# Energy of 4s and 3d states

1. Oct 15, 2013

### Habeebe

1. The problem statement, all variables and given/known data
Explain why, as the atomic number increases, the 4s electronic states fill before the 3d states. The fact that they fill first means they are lower energy. You must explain why they are lower energy.

2. Relevant equations

3. The attempt at a solution
First, I'm aware that 4s is not always lower energy than 3d and they are only asking about the case where previous orbitals have been filled.

Everything I've dug up so far has been kinda vague, mentioning things like penetration and shielding. That doesn't sound like a good answer. More specifically, I've seen penetration described as the electron spending more time closer to the nucleus, however, the electron is really a standing wave, so that doesn't make sense to me.

We've only really discussed the Schrodinger Equation for the hydrogen atom, so that doesn't seem applicable here, but any reasonably simple explanation involving it would be very welcome. Thanks.

2. Oct 16, 2013

### Simon Bridge

What's special about having the lower states filled?

Why not?

Where did you get the idea that the electron is "really" a standing wave?
It is the wavefunction that forms a stationary state - not the electron. "Spending more time close to the nucleus" means more likely to be detected close to the nucleus than far away from it ...

Compare the radial distribution of the 4s and 3d states.
What is the average distance of the electron from the nucleus in these states?

Have you had a look at what shielding may be doing?

What happens to the spacing of atomic energy levels as you increase n?
What does angular momentum do to each energy level?

It is, sort-of, applicable.
The pattern of energy levels for atoms in general is similar to that for hydrogen.

Last edited: Oct 16, 2013
3. Oct 16, 2013

### Staff: Mentor

4. Oct 16, 2013

### Habeebe

I discussed it with my professor, and in conjunction with your replies it makes a lot more sense now. Thanks.