1 (237)

Prévia do material em texto

SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 229
E7C.5(b) Two wavefunctions ψ i and ψ j are orthogonal if ∫ ψ∗i ψ j dτ = 0, [7C.8–254].
In this case the integration is from ϕ = 0 to ϕ = 2π. Let ψ i = exp(iϕ), the
wavefunction with m l = +1, and ψ j = exp(−2iϕ), the wavefunction with m l =
−2. Note that the functions are complex, so ψ∗i = exp(−iϕ). �e integrand is
therefore
ψ∗i ψ j = exp(−iϕ) exp(−2iϕ) = exp(−3iϕ)
and the integral evaluates as
∫
2π
0
exp(−3iϕ)dϕ = (−1/3i) exp(−3iϕ)∣2π
0
= (−1/3i)[
=1
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
exp(−6πi)−
=1
³¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹µ
exp(i0)] = 0
�e identity exp(ix) = cos x + i sin x (�e chemist’s toolkit 16 in Topic 7C on
page 256) is used to evaluate exp(−6πi) = cos(−6π) + i sin(−6π) = 1 + 0 = 1.
�e integral is zero, so the functions are indeed orthogonal.
E7C.6(b) �e normalized wavefunction is ψ(x) = (2/L)1/2 sin(πx/L).�e operator for
position is x̂ = x, therefore the expectation value of the position of the electron
is [7C.11–256]
⟨x⟩ = ∫ ψ∗ x̂ψ dτ = (2/L)∫
L
0
x sin2 (πx/L)dx
�is integral is of the form of Integral T.11 with k = π/L and a = L
= 2
L
⎡⎢⎢⎢⎢⎣
L2
4
− L
4 × π/L
=sin 2π=0
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
sin(2πL
L
)− 1
8 × (π/L)2
⎧⎪⎪⎨⎪⎪⎩
=cos 2π=1
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
cos(2πL
L
)−1
⎫⎪⎪⎬⎪⎪⎭
⎤⎥⎥⎥⎥⎦
= 2
L
× L
2
4
= L/2
Because the probability density ∣ψ(x)∣2 is symmetric about x = L/2, the ex-
pected result is ⟨x⟩ = L/2.
E7C.7(b) �e normalized wavefunction is ψ(x) = (2/L)1/2 sin(πx/L).�e expectation
value of the momentum is ∫ ψ∗ p̂xψ dx, and the momentum operator is p̂x =
(ħ/i)d/dx, therefore
⟨px⟩ = (2/L)∫
L
0
sin(πx/L)p̂x sin(πx/L)dx
= (2ħ/iL)∫
L
0
sin(πx/L)(d/dx) sin(πx/L)dx
Using (d/dx) sin(πx/L) = (π/L) cos(πx/L) gives
⟨px⟩ = (2πħ/iL2)∫
L
0
sin(πx/L) cos(πx/L)dx