Electric properties of material I. and II.
Preparation for test writing
1.Dielectric
1a. Insulator material of low electric conductivity. It contains no mobile charged particle.
2.Give two definitions for electric field strength and effective electric field strength of a parallel plate capacitor.
2a.
Potential difference between two metal sheets of capacitor divided by the distance of the sheets. Effective field strength is proportional to the charge density on one of the capacitor plates.
3. How to characterize homogeneous and inhomogeneous electric field?
3a.By the arrangement of force field lines. When lines run parallel the field is homogeneous. When lines are bent the field is inhomogeneous, a problem occurs at the edge of capacitor.
4. Dipole moment vector and its SI unit of a multi charge system.
4a.
Vectors are in boldface letters: µ, the resultant dipole moment and ri the distance of one of the component charges. SI unit of µ is C/m.
5. What is the vector orientation of µ and E?
5a. Dipole moment vector points from negative to positive charges, while electric field strength vector points toward the negative charges.
6. Polarization.
6a. Polarization can be defined by equation
We add up all the dipole vectors by magnitude and orientation in a volume V and the two quantities are multiplied polarization density or simply polaziationwe get.
Other words: Polarization density, or simplypolarization is thevector fieldthat expresses the density of permanent or inducedelectric dipole momentsin adielectricmaterial
7. Polarizability
7a. At low field strength is linearly dependent on E,
unit α: C2m2J-1
with a proportionality coefficient, α, thepolarizabilitycharacteristic to the molecule, referred to as the deformability of electron cloud. Polarizability change with frequency of electromagnetic radiation is used in Raman spectroscopy.
8. Compare the dipole moments of dichlorobenzene derivatives.
8a. para-dichloro benzene, µ = 0, two vectors with the same magnitude both are localised in a straight line with the contrary direction.
ortho-dichloro benzene has grater moment than meta-dichloro benzene. (Show it by making a drawing.)
9. Compare equations , and
The first eq. refers to nonpolar molecules µ = 0 (Clausius-Mossotti). The second eq. is named after P. Debye. Can be applied polar molecules in a narrow temperature range. If T is high enough the randomizing influence of thermal agitation sets µ = 0.
TheClausius-Mossottirelationconnects therelative permittivityεrof adielectricto thepolarizabilityα of the atoms or molecules constituting the dielectric. The relative permittivity is a bulk (macroscopic) property and polarizability is a microscopic property of matter; hence the relation bridges the gap between a directly-observable macroscopic property with a microscopic molecular property.
10. Polarization types and their distinction
10a. The different polarizations can be distinguished by using up their frequency dependence. 1. Orientation polarization: resonancia achieved in radio frequency and microwave range of EM waves. Under an external electric field, the dipoles rotate to align with the electric field causing orientation polarization to occur. In the higher frequency range there is no resonance between molecules and the field observed.
2.Distortion polarization
Electron and distortion polarization are independent of dipole orientation generated by EM waves. Resonance frequencies (1012 – 1014 s-1).identical the vibrational frequencies that fall into infrared.
3.Electron polarization
The electron polarization is a property of electrons. They lightly follow the frequency of photons in the visible and ultraviolet range. (1014 – 1017 s-1).
Total polarization can be measured in the microwave region, electronic polarization exclusively can be observed in the vis or uv ranges.
11. Explain by using figure below
the relation of phase velocity in the upper and lower medium. Derive the refractive index of two media by using angles.
11a. The velocity ratio of monochromatic light can be phrased by the ratio of refractive indices
when the beam advances from optically less denser to optically denser media.
vincident is greater than vtransmitted because the incident light runs in the less denser medium.
According to the Schnell’s law
nr is referred to the first medium, and has a value greater than one, because .
12. What does the refractive index depend on?
12a.The refractive index depends on the wavelength of light beam and the temperature of the system probed. The wavelength dependence ofis the dispersion. In order to compare the results measured on different Abbe refractometers you need to use identical wavelength, e.g. sodium D-line, λ = .
13. How to give Debye equation in terms of refractive index?
13a. Maxwell law states for nonmagnetic materials:
refractive index squared equals the relative permittivity. Substituting this into Debye equation
we another function applicable to calculate polarizability and dipole moment.
The experimental determination of refractive index easier than that of relative permittivity That is the advantage of this function.
14. What type of interaction in solution is characterized? How does it characterize?
14a. The potential energy vs. distance function is a good tool for characterization of strength of different interactions in solution. The ion – dipole interaction is described by taking the sum of repulsive and attractive interactions.
Starting arrangement of charges and equation:
The pot. energy distance function for ion - dipole interaction:
Boltzmann distribution, Selection rules
Preparation for test writing
1.Give the distribution function for a multicomponent and a binary (two component) system.
1a.
It means that we have j number of components, sitting on εj energy level. Each of them has their own level for waiting a jump to another εj energy level.
The distribution function for a binary system.
2. How does the ratio of change when difference increases?
2a. The increase of difference in the negative exponent decreases the ratio, which means that only a smaller amount of particle is present on the upper level. At the same time, the level distance is also increases.
3. How manyout of every 20 000 molecules is found to be at the first vibration excited state in a time average?
Data:
RT = 298.15·8.314 = 2478.9 Jmol-1, for CO molecules levels Δε=25 kJ mol-1.
We have a system containing 20 000 molecules.
3a.
If we have a system containing 20 000 molecules, than
It means that one out of every 20 000 molecules is found to be at the first vibration excited state in a time average.
4. Calculate the number of microstates for the case eight particles having nine
possible energy levels. Use the table attached here.
Table. The possible arrangements of 8 atoms and 9 allowed energy levels:
Case / n0 / n1 / n2 / n3 / n4 / n5 / n6 / n7 / n8 / Wx
Data 4! = 24, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320.
4a. Let’s fill the boxes by 8 atoms in a way the sum of energies taken up by atoms is 8 units altogether.
Case / n0 / n1 / n2 / n3 / n4 / n5 / n6 / n7 / n8 / Wx / 5 / 1 / 0 / 1 / 1 / 0 / 0 / 0 / 0
The number of microstates,, and in our case
5. Born-Oppenheimer approximation
5a. The quantized energies of a molecule do not interact each other, because of great transition energy differences, they can handle separately. That is why the additivity can be written as
ΔE = ΔEelectronic+ ΔEvibration + ΔErotation
and
ΔEelectronic > ΔEvibration > ΔErotation The inequality tells us the direction of energy
decrease.
6 Transition integral.
6a. The value of transition integral determines a certain transition available or not.
If the transition dipole moment, the moment difference between final and initial state is zero we are unable to measure the energy change of a transition. We say the transition is forbidden.
7. Calculate the energy of 285 nm photon (c = 3 108 m/s, h = 6,63 10-34 Js). Using the Planck equation.
7a. Ephoton = h ν = (hc) / λ
8.Reproduce the energy level diagram of different electronic and vibration transitions.
8a. Superposition principle should be applied.
9.Vibration and rotation selection rule.
9a. Selection rule for vibration is:
where Q is the coordinate of the vibration. For a diatomic: Q = R (bond length). The vibrational transition requires a dipole moment change,
Selection rule for rotation is:
The physical requirement for observation of a pure rotational spectrum is an existing permanent dipole.
There is no dipole moment change when rotation axis coincides the direction of the bond.
File: SP0506
Beer Lambert law, Beer Lambert usage
Preparation for test writing
1. Flux of photons, intensity of photons, transmittance, absorbance
1a. The equation defines flux of photons which crosses an area q, during time t .
Flux can be transformed to intensity , when flux of photons is multiplied by . This intensity refers to a monochromatic source if ν is taken as constant.
The ratio of outgoing and incoming intensities is named as transmittance. This intensity function can be measured directly by spectrophotometer.
Absorbance is a logrithmic intensity ratio, which is concentration, c thickness, l and absorption coefficient, ε dependent.
ε is the molar absorbance if concentration is given in mol dm-3.
Its linear concentration dependence is a good tool to determine concentrations.
The absorbance is wavelength dependent.
2. Draw a concentration dependence of a spectral band. Note the measuring conditions of band series.
2a. One can use the wavelength range where the band intensity is meaningful. Thickness of cuvette should be constant for all the concentration. With increasing concentration the band intensities should be greater and greater.
Draw something like this.
3. What an experimental arrangement can be seen on figure below?
Find the relationship between I0 and I.
3a. The figure is about an experimental arrangement of a monochromatic light intensity decreasing from I0 incoming intensity to I outgoing intensity. The light beam crosses the sample placed in a cuvette. To observe this effect the sample should contain at least one optically active material which absorbs light at the wavelength of incoming beam.
4. Deduce the Beer Lambert law!
4a. The derivative is proportional to a material coefficient, ε the outgoing intensity, I and the concentration of light absorbing material c in the solution. It is given as:
Integrating this equation between limits of intensity, and limits of thickness
we get
After simplification
we name the logarithmic ratio as absorbance. The function form:
Absorbance is directly proportional to the concentration and thickness of solution. The molar absorbance is wavelength and solution matrix dependent.
5. Riboflavine in dilute acetate solution shows a maximum in absorption spectrum at 444 nm.
Calculate the wavenumber in cm-1 and frequency of light at this peak.
Calculate the energy of corresponding electronic transition for one molecule and one mole of riboflavin.
Data: h = 6.626 10-34 Js, c = 3.00 108 ms-1, NA = 6.022 1023, λ = 444 nm
5a. First transform data to proper units:
λ = 444 10-9 m = 444 10-7 cm
the wavenumber
the frequency
For one molecule
For one mole = 269.4552 kJ/mol.
6. Dissociation equilibrium constant, K is given by equilibrium concentrations at dT = 0and dp = 0. . Find K in terms of α, the dissociation degree.
6a. The relationship between the measured concentration of analyte (c) and equilibrium (e.g. [A-]) concentrations
[A-] = [H+] = c∙α and [HA] = (1-α)∙c . Doing the substitutions
7. The optical arrangement and main elements working principle of a double beam spectrophotometer.
7a.
Sources: deuterium lamp UV range, tungsten lamp VIS range
Monochromator: produces monochromatic radiation
Sector mirror: changes beam from reference to sample at a given wavelength
Sample and reference cuvettes:solvent absorbance is eliminated
Detector: (phototube) converts intensity to electric current or voltage
Ratio: is proportional to I/Io
8. Why we use reference system?
8a. Interaction between light and medium. The disturbing interactions should be subtracted by preparing proper reference solution.
Proper reference solution contains all the material in the same concentration as in the sample solution except the optically active material/materials to be studied.
9. Explain the cause the deviation from BL law.
9a. using high analyte concentration:
causing strong electrostatic interactions between molecules,
increases refractive index, consequently photons by-pass the detector, greater absorbance is detected.
if we have a system in chemical equilibrium, equilibrium may shift at high concentrations
10. Give an explanation for the effect of non-monochromatic radiation.
10a. Monochromators do not produce a single wavelength radiation. The “monochromatic” beam contains several wavelegths.
Calibration curve is constructed at band A shows a straight line, while that isconstructed at band B is bent.
Explanation: There is a flat part of absorption band in the vicinity of its maximum (band A), therefore the average wavelength has a small standard deviation. Molar absorbance is hardly shifted at the peak, while there is a great change in molar absorbance at the sides. A small shift in wavelength causes a great change in molar absorbance.
11. Chromophore, red shift
11a. Achromophoreis the part of amoleculeresponsible for its absorbance. For example in acetone CH3- CO -CH3 the carbonyl group. The chromophore is a region in the molecule where the energy difference between two differentmolecular orbitalsfalls within the range of the visible spectrum.
Conjugation of π bonds causes red shift of λmax
Original molecule without conjugation: CH2=CH-CH2-CH2-CH=CH2 λmax =185 nm. Molecule with conjugation: CH2=CH-CH2=CH-CH=CH2 λmax =217 nm
File: SP0708
Luminescence, PhotoelectronSp
Preparation for test writing
1.Singlet and triplet electronic spin states and transitions. Make an arrow diagram and explain it.
1a. Electrons in molecular orbitals are paired, according to Pauli exclusion principle. In ground state the two electrons are paired. In a singlet transition electron preserve original orientation. In a triplet transition one of the electrons flip in spin. Electrons are unpaired, extremely low probability geometry. From singlet transition it can be gained by intersystem crossing.
2. Time domain of fluorescence and phosphorescence.
2a. After irradiation stops the fluorescent intensity abruptly decreases to zero.
Lifetime of fluorescence: 10-5 – 10-9 s.
After irradiation stops
Lifetime of phosphorescence: 10-5 – 10 s.
After irradiation phosphorescent light is visible for a long time. Triplet states, resulting from a spin interconversion process in the excited state, have quite long lifetimes, sometimes to beyond 1 millisecond. That is why phosphorescent light is visible for a long time.
3. Compare in terms of wavelength and energy of absorbed and emitted fotons in the fluorescence process!
3a. The ground singlet state molecule, S0, goes into first excited singlet state, S1,
Fluorescence may appear as an approximate mirror image of the absorption at a lower energy, and hence longer wavelength than the absorption, the difference (in wavenumbers) between the two corresponding bands being known as the Stokes shift.
4. Detail the energy releasing process in phosphorescence!
4a. Phosphorescence may be defined as the emission occurring due to the radiative transition between two states of different spin multiplicity, for example between T1 and S0.
The intersystem crossing process takes place where two electronic states T1 and S1 have the same energy. Relax to ground state with a second flip in spin to satisfy the singlet ground state.
5. The photoionization process.
5a. M + hν = M+ + e-
From monochromatic source a photon ionizes an atom or molecule, depending its energy. In XPS the photon is absorbed by an atom in a molecule, leading to ionization and the emission of a core (inner shell) electron.
In UPS the photon interacts with valence levels of the molecule, leading to ionisation by removal of one of these valence electrons.
6. Apply the energy conservation law to the ionization process according to Koopman’s theorem! Which term of equation is measured?
6a. Photoelectron spectroscopy is based upon a single photon in/electron out process. The law of energy conservation says
The total energy of photon is the sum of the binding energy, BE and the kinetic energy of electron. By measuring the velocity of photon the kinetic energy can be determined. The photon’s energy is well known, therefore binding or ionization energy can be determined.
7. How does electron energy analyser work?
7a. . PES uses monochromatic sources of radiation. E.g. Mg Kα radiation:
hν = 1253.6 eV Sample is placed in a high vacuum chamber (P ~ 10−8 millibar) for excluding other ionization (oxygen, nitrogen or other components from air) processes than the studied one.
The preferred option for photoemission experiments is a concentric hemispherical analyser (CHA) which uses an electric field between two hemispherical surfaces to disperse the electrons according to their kinetic energy. The electric field is increased gradually and the greater the field the higher velocity electron get into the window before the detector.
The higher the velocity measured the lower the ionization energy of photoelectron.
8. Characterize XPS method!
8a. The intensity of the peaks is related to the concentration of the element within the sampled region. The most commonly employed x-ray sources are those giving rise to:
Mg Kα radiation: hν = 1253.6 eV / Al Kα radiation: hν = 1486.6 eVEach element will give rise to a characteristic set of peaks in the photoelectron spectrum at kinetic energies determined by the photon energy and the respective ionization energies. Core ionization energies are characteristic of the individual atom rather than the overall molecule.
The presence of peaks at particular energies therefore indicates the presence of a specific element in the sample under study.
9. Characterize UPS method!
9a. In UPS the photon interacts with valence levels of the molecule, leading to ionisation by removal of one of these valence electrons. After ionization process occurred the electrons remaining rearrange their distribution.
When ejected electron arises from bonding orbital, a vibration term,should be added to energy balance equation.
File: SP0911
Vibrational, Rotational, NMR
Preparation for test writing, (do not forget to explain your notation).
1.Describe the vibration frequency vs. force constant function given for case of a mechanical system, and that of a molecular bond!
1a.
The vibration frequency of both mechanical and molecular system is proportional to force constant, k. The vibration frequency of mechanical system is inversely proportional to the mass of the ball.
For molecular system m is replaced by μ, which is given as .Where m1 and m2 are the vibrating atoms.
2.The quantum mechanical energy term , its description: quantum jump, vibrational energy equation, energy level distances.
2a. The force law F = -kx is valid. The energy levels are equidistant. The change in vibrational quantum number, Δv is restricted to one. The system’s vibrational energy can be increased or decreased by unit amount of energy independently the positions of levels. The vibrational energy difference between two successive levels is given:
3. The anharmonic oscillator model, its descrition: quantum jump, vibrational energy equation, level distances in energy.
3a. To draw a figure is essential.
The energy distance function is an imperfect parabola. The function can be approximated a power series. The change in vibrational quantum number, Δv can take the values ±2, ±3, … The levels are not equidistant.