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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 259
E7F.6(b) �e energy levels are [7F.10–287], E l = ħ2 l(l + 1)/2I, where I is the moment
of inertia.�e minimum energy to start it rotating is the minimum excitation
energy, the energy to take it from the motionless l = 0 to the rotating l = 1
state, ∆E = E1−E0 = (ħ2/2I)(1(1+1)−0(0+1)] = ħ2/I. Evaluating this gives
∆E = (1.0546 × 10−34 J s)2/(3.07 × 10−47 kgm2) = 3.62 × 10−22 J
E7F.7(b) �e energy levels are [7F.10–287], E l = ħ2 l(l + 1)/2I, where I is the moment
of inertia. So that the excitation energy is ∆E = E2 − E1 = (ħ2/2I)[2(2 + 1) −
1(1 + 1)] = 2ħ2/I. Evaluating this gives
∆E = 2(1.0546 × 10−34 J s)2/(3.07 × 10−47 kgm2) = 7.25 × 10−22 J
E7F.8(b) �e energy levels are [7F.10–287], E l = ħ2 l(l + 1)/2I, where I is the moment
of inertia. �e corresponding angular momentum is ⟨l 2⟩1/2 = ħ
√
l(l + 1).
Hence, the minimum energy allowed is 0, through this corresponds to zero
angular momentum, and so rest and not motion. So the minimum energy of
rotation occurs for the state that has l = 1.�e angular momentum of that state
is ⟨l 2⟩1/21 = ħ
√
1(1 + 1 =
√
2ħ =
√
2 × (1.0546 × 10−34 J s) = 1.49 × 10−34 J s .
E7F.9(b) �e diagrams shown in Fig. 7.11 are drawn by forming a vector of length [l(l +
1)]1/2 and with a projection m l on the z-axis. For l = 6 the vector is of length√
42 and has possible projections from −6 to +6 in integer steps on the z-axis;
the vectors are labelled with the value of m l . Each vector may lie anywhere on
a cone described by rotating the vector about the z-axis.
E7F.10(b) Following the pattern shown in Fig. 7F.5 on page 287, the spherical harmonic
Y4,0 is expected to show four angular nodes. �e angle θ = 0 speci�es a point
on the z-axis; this does not describe an angular node or plane.
E7F.11(b) �e real part of the spherical harmonic Y2,+2 is − 12
√
3/π sin2 θ cos 2ϕ. Angular
nodes occur when cos 2ϕ = 0, i.e. at ϕ = π/4, 3π/4, 5π/4, 7π/4 .
�e imaginary part of the same spherical harmonic is − 12
√
3/π sin θ sin 2ϕ.
�is has angular nodes when sin 2ϕ = 0, i.e. at ϕ = 0, π/2, π, 3π/2
E7F.12(b) �e rotational energy depends only on the quantum number l [7F.10–287], but
there are distinct states for every allowed value ofm l , which can range from −l
to +l in integer steps. �ere are 2l + 1 such states, as there are l of these with
m l > 0, l of these with m l