A lot of considerations about the Bohr's Hydrogen Atom

[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:
Rp= proton orbital radius
Re = electron orbital radius
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%

 

[mfb]

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.

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

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.

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

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?

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.

 

[jaumzaum]

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

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.

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

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?

 

[mfb]

Note I said Photon, not proton in:

Oh sorry, I misread that.
Well, in that case: Work is not a useful concept here. Use energy.
The atom emits a photon and loses the corresponding energy.
A reduced energy of the atom corresponds to a higher kinetic energy of the electron, and a lower potential energy.

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?

Quantum mechanics. You have to consider the whole system for the process, as those energy states are a property of the whole system as well.

 

[Nickie]

Thank you for sharing your work and thought process on this problem. It is clear that you have a strong understanding of the Bohr model and the equations involved. Your approach seems sound and the calculations appear to be correct.

In regards to your question about the special name for the waves emitted by the nucleus during atomic electron transitions, these are known as X-rays. They were first discovered by Wilhelm Röntgen in 1895 and are often used in medical imaging and industrial applications due to their ability to penetrate materials. However, in the context of the Bohr model, these X-rays are a result of the high energy transitions of the electron between energy levels and the subsequent emission of photons.

Overall, your work appears to be correct and well thought out. Keep up the good work in your studies of the Bohr model and its applications.