# DeBroglie wavelength of Proton in H atom

1. Feb 25, 2015

### Davephaelon

How would one calculate the de Broglie wavelength of the proton in the hydrogen atom as the electron and proton pirouette around their common center of mass, for the lowest orbital? I believe the center-of-mass of the system would actually be inside the proton due to the great differential in mass between the proton and electron - a ratio of about 1800 to 1. So it might not be an easy calculation.

2. Feb 25, 2015

### MisterX

The de Broglie wavelength corresponds to a particle in a "momentum state" which is a planewave like state. An atom which is at some particular energy level actually has multiple different wavelengths associated with it, So we don't always think of electrons in an atom of having a particular de Broglie wavelength, I think. There might still be situations where that makes sense.

A different length which is useful for describing hydrogen atoms is the Bohr radius, commonly denoted by $a_0$.

3. Feb 26, 2015

### Davephaelon

I just realized this morning the answer to my question. The de Broglie wavelength of the counter-orbiting proton (around the common center-of-mass of the system) in the H atom has to be virtually the same as that of the orbiting electron to a first order approximation. The reason is simple. Imagine two children on a seesaw, one twice as heavy as the other. The balance point between them will have a 2 to 1 ratio, so the heavy child will move at half the velocity as the light child, but being twice as heavy his momentum will be the same as the light child. Same situation in the simplified Bohr version of the Hydrogen atom - the momentum of the counter-orbiting proton has to be the same as the momentum (mv) for the electron. So the formula for the de Broglie wavelength λ = h/mv will produce the same result for both the proton and electron - as the product of mv will be the same for both.

However, something doesn't seem right with this assessment, as it would mean the de Broglie wavelength for the wobbling proton would be approximately 10,000 times the proton's diameter. I'm going to have to think this over some more.

Last edited: Feb 26, 2015
4. Feb 26, 2015

### Staff: Mentor

Did you see MisterX's response above? You are considering a system in which the proton is in an energy eigenstate, and therefore has no meaningful de Broglie wavelength. That's why you can't calculate anything that makes sense.

As a historical note, the notion of de Broglie wavelength that you're using was long ago discarded. It was a good hint and an interesting heuristic on the way to the development of modern ("modern"? we're talking more than 75 years old now) quantum theory, but it never was applicable to systems as complex as a hydrogen atom and is no part of the modern theory.

5. Feb 26, 2015

### Davephaelon

Oh yes I understood what MisterX was driving at; I knew that the basic Bohr model predated, by about a decade, the revolutionary advances by Heisenberg, Schrodinger, and a host of others, that forms the foundation of modern Quantum Mechanics theory. I was aware that the electron's position in its ground state, S-orbital is described by a spherically symmetric region of probability, so that a single orbit/energy is a gross simplification; though a useful heuristic model, as you point out, in early efforts to understand the micro-world.

But, isn't there an equivalence here? By that I mean doesn't the orbiting electron also exist in a eigenstate, where the median value of all its summed energy states corresponds, approximately, to the Bohr radius, or energy? So, wouldn't the median value of all the energy states of the proton also be, approximately, what I indicated above?

6. Feb 26, 2015

### Staff: Mentor

I don't understand - you were asking about calculating the de Broglie wavelength of the proton, and both MisterX and I are saying that is a dated and obsolete concept and that's why you're not getting sensible results out of your calculation. Other than also being dated and obsolete, what does the Bohr model have to do with the de Broglie wavelength of the proton?

The electron isn't orbiting of course... but yes, it is in an energy eigenstate. That means it has a single clearly defined energy, but that's not going to get you any closer to a de Broglie wavelength for the proton.