# Bohr-Sommerfeld model question "Old" Quantum Theory

• RJLiberator
In summary: Taylor series. You can go beyond the first term for the first difference in ## r ##, and beyond the second term for the second difference in ## r ##, etc. ## \\ ## Similarly for ## E_n ##.Yes, that makes sense. Thank you for your help and confirming my answers. In summary, the problem is solved by using the given equations and the "Old" Quantum Theory to find the orbit radius, r_n, and the energy, E_n, in terms of the quantum number, n, and other relevant parameters. The satellite of mass 1000kg in a circular orbit of radius 7000km corresponds to a quantum number of n =
RJLiberator
Gold Member

## Homework Statement

In analogy to the Bohr theory of the hydrogen atom, develop a quantum theory of Earth satellites, obtaining expressions for the orbit radius, r_n, and the energy, E_n in terms of the quantum number n and other relevant parameters. Use the "Old" Quantum Theory. A satellite of mass 1000kg is in a circular orbit of radius 7000 km. To what value of n does this correspond? What is the satellite energy? Determine the differences in the radius, Δr_n, and in the energy ΔE_n, for successive orbits at this radius. (Hint: Differentiate with respect to n and set Δn = 1. Why is this legitimate since n is not a continuous variable?)

## Homework Equations

$$V_{Sat} = \sqrt{\frac{GM_e}{r}}$$

L=angular momentum of satelite:
$$L_{sat} = rm_{sat}v_{sat}$$

T = period
$$T = 2*\pi\sqrt{\frac{r^3}{GM_e}}$$

Bohr–Sommerfeld model,

$$\int_0^T p_rdq_r\ =nh$$
h = Planck's constant = 6.626*10^(-34) J*s

## The Attempt at a Solution

So, based on the above equations and m = 1000kg, r = 7000km, I can find V_s = 7548.6 m/s and the period, T = 5826.6 s.

If I have angular momentum correctly, then that is equal to 5.284*10^(13) kg*m^2/s .

I can easily set up the bounds of integration from 0 to the period, T.
But I've never used the Bohr-Sommerfeld model. I don't know what is meant by the wiki's description of "where p_r is the radial momentum canonically conjugate to the coordinate q which is the radial position and T is one full orbital period."

Once I am able to calculate the integral, it will be easy to solve for n, and thus should be easy to find the Energy.

I believe that we are allowed to treat n like a continuous variable due to how minute it will be compared to the other numbers. n will likely be so small, that it will act like a continuous variable, thus we can differentiate it.

Any help on how to get started with the Bohr-Sommerfeld model in relation to this problem? What is "dq" ?

A simpler form of the Bohr model might be helpful: The results are the same: ## \\ ## 1) ## L=mvr=n \frac{h}{2 \pi} ## (Angular momentum is an integer number times ## h/(2 \pi) ##). ## \\ ## 2) ## \frac{mv^2}{r}=\frac{GMm}{r^2} ## (centripetal force=gravitational force) ## \\ ## 3) ## E_n=\frac{1}{2}mv^2-\frac{GMm}{r} ## (Energy is the sum of the kinetic energy plus the potential energy). ## \\ ## These are 3 equations, and 3 unknowns (##v ##, ##r ##, and ## E_n ##). ## \\ ## You can solve for the unknowns and then compute the required quantities. ## \\ ## Note: The integral ## \int\limits_{0}^{T} p \, dq=mv(2 \pi r)=n h ##, which is the same as ## mvr=n \frac{h}{2 \pi} ##. And you also have ## v=2 \pi r/T ##, but I don't think you need to solve for ## T ##.

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RJLiberator
Hi Charles, thank you kindly for helping me.

I have some follow up questions.

1. Are you sure r is an unknown? The question states that radius = 7000km. I am finding "n", E_n, and v as my unknowns.

2. So, the bohr-sommerfeld statement (that integral) comes down to be simply nh = mv(2*pi*r)?

3. How do you type latex in the middle of the sentence on PhysicsForums like you did? :)

Thank you.

1) ## r ## will be a function of ## n ## and the other constants. They then give you ## r ## so that will tell you what ## n ## is. In these 3 equations, you might not need ## v ##, but you need to solve for ## r ## and ##E_n ## (I could have called it ## r_n ##, and the ## v ## I could have called ## v_n ##). ## \\ ## 2) The answer is yes. ## \\ ## 3) just type ## on both sides of the Latex expression.

RJLiberator
Ah, I see. That makes sense, so using the first equation:

$$mvr = n\frac{h}{2*\pi}$$
We get a function r of n, namely ##r(n)=\frac{n}{mv}\frac{h}{2*\pi}##.

Couldn't I use r = 7,000,000 m to find what v is in the second equation? ##v=\sqrt{\frac{GM}{r}}## and we have r, G (grav. constant), M (mass of earth). This nets us a v of 7548.6 m/s.

With that v and using an r value again of 7,000,000 m, we can find what n is in the first equation which comes out to be a huge 5.0106*10^(47).

RJLiberator said:
Ah, I see. That makes sense, so using the first equation:

$$mvr = n\frac{h}{2*\pi}$$
We get a function r of n, namely ##r(n)=\frac{n}{mv}\frac{h}{2*\pi}##.

Couldn't I use r = 7,000,000 m to find what v is in the second equation? ##v=\sqrt{\frac{GM}{r}}## and we have r, G (grav. constant), M (mass of earth). This nets us a v of 7548.6 m/s.

With that v and using an r value again of 7,000,000 m, we can find what n is in the first equation which comes out to be a huge 5.0106*10^(47).
The second equation gives you ## v=\sqrt{\frac{GM}{r}} ##, so for this case, yes, you can take that shortcut to the answer. ## \\ ## Meanwhile the huge ## n ## does not surprise me. I might try to compute it, and see if I concur.

RJLiberator
Yes, my thoughts when first reading this problem was that n would be very extreme and the changes from the change in n should result in very minute differences. This is what explains why we can differentiate r(n) without n being a continuous variable.

RJLiberator said:
Yes, my thoughts when first reading this problem was that n would be very extreme and the changes from the change in n should result in very minute differences. This is what explains why we can differentiate r(n) without n being a continuous variable.
The arithmetic checked out=I agree with your answers. ## \\ ## Meanwhile ## r_n ## can be written as a Taylor series: ## r(n)=r(n_o)+r'(n_o)(n-n_o)+\frac{1}{2}r''(n_o)(n-n_o)^2+...##. The derivatives get successively smaller. Your explanation is ok too, but I think justifying it with the Taylor series is more mathematical. (Even though ## n ## is integer, you can assume it to be a continuous variable in the formula for ## r(n) ##, and when you put in integers ## n ## and ## n_o ##, the Taylor Series would get you the exact answer for ## r(n) ## if you included all of the terms in the series. In this case you find ## \Delta r=r(n)-r(n_o) ## with just the first derivative term).

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RJLiberator
Charles, thank you for helping me with my problem and even cross-checking your solution with mine.

## 1. What is the Bohr-Sommerfeld model?

The Bohr-Sommerfeld model, also known as the "Old" Quantum Theory, was proposed by Niels Bohr and Arnold Sommerfeld in the early 20th century. It was an attempt to explain the behavior of electrons in atoms, based on the idea that electrons move in orbits around the nucleus. This model was eventually replaced by the more accurate quantum mechanical model.

## 2. How does the Bohr-Sommerfeld model differ from the quantum mechanical model?

The Bohr-Sommerfeld model is a simplified version of the quantum mechanical model. It assumes that electrons move in circular orbits around the nucleus, while the quantum mechanical model considers electrons as waves that exist in a cloud around the nucleus. The quantum mechanical model also takes into account the uncertainty principle, which is not included in the Bohr-Sommerfeld model.

## 3. What was the main contribution of the Bohr-Sommerfeld model?

The Bohr-Sommerfeld model was the first successful attempt to explain the atomic structure and the behavior of electrons in atoms. It introduced the idea of quantized energy levels, where electrons can only occupy certain energy levels and can transition between them by absorbing or emitting energy. This concept laid the foundation for the development of quantum mechanics.

## 4. What are the limitations of the Bohr-Sommerfeld model?

One of the main limitations of the Bohr-Sommerfeld model is that it only works for atoms with one electron, such as hydrogen. It also fails to explain the behavior of heavier atoms and molecules. Additionally, the model does not take into account the spin of electrons, which is a crucial factor in determining the properties of atoms.

## 5. Is the Bohr-Sommerfeld model still relevant today?

The Bohr-Sommerfeld model is no longer the most accurate description of atomic structure, as it has been replaced by the quantum mechanical model. However, it still holds historical significance and is taught in introductory physics courses to provide a foundation for understanding quantum mechanics. Furthermore, the principles and concepts introduced in the Bohr-Sommerfeld model are still relevant and serve as a basis for many modern theories and technologies.

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