Hypothetical Gravitational Atomic Model

Brief Explanation: We’ll explore another question regarding the Electric and Gravitational interactions, just like in our previous post. This time, we’ll apply the Electric and Gravitational Forces to Niels Bohr’s atomic model. We’ll also have to deal with the quantization of the angular momentum of the orbits in Bohr’s model, called Bohr’s Condition.

— — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — —

Question: A student suggested an alternative no Bohr’s Model of the Hydrogen atom in which the proton and the electron have no electric charge and the interaction responsible for the atomic stability is the Gravitational Force.

  1. Determine the ratios of the first stationary orbit on Bohr’s Model and on the student’s gravitational model.
  2. Compare the ionization energies of both models.
  3. Comment on the viability of the gravitational model.

— — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — —

Comprehending: Lets picture both models involved in the question. First of all, Niels Bohr’s Hydrogen Model.

Modelo Bohr.png

As you can see in the picture, the force between the electron and the proton is electric in nature, meaning both particles have a charge. The force is also central, meaning it applies to the center of mass of both particles, creating a centripetal force in the case of the electron orbiting the proton. The distance between the center of mass of both particles is named a0, which is a ‘classical’ name given to what is called Bohr’s Radius.

Now, the gravitational model proposed by the student:

Modelo Grav.png

In this case, both particles are not charged, and their interaction is Gravitational. Nevertheless, the force is central again, creating a centripetal force towards the center of the orbit. The distance of the center of the orbit to its limit is now called r. We are interested in the ratio between this distance and Bohr’s Radius.

As we can see, the two models are very similar in their representation: both have central forces that create the orbit of the electron around the atom. Now lets compare the two forces using two expressions: Newton’s Law of Gravity and Coulomb’s Law.

Força Gravítica.pngLei Coulomb Abreviada.png

(where k = 1/4πε0)

Notice the similarity between the two. Although Gravitation depends on the masses of the objects and the Electric Force on its charges, there is a common factor: the dependence on inverse of the square of the distance. This will prove to be very useful when we go through the exercise.

Now remember what was said about both forces: they are centripetal. Both forces are the only ones that act in their model, becoming the Resulting Force: meaning we can apply Newton’s Second Law (F = m a). In this case it will become:

Lei Newton Centripeta.png

Given that the Electric and Gravitational Forces are centripetal, we’ll just have to equal this expression to the Laws stated above and solve for r.

The velocity is another unknown in this problem. How do we determine it? In order to calculate it, there is some specific knowledge you need to know regarding both hydrogen models. In these models, there is a law that was first derived by Louis de Broglie

Now, for the second question. The comparison of the ionization energies. The logic behind the answer is the following: to ionize the atom we’ll have to give energy to the electron. Exactly how much energy? The answer to that is simple: exactly the energy the electron has. The energy of the electron is divided in two parts: its kinetic and potential energy. The kinetic part is related to its translational motion around the proton. The potential energy part is related to the force that acts in each model. Given that both forces are conservative (meaning they are dependent only of the position of the particle) we have the following relation:

Força e Energia Potencial.png

In integral form, it becomes:

Forma Integral Energia Pot.png

After calculating the potential energy, we just need to sum it to the kinetic energy, obtaining the Mechanical Energy of the electron. That will be the ionization energy of the electron.

— — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — —

Applying:

Lets apply all we did before. First, we have to equal the forces we studied before to the centripetal force equation and solve for r. First, the Electric Force in Bohr’s Model.

Calculo raio bohr.png

(if you want, check this value with v = 2.2E6 m/s. Bohr radius is 5.29E-11 m)

Now, solve for r with the Gravitational Force:

Calculo raio grav.png

 

Now, how do we calculate the speeds of the electrons in both cases? This is where Bohr’s Condition comes in: according to Bohr’s Model, the angular momentum of the electron is quantized according to the following equation:

Quantização do Momento Angular.png

(I will not explain this quantization right now, but I am hoping to make a post about it)

In this equation, the L stands for the angular momentum, me the mass of the electron, v the speed of the electron, r the radius, n is an integer (hence the quantization), and h Planck’s Constant.

Solving for the speed:

Velocidade.png

Now lets apply this to the expressions of the radiuses we derived earlier:

In the Gravitational Model:

Raio Gravítico.png

In the Electric Model:

Raio Bohr.png
(If you want, check Bohr’s radius, just replace the constants in the above expression!)

Now, for the Ratio:

Rácio.png

Replacing the values:

Numeros Racio.png

This means the Gravitational Radius is enormously bigger than Bohr’s Radius! We’ll think about this result later.

Now, for the ionization energy, we apply the integral form of the Potential Energy equation. The thing in this case is that both forces vary with the inverse of the square of the distance, meaning their integration will be the same:

Calculo Energia Potencial.png

In the expression above, A represents the constants related to both Gravity and Electric Force. Replacing this A in both expressions:

Energia Potencial Gravitica.png

Energia Potencial Electrica.png

Now, for the Kinetic Energy. Since we know their speeds:

Energia Cinética.png

From the radius expression we can derive the velocity for both models:

Velocidade.png

Velocidade mod elec.png

Velocidade mod grav.png

And this expression goes into the kinetic energy equation. Because the total energy of the electron is the sum of the kinetic and potential energy:

 

Energia MEcanica Bohr.png

Energia Mecanica Gravitico.png

Again, to compare the value, let’s take the ratio between the two:

Racio Final.png

Now, don’t even try to compute the expression. The numerator is so small that most calculators just say the answer is 0. All we need to know is that Eg<<Ee.

— — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — — —

Conclusion:

This post was quite long, so let’s think briefly evaluate the results. The radius of the Gravitational Model is immensely superior to Bohr’s Model, but how much? Well, let’s compute it, knowing that Bohr’s Radius is 5.29E-11.

Number Crunching.png

This number is absolutely enormous! For comparison, the diameter of the Observable Universe is 8.8E26 m, meaning a single atom would be bigger than the entire Observable Universe! This explains why the ratio is so low, because the energy required to ionize this gravitational atom is very very small.

Once again, Mathematics explains us why this model is impossible!

Note: This post took quite a long time to do and it is very likely that I made some mistakes. If you see anything, please tell me!

Advertisements

One thought on “Hypothetical Gravitational Atomic Model

  1. Pingback: The return of Matiphy! – Matiphy

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s