Section 14.1: Radial Wave Functions for Hydrogenic Atoms
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In Chapter 13, we studied the Coulomb potential as applied to the idealized hydrogen atom. Here we extend this analysis to so-called hydrogenic atoms. Hydrogenic atoms are atoms with only one electron and are therefore highly ionized. We can easily extend the discussion of the radial energy eigenfunctions from Section 13.8 by considering the Coulomb potential, V = −Ze2/r, where Z represents the number of protons. We can therefore make the replacement e2 → Ze2, in the expressions for the Bohr radius and energy equations, which give:
a(Z) = ħ2/μeZe2 = a0/Z , (14.1)
En(Z) = -μe Z2e4/2n2ħ2 = Z2En , (14.2)
where a0 and En are the Bohr radius and the energy levels for the idealized Hydrogen (Coulomb) problem. Notice therefore that the size of hydrogenic atoms decreases linearly as Z increases, and the binding energy increases quadratically as Z increases.
The radial energy eigenfunctions described in Eq. (13.29) change as well, such that
RZ nl(r) = (2Z/na0)3/2[(n − l − 1)! / 2n[(n + l)!]3] e−Zr/nao [(Zr/na0)l + 1/r] vnl(Zr/na0). (14.3)
In the animation, these radial energy eigenfunctions are shown for Z ≤ 16 and n = 1, 2, 3, and 4 with the appropriate (allowed) l values. The quantum numbers are given in spectroscopic notation such that 4f corresponds to n = 4 and l = 3. For the radial energy eigenfunction, notice how the number of crossings is related to the quantum numbers n and l. You should see that the number of crossings is n − l − 1. Also note how the spatial extent of the radial energy eigenfunctions decrease as Z increases according to Eq. (14.1).
In addition to RZ nl(r), for the same range of Z values and the same quantum numbers, R2Z nl(r) and the probability density, PZ nl(r) = R2Z nl(r)r2, are also shown. You can change the start and end of the integral for R2Z nl(r) and R2Z nl(r)r2 as well as change the range plotted in the graph by changing values and clicking the button associated with the state in which you are interested. You should quickly convince yourself that
∫ R2Z nl(r)r2 dr = ∫ u2Z nl(r) dr = 1 , [integrals from 0 to infinity] (14.4)
and that indeed PZ nl(r) = R2Z nl(r)r2.