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

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1 Chem 44 - Homework due ondy, pr. 8, 4, P..

2 . Put this in eq 8.4 terms: E m = m h /m e L for L=d The degenery in the ring system nd the inresed sping per level (4x bigger) mkes the sping between the HOO nd LUO bigger thn the regulr sping in the liner se. 4.

3 5. This nswer goes beyond the ourse, uses group theory. I think tht both u g nd u g re llowed in simpler piture fitting your experiene. The simpler piture works for hetero tom ditomis.

4 . Just think of it more simply, O is triplet so S =, if no hnge in spin, when seprte to O, must be triplets (S=) or singlets (S=) sine two triplets n ombine S= s +s, s +s -, s -s =,, nd one of these hs net S=; similrly triplet nd singlet n give totl S==+. 7. Here it is importnt to tke into ount tht the next expnsion term is (L log) ~ nd it must be smll ompred to L log~. For tht to hppen, need <. for % ury or <. for.% ury.

5 8. P9.) Clulte the trnsition dipole moment, mn z * m z n d, where µ z = er os for trnsition from the s level to the p z level in H. Show tht this trnsition is llowed. The integrtion is over r,, nd. Use r,, The trnsition is llowed if the trnsition dipole moment is non-zero: μ mn z ψ m τμ ψ τ z n r e r os for the p z wve funtion. We use the wve funtions for hydrogen s orbitl: ψ e π r nd the wve funtion for the p z orbitl given bove. The trnsition dipole moment is then: μ mn z ψ π - er Cos θ ψ dτ π oing the integrls one t time: π dφ ; π Cos θsin θdθ ; π Cos π r -r -r θsin θdθ dφ - er r e e dr 4 r -r -r -r -r -r -r r e 4 - r e e er r e e dr e e e dr dr μ mn z e -r -r 4 e r e π e 5 dr e 4 e 4 e r Exp r 4 π e dr 4 π e 4 e 4 4! 4 π π e 4

6 9. P9.9) The bsorbne of nulei id solution t nm is lled. The O (i.e., optil density) is the mount of nulei id in volume of. ml in.-m pth length ell for whih =.. How mny moles of nuleotide re ontined in. ml solution of double-strnded nuleotide for whih =.5, ssuming the extintion oeffiient per nuleotide is 7. m. Express this quntity in O s. ssume. m pth length. Solving λ I log I for the onentrtion,, gives: Therefore the onentrtion of nulei id in the solution is:.5 7 m m 4 mol.574 l Consequently, ml of nulei id solution ontins moles. This onentrtion orresponds to n λ =.5,. P9.) Beuse of intertions between trnsition dipoles of the onstituent nuleotides, the extintion oeffiient for single strnd polynuleotide is not simply the sum of the extintion oeffiients for the individul nuleotides. These dipole-dipole intertions depend on /r where r is the distne between bses, so for the purpose of lulting the extintion oeffiient for single-strnded polynuleotide, only nerest neighbor intertions need be onsidered. For hypothetil polynuleotide strnd GpCpUp...pG the extintion oeffiient is (GpCpUp...pG) = [(GpC) + (CpU) +(pg)] [(Cp) + (Up) +... (p)] Interting nuleotides pirs in single strnd re indited by XpY, where p represents the phosphte group tht joins the nuleotides X nd Y. In the eqution bove (pg), (pc), et., re extintion oeffiients for omponent dinuleotide phosphtes per mole of nuleotide. Hene they re ounted twie, whih ounts for the in the expression bove. To orret for this ft, the extintion oeffiients of the individul nuleotides, exept for the terminl nuleotides, re subtrted. The preeding eqution gives good greement with experimentl extintion oeffiients for nd single strnds. Consider the tbles of extintion oeffiients per nuleotide ( m ) t nm, 98K, nd ph 7 for nuleotides. p Cp Gp Up p pc pg pu Cp CpC CpG CpU Gp GpC GpG GpU Up UpC UpG UpU Clulte the extintion oeffiient for the single strnd pcpgpupuppgpu t 98K nd ph 7. b. Clulte for.5 4 solution of the in prt for pth length of. m.. epet the lultions in prt nd b for the single strnd GpCpUpUpp. ssume pth length of. m.

7 . pcpgpupuppgpu pc CpG GpU UpU Up pg GpU Cp Gp Up Up p Gp pcpgpupuppgpu m m 4 pcpgpupup pgpu 8.7 m b. s in P9.9, solving for the onentrtion,, gives: m 4-4 m.8. GpCpUpUpp GpC CpU UpU Up p Cp Up Up p GpCpUpUpp ( ) 7.4 m m.5.m =. (still out of rnge of spetrometer!!) m m ). P9.) For solutions omposed of more thn one bsorbing speies the bsorbnes t given wvelength re dditive. For two bsorbing speies nd the bsorbnes of the individul speies t re dditive: Therefore, the onentrtions of the two speies n be obtined by mking bsorbne mesurements t two different wvelengths nd, nd using known vlues of the four extintion oeffiients.,,,. erive expressions for the onentrtions nd in terms of the extintion oeffiients,,,, the pth length l, nd the bsorbnes nd. Y b. For tyrosine (Y) the extintion oeffiients t 4 nd8 nm re 4 Y 8, m nd 5 m. For tryptophn (W) the orresponding extintion oeffiients re W 4 9 m W nd 8 58 m. solution of tyrosine nd tryptophn hs bsorbnes of 4 =.5 nd 8 =.. Clulte the onentrtions of tyrosine nd tryptophn. ssume pth length of. m. From: follows: ;, where nd depit two different wve lengths, nd nd designte two different bsorbing speies. Solving for nd yields: nd

8 ombining these two equtions gives: b) With: tyrosine (Y) speies with: Y 4 m Y 8 m 5 tryptophn (W) speies with: W 4 m 9 W 8 m nd. 8, the onentrtion of tyrosine (Y) n be lulted s: m m nd: P9.5) s explined in Setion 9., FET n be used to determine distnes between fluoresent hromophores in mromolules, thus providing informtion on mromoleulr onformtion. The effiieny of energy trnsfer E t in FET experiment is given by Et r where r is the distne between the donor () nd eptor () hromophores. The exittion trnsfer n be determined from fluoresent life times Et where is the fluoresent life time of the donor in the bsene of the eptor nd + is the life time of the donor in the presene of the eptor.

9 Consider protein lbeled with donor nphthyl group nd n eptor dnsyl group. The fluoresene lifetime for the nphthyl group in the protein is ns. When dnsyl is dded to the protein the life time of the nphthyl group dereses to 8 ns. Clulte the distne r between the nphthyl nd dnsyl hromophores ssuming the Förster rdius = 4 Å. Using: r E t nd solving for r gives: r r r 4 ns 8ns r 4.9 r Extr Problems..

10 . 4.

11

12

13

14 9..

15 .. P9.8) n importnt biologil pplition of bsorption spetrosopy is the determintion of the onentrtions of solutions of nulei ids. The * eletroni trnsitions within the purine nd pyrimidine bses of nulei ids hve bsorption mxim ner wvelength of nm. ssume the extintion oeffiient of nulei id is. 4 m t nm. If the onentrtion of nulei id solution is 5. 4, lulte the bsorbne of this solution in. m ell t nm. I λ log I m m 5

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