# Why the energies of orbitals in the same subshell decrease with increase of atomic number?

Why the energies of orbitals in the same subshell decrease with increase of atomic number? For eg., energy of 2s orbital of hydrogen atom is greater than energy of 2s orbital of lithium atom. What is the reason?

• Hydrogen doesn't have a '2s orbital' or a 2s subshell. It only has a single s orbital in the 1s subshell. – user60221 Apr 29 '18 at 15:26
• @KianStevens an excited state of hydrogen can have the electron promoted to the 2s orbital. – Tyberius Apr 29 '18 at 16:03
• @Tyberius Maybe so. I wasn't on about excited states however. – user60221 Apr 29 '18 at 16:13
• @KianStevens I guess my point is just that, whether experimentally or through the Schrodinger equation, you can measure the energy of other orbitals of hydrogen. In some sense, hydrogen is the only atom that has a 2s orbital. – Tyberius Apr 29 '18 at 16:35
• I think the point is that the atom "has" a 2s orbital, but in the ground state there's just no electron in that state. – Mark Wolfman Apr 30 '18 at 0:12

Long answer: The energy of the orbital is driven by the Hamiltonian operator $\hat{H} = \hat{T} + \hat{V}$ where $\hat{V}$ is the potential energy. For a simple 1-electron atom/ion, the potential energy is the electrostatic attraction between the positively charged nucleus $Q$ and the negative electron $q$: $V = \frac{1}{4\pi\epsilon_0} \frac{Qq}{r}$. Lithium has 3 protons in the nucleus versus the 1 proton for hydrogen, so $Q$ is three times higher and the potential energy of the 1s orbital in $\ce{Li^{2+}}$ is roughly three times more negative than the 1s orbital in $\ce{H}$ (remember $q$ is negative).