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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 225
E7B.7(b) A function is an acceptable wavefunction if it: (1) is not in�nite over a �nite
region; (2) is single-valued; (3) is continuous; (4) has a continuous �rst deriva-
tive. Note that for a particle on a ring, the condition to be single valued implies
the function must be periodic over 2π, meaning that the function ‘joins up’
with itself as it goes through 2π. Expressed mathematically this condition is
ψ(ϕ) = ψ(ϕ + 2π), where ϕ is the angle.
(i) cos ϕ is periodic over 2π and satis�es all the other conditions so it is an
acceptable wavefunction.
(ii) sin ϕ is periodic over 2π and satis�es all the other conditions so it is an
acceptable wavefunction.
(iii) cos(0.9ϕ) is not periodic over 2π (in fact it is periodic over (2π)/0.9).�e
function does not ‘join up’ a�er the angle goes through 2π and therefore
it is not an acceptable wavefunction.
E7B.8(b) �e normalized wavefunction isψ(x) = (2/L)1/2 sin(3πx/L), and so the prob-
ability density is P(x) = ∣ψ(x)∣2= (2/L) sin2 (3πx/L).�is ismaximizedwhen
sin2(3πx/L) = 1, and so when sin(3πx/L) = ±1. �ese values occur when
3πx/L = π/2, 3π/2, 5π/2 and hence x = L/6, L/2, 5L/6 .
Nodes occur when the wavefunction goes through zero: sin(3πx/L) = 0. �e
sine function is zero when 3πx/L = π, 2π, hence x = L/3, 2L/3 . �e wave-
function goes to zero at x = 0 and x = L, but these do not count as nodes as the
wavefunction does not pass through zero.
Solutions to problems
P7B.2 (a) �e task is to �nd N such that ψ = N cos(ϕ) satis�es the normalization
condition [7B.4c–248], ∫ ψ∗ψ dτ = 1. In this case the integration is over ϕ
and the range is 0 to 2π; the function is real so ψ∗ = ψ. Using the identity
sin2 x + cos2 x = 1 the integrand is expressed as cos2 ϕ = 1 − sin2 ϕ, and
then the integral is evaluated using Integral T.2
N2 ∫
2π
0
(1 − sin2 ϕ)dϕ = N2 [ϕ∣2π
0 − (2π/2) + (1/4) sin(4π)] = πN2
Setting this equal to 1 gives N = (π)−1/2 .
(b) �e task is to �ndN such thatψ = N sin(m lϕ) satis�es the normalization
condition [7B.4c–248], ∫ ψ∗ψ dτ = 1. In this case the integration is over
ϕ and the range is 0 to 2π; the function is real so ψ∗ = ψ. �e integral is
evaluated using Integral T.2
N2 ∫
2π
0
sin2(m lϕ)dϕ = N2[(2π/2) − (1/4m l)
=0
³¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹·¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹µ
sin(4m lπ)] = πN2
where sin nπ = 0 for integer n is used. Setting the result equal to 1 gives
N = (π)−1/2 .