Solving Gauss' Law Problem with J.J. Thomson Model of Hydrogen Atom

In summary, the conversation discusses an incorrect model of the hydrogen atom proposed by J.J. Thomson, which suggested a positive cloud of charge with a negative point charge at the center. The conversation then goes on to discuss using Gauss' Law to show that the electron is in equilibrium at the center and would experience a restoring force when displaced. It is also shown that the force constant K is equal to ke^2/R^3 and an expression for the frequency of simple harmonic oscillations is found. The conversation ends with a discussion on how to show that F = -Kr without first finding the electric field.
  • #1
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1
I'm having trouble with the following problem:

An early (incorrect) model of the hydrogen atom, suggested by J.J. Thomson, proposed that a positive cloud of charge +e was uniformly distributed throughout the volume of a sphere of radius R, with the electron an equal-magnitude negative point charge -e at the center. (a) Using Gauss' Law, show that the electron would be in equilibrium at the center and, if displaced from the center a distance r < R, would experience a restoring force of the form F = -Kr, where K is a constant. (b) show that K = ke^2/R^3. (c) Find an expression for the frequency f of simple harmonic oscillations that an electron of mass m would undergo if displaced a short distance (<R) from the center and released. (d) Calculate a numerical value for R that would result in a frequency of electron vibration of 2.47 x 10^15 Hz, the frequency of the light in the most intense line in the hydrogen spectrum.

The second half of Part a) is where I'm having most of my trouble.

a) Using Gauss' law I can show that the electric field at the surface of the sphere is 0. Therefore the electron is in equilibrium. I'm not sure how to show that F restoring = -Kr without first doing part b) (see below)

b) If I move the electron to position r. I can show that the E-field from the positive cloud of charge at that point is (ke^2/R^3)r so the K in part a) = ke^2/R^3

Any comments or suggestions would be greatly appreciated.

Thanks.
 
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  • #2
Originally posted by discoverer02
Using Gauss' law I can show that the electric field at the surface of the sphere is 0. Therefore the electron is in equilibrium.

That doesn't prove that the electron is in equilibrium, since the electron is not at the surface of the sphere: it is in the center.

Symmetry alone should tell you that the electron has to be in equilibrium at the center, but you can also formally derive that there is no force at the center using Gauss's law ... if you want to determine whether the equilibrium is stable, Gauss's law will tell you the sign of the field a small distance from the center, and thus the direction of the force. Of course, it will also tell you the exact force and the constant K as you note, but perhaps they are speaking qualitatively.

If I move the electron to position r. I can show that the E-field from the positive cloud of charge at that point is (ke^2/R^3)r

You mean the electric force, not the electric field. Your derivation of K is correct.
 
  • #3
You're right, since r = 0 at the center of the positively charged cloud the E-field of the cloud there is zero so the -e is in equilibrium there.

I'm now kind of confused by part b). At a distance r from the center the pos. cloud's E-field is (ke/R^3)r and not (ke^2/R^3), so I guess I'm not seeing how K becomes ke^2/R^3.
 
  • #4
F = -Kr
E = (ke/R3)r
F = (-e)E
 
  • #5
"
a) Using Gauss' law I can show that the electric field at the surface of the sphere is 0. " Was that a typo? Surely you meant to say "at the center of the sphere"?

Move the electron from the center of the sphere. The total force from the part of the sphere beyond the electron is 0 (the field at any point inside a hollow sphere is 0) so you can ignore all of the charge past r. What is the total charge inside r? Now think of that charge as concetrated at the center.
 
  • #6
Yes, I made a mistake. I should only be concerned with what's going on at the center of the sphere.

Is there some way to show that F = -Kr without first finding the E-field of the positively charged cloud at r and then finding F = -eE?
It sounds like the problem is looking for something like this.

Anyway, other than that, I've got the rest of the problem figured out.

Thanks to all for the help.
 
  • #7
Originally posted by discoverer02
Is there some way to show that F = -Kr without first finding the E-field of the positively charged cloud at r and then finding F = -eE?

You could do it by directly integrating differential force due to an infinitesimal charge over the volume of the sphere, by superposition. But any solution involving Gauss's law will solve for the field, from which you find the force, and it looks to me like they are asking for a Gauss law solution to (b).
 

1. What is Gauss' Law and what does it have to do with the J.J. Thomson model of the hydrogen atom?

Gauss' Law is a fundamental law in electromagnetism that relates the electric field to the distribution of electric charges. The J.J. Thomson model of the hydrogen atom is a simplified model of an atom that can be used to solve Gauss' Law problems.

2. How does the J.J. Thomson model of the hydrogen atom differ from the more commonly known Bohr model?

The J.J. Thomson model of the hydrogen atom is based on a positively charged sphere with embedded electrons, while the Bohr model has a nucleus with orbiting electrons in discrete energy levels. The J.J. Thomson model is simpler and easier to use for solving Gauss' Law problems.

3. Can the J.J. Thomson model of the hydrogen atom be used to solve Gauss' Law problems for other elements?

Yes, the J.J. Thomson model can be used to solve Gauss' Law problems for any element. However, it is most accurate for hydrogen, as it does not take into account the more complex structure of other atoms.

4. How does the J.J. Thomson model of the hydrogen atom account for the electron's charge?

The J.J. Thomson model assigns a negative charge to the electrons embedded in the positively charged sphere. This creates an overall neutral atom, as the positive and negative charges cancel each other out.

5. Are there any limitations to using the J.J. Thomson model for solving Gauss' Law problems?

Yes, the J.J. Thomson model is a simplified model and does not take into account the full complexity of atoms. It may not yield accurate results for more complex systems or for atoms with multiple electrons.

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