What Happens to the Effective Potential in a Hydrogen Atom for l>0 and r<r(min)?

alfredbester
Messages
38
Reaction score
0
Basic SE, Hydrogen atom I'm given the basicc equation for the effective potential I've worked out the bohr radius (a)

Q. Describe what happens to the effective potential for orbital angular momentum l>0 and electron radius r < r(min)

r(min) = l(l+1)a

I'm thinking that the effective potential will become positive which would indicate that the electron will repel but am having a hard time seeing how positives and negatives can repel.

My answer (2nd year exam). Would probably be.

The effective potential will become positive which indicates a negative charge is repelled by a postive one. This is not a physically possible solution,.
 
Physics news on Phys.org
Hi alfred,

I think you would find it helpful to make a little sketch of the effective potential. One thing you will see is that the effective potential does not immediately become positive after the minimum is passed. In fact, you know that the sign of the potential doesn't determine whether the force is attractive or repulsive. Remember that you can always add a constant to the potential without changing the physics. Something does change, however, when you cross the minimum of the potential. Hint: how does the potential relate to the force?

Also, the effective potential includes the centrifugal barrier (the angular momentum piece) which can lead to an effective repulsion. What is the physical origin of this seeming repulsion?
 
Last edited:
Thanks, the distance derivative of the potential is the Force (F = - dU/dr) so the force is attractive since r<rmin has negative gradient.

The second part I think, as r is decreasing the velocity needs to increase to conserve angular momentum so the centrifrugal (outward) force (proportional to v-squared) increases.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top