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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 245
integer n; it follows that k = (2n + 1)π/L. Negative values of n do not
generate distinct solutions: for example if n = −2, k = −3π/L and the
wavefunction is cos(−3πx/L). However, because cos(−θ) = cos(θ) this
wavefunction is cos(3πx/L), which is that for n = 1. All negative values
of n therefore simply duplicate solutions given by positive values of n. In
summary
ψn(x) = cos([2n+1]πx/L) En = h2(2n+1)2/8mL2 n = 0, 1 . . .
It follows that the three lowest energy wavefunctions areψ0 = cos(πx/L),
ψ1 = cos(3πx/L), and ψ3 = (cos 5πx/L); these are depicted in Fig. 7.6
−0.4 −0.2 0.2 0.4
−1.0
−0.5
0.5
1.0
x/L
ψn
ψ0
ψ1
ψ2
Figure 7.6
(c) Substituting ψ(x) = sin k′x into the Schrödinger equation gives
−(ħ2/2m)(d2/dx2)(sin k′x) = −(ħ2/2m)(−k′2 sin k′x) = E sin k′x
�e energy is therefore E = ħ2k′2/2m. �e boundary conditions are
sin(±k′L/2) = 0, which is satis�ed by k′L/2 = n′π for integer n′. Negative
values of n′ do not generate distinct wavefunctions and n′ = 0 is not
permitted as the wavefunction is zero everywhere. In summary
ψ′n′(x) = sin(2n′πx/L) En′ = h2(2n′)2/8mL2 n′ = 1, 2 . . .
�ewavefunctionswith the three lowest energies are thusψ′1 = sin(2πx/L),
ψ′2 = sin(4πx/L), and ψ′3 = sin(6πx/L); these are depicted in Fig. 7.7.
(d) Expressed as multiples of h2/8mL2 the cosine solutions have energies 1,
9, 25, . . . , and the sine solutions have energies 4, 16, 36, . . . .�ese are the
same set of energies as for the box between 0 and L, n2h2/8mL2.
(e) �e normalized cos wavefunctions are ψ = N cos[(2n + 1)πx/L], and
are normalized when ∫
L/2
−L/2 ψ∗ψ dx = 1. �e integral is evaluated by �rst
using the identity cos2 x + sin2 x = 1 and then Integral T.2
N2 ∫
L/2
−L/2
cos2[(2n + 1)πx/2L]dx = N2 ∫
L/2
L/2
1 − sin2[(2n + 1)πx/2L]dx
= N2 x − N2{x/2 − (L/2(2n + 1)π) sin[(2n + 1)πx/L]}∣−L/2L/2 = N2(L/2)