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.