Atomic & Rotational Spec Flashcards

Question

Where does the selection rule of Δm_l = 0, +/- 1 for a H-atom spectra arise from?

Answer 1

Due to the helicity of the photon being σ = +/- 1

Answer 2

Electron repulsion Σ_{i =!j} V_ij term makes it insolvable As in general need 3N spatial and N spin coordinates Solved using orbital approx

Answer 3

Assume ψ_space is product of n, one-e^- wavefunctions/orbitals Each orbital has [(-hbar²/2m_e) \* ∇_i² + V_iN + average(Σ_i=!j V_ij)]\*φ(r_i) = E\*φ(r_i) average of sum means that e- i experiences mean field of all other electrons

Answer 4

ψ_space should be a linear combination of orbitals with define symmetry wrt e- permutation Is not so doesn't satisfy Pauli's principle Also neglects spin and correlation

Answer 5

Energy level now depends on n and l for alkali metals the ns orbital experiences more attraction as higher Z and penetrating so lower in E

Answer 6

Core excitations at higher energies

Answer 7

Zeff - change in the Z Quantumm defect - changes the n quantum number

Answer 8

Sharp: ns -\> np Principal: np -\> ns Diffuse: nd -\> np Fundamental: nf -\> nd

Answer 9

Δn is unrestricted Δl = +/- 1 Δm_l = 0, +/- 1 Δj = 0, +/- 1 ΔS = 0

Answer 10

Spin orbit coupling - increases as atomic number does Orbital AM **_l_** couples to spin AM **_s_** to give a total AM **_j_** **_j_** = **_s_**+ **_l_**

Answer 11

Range given by Clebsch-Gordon series j = l+s, l+s-1, ..., | l-s|

Answer 12

**_l_** \* **_s_** = 1/2 \*(**_j_² - _l²_ - _s²_)** from **_j_ = _l_ + _s_** Sub in eigenvalues: **_l_** \* **_s_** = hbar²/2 \* [j(j+1) - l(l+1) - s(s+1)] E of a given level: E = 1/2 hcA \* [j(j+1) - l(l+1) - s(s+1)] where A is spin-orbit coupling constant proportional to Z⁴/n³l³

Answer 13

line at specific l and s splits to two lines for different j values lines at j values split to 2j+ 1 due to different mj values ( when in absence of external fields)

Answer 14

Conservation of angular momentum NOTE: cannot go from 0 to 0 as absorption of photon gives angular momentum

Answer 15

AM **_L_** and total spin AM **_S_** arise from additions of possible **_l_i_** or **_s_i_** from Clebsch-Gordon Couple to give total AM **_J_** = L+S, L+S-1, ..., |L-S| Coupling is the different values of J which can be seen

Answer 16

Fermions have 1/2 integer spin Bosons have integer spin

Answer 17

Wavefn must be anti-symmetric wrt exchange of two identical fermions and symmetric wrt exhange of bosons

Answer 18

α(1)α(2), β(1)β(2) - both are symmetric α(1)β(2), α(2)β(1) - no defined symmetry so must take linear combination 1/sqrt2[α(1)β(2) +/- α(2)β(1)] where + is symm and - is anti First two and + linear combination form a triplet which are symm Final - linear combination is a singlet

Answer 19

One e- spin up and the other down Cancellation to give **_S_** = 0, S = 0, Ms = 0, a single arrangement

Answer 20

Three arrangements with S = 1 Ms = 0 - one up one down spin Ms = +1 - two up spin Ms = -1 - two down spin

Answer 21

ψ = (1/sqrt2) [φ_1s(1)φ_2s(2) +/- φ_1s(2)φ_2s(1)] This linear combination is to give symmetry, + is symm and - is anti Triplet has symm spin wavefn so must have ψ^- Singlet has antisymm spin wavefn so must have ψ+

Answer 22

P^(ψ^-(1,2)) = -ψ^-(2,1) when r1 = r2 then ψ^-(1,1) = -ψ^-(1,1) so ψ^-(1,1) = 0 and 2e- cannot exist at same place however for ψ⁺ there is a max

Answer 23

e- in the triplet have a 0 prob of being in same location - seen as a dip in graph e- in the singlet have a maximum in graph when in same location, so more repulsion Means triplet is lower in energy overall

Answer 24

Singlet to triplet or vice versa transition forbidden As every triplet state is lower than corresponding singlet

Answer 25

Configuration: number of e- in each orbital Terms: different L & S config (atomic symbols in ^SL), from spin correlation Levels: different J levels of terms, from magnetic coupling of total spin and total orbital AM States: all 2J+1 possibilites of the mj

Answer 26

Pauli Exclusion principle - cannot have identical quantum numbers (q.n.) E.g. ³D for Carbon has L=2 and M_L = 2 compenent, meaning m_l1 = m_l2 = 1 and S = 1 so M_s=1 and so m_s1= m_s2 = 1/2, same l and n so has same q.n.

Answer 27

Table of every combination of m_s and m_l Can eliminate all terms of different Σm_l values

Answer 28

Term with largest S is lowest in energy For given S the term with largest L is lowest in energy When several levels: if subshell less than half full then lowest J level is lowest in energy but if more than half full then highest J level Assumes spin correlation \>\> orbital am coupling \>\> spin-orbit coupling

Answer 29

Only for ground states of atoms Valid for period 1-2 and some TM

Answer 30

At large Z spin-orbit coupling is strong then j-j coupling As spin prefers to couple with its own orbital a.m. to give total j

Answer 31

**_j_**_i - total a.m. of individual e- orbital a.m. and spin a.m. couple **_J_** - sum of **_j_**_i

Answer 32

Applicable when S=0, so is a singlet External field **_B_** lifts degeneracy of M_L components which leads to splitting in the spectrum

Answer 33

Classically: E = -**m\*B** = -γ_eL_zB QM: E= -γ_eM_LB\*hbar = μ_BM_LB where γ_e = -e/2m and μ_B (bohr magneton) = -γ_e\*hbar

Answer 34

ΔM_L = 0, +/- 1 Single signal therefore splits

Answer 35

When atom of S =! 0 is subjected to an external field Causes splitting into 2J+1 levels, and the energy of which is dependent on J, L and S

Answer 36

E = -γ_e(**_L_** + 2**_S_**).**_B_** Only need to consider projection of **_L_** & **_S_** on **_B_** **_L_**.**_B_** = (**_L_**.**_J**_)(_**B**_._**J_**)/**_J_**² E = -γ_eg_J**_B**_._**J_** = -γ_eg_JM_JB\*hbar So proportional to M_J and B, not degen splitting so numerous peaks for each ΔM_J where Lande factor g_J (J,L,S) = 1 + [J(J+1) + S(S+1) - L(L+1)]/2J(J+1)

Answer 37

10cm to 1mm wavelength In the microwave

Answer 38

I = Σ_i m_ir_i² where m_i is mass of ith particle and r_i is perpendicular distance from axis Rotational equivalent of mass

Answer 39

J = I\*ω E = J²/2I = 0.5\*I\*ω² (where I is inertia)

Answer 40

Rigid rotor represents particle on a sphere

Answer 41

Purely kinetic energy H^ψ = [-(hbar)²/2μ]\*∇²ψ = Eψ Where ∇² = δ²/δr² + (2/r)(δ/δr) + (1/r²)\*Λ² in spherical polar coordinates Cyclic boundary conditions that ψ(φ+2π) = ψ(φ) Gives Spherical harmonics as the solutions

Answer 42

E_J,Mj = j^²/2I = J(J+1)\*hbar²/2I = BJ(J+1) with rotational q.n. J = 0,1,2,3... rotational a.m. projection q.n. m_J = 0, +/- 1, +/- 2... +/- J B is rotational constant in joules B = hbar²/2I

Answer 43

Given as wavenumbers or rotational terms B~ = B/hc = h/8π²cI = h/ h/8π²cμR² No zero point energy associated with rotation - get more widely spaced with increasing J As B~ α 1/μR² for diatomics relate this to bond lengths \<1/R²\>

Answer 44

Reduced mass does change so B~ ratio changes

Answer 45

Each J level has 2J+1 degenerate states Arises from projection quantum number M_J

Answer 46

From Boltzmann distribution, n_iα g_j\*exp(-E_j/kT) g_j = 2J+1 and E_J = hcB~J(J+1) Most populated when dn_i/dJ = 0

Answer 47

Energy relative to ZPE, NJ/N0 = (2J+1)*exp(-E/kT) dn_i/dJ = 0 dn_i/dJ = [2-(2J+1)²\*hcB~/kT} \* exp[-hcB~J(J+1)/kT] = 0 J_max = Sqrt[kT/2hcB~] - 1/2

Answer 48

As temp increases the more are in a higher most populated rot J-level Higher B, and hence spacing, then lower the most populated rot J-level

Answer 49

ψ_ml (φ) = Exp(im_lφ)/Sqrt[2π] Where m_l = +/- Sqrt[2EI]/hbar Cyclic boundary conditions restrict to m_l = 0, +/- 1, +/- 2, +/- 3 Real part plotted shows symmetry, odd & even J levels have opposite parity

Answer 50

Parity - symmetry wrt inversion Parity = (-1)^J where J is the J level occupied

Answer 51

Properties required to do a form of spectroscopy

Answer 52

Must posses a permanent dipole moment

Answer 53

ΔJ = +/- 1 : Due to conservation of a.m. as need to change parity ΔK = 0 : required for non-linear or polyatomic

Answer 54

F_J = E_J / hc = B~J(J+1) where B~ = B/hc

Answer 55

v~ = F_J+1 - F_J = B~(J+1)(J+2) - B~J(J+1) v~ = 2B~(J+1) Equally spaced lines with separation of 2B

Answer 56

As J increases the gap between levels increases due to J² dependence of rotational eigenvalues

Answer 57

Info on geometry For diatomic molecules it gives bond length

Answer 58

Population of lower levels and J dependence of transition strength See that population of many levels as k_BT large wrt rotation energy

Answer 59

Moments of inertia about three mutually perpendicular axes through centre of mass Called axis a,b and c so I_a \< I_b \< I_c with I_c - max

Answer 60

Molecule with zero dipole moment so doesn't appear in spectra I_a = I_b = I_c E.g. T_d, O_h

Answer 61

Spherical tops - identical I_i Symmetrical tops - two identical I_i Asymmetrical tops - all unique I_i

Answer 62

Prolate tops - long and thin molecules with I_a b = I_c Oblate tops - discus type molecules with I_a = I_b \< I_c

Answer 63

Constant for each moment of inertia for the three axes A~ \> B~ \> C~ as I_c is the largest Cannot relate to bond lengths individually as includes angles and lengths

Answer 64

Projection quantum number for projection on the unique a axis Where a is body-fixed axis in prolate, and c in oblate

Answer 65

J_Ka - J is total a.m. and K_a is projection q.n. Each level has 2J+1 degen (from M_J) and if K\>0 has two fold degen of +/-K

Answer 66

F_J,K = B~J(J+1) + (A~-B~)K² So J = 0, 1, 2, 3.... and K = 0, +/- 1, +/- 2,..., +/- J

Answer 67

|**J**| = Sqrt[J(J+1)] \* hbar J_a = K \* hbar , is what extent it rotates wrt principle axis

Answer 68

When K≈J then rot a.m. is nearly parallel to principle axis K = 0 then rot a.m. is perpendicular to principle axis

Answer 69

J_Kc where J is total a.m. and Kc is projection q.n. (on unique axis, c) Each level has 2J+1 degen (from M_J) and when K\>0 two-fold extra degen (+/- K)

Answer 70

F_J,K = B~J(J+1) + (C~-B~)K² C~ \<= B~ so term is -ve

Answer 71

NONE, K =! M_J K refers to projection on a body-fixed axis M_J is a projection on a space-fixed axis

Answer 72

Prolate: (A~-B~)K² \> 0 so for a given J the energy increases with K Oblate: (C~-B~)K² \< 0 so for a given J the energy decreases with K

Answer 73

I_a = 0 and so A~ = ∞ As only K = 0 F_J = B~J(J+1) and J = 0,1,2,3...

Answer 74

A~=B~=C~ F_J = B~J(J+1) where J = 0,1,2,3... Degen = (2J+1)² No pure rotational spec as no dipole moment

Answer 75

Wihtin rigid rotor approx they are the same Have equally spaced lines with separation = 2B~ So no info on unique axis

Answer 76

Bonds stretch during rotation So inertia and rot constants change with J Treat as perturbation to rigid rotor as terms not large in molecules at room T

Answer 77

F(J) = B~J(J+1) - D~J(J+1)² Where D~ is centrifugal distortion constant in cm^-1 D~ = 4B~³/ω_e² and ω_e is vibrational frequency (stronger bonds cause lesser distortion)

Answer 78

Values are smaller than expected - more prominent as J increases Transitions occur at v~(J) = F(J+1) - F(J) = 2B~(J+1) - 4D~(J+1)³

Answer 79

v~(J) = 2B~(J+1) - 4D~(J+1)³ Plot (J+1)² on x-axis and v~(J)/(J+1) Gives intercept = 2B~ and gradient = -4D~

Answer 80

v~ = F(J+1, K) - F(J,K) v~ = 2(B~-D~_JKK²)(J+1) - 4D~_J(J+1)³

Answer 81

Each J levels has 2J+1 which are degenerate due to space quantization When there is a field (**_E_** or **_B_**) the degenercy is lifted

Answer 82

Electric dipole μ interacts with an applied field, results in an interacting energy **_μ_**.**_E_** = -μEcosθ Effects energy levels of symmetrical tops

Answer 83

μ_J = μ\*cosα = μ\*K/Sqrt[J(J+1)] where μ_Jis projection along **_J_** External field making an angle β to J which leads an interaction energy E_stark = -μ_J\*E\*cosβ = -μEKM_J/[J(J+1)] where E is electric field, and M_J means there are different values

Answer 84

E_stark α |**_J_**|^-2

Answer 85

As depends on M_J it causes distinct energy levels Has larger splitting when J is smaller

Answer 86

Selection rule: ΔM_J = 0 which will cause splitting E.g. So if from J=1 to 2 there is one peak, when field applied will get 3 separate peaks μE/3 in energy apart for M_J 1, 0, -1 of the J=1 Equal in intensity and only 3 as can only do three transitions from J=1 to 2 to keep selection rule

Answer 87

**_I_** is the a.m. and I is the quantum number

Answer 88

Mass number Even: I is integral (0,1,2,3..) nuclei are bosons Mass number Odd: I is 1/2 integral (1/2, 3/2, ...) nuclei are fermions

Answer 89

Nuclear spin causes magnetic moment which interacts with external magnetic fields & internal magnetic field (gives small splittings) Determines whether rotlevels in symm molecules actually exist - due to nuclear spin stats

Answer 90

ψ_tot = ψ_elψ_vibψ_rotψ_ns ψ_tot if boson is symm, if fermion is antisymm Consider ground state of molecule, and then work out symmetry of each part, ψ_eland ψ_vibare constant Gives several combination, find weighting of them which is shown in spectra

Answer 91

of sym ψ_ns / # of antisym ψ_ns = (I + 1) / I See the ratio of this or lack of some in the spectra

Answer 92

E = -(μEKM_J)/[J(J+1)] define for J and K then will have splitting to M_J levels

Answer 93

Larger splitting between levels when J small

Answer 94

Introduces selection rule ΔM_J = 0 Leads to splitting in spectrum to number of M_J levels in the lower state

Answer 95

Find term symbol for the molecule (e.g. H₂ has ¹Σ_g⁺) Usually symmetric but need to check, O₂ is the normal exception

Answer 96

All v=0 vib levels are symmetric

Answer 97

Even J are symmetric, odd J levels are anti from (-1)^J

Answer 98

Nuclear spin Even mass no then I is an integer and a boson (symmetric) Odd - I is a half-integer and so fermion (antisymm)

Answer 99

Even mass - boson, must be symm Odd mass - fermion, must be anti

Answer 100

Find mass number - this determines ψ_total and ψ_ns Find symmetry of elec state to give ψ_el Then find combinations to give correct ψ_total, remember that NS must be correct for the compound!

Answer 101

F = -k_Fx PE: V(x) = -∫F dx = (k_Fx²)/2

Answer 102

G_v = (v+1/2)ω_e where v is vib qn and ω_e is vibrational const

Answer 103

ω_e = ν_vib/c~ = (1/2πc~) Sqrt[k_F/μ] where : ν_vib is the classical freq of oscillation k_F is the force const μ is the reduced mass

Answer 104

E = hc~G~(v+1/2) E = hν(v+1/2) where ν is freq E = hbarω_e(v+1/2)

Answer 105

E = (1/2)hν where ν is freq

Answer 106

Equally spaced non-degen energy levels

Answer 107

As q increases the prob density becomes more classical Penetration into classically forbidden regions

Answer 108

H^ = -(1/2)(d²/dq²) + (1/2)q² where q is vib coord, is 0 at eqm bond length

Answer 109

ψ₀ = exp(-q²/2)

Answer 110

Dipole moment that changes during vibration TDM =! 1

Answer 111

Δv = +/- 1 where v is rot qn

Answer 112

R₂₁ =! 0 R₂₁ = (dμ/dq) ∫ψ`*`qψ dq (dμ/dq) =! 0, dipole changes - gross selec rules ∫ψ`*`qψ dq =! 0, leads to Δv=+/- 1 which is specific selec rule

Answer 113

v~ = ω_e One line in spectrum

Answer 114

As bond stretches the bond gets weaker so slope decreases to limit

Answer 115

V(R) = D_e[ 1-exp(-β(R-R_e))]² Gives more accurate than HO

Answer 116

G~ = (v+1/2)ω_e - (v+1/2)²ω_ex_e where v=0,1,2,... v_max where x_e is the anharmonicity constant

Answer 117

x_e = ω_e/4D_e where D_e is the dissociation energy

Answer 118

D₀ = D_e - ZPE where D_e is dissociation energy to 0

Answer 119

Δv = +/- 1, (+/-2, +/-3, etc. are weaker)

Answer 120

Fundamental (0->1) 1st overtone (0->2) 1st hot band (1->2)

Answer 121

Rot transition also occurs E_tot = E_vib + E_rot

Answer 122

Δv = +/- 1, (+/- 2, +/- 3, etc weaker) ΔJ = +/- 1

Answer 123

For rot vib spec: when ΔJ = +1, transitions give rise to R-branch with higher wavenumbers when ΔJ = -1, transitions give rise to P-branch with lower wavenumbers

Answer 124

S~ = (v+1/2)ω_e - (v+1/2)²ω_ex_e + B~J(J+1) where only last term is from rot

Answer 125

Signal when ΔJ=0 Absent unless additional angular momentum (from elec a.m.) or degen bending mode present

Answer 126

Spacing in the branch = 2B~

Answer 127

Central gap = 4B~

Answer 128

ω₀ = ω_e - 2ω_ex_e

Answer 129

B~ α 1/μR² Treat as `<`1/R²>, which decreases with v

Answer 130

B~₀ > B~₁ > B~_v>1 B~_v = B~_e - α(v+1/2)

Answer 131

Use transitions with common upper/lower levels to determine rot const for each level

Answer 132

At high J the wavenumber separation of adj transitions can change sign Causes transitions to bunch up

Answer 133

Independent, harmonic vibrations of polyatomic molecules These vibrations must: leave centre of mass unmoved, involve all atoms moving in phase, and transform as an irrep of molecular point group

Answer 134

Means all atoms moving in phase and going back to original position at the same time

Answer 135

N isolated atoms have 3N dof

Answer 136

Linear molecule = 3 Non-linear = 3

Answer 137

Linear = 2 Non-linear = 3

Answer 138

Linear = 3N-5 Non-linear = 3N-6

Answer 139

Perform symm operations on stretches 1 if unchanged and -1 if swapped

Answer 140

ψ₀ transforms as totaly symmetric irrep of relevant point group

Answer 141

ψ₁ has the same symmetry as the normal mode This is because has same symm as normal coord q (which is a linear function)

Answer 142

ψ₂ transforms as totally symm IR, as proportional goes via q² ψ₃ transforms as same symm of normal mode, as proportional to q+q³ when degen modes then use direct product tables

Answer 143

New one: Δv_i = +/- 1 Dipole moment must change with vib (standard one)

Answer 144

TDM separates across x,y, and z Each is scalar, and at least 1 of them must transform as totally symm irrep Check if direct prod transform as totally symmetric, with v' being final

Answer 145

Transforms as x,y,z from character table

Answer 146

Parallel - if transition moment of a mode is parallel to symmentry axis Perp - same as above but perpendicular

Answer 147

Symm of normal mode (and therefore v=1) is the same as x,y,z Due to the starting one being totally symm

Answer 148

Simultaneous excitation of both symm and asymmetric stretches e.g. v₁=0, v₃=0 -> v₁=1, v₃=1 Direct prod of how v₁ and v₃ gives how they transform overall

Answer 149

Parallel (Σ-Σ) - ΔJ=+/- 1, so P & R branches Perpendicular (Σ-Π) - ΔJ=0,+/- 1 (Q,P, and R branches)

Answer 150

Occurs due to additional angular momentum from degen bending mode

Answer 151

ψ_total(r,Q,θ) = ψ_el(r) ψ_vib(r) ψ(θ) Assumption: single ψ_el for a given config for a given elec config as separable to nuc coord

Answer 152

1 value for PE at specific separation so e- can arrange in lowest E config quickly

Answer 153

Classified wrt angular momentum around internuclear axis, λ λ analogous to mI in atoms

Answer 154

m_I = 0, combine to give σ,σ`*` (λ=0) where σ means cylindrical symm about the internuclear axis

Answer 155

Each has m_I = +/- 1, combine to give π,π`*` (λ = +/- 1) where π means antisymmetry about internuclear axis

Answer 156

Λ = Σ λ_i λ is for each e- Λ =0 gives Σ Λ =+/- 1 gives Π Λ =+/- 2 gives Δ

Answer 157

^2S+1Λ_Ω where Λ = total angular momentum Ω = |Λ + Σ| and Σ = projection of S on internuclear axis

Answer 158

Homonulcear u - antisymm g - symm

Answer 159

g x g = g u x u = g g x u = u x g = u

Answer 160

For sigma terms It is with respect to reflection in a plane containing the internuclear axis

Answer 161

Singlet: ψ_spin is anti, so ψ_space must be symm Therefore g,+ states Triplet must be paired with g,- states

Answer 162

BO Approx allows for separation: ψ_tot = ψ_{el_{(r) ψ_{vib_{(R)

TDM therefore required elec and vib part to not equal to 0}}}}

Answer 163

ΔΛ = 0, +/- 1 ΔS = 0 ΔΣ = 0 when they exist: g <-> u + <-> +, and - <-> -

Answer 164

Square of vib overlap integral (from square of TDM) Deals with vibrational changes when transition occurs

Answer 165

R unchanged as: Elec transitions take place on such a short timescale that the nuclei reamin frozen

Answer 166

Probability of a transition between elec states is governed by square of overlap integral of two vib wavefunctions No selection rule governing allowed vib changes which accompany the elec transition

Answer 167

Short progression: when similar bonding state (e.g. bond to bond) then 0->0 more likely Long progression: then different bonding (e.g. bond to anti) then 0->excited states

Answer 168

Do dGv/d(v+1/2) = ω_e - 2ω_ex_e (v+1/2) at the dissociation limit (v+1/2)_max ->0 gives: G~_max = D_e = ω_e²/(4ω_ex_e)

Answer 169

Plot ΔG as a function of (v+1/2) Area under plot gives dissociation energy At high v, then true curve falls off and so linear extrapolation performed

Answer 170

Only larger changes in rot const observed So band heads are observed in elec, and can be in either branch This is because `<`R²> change can be large depending on bonding character of final and initial state Large change in R_e means B'<`

Answer 171

When lines coalesce, dv~/dJ = 0 Can be in either branch (but R more likely) In R branch: dv~/dJ = (B' + B'') + 2(B'-B'')(J+1) In P branch: dv~/dJ = -(B' + B'') + 2J(B'-B'') therefore (J+1)_head = - (B'+B'')/2(B'-B'') J_head = (B'+B'')/2(B'-B'')

Answer 172

When transition has ΔΛ > 0, then additional angular momentum can cause it Coudl alternatively form vib mode

Answer 173

Records ionization energies for removal of e-s Provides measure of energy of MOs

Answer 174

1. Excite with fixed λ (above IE) 2. Measure KE of ejected photo-e^-

Answer 175

overall: hv = I + E_M+ + KE_e- + KE_M+ where I is ionisation energy, and E_M+ is the internal energy of the ion KE_e- >>> KE_M+, due to M being sig heavier so less KE when e- leaves therefore make assumption KE_e- = hv - I - E_M+

Answer 176

Franck-Condon Principle Removal of strongly bond e- results in substantial red in bonding character Therefore can occupy various vib energy levels of excited state

Answer 177

Short - weakly anti/bonding e- Long - strongly anti/bonding e-

Answer 178

Photons scattered by molecules Weak but still occur

Answer 179

Rayleigh scattering - elastic, leaves molecule in same state Raman scattering - inelastic, leaves molcules in different quantum state (rot or vib)

Answer 180

I = I₀ [8πNα²/λ⁴R²] (1+cos²θ) where N is number of scatterers α is polarisability, and R is distance between scatterer and observer

Answer 181

When the dipole scatterer << λ of light

Answer 182

I α R^-2, inverse square law! I α λ^-4, sig more effective with shorter wavelength of light

Answer 183

Scattered photon has different energy (frequency, wavelength)

Answer 184

Result of raman scattering: Stokes lines - when scattered photon has lost energy Anti-Stokes - when scattered photon has gained energy

Answer 185

Rayleigh strongest scattering

Answer 186

Raman active requires a molecule to have anisotropic polarisability where anisotropic = different polarising in one plane compared to another

Answer 187

Polarisability is isotropic - same on each plane Always is in atoms and some molecules

Answer 188

Anisotropic as the molecule rotates the polarisability presented in E field changes

Answer 189

ΔJ = 0,+/- 2 where 0 is from Rayleigh, + is from stokes , and anti being -ve 2 as is an effective 2-photon process

Answer 190

When anisotropic molecule rotates the polarisability presented to E field changes This means induced dipole modulated by rotation Results in rotational transitions occuring

Answer 191

Due to nuclear spin-stats

Answer 192

Missing lines may be present in spectra e.g. O₂ and CO₂ have alternate lines missing (8B~ instead of 2B~ in rot or 4B~ in rot raman)

Answer 193

Polarizability must change during vibration Therefore: normal mode must transform with same symm as quadratic forms (x², xy, etc.)

Answer 194

Δv = +/- 1 +ve is stokes and -ve is anti

Answer 195

v>0 are weakly populated

Answer 196

Vib mode may either be Raman or IR active but not both (only when centre of inversion symm)

Answer 197

IR - transforms as x,y,or z Raman - transforms as quadratic term (e.g. z², x²+y², etc)