# Atomic energy levels Silicon

1. Jan 12, 2015

### Matt atkinson

1. The problem statement, all variables and given/known data
Silicon has the configuration [Mg] $3p^2$.
Explain why there are more L, S, J levels for the $3p4p$ configuration
than in the $3p^2$ configuration.

2. Relevant equations

3. The attempt at a solution
My thought is because in the $3p^2$ subshell you have less variations of $M_l$ and $M_s$ because both electrons cannot have the same quantum numbers, wheres with the $3p4p$ level, they could both have $M_l=+1$ and $M_s=+1$ because the $n$ the principle quantum number is different?

Last edited: Jan 12, 2015
2. Jan 12, 2015

### Staff: Mentor

You're on the right track. But there are more states than just the one you cited ($M_l=+1$ and $M_s=+1$).

3. Jan 13, 2015

### Matt atkinson

Yes, i understand that there would twice (?) as many states in the excited level because the electrons would be free to have any $m_l$ or $m_s$, but the states for the $3p^2$ level are restricted by the Pauli principle.

4. Jan 13, 2015

### Staff: Mentor

Not twice, because it is only the states where the two electrons have the same spin in the same orbital that have to be discarted. But I think you get the idea.

5. Jan 13, 2015

### Matt atkinson

Ah okay thankyou!