Spectroscopy (Topics 8,10,12,14,15)

Question 1

What does the de Broglie relation allow us to conclude?

Answer 1

Particles with low linear momentum have long wavelengths and vice versa

Question 2

Which type of transitions in a molecule is likely to be induced by photons with a wavelength in the microwave region?

Answer 2

Vibrational

Question 3

What is spectoscopy?

Answer 3

the study of the interaction of electromagnetic
radiation with matter
How and why things scatter, absorb, or emit light
Exchange of energy between radiation and matter

Question 4

Electromagnetic radiation is

Answer 4

Light
Oscillating electric and magnetic fields that propagate as a wave

Question 5

c =

Answer 5

speed of light : 2.998 x 10^8 ms^-1
𝜆 (wavelength (m)) x 𝜈 (frequency (Hz = s^-1)

Question 6

𝜆 =

Answer 6

The wavelength at which a substance has it’s strongest photon absorption
ℎ (6.626 x 10^-34 Js) / 𝑚 (mass(kg)) x 𝑣 (velocity (ms^-1))

Question 7

angular frequency 𝜔

Answer 7

2 𝜋 𝜈 (frequency (s^-1)
2 𝜋 / T (s)

Question 8

𝜈 (1D)

Answer 8

𝑛𝑐/2𝐿
1 / 2𝜋 (𝑘𝑓 / 𝑚𝑒𝑓𝑓)^1/2

Question 9

𝜈~ (wavenumber (cm^-1) =

Answer 9

𝜈 (s^-1) / c (2.998 x 10^8 ms^-1)
1 / 2𝜋c (𝑘𝑓 / 𝑚𝑒𝑓𝑓)^1/2

Question 10

𝑛 (1D) =

Answer 10

2𝐿𝜈/𝑐

Question 11

𝑛𝑥^2 + 𝑛𝑦^2 + 𝑛𝑧^2 (3D)

Answer 11

4𝐿^2𝜈^2 / C^2

Question 12

Infrared radiation frequency

Answer 12

n ~10^12 – 10^13 s-1

Question 13

Visible light radiation frequency

Answer 13

n ~10^14 s-1

Question 14

Planck Constant ℎ =
ℏ =

Answer 14

6.626 x 10^-34 Js
ℏ / 2 pi

Question 15

Planck-Einstein Relation:

Answer 15

𝐸 (J) = ℎ (6.626 x 10^-34 Js) x 𝜈 (frequency (s^-1))

Question 16

𝐸 = 𝑚𝑐^2
𝐸^2 =
So for a photon, 𝐸 =

Answer 16

𝐸^2 = (𝑚𝑐^2)^2 + (𝑝𝑐)^2
𝐸 = 𝑝𝑐
𝐸 (energy (J))
𝑚 (mass (kg))
𝑝 (momentum (kg m s^-1)
𝑐 (speed of light (2.998 x 10^8 ms^-1))

Question 17

Einstein mass-energy relation

Answer 17

E = mc^2

Question 18

de Broglie Wavelength

Answer 18

lambda (wavelength (m)) = ℎ (6.626 x 10^-34 Js or kg m^2 s^-1 / 𝑚 (mass(kg)) 𝑣 (velocity(m s^-1))

Question 19

How do you define a peak?

Answer 19

Peak position
Peak intensity
Peak breadth

Question 20

delta E =

Answer 20

E1 - E0

Question 21

Bohr condition

Answer 21

Photon energy ℎ𝜈 must be equal to delta E in order for a transition to occur

Question 22

𝝉 (excited state lifetime)=

Answer 22

1 / A1->0

Question 23

Peak intensity information

Answer 23

Population
Transition Probability
State Degeneracy
Experimental: Path length of sample

Question 24

Boltzmann expression

Answer 24

𝑁(𝐽) / 𝑁= 𝑔(𝐽) 𝑒^− (𝐸(𝐽) 𝑘𝑇)
𝐸(𝐽) = 𝐵~ℎ𝑐𝐽(𝐽 + 1)
𝑔(𝐽) = 2𝐽 + 1
(J) subscript

Question 25

N upper / N lower =

Answer 25

𝑒−^(∆𝐸 𝑘𝑇)

Question 26

At room temp (298K), kT =

Answer 26

200 cm^-1

Question 27

Most critical factor in determining selection rules and line intensities

Answer 27

Transition dipole moment Measure of the electric dipole moment associated with movement of charge from its initial state (state 0) to its final state (state n)

Question 28

Gross Selection rule

Answer 28

In order for a transition to occur, it must have a nonzero transition dipole moment In order for an atom or molecule to absorb/emit a photon at a specific frequency, it must possess (at least transiently) a dipole oscillating at that frequency

Question 29

𝐴 = 𝜀𝐿𝑐 units

Answer 29

Absorbance (unit less) = molar extinction coefficient (M^-1 cm^-1) x sample thickness (often 1cm) x concentraTion (M)

Question 30

Commonly used versions of Beer-Lambert Law 𝐴 = 𝜀𝐿𝑐

Answer 30

log (𝐼0/𝐼) = 𝜀𝐿𝑐 log (𝐼0/𝐼) = 𝜎𝐿[𝑁] = 𝛼𝐿 𝐼 = transmitted light intensity 𝜀 = molar extinction coefficient (M^-1 cm^-1) 𝐿 = sample thickness (often 1cm) 𝑐 = concentration (M) 𝜎 = absorption cross section [𝑁] = Concentration (molecule cm^-3) 𝛼 = absorption coefficient

Question 31

Heisenberg Uncertainty Δ𝐸Δ𝑡 >/=

Answer 31

ℎ / 2𝜋

Question 32

Microwave wavelength

Answer 32

1cm - 100 um

Question 33

Infra-red wavelength

Answer 33

1um - 100 um

Question 34

Visible light wavelength

Answer 34

400 - 700 nm

Question 35

Order of energies for 3 spectroscopies

Answer 35

Delta E elec >> Delta E vib >> Delta E rot

Question 36

Rotational energies are in ... region of electromagnetic spectrum (microwave)

Answer 36

1 – 100 cm^-1

Question 37

𝐼 (moment of inertia) =

Answer 37

Sum of (𝑖) 𝑚𝑖𝑟𝑖^2 𝑖 is subscript 𝜇𝑟^2

Question 38

𝐹𝐽 = 𝐽 is subscript 𝐹~(𝐽) = (rotational energy levels)

Answer 38

𝐵𝐽(𝐽 + 1) 𝐵~𝐽 (𝐽 + 1) or 𝐵~𝐽 (𝐽 + 1) - 𝐷~(𝐽)𝐽^2(𝐽 + 1)^2 (𝐽) is subscript

Question 39

𝐵 (cm^-1)= 𝐵~ (rotational constant (cm^-1) =

Answer 39

ℏ^2 / 2𝐼 ℏ / 4𝜋𝑐𝐼 ℎ / 8𝜋^2c𝐼

Question 40

𝜈~ (wavenumber (cm^-1) (𝐽 + 1 ← 𝐽) =

Answer 40

2𝐵~(𝐽 + 1) 2𝐵~(𝐽 + 1) - 4𝐷~(𝐽) (𝐽 + 1)^3 First J subscript

Question 41

What will will result in a "red shift"

Answer 41

Increase in the wavelength of a wave with respect to the detector Decrease in the wavenumber of a wave with respect to the detector Decrease in the frequency of a wave with respect to the detector

Question 42

For the Doppler broadening of a peak, what affects the width of the observed peak?

Answer 42

The temperature and the peak position

Question 43

What function can be used to describe a peak in a spectrum?

Answer 43

Gaussian function

Question 44

What will cause deviationes in the linearity of the Beer-Lambert law?

Answer 44

Changes in the refractive index at high analyte concentration Fluorescence or phosphorescence of the sample Scaterring of light due to particulates in the sample

Question 45

The lifetime broadening of a spectral line arises from

Answer 45

The Heisenberg uncertainty principle

Question 46

What are considered absorption and emission processes in spectroscopy

Answer 46

Spontaneous emission and stimulated absorption

Question 47

The intensity of the spectroscopy transition can be predicted from

Answer 47

The population of the initial state and degeneracy

Question 48

In the Beer-Lambert law, the graphical representation of -log(10) of transmittance versus concentration of a solution can be fitted using

Answer 48

a linear regression

Question 49

𝑉 (Potential energy (J))

Answer 49

𝑉 = 1/2 𝑘𝑓 𝑥^2 𝑘𝑓 = spring force constant (Nm^-1) 𝑓 is subscript 𝑥 = 𝑟 − 𝑟𝑒 𝑟𝑒 = distance away from equilibrium (m) 𝑒 is subscript

Question 50

𝐺~(v) = (vibrational energy levels)

Answer 50

(v + 1/2) 𝜈~ (v + 1/2) 𝜈~ - (v + 1/2)^2 𝑥𝑒 𝜈~ (anharmonic) 𝑥𝑒 = Unitless anharmonicity constant, 𝑒 subscript

Question 51

𝐸v = (vibrational energy)

Answer 51

(v + 1 / 2) ℏ𝜔

Question 52

𝜔 =

Answer 52

(𝑘𝑓 / 𝑚𝑒𝑓𝑓)^1/2

Question 53

Gross Selection Rule

Answer 53

Dipole must change with displacement

Question 54

From selection rules, transitions are only allowed for... (P, Q, R)

Answer 54

𝚫𝐯 = ±𝟏 𝚫J = ±𝟏 P: 𝚫 J = –1 Q: 𝚫 J = 0 R: 𝚫 J = +1

Question 55

𝑆~ (v, 𝐽 ) = (transition energy levels)

Answer 55

𝐺~(v) + 𝐹~(𝐽) = (v + 1/2) 𝜈~ + 𝐵~𝐽(𝐽 + 1)

Question 56

Δ𝑆~𝑃 (v, 𝐽) 𝑃 is subscript

Answer 56

𝜈~𝑃(𝐽) = 𝜈~ - 2𝐵~𝐽 𝑃 is subscript 𝜈~ − (𝐵~1 + 𝐵~0)𝐽 + (𝐵~1 - 𝐵~0)𝐽^2

Question 57

Δ𝑆~𝑄 (v, 𝐽) 𝑄 is subscript

Answer 57

𝜈~𝑄(𝐽) = 𝜈~ 𝑄 is subscript 𝜈~ + (𝐵~1 - 𝐵~0)𝐽(𝐽 + 1)

Question 58

Δ𝑆~𝑅 (v, 𝐽) 𝑅 is subscript

Answer 58

𝜈~𝑅 (𝐽) = 𝜈~ + 2𝐵~(𝐽 + 1) 𝑅 is subscript 𝜈~ + (𝐵~1 + 𝐵~0)(𝐽 + 1) + (𝐵~1 - 𝐵~0)(𝐽 + 1)^2

Question 59

Rotational spectroscopy rotors

Answer 59

rigid rotor and non-rigid rotor

Question 60

Vibrational spectroscopy oscillators

Answer 60

harmonic oscillators and anharmonic oscillators

Question 61

𝐵v =

Answer 61

𝐵e − 𝛼e(v + 1/2) e subscript 𝐵e: Rotational constant at the equilibrium structure 𝛼: Constant, reflects shape of potential energy curve

Question 62

Combination difference to the same J level

Answer 62

𝜈~𝑅 (𝐽-1) - 𝜈~𝑃(𝐽+1) = 4 𝐵~0 (𝐽 + 1/2)

Question 63

Combination difference from the same J level

Answer 63

𝜈~𝑅 (𝐽) - 𝜈~𝑃(𝐽) = 4 𝐵~1 (𝐽 + 1/2)

Question 64

Number of vibrational normal nodes

Answer 64

3N – 5 for a linear molecule and 3N – 6 for a nonlinear molecule

Question 65

ΔG~ v = 0 → 1 (fundamental):

Answer 65

𝐺~(v + 1) - 𝐺~(v) = (v + 3/2) 𝜈~ - (v + 3/2)^2 𝑥𝑒 𝜈~ - (v + 1/2) 𝜈~ + (v + 1/2)^2 𝑥𝑒 𝜈~ (using G~ equation) = 𝜈~ - 2(v + 1) 𝑥𝑒 𝜈~

Question 66

ΔG~ v = 0 → 2 (1st overtone)

Answer 66

𝐺~(v + 2) - 𝐺~(v) = (v + 5/2) 𝜈~ - (v + 5/2)^2 𝑥𝑒 𝜈~ - (v + 1/2) 𝜈~ + (v + 1/2)^2 𝑥𝑒 𝜈~ (using G~ equation) = 2𝜈~ - 2(2v + 3) 𝑥𝑒 𝜈~

Question 67

Which one of the following functions best describes the simple harmonic oscillator?

Answer 67

y = Ax^2

Question 68

At which frequency is the R(0) transition located?

Answer 68

𝜈 + 2B

Question 69

In the simple harmonic oscillator model for a diatomic molecule, steeper slopes of the parabola indicates

Answer 69

stronger bond between the atoms in the molecule

Question 70

The first P-branch transition is

Answer 70

P(0)

Question 71

Parameters that can affect the full-width-at-half-maximum of an observed optical transition in a spectrum?

Answer 71

Temperature The lifetime in an excited state The phase of the medium Atomic mass of the species

Question 72

Parameters that can affect the intensity (height) of an observed peak in a microwave spectrum?

Answer 72

The instrument resolution The degeneracy of the rotational energy levels involved in the transition The temperature The rotational transition dipole moment

Question 73

𝐵~𝑒

Answer 73

Rotational constant for the equilibrium bond length of the molecule. It corresponds to the rotational constant when the molecule is in its lowest vibrational state, assuming no vibrational excitation. It provides the most fundamental measure of the molecule's moment of inertia in its equilibrium configuration

Question 74

𝐵~0

Answer 74

the rotational constant for the molecule when it is in its vibrational ground state (v=0). It accounts for the slight increase in bond length due to zero-point vibrational motion even in the ground state. The value of 𝐵~0 is typically slightly less than 𝐵~𝑒 because the average bond length is longer due to the vibrations

Question 75

𝐵~1

Answer 75

This is the rotational constant for the first vibrational excited state (v=1). The bond length further increases when the molecule is vibrationally excited, leading to an even larger moment of inertia and thus a smaller rotational constant compared to 𝐵~0

Question 76

Fundamental band

Answer 76

the transition from the ground vibrational state (v=0) to the first excited vibrational state (v=1) in a molecule. This is the most basic and usually the most intense absorption band observed in vibrational spectroscopy.

Question 77

Hot bands

Answer 77

transitions that originate from vibrationally excited states higher than the ground state. These transitions occur when a molecule that is already in an excited vibrational state (v=1, v=2, etc.) absorbs additional energy and transitions to an even higher vibrational state (v=2→v=3, v=3→v=4, etc.) energy difference of hν