the protons even stay a little close to each other. They would just strongly repel)