# A lot of considerations about the Bohr's Hydrogen Atom

1. Mar 2, 2013

### jaumzaum

I'm studying the Bohr's hydrogen atom and my teacher gave us a challenge question. When I was working in the problem I've got a couple of other questions that I don't know the answer.

The initial problem was the following:
Today we know the electrons are not the only particles moving inside the atom. The nucleus is moving too, but with a very smaller radius. In the Bohr hydrogen atom we can say the nucleus describes a circumference with the same angular velocity of the electron, and they are always "diametrically" opposed. The sum of the angular momentum of the nucleus and the electron is a integer multiple of h/2 π. Calculate the percentage deviation of the new Rydberg constant in this case.

OK, let's begin
Let:
Vp = proton orbital velocity
Ve = electron orbital velocity
w = angular velocity of electron/proton
Mp = proton mass
Me = electron mass
q = charge of electron/proton
Rh = Rydberg constant
c = speed of light
ε = vacuum permittivity

We have:
Me w Re² + Mp w Rp² = n h/2π (I)
Me w² Re = Mp w² Rp = q²/4πε(Re+Rp)² (II)

From (II) Rp = Me Re/Mp
From (I) w Re² Me(Me+Mp)/Mp = n h/2π (squaring)
Re4 w² Me² (Me+Mp)²/Mp² = n²h²/4π² (substituting w)
Re4 Me² (Me+Mp)²/Mp² . q²/4πε(Re+Rp)² Me Re = n²h²/4π²
But (Re+Rp)² = Re² (Me+Mp)²/Mp²
Doing all the calculations,
Re = h² ε/π q² Me . n²
Rp = h² ε/π q² Mp . n² (that's because we don't wee the nucleus orbiting, its radius is 1836x smaller than the electron 's

Finding w:
w = Me Mp/(Me+Mp) . q4 π/2h³ε² . 1/n³

Now we have to calculate the energy of the electron.
But I don't think we can do Me Ve²/2 - q²/4πε (Re+Rp), because q²/4πε (Re+Rp) stand for the potential energy of the set electron+proton, and not only for the electron. The difference in q²/4πε (Re+Rp) will be responsible for a difference in electron AND proton kinetic energy. Is it right?

I did the following:
The difference of the electron kinetic energy is equal to the resultant work done on it. The work is negative for the electric force and positive for the photon.
So:

The variation in kinetic energy between levels n1 and n2 is:
ΔMe Ve²/2 = -Mp²/(Me+Mp)² . q4 Me/8ε²h² . (1/n1² - 1/n2²)

The work done by the electric force is
-2Mp ²/(Me+Mp)² . q4 Me/8ε²h² (1/n1² - 1/n2²)

So hc/ λ = q4 Me/8ε²h² (1/n1² - 1/n2²) . Mp²/(Me+Mp)²
And 1/λ = Rh (1/n1² - 1/n2²) Mp²/(Me+Mp)²

So the deviation is 1-Mp²/(Me+Mp)² = Me/(Me+Mp) = 1/1837 = 0.05443%.

Have I done it right?

Besides, the 1/ λ for the proton can be calculated too
1/λ' = 1/λ . (Me/Mp)
λ' = 1836 λ
This equation says in every atomic electron transition in the hydrogen atom, the nucleus emits waves in the range of 10-4 to 10-3 m. Is it right? Do this waves have any special name?

--------CORRECTION---------
Deviatio n = 1-Mp²/(Me+Mp)² = Me(Me + 2Mp)/(Me+Mp)² = 3673/1837² = 0.1088%

Last edited: Mar 2, 2013
2. Mar 2, 2013

### Staff: Mentor

This is wrong. Today we know that both do not "move" at all.

There is no work done, you cannot switch between a proton at rest and a moving proton like that.

Radiation is emitted by the whole atom, in a single photon with the modified energy. There is no "wave from the proton".

The problem is very similar to the classical Kepler problem and the transition from a 2-body system to a 1-body system.

3. Mar 2, 2013

### jaumzaum

I didn't understand this, can you explain it better?
Note I said Photon, not proton in:
"work" is positive for the photon
What I mean is that the hf the photon gives to the electron contributes positive for its kinetic energy and the work by the electric force contributes negative to it.

So the energy that would be emitted by the nucleus only add to the energy ithat would be emitted by the electron only, in a single photon? Why this happens?

4. Mar 3, 2013