1 (251)

Prévia do material em texto

SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 243
P7D.4 �e wavefunction with n = 2 is ψ2(x) = (2/L)1/2 sin(2πx/L), and this is used
in [7C.11–256], ⟨Ω⟩ = ∫ ψ∗Ω̂ψ dτ, to compute the expectation value for the
cases Ω̂ = p̂x = (ħ/i)d/dx and Ω̂ = p̂2x .�e e�ect of p̂x on the wavefunction is
p̂xψ2 = (ħ/i)d/dx [(2/L)1/2 sin(2πx/L)] = (ħ/i)(2/L)1/2(2π/L) cos(2πx/L)
and the e�ect of p̂2x is found by applying p̂x to this result
p̂2xψ2 = (ħ/i)d/dx [(ħ/i)(2/L)1/2(2π/L) cos(2πx/L)]
= ħ2(2/L)1/2(2π/L)2 sin(2πx/L)
Hence
⟨px⟩ = ∫
L
0
ψ∗2 p̂xψ2dx = (ħ/i)(2/L)(2π/L)∫
L
0
sin(2πx/L) cos(2πx/L)d x
�is integral is of the form of Integral T.7 with k = 2π/L, a = L
⟨px⟩ = (ħ/i)(2/L)(2π/L) × (L/4π) sin2(2πL/L) = 0
⟨p2x⟩ = ∫
L
0
ψ∗2 p̂
2
xψ2 dx = ħ2(2/L)(2π/L)2 ∫
L
0
sin2(2πx/L)dx
�is integral is of the form of Integral T.2 with k = 2π/L, a = L
⟨p2x⟩ = ħ2(2/L)(2π/L)2 × [L/2 − (L/8π) sin(4πL/L)] = 4π2ħ2/L2
P7D.6 �e wavefunction of the state with quantum number n, in the range 0 ≤ x ≤ L,
is ψn(x) = (2/L)1/2 sin(nπx/L).
(a) �e probability density is Pn(x) = ∣ψn(x)∣2 = (2/L) sin2(nπx/L) which
is symmetric about x = L/2, meaning that Pn(L/2+ x) = Pn(L/2− x) for
all x. Hence, the particle is equally likely to be at position x as at L−x, and
so the average positionmust be in themiddle of this range, i.e. ⟨x⟩ = L/2 .
(b) �e particle has kinetic energy (only) when it is in the box, but as it
is constrained to be within the box it must move back and forth at a
constant speed between the two in�nite walls. �erefore the probability
of the particle travelling with momentum +p (to the right) is equal to the
probability of it travelling with momentum −p (to the le�), and so the
average value of the momentum must be 0 .
(c)
⟨x2⟩ = ∫
L
0
ψ∗n x̂
2ψn dx = (2/L)∫
L
0
x2 sin2(nπx/L)dx
�e integral is of the form of Integral T.12 with a = L and k = nπ/L
⟨x2⟩ = (2/L)[L3/6 − {L3/4πn − L3/8n3π3} sin(2nπL/L)
− {L3/4n2π2} cos(2nπL/L)] = L2(1/3 − 1/2n2π2)