Recent content by FloridaGators

Thank you for helping. Do you mind my asking what your search inquiry in google was to find that?

The Hamiltonian for a Hydrogen atom in Cartesian Coordinates (is this right?): \hat{H} = - \frac{\bar{h}^2}{2m_p}\nabla ^2_p - \frac{\bar{h}^2}{2m_e}\nabla ^2_e - \frac{e^2}{4\pi\epsilon _0r} In Spherical Coordinates do I just use: x=r sin θ cos φ, y = r sin θ sin φ, and z = r cos θ?

Ya i actually just read that, but I still don't understand. You have a state where electron 1: l = 0 s = +1/2 electron 2: l = 1 s = -1/2 Here L = 1, and S = 0 So doesn't this correspond to a state of 1P1.. why is this term not included?

Hund's rules would just tell me which state is the lowest in energy of all the possible ones right? But the ones I listed are not Pauli forbidden states right and they are all possible?

Term Symbol for Carbon Total Angular Momentum 1. Write the term symnbols for carbon atom He2s22p2 which has two equivalent p electrons taking into account the Pauli exclusion principle. 2... 3. Since the carbon has two equivalent p-electrons, we know the l1=1 and that l2=1 as well. Also the...

oh ya, duh, my mistake, how about the 2nd one. the 2nd one is right though? 5/2 and 3/2?

ah! so for (1) does that mean J can equal 0 or 1 and for (2) J = 5/2 or 3/2? is that all the values for J? Thanks for the help by the way!

ahh! then for (1) does that mean S = 1, and since it is a P orbital it is L = 1 Hence J = 2. Is this all the values that J can attain?"

1. List all the possible values of total angular momentum for the following 1. 3P 2. 2D Homework Equations j = s + lThe Attempt at a Solution is this problem as simple as, for 1. for 3p, n = 3, l = 0,1,2 since it's a p orbital, l = 1 thus j = l + s, all possible values are 3/2, and 1/2 but...

[b]1. http://img10.imageshack.us/img10/8602/232cw.jpg /b][/URL] Homework Equations http://img199.imageshack.us/img199/484/232ag.jpg sorry about the insert pictures, i don't know how to type up the eqns easily The Attempt at a Solution Now I know that outside the box the Schrödinger equation...