## Quantum numbers & orbitals

$H_{\psi }=E_{\psi }$

1-D: $E_{TOTAL}\psi (x)=E_{k}\psi (x)+V(x)\psi(x)=-\frac{h^{2}}{8\pi ^{2}m}\frac{d^{2}\psi(x)}{dx^{2}}+V(x)\psi(x)$

Alicia Yu 1A
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### Quantum numbers & orbitals

Professor Lavelle said that during lecture 2s and 2p are the same in a Hydrogen atom. Why is that?

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### Re: Quantum numbers & orbitals

The 2s subshell and the 2p subshell have the same principal quantum number, 2. When we solve the Schrodinger equation for one electron and one nucleus (hydrogen or any one-electron ion such as C5+) we get an equation for the energy depending only on the principal quantum number, n. We have seen this equation, $E=\frac{-hR_{H}}{n^{2}}$ (multiplied by the atomic number squared, Z2, which is 1 for hydrogen). When we try to solve the Schrodinger equation for multi-electron atoms, we can get approximate solutions, but the energy will depend on more than just n because there are other electrons repelling the electron we're considering, and they affect different subshells to different extents (shielding and penetration).

In short, other electrons in the atom/ion change the energies of the subshells relative to each other (which won't occur in hydrogen or one-electron ions).