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Mechanical Kinetic Energy vs Electrical Potential Energy

  • Thread starter jg370
  • Start date
  • #1
jg said:
The problem I am working on is as follows:

Consider an atom with a single electron in orbit of radius r about a nuclear charge +Ze.

As requested, I have dertermine that the electric potential energy of the nucleus at the electron position and the potential energy of the electron are respectively:

[tex]V(r) = \frac{1}{{4\pi\varepsilon_o}}\frac{Ze}{r}[/tex]


[tex]PE(elect) = \frac{Ze}{{4\pi\varepsilon_o r}}[/tex]

Next, I am asked to use the Coulomb law for interaction between the orbiting electron and the nucleus and write Newton's second law of motion for the orbiting electron.

For this part of the problem, I have come up with:

[tex]F(elect) = F(cent)[/tex], which implies,

[tex]-\frac{Ze}{{4\pi\varepsilon_o r}} = m\frac{v^2}{2}[/tex]

My difficulty arises as I am asked to show that:

[tex]KE = -\frac{1}{2}PE[/tex]

Could someone give me a hint how to proceed with this part of the problem?

Thank you kindly,


Answers and Replies

  • #2

Well you're very close to the answer. You've got a tiny mistake here (on the RHS):

[tex]-\frac{Ze}{{4\pi\varepsilon_o r}} = m\frac{v^2}{2}[/tex]

If you fix it and then write the general equation for KE, I think you'll see the answer.
  • #3
As jdavel said, check the expression on the RHS. This sort of 'coding' error (if you will) left me stumped on a problem which was just recently cleared up for me here. I better start getting in the habit of proof-reading my work. I suggest you do the same.
  • #4
By the way, the result in this problem is called the Virial Theorem.