Midterm/Exam one (E1-E5) Flashcards

Question

# E2 Quantum yield

Answer 1

the fraction of excited state molecules that return to the ground state via fluorescence

Answer 2

increasing T decreases quantum yield because more molecules remain excited or lose energy via non radiative processes

Answer 3

a molecule in the ground state of an excited level transitions to a higher level of a lower energy excited state while *conserving* electron spin | opposite of intersystem crossing

Answer 4

The transfer of energy of excited molecules to the environment via collisions

Answer 5

A molecule will take the shortest pathway from the excited state to the ground state and these may be quicker

Answer 6

If the detector were to be placed head on from the source light it would measure the radiation of the source light and the sample resulting in a net positive error.

Answer 7

An increase in volume of the sample increases the sensitivity, if the cuvette is placed horizontally instead of vertically the volume is increased

Answer 8

The change in energy from the ground state to the excited state is larger than that for the excited state to the ground state because of energy lost by vibrational relaxations

Answer 9

non radiative relaxation mechanism where electrons in the ground level of an excited state transtion to a higher level of a lower energy excited state *without* conserving electron spin | opposite of internal conversion

Answer 10

often done by halides, reduces the intensity of fluorescence signal

Answer 11

False, quenching abilites for halides increase as size increases

Answer 12

Basicity | pH > 7

Answer 13

They can hinder quick proton transfer mechanisms that occur faster than fluorescence

Answer 14

The Perkin-Elmer instrument has a Xe bulb as opposed to an LED source that can excite molecules at a number of different wavlengths as oppsed to the one emission of the LED bulb

Answer 15

Fo/F = 1 + Ksv[halide] | y= mx + b

Answer 16

F = kC - fluorescence is proportional to concentration | C= concentration k = product of constants

Answer 17

- more intense than raman scattering - emitted photons have the same energy as the absorbed photons - elastic

Answer 18

- less intense signals than rayleigh scattering - emitted photons have a different wavelength than absorbed photons - inelastic

Answer 19

- a kind of raman scattering - emitted photons have less energy than the absorbed photons (loss of E) - favored over anti stokes

Answer 20

- kind of raman scattering - emitted photons have more energy than absorbed photons (gain in E) - less favored than stokes scattering

Answer 21

to be raman active there must be a change in the polarizability of the molecule | ex: induced dipole moment

Answer 22

to be IR active the dipole moment of the molecule must change (change in charge distribution)

Answer 23

Raman signals are very weak compared to Rayleigh signals, a sufficent noise to signal ratio must be seen and that is created by a high powered light source (eg laser)

Answer 24

= (1/ x nm - 1/y nm) 1*10^7 | x = excitation wavelength and y = wavelength of peak

Answer 25

The wavelength was too close to the emission wavelength causing the solution to fluoresce and wash out any other signal | this is why the wavelength cannot be too small/ too high energy

Answer 26

IR spectra often have an overlap of signals, Raman peaks are less cluttered and feature less overlap of signals

Answer 27

aqueous samples can be used with Raman, for IR samples must be dried extensivley before analysis

Answer 28

- sample is placed on a textured metal or in a collidal solution and plasmons produced by the laser enhance the Raman signal - can be used for highly diluted samples because of signal enhancement

Answer 29

- A flame source has better repetition of sample delivery (more precise) but effiency of atomization is low because of the temperature - An electrothermal source has high effiency of atomization due to temperatures but poor repetition of sample delivery (less precise)

Answer 30

if droplets of sample are too large they cannot be carried by the combustion gases to be atomized and concentration of solution is lower than the initial sample

Answer 31

1. the solvent is removed and the sample is vaporized by a flame or electrothermal source 2. the vaporized sample is exposed to increased temperatures to create free atoms and ions

Answer 32

Nstar/N = (gstar/g)e^(-delta E/kT)

Answer 33

We want to maximize N star because that means there are more atoms in the excited state

Answer 34

False, a small delta E and large T are favored - increases sensitivity

Answer 35

they have lower surface tensions and allow for smaller droplets to form, improves atomization efficency esp in flame

Answer 36

the bandwidth is much smaller, allows for increased sensitivity and measurement - also the hollow cathode lamp emits at the exact wavelength of the metal of interest

Answer 37

an electric current creates anions and cations within a gas, cations collide with the negative cathode that then emits metal atoms that emit light that serve as the source light for the instrument | M --- M star + hv

Answer 38

A releasing agent forms compounds with species that cause chemical interference and allow the atom of interest to be released for detection | Ex: La3+ for phosphate and aluminum

Answer 39

A new one must be used for each individual metal that needs to be studied

Answer 40

A standard addition method would require far more samples | ex: six unknowns would require 36 samples in total to be made

Answer 41

The concentration of CaCO3 in water - we used tap water and assed its hardness by converting concentration of Ca to CaCO3 using molar masses

Answer 42

strontium had the highest average signal which is to be expected considering its low ionization energy - signal increased as you moved down the column (ionization energy trend)

Answer 43

The LOD in ppb for the flame instruments were thousands of times more than the ICP-OES - low limit of detection notes increased sensitivity for dilute samples

Answer 44

standard signal/ ppb of standard = unknown signal/ppb of species unknown - use molar masses to convert

Answer 45

- atomization is the conversion of a sample into free atoms and ions by a thermal energy source - excitation is the transition of an electron from a lower level ground state to a higher energy level excited state by energy input (whether Eth or light/ radiation)

Answer 46

less, so transitions with more atoms returning to the ground state are more favorable

Answer 47

Self absorption occurs most often in flame spectroscopy where excited atoms at the center of the flame emit radiation that is then absorbed by atoms in the cooler part of the flame so a lower emission signal than expected is recorded

Answer 48

this is coupling emission with mass spec, allowing for analysis of structure and qualitiative analysis concerning concentrations - also has very low detection limits

Answer 49

alkali metals (earth and non) - low ionization energies

Answer 50

all metals but transition metals must be used with ICP-OES bc of higher ionization energies | remember plasma has a higher T

Answer 51

this instrument had a potassium filter attatched

Answer 52

Lower slope and flattening of the curve because of false decrease in absorbance | practice exam Q 1

Answer 53

The colored particles in the added soda absorbed the radiation of quinine due to it being a complimentary color | practice exam Q 4

Answer 54

One of the constants that is a part of the fluorescence equation is Po (the source light) which increases signal, but absorbance is part of a ratio so increased Po would have no effect | practice exam Q 5

Answer 55

Creates a magnetic field that creates the hot argon plasma as it moves in a circular motion | practice exam Q 12

Answer 56

add the numbers | ex: 1:1 = a DF of 2

Answer 57

V/Vo | practice exam Q 17

Answer 58

**electronic** only