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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 351
[dye]/(mol dm−3) 0.0010 0.0050 0.0100 0.0500
T/% 68 18 3.7 1.03 × 10−5
T 0.68 0.18 0.037 1.03 × 10−7
ε/(dm3mol−1 cm−1) 670 596 573 559
�e molar absorption coe�cient is clearly not independent of the concentra-
tion; this may indicate that the dye molecules are interacting with one another
at higher concentrations.
E11A.6(b) �e transmittance T is the ratio I/I0, hence the Beer–Lambert law [11A.8–421]
can be written T = I/I0 = 10−ε[J]L . It follows that logT = −ε[J]L and hence
ε = −(logT)/[J]L. With the given data, T = 0.29 and L = 0.500 cm, the molar
absorption coe�cient is calculated as
ε = −(log 0.29)/[(18.5×10−3moldm−3)×(0.500 cm)] = 58.1... dm3mol−1 cm−1
�e molar absorption coe�cient is therefore ε = 58 dm3mol−1 cm−1 . For a
path length of 0.250 cm the transmittance is
T = 10−(58.1... dm
3 mol−1 cm−1)×(0.250 cm)×(18.5×10−3 mol dm−3) = 0.538...
Hence T = 54% . Note the conversion of the concentration to mol dm−3.
E11A.7(b) �e ratio of the incident to the transmitted intensities of light a�er passing
through a sample of length L, molar concentration [J], and molar absorption
coe�cient ε is given by [11A.8–421], T = I/I0 = 10−ε[J]L . It follows that logT =
−ε[J]L, which rearranges to give L = −(logT)/ε[J].
�e missing piece of information (in fact misplaced to Exercise E11A.7(a)) is
that the concentration of the absorber is 10 mmol dm−3. �e light intensity is
reduced to half when T = 0.5, hence the thickness of the sample is calculated
as
L = −(log 0.5)/[(30 dm3mol−1 cm−1) × (0.010 mol dm−3)] = 1.0 cm
�e light intensity is reduced to one tenth when T = 0.1
L = −(log 0.1)/[(30 dm3mol−1 cm−1) × (0.010 mol dm−3)] = 3.3 cm
E11A.8(b) �e integrated absorption coe�cient is given by [11A.10–423],A = ∫band ε(ν̃)dν̃,
where the integration is over the band, and ν̃ = λ−1 is the wavenumber. �e
band extends from 167 nm, corresponding to ν̃ = (167×10−7 cm)−1 = 5.98...×
104 cm−1, peaking at 200 nm, ν̃ = (200 × 10−7 cm)−1 = 5.00 × 104 cm−1, and
ending at 250 nm, ν̃ = (250 × 10−7 cm)−1 = 4.00 × 104 cm−1.