Ⅰ Pre-laboratory assignment1. Introduction summaryIn this experiment, we use the method of titration to analyze hard water. Hardening ions are result of acidic rainwater, which reacts with slightly soluble carbonate salts of calcium and magnesium to dissolve it. Of the hardening ions, calcium and magnesium ions are dominant. Though most of the hardening ions come from minerals deposited in the ground, other sources such as factory waste may supply hardening ions.
Lifetime of Fluorescence 202***** 이** Background Luminescence is observed as a direct consequence of electrons being excited and relaxed to the ground state. After being excited to a higher energy level, electrons can undergo various pathways. For an electron in the S0 state, initial excitement using laser can make it go to the S1 state or higher. Electrons that are excited to higher vibrational state are quickly relaxed to the lowest vibrational level of S1, also known as internal conversion. Here, the electron can directly relax into S0 state, which results in fluorescence, or it may first undergo intersystem crossing to the triple state and then phosphoresce to S0 state. Such change of energy and spin state of electrons can be depicted using a Jablonski diagram. The lifetime of fluorescence is the average time that molecules spend in the excited state before returning to the ground state. Lifetime of a fluorophore is a very important feature in describing its characteristics. Lifetihe rate of collision with quenching agents, energy transfer and excited state reactions. It also informs us about the behavior of the fluorophore within its environment ADDIN EN.CITE Lakowicz19995[1]556Lakowicz, Joseph R.Principles of fluorescence spectroscopyxxiii, 698 p.2ndFluorescence spectroscopy.1999New YorkKluwer Academic/Plenum03*************9153420106761LSA QD96.F56.L34 1999SCI QD96.F56.L34 1999852 Hayden Library. QD96.F56.L34 1999Library Storage Annex QD96.F56.L34 1999[1]. However, the typical fluorescence lifetime is 10ns, and can be very difficult to measure. In this experiment, time-correlated single-photon counting (TCSPC) is used for the detection of photons emitted from fluorescence. Basically, laser pulse excites the molecule, which then emits photons to be detected by the photomultiplier tube (PMT) detector. However, conditions such as pulse repetition rate is adjusted so that only one photon at a time is detected. Then the photons are recorded in a histogram by the titation and emission. Such method of recording gives several advantages. One is that there is almost no loss of photons in the recording process. Provided that the timing electronics are fast enough, the photons that arrive to the detector are counted in the photon distribution. Therefore, TCSPC has essentially ideal counting efficiency ADDIN EN.CITE Becker20156[2]665Becker, WolfgangIntroduction to Multi-dimensional TCSPCSpringer Series in Chemical Physics1-632015Springer International Publishing0172-6218https://dx.doi.org/10.1007/978-3-319-14929-5_110.1007/978-3-319-14929-5_1[2]. One of many factors that can affect the fluorescence intensity is quenching. Quenching refers to any processes that reduces fluorescence intensity. Two main mechanisms of quenching are dynamic quenching, which is due to collision from diffusion, and static quenching due to complex formation. There are many mechanisms that explain how quenching arises, and it is thought that those mechanisms all work together. s involve electron exchange, photoinduced electron transfer and excited state reactions. Also, Förster resonance energy transfer may contribute to diminishing the fluorescence intensity. An interesting thing to consider would be the difference between quenching and resonance energy transfer. While both affect the fluorescence intensity, the main discrepancy between the two is the distance between the fluorophore and the quencher or acceptor molecule. Resonance energy transfer is a process in which an electron of the acceptor molecule is excited as the electron of the donor molecule, which in this case would be the fluorophore, is relaxed. Then the acceptor molecule is relaxed, during which it may fluoresce or dissipate the energy as heat. This process can occur even when the two molecules are far apart from each other. However, quenching, requires for the fluorophore and quencher to directly contact each other at the van der Waals radii. Due to this distinct distant requirements, the trent information ADDIN EN.CITE Lakowicz19995[1]556Lakowicz, Joseph R.Principles of fluorescence spectroscopyxxiii, 698 p.2ndFluorescence spectroscopy.1999New YorkKluwer Academic/Plenum03*************9153420106761LSA QD96.F56.L34 1999SCI QD96.F56.L34 1999852 Hayden Library. QD96.F56.L34 1999Library Storage Annex QD96.F56.L34 1999[1]. In this experiment, we aim to measure fluorescence lifetime of coumarin, riboflavin and flavin adenine dinucleotide (FAD). Coumarin was prepared with ethanol as solvent, and riboflavin and FAD were prepared as aqueous solutions. The pH for riboflavin solution was arranged as 1, 4, 7, 10 and 13, while the pH of FAD solution was 7. The solutions were excited using diode laser at 375 nm, and the intensity was measured using TCSPC. Results Sample solution Concentration Max emission wavelength (nm) Lifetime (ns) Lifetime from literature (ns) Coumarin 460 460 1.00 3.1 Riboflavin pH 1 540 0.699 0.6 0.05(a) 4 540 4.87 5.11 0.05(b) 7 540 4.64 5.06 0.08 10 540 4.07 5.72
In this experiment, atomic force microscopy (AFM) is used for the determination of force constant of a common elastomer polydimethylsiloxane (PDMS), and the relationship between the force constant and curing agent ratio. Common AFMs are consisted of mainly three parts: cantilever with a sharp tip, scanner and a feedback control system. Basically, interaction between the atoms on the surface of the sample and the tip of cantilever interact, which causes the cantilever to bend. The displacement is measured be optical lever technique. Whether the interaction is repulsive or attractive is dependent on the distance between the tip and the sample.
Exp. 3 Computational Chemistry and Quantum Tunneling Group number: Student ID: Name: Compounds E pyramid ( Hartree ) E planar ( Hartree ) E a (kcal/ mol ) NMe 3 -173.3979586864 -173.3880475076 6.219359923909 PMe 3 -458.6708448779 -458.5995986567 44.70768833887 The effect of the nature of the inverting atom on the activation energy of inversion The activation energy for PMe 3 is much larger than that for NMe 3 . In this comparison, the only difference between the two compounds is the size of the central atom. Phosphorus is much larger than nitrogen. This provides a bigger barrier for the lone pairs to tunnel through. Also, since the bond length between the central atom and carbon is longer, the distance the carbon needs to travel for inversion is longer. This may also be the cause for larger activation energy. Compounds E pyramid ( Hartree ) E planar ( Hartree ) E a (kcal/ mol ) PMe 3 -458.6708448779 -458.5995986567 44.70768833887 PEt 3 -575.8798142843 -575.8133623713 41.6992138795 The ent steric bulkiness on the activation energy of inversion In this case, the activation energy for PEt 3 is smaller than that for PMe 3 . Although the ethyl group is bulkier than the methyl group, it affects the pyramidal state more than it does the planar state. Therefore, the pyramidal state is destabilized in triethylphosphine , which is the main cause for the lowering of activation energy. Compounds Pyramidal E HOMO ( Hartree ) Planar E HOMO ( Hartree ) Δ E HOMO (kcal/ mol ) NMe 3 -0.1924 -0.1529 24.78663 PMe 3 -0.2081 -0.1028 66.07676 PEt 3 -0.2074 -0.1092 61.62144 The HOMO energy for pyramidal and planar amines and phosphines The HOMO of each molecule contains the lone pair electron. For the pyramidal case, the lone pair would occupy the sp 3 orbital, and in the planar case, it would occupy the p orbital. Energy change was greater in the case of phosphines, which can be explained using the same logic aforementioned. Also, similar to the activation energy, triethylphosphine showedOMO energy than in the case of trimethylphosphine . Compounds E pyramid ( Hartree ) E planar ( Hartree ) E a (kcal/ mol ) PMe 3 -458.6708448779 -458.5995986567 44.70768833887 PMe 2 F -518.2948506125 -518.1960949214 61.97014501295 The effect of the substituent electronegativity on the activation energy of inversion In this case, one of the methyl groups was changed into fluorine, which has a higher electronegativity than carbon. The electronegative fluorine atom withdraws electrons from phosphorus, and the lone pair on the phosphorus has a higher s character. Since the lone pair occupies the p orbital in the planar state, electronegative atoms tend to destabilize the planar state, and the activation energy is elevated. Compounds E pyramid ( Hartree ) E planar ( Hartree ) E a (kcal/ mol ) PMe 3 -458.6708448779 -458.5995986567 44.70768833887 PMe 2 Ph -649.2295029511 -649.1701302200 37.25695922027 PMe 2 ( SiH 3 ) -708.7767432065 -708.7318371311 28.17899377245 The effect of conjugation on tf inversion Conjugation may also affect the activation energy. Delocalization of the lone pair on the central atom can stabilize the transition state, leading to decreased activation energy. For dimethylphenylphosphine , the aryl group substituent forms a conjugated system that can delocalize the lone pair on the phosphorus atom. For the third case, silicon has an empty d orbital that can take up the lone pair. Therefore the transition state is stabilized and the activation energy is lowered. Infinite potential barrier Show that the energy levels are two-fold degenerate. In the figure below, two wells can be seen in blue, and the wave functions are formed in each well at the same energy and not divided into two (marked in red circle), which indicates two-fold degeneracy Show that the states are localized at two potential wells of length a. It is easily seen in the graph, since the wavefunctions exist only in the well, formed by the blue line, and not in between. Infinite potential barrial barrier Show that the energy levels are split, compared with the infinite barrier case. Show that the tunneling frequency gets higher in the higher vibration level. It can be seen in the graph that the energy levels are split through parts such as the one marked in blue circle. This is different from the infinite barrier case. If it was twofold degenerate, the two lines should not have been divided, but rather be shown as one line. Show that the wavefunctions do not vanish on the interval – b+a /2 ≤ x ≤ b-a/2. Upon observation of the graph, the black lines that indicate the wavefunctions are not flat and there are wavefunctions in the given interval such as in the case of A = 0.5, where the wavefunction in the indicated interval is marked in red circle. A=0.25 A=0.5 A=0.75 Reference 1.Montgomery, C.D., Factors Affecting Energy Barriers for Pyramidal Inversion in Amines and Phosphines: A Computational Chemistry Lab Exercise. Journal of Chemical Education, 2013. 90 (5): p. 661-664. {}
In this experiment, cyclic voltammetry, along with chronocoulometry was used to distinguish the electron transfer mechanism between outer-sphere and inner-sphere electrode reaction. With chronocoulometry, the electrochemical area of the electrode was determined first using [Fe(CN)6]3- whose diffusion coefficient was known. Using the obtained area, diffusion coefficients for the other two electrolytes were determined. Applying the diffusion coefficient values, rate constant for each redox reactions were calculated from the peak separation of cyclic voltammetry. From the fact that the rate constant for outer-sphere electrode reaction does not depend much on the electrode, comparing the rate constant for glassy carbon and platinum electrode showed that reaction [Fe(CN)6]3- is outer-sphere, while the mechanism for the other two is inner-sphere.