Chapter 15

Question 1

15-1 Explain the difference between a fluorescence emission spectrum and a fluorescence excitation spectrum. Which more closely resembles an absorption spectrum?

Answer 1

In a fluorescence emission spectrum, the excitation wavelength is held constant and the emission intensity is measured as a function of the emission wavelength. In an excitation spectrum, the emission is measured at one wavelength while the excitation wavelengths are scanned. The excitation spectrum closely resembles an absorption spectrum since the emission intensity is usually proportional to the absorbance of the molecule.

Question 2

Define fluorescence F i t p b w a m i a e s, r t a g s o t s s m w t e o a p.

Answer 2

F i t p b w a m i a e s, r t a g s o t s s m w t e o a p. Fluorescence is the process by which a molecule in an excited state, relaxes to a ground state of the same spin multiplicity with the emission of a photon.

Question 3

Define phosphorescence P i t p b w a m i a e s, r t a g s o a d s m w t e o a p.

Answer 3

P i t p b w a m i a e s, r t a g s o a d s m w t e o a p. Phosporescence is the process by which a molecule in an excited state, relaxes to a ground state of a different spin multiplicity with the emission of a photon.

Question 4

Define resonance fluorescence. R f i o w a e s e r o t s f a u t c t e.

Answer 4

R f i o w a e s e r o t s f a u t c t e. Resonance fluorescence is observed when an excited species emits radiation of the same frequency at used to cause the excitation.

Question 5

Define singlet state A s s h n u e (S= 0), s m = 2+1 = 1.

Answer 5

A s s h n u e (S= 0), s m = 2+1 = 1. A singlet state has no unpaired electrons (S= 0), spin multiplicity = 2S+1 = 1.

Question 6

Define triplet state A t s h t u e, s m = 3

Answer 6

A t s h t u e, s m = 3 A triplet state has two unpaired electrons, spin multiplicity = 3

Question 7

Define vibrational relaxation V r i t p b w a m l v e w e a p.

Answer 7

V r i t p b w a m l v e w e a p. Vibrational relaxation is the process by which a molecule loses vibrational energy without emitting a photon.

Question 8

Define internal conversion I c i t i p i w a m c t a l e s w e r.

Answer 8

I c i t i p i w a m c t a l e s w e r. Internal conversion is the intermolecular process in which a molecule crosses to a lower electronic state with emitting radiation.

Question 9

Define external conversion E c i a r p i w a m l e e w t t e t t s o a s.

Answer 9

E c i a r p i w a m l e e w t t e t t s o a s. External conversion is a radiationless process in which a molecule loses electronic energy while transferring that energy to the solvent or another solute.

Question 10

Define intersystem crossing I c i t p i w a m i o s s c t a s s w n t s t e (e.g., s → t).

Answer 10

I c i t p i w a m i o s s c t a s s w n t s t e (e.g., s → t). Intersystem crossing is the process in which a molecule in one spin state changes to another spin state with nearly the same total energy (e.g., singlet → triplet).

Question 11

Define predissociation P o w a m c f a h e s t a u v l o a l e s i w t v e i g e t r t b.

Answer 11

P o w a m c f a h e s t a u v l o a l e s i w t v e i g e t r t b. Predissociation occurs when a molecule changes from a higher electronic state to an upper vibrational level of a lower electronic state in which the vibrational energy is great enough to rupture the bond.

Question 12

Define dissociation D o w r p a m d t a s w s v e f a b t b.

Answer 12

D o w r p a m d t a s w s v e f a b t b. Dissociation occurs when radiation promotes a molecule directly to a state with sufficient vibrational energy for a bond to break.

Question 13

Define quantum yield Q y i t r o p e t p a.

Answer 13

Q y i t r o p e t p a. Quantum yield is the ratio of photons emitted to photons absorbed.

Question 14

Define chemiluminescence. C i a p b w r i p a a r o a c r.

Answer 14

C i a p b w r i p a a r o a c r. Chemiluminescence is a process by which radiation is produced as a result of a chemical reaction.

Question 15

Why is spectrofluorometry potentially more sensitive than spectrophotometry? F s, t a s, F, i p t t s i P0 a t t s. I s, t a, A, i p t t r o P0 t P. I P0 o t t s t P0 p a c i i P o t s t P. T t r d n c. A a r, t s o f c b i b i P0 o t s, b t t o a d n c.

Answer 15

F s, t a s, F, i p t t s i P0 a t t s. I s, t a, A, i p t t r o P0 t P. I P0 o t t s t P0 p a c i i P o t s t P. T t r d n c. A a r, t s o f c b i b i P0 o t s, b t t o a d n c. For spectrofluorometry, the analytical signal, F, is proportional to the source intensity P0 and the transducer sensitivity. In spectrophotometry, the absorbance, A, is proportional to the ratio of P0 to P. Increasing P0 or the transducer sensitivity to P0 produces a corresponding increase in P or the sensitivity to P. Thus the ratio does not change. As a result, the sensitivity of fluorescence can be increased by increasing P0 or transducer sensitivity, but the that of absorbance does not change.

Question 16

Answer 16

(a) Fluorescein because of its greater structural rigidity due to the bridging –O– groups.
(b) o,o’-Dihdroxyazobenzene because the –N=N– group provides rigidity that is absent
in the –NH–NH– group.

Question 17

15-9. Quinine in a 1.664-g antimalarial tablet was dissolved in sufficient 0.10 M HCI to give 500 mL of solution. A 20.00-mL aliquot was then diluted to 100.0 mL with the acid. The fluorescence intensity for the diluted sample at 347.5 nm provided a reading of 245 on an arbitrary scale. A standard l00-ppm quinine solution registered 125 when measured under conditions identical to those for the diluted
sample.

Calculate the mass in milligrams of quinine in the tablet.

Answer 17

100 ppm quinine yields 125 arbitrary units, the sample yields 245 a.u. so it is 196 ppm. This is a 5x (20 mL diluted to 100 mL) diluted, so the 500 mL solution is 196 ppm x 5 or 980 ppm. Since 1 ppm is 1 mg/L, the sample is 980mg/L or 490 mg per 0.5 L.