Solutions
253
Traveling along a closed path C, γ returns to its initial value γ i if C does not contain
the degeneracy point, while it goes from γ i to γ i ± 2π if C contains the degeneracy
point (see Fig. 5.5). In either case both |ϕ 1 and |ϕ 2 retain their sign.
Problems of Chap. 6
6.1 We calculate the cross section from Eq. (6.4):
σ bim =
πμ
8K B T
1/2
K =
π N A μ
8RT
1/2
K = 5.8 · 10
−22 m
2
= 0.058 Å
2
.
The reduced mass μ was computed from the molecular masses:
μ =
208.22 · 79.10
208.22 + 79.10
10
−3
N A
kg =
0.0573
N A
kg .
The hard sphere cross section can be evaluated from the molecular volumes. For
anthraquinone V mol = 208.22 · 10
24
/(N A ρ) = 264 Å
2 . For pyridine V mol = 79.10 ·
10
24
/(N A ρ) = 133 Å
2 . The respective radii are then R mol = (3V mol /4π)
1/3
= 4.0
Å and 3.2 Å, respectively. Then, σ = 160 Å
2 , about 3000 times the real one. This
means it takes an average of about 3000 collisions with pyridine molecules to quench
an excited anthraquinone molecule.
6.2 In gas phase the overall rate of collisions is given by Eq. (6.2). The number of
collisions a single X molecule undergoes per unit time is computed putting the number density C X = 1. In the formula we have: μ XY = 100 /(2 · 10
3 N A ) = 0.05/N A
kg and C Y = N A P/(RT ) = 2.45 · 10
25 . So the collision rate is 4.35 · 10
8 s
−1 and
the average time between two collisions is its inverse, i.e., 2.3 ns.
In solution the rate for hard sphere encounters is given by Eq. (6.10). The number density of Y is 1000 N A molec/m
3 , and the concentration of X is put equal
to 1. The diffusion coefficients, multiplied by Avogadro’s number, are N A D =
R gas T /(6πηR) = 3.31 · 10
14 m
2 /s. If R int = R X + R Y , the rate of encounters is
6.7 · 10
9 s
−1 and the average time between two encounters is 0.15 ns.
6.3 FRET requires the fluorescence spectrum of the donor, at λ ≥ λ 00 (S 1 ), to overlap
with the absorption spectrum of the acceptor, λ ≤ λ 00 (S 1 ). So λ 00 (S 1 ) must be shorter
for the donor than for the acceptor. Viable donor–acceptor pairs are: A-B, A-C, A-D,
A-E, B-C, B-D, B-E, C-D, C-E, E-D.
For triplet sensitization the same rule holds for λ 00 (T 1 ) and the viable pairs happen
to be the same.
For singlet fission we want 2(E T 1 − E S 0 ) ≤ E S 1 − E S 0 , which translates into
2λ 00 (S 1 ) ≤ λ 00 (T 1 ). This relationship is obeyed by C and D. Actually C and E are
very close to the limit. Triplet–triplet annihilation requires just the opposite, i.e.,
2(E T 1 − E S 0 ) ≥ E S 1 − E S 0 , so in principle A and B should do it, while low yields
may be obtained for C and E.
253
Traveling along a closed path C, γ returns to its initial value γ i if C does not contain
the degeneracy point, while it goes from γ i to γ i ± 2π if C contains the degeneracy
point (see Fig. 5.5). In either case both |ϕ 1 and |ϕ 2 retain their sign.
Problems of Chap. 6
6.1 We calculate the cross section from Eq. (6.4):
σ bim =
πμ
8K B T
1/2
K =
π N A μ
8RT
1/2
K = 5.8 · 10
−22 m
2
= 0.058 Å
2
.
The reduced mass μ was computed from the molecular masses:
μ =
208.22 · 79.10
208.22 + 79.10
10
−3
N A
kg =
0.0573
N A
kg .
The hard sphere cross section can be evaluated from the molecular volumes. For
anthraquinone V mol = 208.22 · 10
24
/(N A ρ) = 264 Å
2 . For pyridine V mol = 79.10 ·
10
24
/(N A ρ) = 133 Å
2 . The respective radii are then R mol = (3V mol /4π)
1/3
= 4.0
Å and 3.2 Å, respectively. Then, σ = 160 Å
2 , about 3000 times the real one. This
means it takes an average of about 3000 collisions with pyridine molecules to quench
an excited anthraquinone molecule.
6.2 In gas phase the overall rate of collisions is given by Eq. (6.2). The number of
collisions a single X molecule undergoes per unit time is computed putting the number density C X = 1. In the formula we have: μ XY = 100 /(2 · 10
3 N A ) = 0.05/N A
kg and C Y = N A P/(RT ) = 2.45 · 10
25 . So the collision rate is 4.35 · 10
8 s
−1 and
the average time between two collisions is its inverse, i.e., 2.3 ns.
In solution the rate for hard sphere encounters is given by Eq. (6.10). The number density of Y is 1000 N A molec/m
3 , and the concentration of X is put equal
to 1. The diffusion coefficients, multiplied by Avogadro’s number, are N A D =
R gas T /(6πηR) = 3.31 · 10
14 m
2 /s. If R int = R X + R Y , the rate of encounters is
6.7 · 10
9 s
−1 and the average time between two encounters is 0.15 ns.
6.3 FRET requires the fluorescence spectrum of the donor, at λ ≥ λ 00 (S 1 ), to overlap
with the absorption spectrum of the acceptor, λ ≤ λ 00 (S 1 ). So λ 00 (S 1 ) must be shorter
for the donor than for the acceptor. Viable donor–acceptor pairs are: A-B, A-C, A-D,
A-E, B-C, B-D, B-E, C-D, C-E, E-D.
For triplet sensitization the same rule holds for λ 00 (T 1 ) and the viable pairs happen
to be the same.
For singlet fission we want 2(E T 1 − E S 0 ) ≤ E S 1 − E S 0 , which translates into
2λ 00 (S 1 ) ≤ λ 00 (T 1 ). This relationship is obeyed by C and D. Actually C and E are
very close to the limit. Triplet–triplet annihilation requires just the opposite, i.e.,
2(E T 1 − E S 0 ) ≥ E S 1 − E S 0 , so in principle A and B should do it, while low yields
may be obtained for C and E.