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258 7 QUANTUM THEORY
(i) With m l = 1, nodes in the real part occur at π/2, 3π/2 , and in the imag-
inary part at 0, π .�ere are 2 nodes in each of the parts.
(ii) With m l = 2, nodes in the real part occur at π/4, 3π/4, 5π/4, 7π/4 , and
in the imaginary part at 0, π/2, π, 3π/2 .�ere are 4 nodes in each of the
parts.
E7F.3(b) �e normalization condition is ∫ ψ∗m lψm l dτ = 1. In this case the integral is
over ϕ in the range 0 ≤ ϕ ≤ 2π, and the wavefunction is ψm l = N cosm lϕ,
hence
N2 ∫ ψ∗m lψm l dτ = N2 ∫
2π
0
cos2(m lϕ)dϕ
�e integral is evaluated by using the identity cos2 θ+sin2 θ = 1, hence cos2 θ =
1 − sin2 θ, and then using Intergral T.2 with a = 2π and k = m l
N2 ∫
2π
0
[1 − sin2(m lϕ)]dϕ = N2 ∫
2π
0
dϕ − N2 ∫
2π
0
sin2(m lϕ)dϕ
= 2πN2 − N2 [ 12 × 2π − (1/4m l) sin(2πm l)]
Because sin(2πm l) = 0 for integer m l it follows that the integral evaluates to
πN2 and hence the normalization factor is N = π−1/2 .
E7F.4(b) �e integral to evaluate is ∫
2π
0 cos(m lϕ) cos(m′
lϕ)dϕ. Using the identity given
in the question, the integral can be rewritten as
1
2 ∫
2π
0
cos[(m l +m′
l)ϕ] + cos[(m l −m
′
l)ϕ]dϕ
= 1
2
sin[(m l +m′
l)ϕ]
m l +m′
l
+
sin[(m l −m′
l)ϕ]
m l −m′
l
∣
2π
0
= 1
2
(
sin[2π(m l +m′
l)] − sin 0
m l +m′
l
+
sin[2π(m l −m′
l)] − sin 0
m l −m′
l
)
= 0
Hence these two wavefunctions are orthogonal.
E7F.5(b) �e energy levels on a particle on a ring are [7F.4–283], Em l = m2l ħ
2/2I where
I is the momentum of inertia of the system, I = mr2, see�e chemist’s toolkit
20 in Topic 7F on page 282. �is is equal to the classical average energy when
1
2 kT = m2l ħ
2/2I. Here kT = (1.3806×10−23 JK−1)×(298 K) = 4.11...×10−21J.
Solving this for ∣m l ∣ gives
∣m l ∣ = ( kTmr
2
ħ2
)
1/2
= [(1.6726 × 10−27 kg) × (100 × 10−12 m)2(4.11... × 10−21J)]1/2
1.0546 × 10−34 J s
= 2.49
As m l must be integral the closest level is ∣m l ∣ = 2.