A student hands in the following work on the lowest hydrogen state

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Atomic, Molecular & Optical Physics · One-electron atoms and the origins of atomic structure

Quantum Mechanics: From the Photon to the Hydrogen Atom, Ch The hydrogen atom

Problem

A student hands in the following work on the lowest hydrogen state, for which the series factor is w=1w=1 and l=0l=0, so that R(r)=A e−r/a0R(r) = A\,e^{-r/a_0} with AA real and positive.“Normalising: ∫0∞∣R∣2 dr=A2∫0∞e−2r/a0dr=A2a0/2=1\int_0^\infty |R|^2\,dr = A^2\int_0^\infty e^{-2r/a_0}dr = A^2a_0/2 = 1, so A=2/a0=6.15 nm−1/2A = \sqrt{2/a_0} = 6.15\ \mathrm{nm}^{-1/2}. Also, the series stops at jmax⁡=0j_{\max}=0, so there are no radial nodes; the number of nodes is what nn counts, so this state has n=0n=0 and E=−13.6 eV/02E=-13.6\ \mathrm{eV}/0^2, which is infinite. Presumably one just declares nn to start at 11 by convention.”Take a0=0.0529a_0 = 0.0529 nm. ((a)) Identify the two mistakes. ((b)) Normalise RR correctly and give AA in nm−3/2^{-3/2}. You may use ∫0∞xqe−xdx=q!\int_0^\infty x^qe^{-x}dx = q!. ((c)) Give the correct value of nn for this state, and its energy in eV.

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