Chem 344 - Homework 11 – due Monday, Apr. 28, 2014, 2 PM

1.

2.

3.

4.

The degeneracy in the ring system and the

increased spacing per level (4x bigger) makes the

spacing between the HOMO and LUMO bigger

than the regular spacing in the linear case.

Put this in eq 8.4a terms: Em = m2h

2/2meL

2 for L=6d

5.

This answer goes

beyond the course,

uses group theory.

I think that both

ug and ug

are allowed in a

simpler picture

fitting your

experience. The

simpler picture

works for hetero

atom diatomics.

6.

7.

Just think of it more simply, O2 is

triplet so S = 1, if no change in spin,

when separate to 2O, must be

triplets (S=1) or singlets (S=0) since

two triplets can combine S= s1+s1,

s1+s1-1, s1-s1= 2,1,0 – and one of

these has net S=1; similarly triplet

and singlet can give total S=1=0+1.

Here it is important to take into account that

the next expansion term is (L log10)2~A

2 and it

must be small compared to L log10~A. For

that to happen, need A<0.1 for 1% accuracy or

A<0.01 for 0.01% accuracy.

8. P19.6) Calculate the transition dipole moment, z

mn m

* z n d , where µz = –er cos

for a transition from the 1s level to the 2pz level in H. Show that this transition is allowed. The integration is

over r, , and . Use

210 r ,, 1

32

1

a0

3 2r

a0

er 2a0 cos for the 2pz wave function.

The transition is allowed if the transition dipole moment is non-zero:

0τψ μ τψμ nzm

mn

z

We use the wave functions for a hydrogen 1s orbital:

0ar2

3

0

100 ea

1

π

1ψ

and the wave function for the 2pz orbital given above. The transition dipole moment is then:

dreea

rrer -dφdθθSin θCos

a

1

π

1

a

1

π32

1

dτ ψ θCoser - ψμ

00 ar-

a 2r-

0

2π

0 0

2

π

0

22

3

0

23

0

100210

mn

z

Doing the integrals one at a time:

2dφ

2π

0

;

3

2dθθSin θCos

π

0

2 ;

0

ar-

a 2r-

4

00

ar-

a 2r-

0

4

0

ar-

a 2r-

0

2 dreera

edree

a

redree

a

rrer - 000000

4

0

5

0

0

5

00

00 00

4

00

ar-

a 2r-

4

0

a e 243

3224a

243

3224

a

e

2a

1

a

1

!4

a

edr

2a

1

a

1rExpr

a

edreer

a

e00

00

4

03

0

4

0

23

0

23

0

mn

z

a e 2 243

128a e

243

2484

a e 2433

32 32 24

a π 32

π 4a e

243

32242π

3

2

a

1

π

1

a

1

π 32

1μ

9. P19.9) The absorbance of a nucleic acid solution at 260 nm is called A260. The OD (i.e., optical density) is

the amount of nucleic acid in a volume of 1.00 mL in a 1.00-cm path length cell for which A260 =1.00. How

many moles of nucleotide are contained in a 1.00 mL solution of a double-stranded nucleotide for which

A260 = 2.50, assuming the extinction coefficient per nucleotide is 7.000 103 M

–1 cm

–1. Express this

quantity in OD’s. Assume a 1.00 cm path length.

Solving εcI

IlogA 0

λ

for the concentration, c, gives:

ε

Ac

Therefore the concentration of nucleic acid in the solution is:

cmcmM 1107

2.5c

113

l

mol4105714.3c

Consequently, 1 ml of nucleic acid solution contains 3.5714 10-7

moles. This concentration corresponds

to an Aλ = 2.5,

10. P19.10) Because of interactions between transition dipoles of the constituent nucleotides, the extinction

coefficient for a single strand polynucleotide is not simply the sum of the extinction coefficients for the

individual nucleotides. These dipole-dipole interactions depend on 1/r3 where r is the distance between

bases, so for the purpose of calculating the extinction coefficient for a single-stranded polynucleotide, only

nearest neighbor interactions need be considered. For a hypothetical polynucleotide strand GpCpUp ...ApG

the extinction coefficient is

(GpCpUp ...ApG) = 2 [(GpC) + (CpU) +(ApG)]– [(Cp) + (Up) + ... (Ap)]

Interacting nucleotides pairs in a single strand are indicated by XpY, where p represents the phosphate

group that joins the nucleotides X and Y. In the equation above (ApG), (ApC), etc., are extinction

coefficients for component dinucleotide phosphates per mole of nucleotide. Hence they are counted twice,

which accounts for the 2 in the expression above. To correct for this fact, the extinction coefficients of the

individual nucleotides, except for the terminal nucleotides, are subtracted. The preceding equation gives

good agreement with experimental extinction coefficients for DNA and RNA single strands.

Consider the tables of extinction coefficients per nucleotide (M–1

cm–1

) at 260 nm, 298K, and pH 7 for

RNA nucleotides.

Ap Cp Gp Up ApA ApC ApG ApU CpA CpC

15.34

103

7.60

103

12.16

103

8.70

103

13.65

103

10.67

103

12.79

103

11.42

103

10.67

103

7.52

103

CpG CpU GpA GpC GpG GpU UpA UpC UpG UpU

9.39

103

8.37

103

12.92

103

9.19

103

11.43

103

10.96

103

12.52

103

8.90

103

10.40

103

10.11

103

a. Calculate the extinction coefficient for the single strand RNA ApCpGpUpUpApGpU at 298K and pH 7.

b. Calculate A260 for a 1.50 10–4

M solution of the RNA in part a for a path length of 1.00 cm.

c. Repeat the calculations in part a and b for the single strand RNA GpCpUpUpApA. Assume a path length of 1.00

cm.

a.

GpApUpUpGpCp

GpUApGUpAUpUGpUCpGApC

2ApGpUApCpGpUpUpε

113

113

cmM1012.1615.3410.2110.2112.167.60

cmM1010.9612.7912.5210.1110.969.3910.672ApGpUApCpGpUpUpε

114 cmM10712.8ApGpUApCpGpUpUpε

b. As in P19.9, solving εcA260 for the concentration, c, gives:

13.0681.00cmM101.510712.8εcA -4114

260 cmM

c.

ApUpUpp

ApAUpAUpUCpUpC

C

G2AGpCpUpUpApε

113

113

113

104.67

10)34.1570.870.860.7(

1065.1352.1211.1037.819.92AGpCpUpUpApε

cmM

cmM

cmM

1.00cmM101.5104.67εcA -4113

260 cmM = 10.11 (still out of range of spectrometer!!)

11. P19.12) For solutions composed of more than one absorbing species the absorbances at a given

wavelength are additive. For two absorbing species M and N the absorbances of the individual species at

are additive:

A A

M A

N

McM

NcN

McM

NcN

Therefore, the concentrations of the two species can be obtained by making absorbance measurements at

two different wavelengths 1 and 2, and using known values of the four extinction coefficients .

1

M, 2

M, 1

N, 2

N

a. Derive expressions for the concentrations cM and cN in terms of the extinction coefficients

1

M, 2

M, 1

N, 2

N , the path length l, and the absorbances A1

and A2

.

b. For tyrosine (Y) the extinction coefficients at 240 and280 nm are 240

Y 11,300 M

1cm

1 and

280

Y 1500 M

1cm

1 . For tryptophan (W) the corresponding extinction coefficients are

240

W 1960 M

1cm

1 and 280

W 5380 M

1cm

1 . A solution of tyrosine and tryptophan has absorbances

of A240 = 0.350 and A280 = 0.226. Calculate the concentrations of tyrosine and tryptophan. Assume a path

length of 1.00 cm.

a)

From: M N M N M N

M N M NA A A c c c c

follows: N

N

1M

M

1

N

1

M

11 cεcεAAA ; N

N

2M

M

2

N

2

M

22 cεcεAAA ,

where 1 and 2 depict two different wave lengths, and N and M designate two different absorbing species. Solving

for cN and cM yields: MN

2

M

2

N

2

2N c

ε

ε

ε

Ac

and NM

1

N

1

M

1

1M c

ε

ε

ε

Ac

combining these two equations gives:

NM

1

N

1

M

1

1

N

2

M

2

N

2

2N c

ε

ε

ε

A

ε

ε

ε

Ac

NM

1

N

1

N

2

M

2

M

1

1

N

2

M

2

N

2

2N c

ε

ε

ε

ε

ε

A

ε

ε

ε

Ac

M

1

1

N

2

M

2

N

2

2NM

1

N

1

N

2

M

2N

ε

A

ε

ε

ε

Ac

ε

ε

ε

εc

M

1

1

N

2

M

2

N

2

2

M

1

N

1

N

2

M

2N

ε

A

ε

ε

ε

A1

ε

ε

ε

ε1c

M

1

N

1

N

2

M

2

M

1

1

N

2

M

2

N

2

2

N

ε

ε

ε

ε1

ε

A

ε

ε

ε

A1

c

b)

With: tyrosine (Y) → species N with: 11Y

240

N

1 cmM 11300εε

11Y

280

N

2 cm1500Mεε

tryptophan (W) → species M with: 11W

240

M

1 cm1960Mεε

11W

280

M

2 cmM3805εε

0.350AA 2401 and 0.226AA 2802 ,

the concentration of tyrosine (Y) can be calculated as:

111

M

1

N

1

N

2

M

2

M

1

1

N

2

M

2

N

2

2

N

1960

11300

1500

53801

1960

0.350

1500

5380

1500

0.226

1

1

ε

ε

ε

ε1

ε

A

ε

ε

ε

A1

c

cmMcm

678.19

10898.4c

4

N

M5

N 10489.2c

and: M5

NM

1

N

1

M

1

1M 10489.2

1960

11300

1960

0.350c

ε

ε

ε

Ac

M5

M 10507.3c

12. P19.15) As explained in Section 19.11, FRET can be used to determine distances between fluorescent

chromophores in macromolcules, thus providing information on macromolecular conformation. The

efficiency of energy transfer Et in a FRET experiment is given by

Et R0

6

R06 r6

where r is the distance

between the donor (D) and acceptor (A) chromophores. The excitation transfer can be determined from

fluorescent life times

Et 1DA

D

where D is the fluorescent life time of the donor D in the absence of the

acceptor A and D+A is the life time of the donor in the presence of the acceptor.

Consider a protein labeled with a donor naphthyl group and an acceptor dansyl group. The fluorescence

lifetime for the naphthyl group in the protein is 23 ns. When dansyl is added to the protein the life time of the

naphthyl group decreases to 18 ns. Calculate the distance r between the naphthyl and dansyl chromophores

assuming the Förster radius R0 = 34 Å.

Using:

66

0

6

01rR

RE

D

DAt

and solving for r gives:

D

DA

RrR

1

6

066

0

1

1

1

1

6

0

6

0

6

06

D

DA

D

DA

RRR

r

061

1

1Rr

D

DA

A341

23ns

18ns1

1r 6

A09.42r

Extra Problems

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12. P19.8) An important biological application of absorption spectroscopy is the determination of the

concentrations of solutions of nucleic acids. The * electronic transitions within the purine and

pyrimidine bases of nucleic acids have absorption maxima near a wavelength of 260 nm. Assume the

extinction coefficient of a nucleic acid is 1.00 104 M

–1cm

–1 at 260 nm. If the concentration of a nucleic

acid solution is 5.0010–4

M, calculate the absorbance of this solution in a 1.00 cm cell at 260 nm.

εcI

IlogA 0

λ

1.00cmM105cmM

1101A 4-4

260

5A260