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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 275
Solutions to problems
P8A.2 Twowavefunctions, ψ1 andψ2 are orthogonal if the integral ∫ ψ∗1ψ2 dτ is equal
to zero. In spherical polar coordinates the range of integration is r = 0 to∞,
θ = 0 to π and ϕ = 0 to 2π; with a volume element is r2 sin θ dr dθ dϕ.
(i) Hydrogenic 1s and 2s orbitals take the form ψ1s = R1,0(r)Y0,0(θ , ϕ) and
ψ2s = R2,0(r)Y0,0(θ , ϕ). Because the angular function is the same for
the two orbitals, the orthogonality must arise from the radial parts, so
only these need be considered further.�e radial functions are R1,0(ρ) =
N1,0e−ρ/2, where ρ = 2Zr/a0 and R2,0(ρ′) = N2,0(2 − ρ′)e−ρ′/2 where
ρ′ = Zr/a0. �e latter is rewritten in terms of ρ by noting that ρ′ = ρ/2:
R2,0(ρ) = N2,0(2− ρ/2)e−ρ/4.�e relevant integral is conveniently taken
over ρ rather than r. Noting that r2 = ρ2(a0/2Z)2
∫
∞
0
ρ2(a0/2Z)2R1,0R2,0 dρ =
C
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
(a0/2Z)2N1,0N2,0
× ∫
∞
0
e−ρ/2 (2 − ρ/2)e−ρ/4 ρ2 dρ
= C ∫
∞
0
(2ρ2 − ρ3/2)e−3ρ/4 dρ
= 2C [ 2!
(3/4)3
] − 1
2C [ 3!
(3/4)4
]
= 2C × 128
27
− 1
2C ×
1536
81
= 0
�e integrals are evaluated using Integral E.3 with k = 3/4 and the rele-
vant value of n. �e integral is zero and therefore the 1s and 2s orbitals
are orthogonal .
(ii) It is explained in Section 8A.2(g) on page 313 that the px and py orbitals
are proportional to x and y respectively. �erefore, in Cartesian coordi-
nates, the integral of the product of these two functions is proportional
to ∫
∞
−∞ ∫
∞
−∞ ∫
∞
−∞ xy dx dy dz. �e integrand is an odd function of both
x and y, so when evaluated over a symmetrical range the result is zero.
�ese orbitals are therefore orthogonal .
Alternatively, consider the angular parts of the orbitals given in [8A.21–
314]: px ∝ sin θ cos ϕ and py ∝ sin θ sin ϕ. �e product of these two
contains the term (cos ϕ sin ϕ) which is equal to 12 sin 2ϕ. �e integral
of this function over the range ϕ = 0 to 2π is zero (the integral over two
complete sine waves).�erefore the functions are orthogonal.
P8A.4 Ionization of He+ and Li2+ both occur from the ground state, with n = 1. �e
ionization energy is the energy needed to take the electron from its original
state to the state with n = ∞, which has energy zero by de�nition. �erefore
the ionization energy is simply minus the energy of the ground state. Using
[8A.13–308] follows that I = hcZ2R̃N. �e ratio of the ionization energies of
Li+, Z = 3, and He+, Z = 2, is therefore ILi/IHe = (3/2)2(R̃Li/R̃He).