Exclusion principal

1. Aug 5, 2008

learning_phys

why is it that the p subshell in a hydrogen atom can hold 6 electrons when the pauli exclusion principle states that no two identical fermions may occupy the same quantum state.

we have two electrons with opposite spins, how do 4 other electrons occupy the p-subshell?

2. Aug 5, 2008

will.c

Remember that we have several numbers determining the quantum state. In the p shell, there are three distinct configurations for the orbital angular momentum, and two spin states for each electron.

3. Aug 5, 2008

learning_phys

p shell correspond to l=1 (orbital angular momentum)

i think i figured it out, it's the three magnetic quantum number that allows the 2x3=6 states

does the subshell (orbital angular momentum quantum number) contribute to the energy?

if not, then the energy degeneracy (for n=2) in a full 2p and 2s subshell would 6+2=8 right? 6 from the p subshell and 2 from the s subshell?

Last edited: Aug 5, 2008
4. Aug 5, 2008

Redbelly98

Staff Emeritus
There are 3 orbital angular momentum states for p shells.

3 orbital states, with 2 spin states in each, gives 6 possible states.

5. Aug 5, 2008

learning_phys

lets let the principal quantum number = 2 (n=2)
this corresponds to an azimuthal quantum number of l=0 and l=1
l=0 is the s subshell, l=1 is the p subshell
for each subshell, there is a corresponding magnetic quantum number
for l=1, the corresponding magnetic quantum number is m=-1,0,1
i am assuming that this is what you mean by the 3 orbital angular momentum states (m=-1,0,1)

back to the energy degeneracy question, the energy quantization of an electron in a given shell (principal quantum number) is given by $$E_n=-\frac{Z^2 R_E}{n^2}$$ which means all the electrons in the n=2 shell will have the same energy right?

so in the n=2 shell, there is the s subshell and p subshell... which means there can be 8 electrons... so this means the degeneracy is 8 fold right?

6. Aug 5, 2008

Redbelly98

Staff Emeritus
Yes. Sounds like you've got it.

7. Aug 5, 2008

learning_phys

when an element (say a potassium atom or some other atom) is in its ground state, is it always true that the number of electrons is equal to the number of protons?

8. Aug 6, 2008

Redbelly98

Staff Emeritus
The electrons and proton numbers are equal for neutral atoms, and not equal for charged ions. The ground state is irrelevant.

9. Aug 6, 2008

learning_phys

http://grephysics.net/ans/8677/30

check out letter D.

"Because the problem states that potassium is in the ground state, the atomic number is the same as the number of electrons in the configuration. The sum of the superscripts is 19."

10. Aug 6, 2008

Redbelly98

Staff Emeritus
I think I see what they are saying. They seem to take "the ground state" of an atom to be its neutral (i.e. not ionized) ground state by definition. So yes the electron and proton numbers would be equal in that case.

However, one could also speak of the "ground state" of K+, a positively charged potassium ion. In this case, the numbers of electrons and protons are not equal. I believe one then speaks of a potassium ion, and not a potassium atom.

Moreover, it's not necessary for an atom to be in the ground state in order to be neutral. It could be in an excited state, and as long as it is not ionized there are still equal numbers of protons and electrons.