N.8 Why the s states have the least energy

The probability of being found near the nucleus, i.e. the origin, is determined by the magnitude of the relevant hydrogen wave function $\vert\psi_{nlm}\vert^2$ near the origin. Now the power series expansion of $\psi_{nlm}$ in terms of the distance $r$ from the origin starts with power $r^l$, (D.8). For small enough $r$, a p, (i.e. $\psi_{n1m}$), state involving a factor $r$ will be much smaller than an s, ($\psi_{n0m}$), state without such a factor. Similarly a d, ($\psi_{n2m}$), state involving a factor $r^2$ will be much less still than a p state with just single factor $r$, etcetera. So states of higher angular momentum quantum number $l$ stay increasingly strongly out of the immediate vicinity of the nucleus. This reflects in increased energy since the nuclear attraction is much greater close the nucleus than elsewhere in the presence of shielding.