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276 8 ATOMIC STRUCTURE AND SPECTRA
�e Rydberg constants for Li and He are in principle di�erent because the
reduced masses of the atoms are di�erent. However, this di�erence is too small
to have a signi�cant e�ect on the result of the present calculation, given the
precision of the data.�erefore, it is assumed that R̃Li/R̃He = 1 and hence
ILi = IHe(3/2)2 = (54.36 eV) × (3/2)2 = 122.3 eV
P8A.6 From [8A.21–314], the wavefunctions for the px and py orbitals can be written
asψpx = r f (r) sin θ cos ϕ andψpy = r f (r) sin θ sin ϕ.�en, using the identities
cos x = (eix + e−ix)/2 and sin x = (eix − e−ix)/2i, these wavefunctions can be
rewritten
ψpx = (2)−1r f (r) sin θ(eiϕ + e−iϕ) ψpy = (2i)−1r f (r) sin θ(eiϕ − e−iϕ)
�e form of the operator l̂z is given in [7F.5b–284], l̂z = (ħ/i)∂/∂ϕ. To deter-
mine whether or not ψpx is an eigenfunction of l̂z , the operator is allowed to
act on the function
ħ
i
∂
∂ϕ
1
2
r f (r) sin θ(eiϕ + e−iϕ) = ħ
i
1
2
r f (r) sin θ(ieiϕ − ie−iϕ)
It is evident that the e�ect of the operator is not to regenerate the original
function times a constant, so ψpx is not an eigenfunction of the operator. A
similar calculation shows that the same is true for ψpy .
However, the functions e±iϕ are eigenfunctions of l̂z , (ħ/i)∂/∂ϕ e±iϕ = ±ħe±iϕ .
Using the identities e±ix = cos x ± i sin x, suitable combinations of ψpx and ψpy
are found which are proportional to e±iϕ , and hence which are eigenfunctions
of l̂z .�ere are two such combinations
ψpx ± i × ψpy = r f (r) sin θ cos ϕ + i × r f (r) sin θ sin ϕ = r f (r) sin θ e±iϕ
P8A.8 �e expectation value of 1/r is given by ⟨1/r⟩ = ∫ ψ∗r−1ψ dτ.�ewavefunction
can bewritten as a product of a radial part and an angular partψ = R(r)Y(θ , ϕ).
As 1/r is a function of r only, the integral over the angles can be evaluated
separately, and because the Y(θ , ϕ) are normalized with respect to integra-
tion over the angles the integral simpli�es to ⟨1/r⟩ = ∫
∞
0 R(r)2(1/r)r2 dr =
∫
∞
0 R(r)2r dr.
(a) For a 1s orbital the radial function is R1,0(r) = 2(Z/a0)3/2e−Zr/a0 (Ta-
ble 8A.1) and so
⟨r−1⟩ = 4(Z/a0)3 ∫
∞
0
re−2Zr/a0dr = 4(Z/a)3[1!/(2Z/a0)2] = Z/a0
where the integral is evaluated using Integral E.3 with n = 1 and k =
2Z/a0.