CHEM 343: Problem Set #4 (Spectroscopy)

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1 CHEM 343: Problem Set #4 (Spectroscopy) 1) What is the energy, in ev, of UV radiation at 250 nm? What about Visible radiation at 550 nm? hc a) Use the expression E = = hv. Where c is the speed of light, h is Plank s constant, and lambda is in m λ if c is in m/s. 2) Information: Visible radiation is typically considered to be in the range of 400 nm to 700 nm, and UV radiation is considered to be anything below 400 nm, down to around 10 nm for what is called vacuum UV. 3) What is meant by quantization in spectroscopy? a) For example, where an electron is excited from the ground state to an excited state in one jump. 4) What are the four possible electronic excitations? a) Look at class notes for these. The four transitions are σ to σ*, π to π*, n to σ*, and n to π*. 5) What is the difference between a photometer and a spectrophotometer? a) A spectrophotometer is designed to acquire a complete absorption spectrum of the compound of interest. 6) What is the difference between a fluorometer and a spectrofluorometer? a) A spectrofluorometer is designed to acquire a complete emission spectrum of the compound of interest at a specific wavelength of excitation. 7) What are the components of a typical spectrophotometer? a) A source, typically one with UV and Visible wavelength output, a monochromator, the sample, then the detector, which is coupled to an electronic out system, i.e., op amps to amplify the signal output. 8) Why must the sample be irradiated with a monochromatic light with a conventional spectrophotometer or photometer? Is this true for a system based on the photodiode array? a) With a conventional spectrophotometer, such as one based on PMT detection, there is no method to discern between different wavelengths of light that impinge on the PMT. Therefore, it is important to discriminate those wavelengths before the sample. In this way, when the specific photons strike the PMT, we will know what small range of wavelengths were present at that time. Conversely, a system based on diode array detection does not require wavelength selection since all wavelength elements, with the consideration of bandwidth (2 nm between wavelength elements), impinge on the detector simultaneously, and that the electronics are fast enough to retrieve the data stored on the diode chip. In this way, a complete spectrum can be acquired in one pulse. Therefore, a monochromator is not required with a diode array detector. b) Figure 1: A diode array detector chip

2 Figure 2: A diode array system. Note that the violet light is most reflected see the order of light waves along the diode in the next figure. Figure 3: What the diode array "sees" following dispersion from the grating. 9) Name two radiation sources. a) Tungsten and mercury lamp 10) Name two wavelength selectors. a) Monochromator and prism 11) Describe a photomultiplier tube. a) See Figure 4

3 hν b) Figure 4: PMT cascade of electron amplification 12) Why can one find baffles in a monochromator? a) Baffles help prevent detection of unwanted light streams. 13) What is a grating? a) See class notes 14) How are the unwanted orders of photoradiation eliminated from the monochromatic light? a) With the use of absorption filters. Say the unwanted orders are less than 400 nm and the wanted wavelengths are greater than 400, then use a filter that will absorb anything less than 400 nm. 15) What first order wavelength will be realized if the incident angle to a grating is 5, the reflected angle is 50, and the groove density is 1400/mm? d is λ per blaze ( nm ) blaze n is the order (think interference) i is the angle of incident radiation d a) Use the equation λ = ( sin i + sin r) r is the angle of reflective radiation. If you could observe the n dispersed light at 50, then it would be 609 nm. 16) What is the effect of increasing the focal plane length on the resolution of the light incident upon the exit slit in a Czerny-Turner monochromator? a) If you increase the focal plane length of the light incident on the exit slit, you will increase the resolution of light at that point. 17) What is the effect of increasing the focal plane length on the light intensity incident upon the exit slit in a Czerny-Turner monochromator? a) The problem is that increasing the resolution concurrently comes with the negative effect that the intensity of light has now decreased which can create problems with detection. 18) If you decrease the entrance and exit slits in a Czerny-Turner monochromator by a factor of 2, what will be the effect on the resolution of a desired wavelength? 1 a) λ = w D. If you decrease w by 2, then you increase the resolution by 2. 19) What is the difference between a monochromator based on an Echellette grating and one based on an Echelle grating? The table below provides a nice summary of the differences between these gratings.

4 a) b) Figure 5: Echelle grating system for ICP 20) At what absorbance (or absorbance range) is the noise in a spectrophotometer or photometer at a minimum?

5 a) 21) What is an isosbestic point? a) The point at which two species in the same solution have the same extinction coefficient.

6 b) 22) How could you increase the resolution of a spectrum in a spectrophotometer or photometer? a) Decrease the slit widths. 23) What is the effect of temperature on an electronic transition in UV-vis spectroscopy and how could you use this phenomenon for analytical purposes? a) A decrease in temperature will decrease the population of electrons in the excited state. You could in fact decrease the temperature to assist in better defining the electronic absorption of a molecule without the influence of rotational and vibrational effects. Look at the two spectra below for a clearer description of what I am writing about. b) Figure 6: Room temperature spectrum for benz[a]anthracene in heptane

7 c) Figure 7: benz[a]anthracene in heptane at 15 K 24) In molecular fluorescence, why are the reasons for the energy losses that account for the radiative transitions to occur at lower energy? Hint: quantum yield. k f a) φ =, kf: fluorescence k f + ki + kec + kic + k pd + kd ki: intersystem crossing kec: external conversion kic: internal conversion kpd: predissociation kd: dissociation Each of these processes competes with the rate of fluorescence, kf. 25) How can fluorescence spectroscopy be so much more sensitive than absorption spectroscopy? a) Since almost no stray light can impinge on the PMT when the compound of interest (that fluoresces) is not present, the background is extremely clean and free of unwanted signal. This in turn increases the signal/noise ratio of the system and permits incredibly low sensitivity to target (fluorescent) compounds. 26) What is quenching in a calibration curve for a fluorescent system? a) F = K C Where F is the fluorescent intensity, K is a constant when the source power is constant, and C is the analyte concentration Note that both self-quenching and self-absorption can occur (and thus result in non-linearity) Self-quenching: collisions between excited species at high analyte concentrations Self-absorption: when there is an overlap of emission and excitation energy bands. Note that the equation indicates that an increase K will realize an increase in F. In fact, the use of Xe lamps, and quite importantly, LASERS, yields intense light sources that do provide this analytical advantage.

8 27) Describe how a fluorometer works. a) Figure 8: A fluorometer a) Figure 9: A spectrofluorometer 28) How can one realize phosphorescence? a) One has to realize an intersystem crossing. It is for this reason that electronic relaxation can take several seconds and occur at significantly longer wavelengths compared to the energy of excitation. An intersystem crossing is actually forbidden and as such quite unfavorable, but when

9 it occurs, you could use it for analytical purposes since it is so unique and simultaneously extremely sensitive since there is no shortage of signal. 29) How does an ICP work? What is the input signal for an ICP. Hint: think about emission from excited state. a) Please see Skoog for a description.

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