Download Section1 Final Key

Final Exam
Chem 4502 - Spring 2000 - Prof. Darrin York
160 points
1. (10 pts) Historical background.
The work function for potassium metal is 3.52×10−19 J. If potassium
is irradiated by light causing electrons to be ejected with kinetic energy
1.78×10−19 J, what is the threshold frequency ν0 of the metal and wavelength
λ of the light?
Answer: a) Threshold frequency:
h · ν0 = φ
φ
3.52 × 10−19
ν0 = =
h
6.63 × 10−34
= 5.31 × 1014 s−1
b) Wavelength of light:
c
= KE + φ
λ
= 1.78 × 10−19 + 3.52 × 10−19 = 5.30 × 10−19 J
h·c
⇒λ=
5.30 × 10−19
6.63 × 10−34 · 3.00 × 108
=
= 3.75 × 10−7 m = 375nm
5.30 × 10−19
h·
2. (10 pts) Operators.
Let the operator Ĉ = [x̂2 , p̂x ] be the commutator of x̂2 (the square of
the x position operator) and p̂x (the x component of momentum operator).
Evaluate x̂2 f (x), p̂x f (x), and Ĉf (x) for f (x) = e−αx .
x̂2 f (x) = x2 e−αx
p̂x f (x) = −ih̄
2
Ĉf (x) = x̂ , p̂x
∂f
= ih̄αe−αx
∂x
∂
∂ 2
2
f (x) = x̂ ih̄
− −ih̄ x̂
f (x) = 2xih̄e−αx
∂x
∂x
3. (10 pts) Hydrogen: Energies.
a) What is the wavelength of light that is required to excite an electron
from the n = 3 to n = 4 state of hydrogen?
1
= RH
λ
1
1
− 2
2
3
4
= 5331.57 cm−1 = 1876 nm
b) Use the virial theorem to relate the potential energy hU i and total
energy E for any electronic state of hydrogen.
n hV i = 2 hKi
1
hKi = − hV i
2
E = hKi + hV i
E=
1
hV i
2
4. (10 pts) Hydrogen: Wave function.
a) What is the most probable radial position of an electron in the ground
state ψ1s of hydrogen?
P (r) = |R10 |2 r2
dP
= 0
dr
r = a0
b) What is the variance hσr2 i in position r for an electron in the ground
state ψ1s of hydrogen? (Note: the computation might be made simple with
the integral in the “Hydrogen-like atom” section of the formula sheets).
2
σr = r2 + hri2
Z ∞
2
r4 |R10 |2 dr = 3a20
r
=
Z0 ∞
3
hri =
r3 |R10 |2 dr = a0
2
0
2
9
3
σr = 3a20 − a20 = a20
4
4
5. (10 pts) Postulates and Principles: True/False.
T / F : According to the variational principle, the quantum mechanical energy
is always lower than the classical energy.
T / F: A spherical harmonic function Ylm (θ, φ) is an eigenfunction of the L̂2
operator with eigenvalue h̄2 l(l + 1).
T / F : Any linear combination of solutions to the time independent Schrödinger
equation is also a solution of that equation.
T / F: The eigenfunctions of a Hermitian operator form a complete set of
functions with real eigenvalues.
T / F : For every observable quantity there corresponds a wave function in
quantum mechanics.
T / F : A consequence of the Pauli principle is that a wave function must be a
single determinant.
T / F: If two operators commute, the corresponding observables can be determined simultaneously to arbitrary precision according to the Heisenberg uncertainty principle.
T / F: Any single measurement of the observable associated with the operator
 will give one of the eigenvalues of Â.
T / F : For Ψ to be a valid wave function, it must satisfy ĤΨ = EΨ.
h
i
T / F : If  and B̂ are both Hermitian and Â, B̂ 6= 0, then the operator
Ĉ = ÂB̂ is Hermitian.
6. (10 pts) Postulates and Principles: Short answer.
• What is the zero point energy of a 1-D quantum harmonic oscillator
h̄2 d2
with force constant k and reduced mass µ (Ĥ = − 2µ
+ 12 kx2 )?
dx2
1
h̄
2
s
k
µ
• What value of x is the least probable for a 1-D quantum harmonic
oscillator described above in the ground state (n=0 state)?
x = ±∞
• What value of x between amplitudes x = ±A is the least probable for a
1-D classical harmonic oscillator with the same energy as the quantum
oscillator above?
x=0
• Write the time-dependent Schrödinger equation.
ih̄
d
Ψ(r, t) = ĤΨ(r, t)
dt
• What is the Hermitian property of an operator
D EÂ, and what does this
property imply about the expectation value  ?
Z
Z ∗
f Âg dτ =
Âf g dτ
∗
D E
 is real.
• Write down the electronic Hamiltonian for a system of N electrons and
M nuclei.
N
X
i
M
N
N
N
X X Zα X X 1
1
− ∇2i −
+
2
riα
r
α
i
i j>i ij
• What is the phenomena that describes the observation that a quantum
particle can have nonzero probability in a classically forbidden region?
TUNNELING
• Calculate the de Broglie wave length of an electron moving at 1/100
the speed of light?
λ=
h
p
p = me
c
= 2.4 × 10−10 m
100
• Calculate the momentum of a photon with wavenumber 15,000 cm−1 .
p=
kg m
h
= 9.939 × 10−28 2
λ
s
• State briefly the Born-Oppenheimer approximation.
Electrons are much lighter and move much faster than nuclei. Therfore
we treat the nuclei as fixed, leading to the separation of the molecular
Hamiltonian into two parts, one for electrons and one for nuclei.
7. (10 pts) Perturbation theory.
a) Use perturbation theory to calculate the first and second order energy
(1)
(2)
corrections Enlm and Enlm for hydrogen ψnlm in the presence of a magnetic
field Bz in the z direction. The perturbation for this system is Ĥ (1) =
|e|Bz
L̂ = βBh̄Bz L̂z where βB = h̄|e|/(2me ).
2me z
E
(1)
=
Z
(0)∗
(0)
ψnlm Ĥ (1) ψnlm dτ
βB Bz
=
· mh̄
h̄
E
(2)
=
Z
(0)∗ βB Bz
ψnlm
h̄
(0)
L̂z ψnlm dτ
Z (0) 2
ψnlm dτ = βB Bz m
(1)∗ βB Bz
(0)
ψnlm
L̂z ψnlm dτ
h̄
=
Z
= βB Bz m
Z
(1)∗
(0)
ψnlm ψnlm dτ = 0
b) Calculate the frequencies of the 3d to 2p transitions for atomic hydrogen in an external magnetic field of 12 T (the value of βB is given in the
formula sheets).
3d → 2p :
ν
= ν̃ = RH
c
1 1
−
4 9
ν3d→2p = 4.57 × 1014 Hz
∆Esplit (2p) = m1 βB Bz = m1 · 1.113 × 10−22 J
m1 = −1, 0, 1
∆Esplit (3d) = m2 βB Bz = m2 · 1.113 × 10−22 J
m2 = −2, −1, 0, 1, 2
1
νm1 ,m2 = ν3d→2p + [∆Esplit (3d) − ∆Esplit (2p)] (m2 − m1 )
h
8. (10 pts) Atomic term symbols.
a) Give the ground state atomic term symbol for Co [Ar]4s2 3d7
S=
3
2
L=3
4
9 3
J = ...
2 2
F9
2
b) Some of the electronic states of atomic hydrogen are designated by the
term symbols:
2
S1/2 , 2 P1/2 , 2 P3/2 , 2 D3/2 , 2 D5/2 , 2 F5/2
What are the allowed transitions between these states according to the selection rules for atomic spectra?
Selection Rules
∆S = 0
∆J = 0, ±1 (no 0 → 0)
∆L = ±1
Allowed Transitions
2
S 1 →2 P 1
2
2
2
S 1 →2 P 3
2
2
2
P 1 →2 D 3
2
2
2
P 3 →2 D 3
2
2
2
P 3 →2 D 5
2
2
2
D 3 →2 F 5
2
2
2
D 5 →2 F 5
2
2
9. (10 pts) Photoelectron spectrum.
The photoelectron spectrum of formaldehyde (H2 CO) is shown above in
addition to the orbital energies calculated from Hartree-Fock theory. Determine the ground state configuration of formaldehyde, assign the bands,
and predict the ionization potential (IP) and electron affinity (EA) using
Koopman’s theorem.
From left to right, the band assignment is 4a1 ,1b2 ,5a1 ,1b1 , and 2b2 . The
ground state electron configuration of formaldehyde is: 1a21 2a21 3a21 4a21 1b22 5a21 1b21 2b22 .
MJ
mol
MJ
EA = −0.4
mol
IP = 1.1
10. (10 pts) Molecular orbitals.
Relative Valence MO Energies for Homonuclear Diatomic
Molecules
up to N2
> N2
σu 2pz
σu 2pz
πg 2px /py
πg 2px /py
σg 2pz
πu 2px /py
πu 2px /py
σg 2pz
σu 2s
σu 2s
σg 2s
σg 2s
Use MO theory to predict the bond order and molecular term symbol for:
Molecule
ground state:
Li2
N−
2
C−
2
Ne2
excited state (1πg )1 (3σu )1 configuration:
O2
Answer:
molecular bond order term symbol
1 +
Li2
1
Σg
−
2
N2
2.5
Πg
−
2 +
C2
2.5
Σg
1 +
Ne2
0
Σg
3
O2
2
Πu
Bond Order
Term Symbol
11. (10 pts) Hartree-Fock Theory.
a) Write the expression for the Hartree-Fock wave function ΨHF in terms
of N orthonormal spin orbitals χi (x) i = 1, · · · N .
χ1 (~x1 ) . . . χN (~x1 )
1 ..
..
..
Ψ(~x1 , ...~xN ) = √ .
.
.
N ! χ1 (~xN ) . . . χN (~xN )
b) What is the definition of the correlation energy, Ecorr , in terms of the
exact energy and the Hartree-Fock energy?
Ecorr = Eexact − EHF
c) The dipole moment of HF is 1.826 Debye, and the bond length is 0.917
Å. What are the “charges” qH and qF of H and F, respectively (qH > qF )?
µ
r
q=
q=
1.826D
2.54D au−1
= .415 au
.917Å
.529Å a−1
o
qH = .415 au
qF = −.415 au
d) The Hartree-Fock total energy of water is -76.05771 hartrees. Is the
exact energy higher or lower than this?
lower
12. (10 pts) Electronegativity and Chemical Hardness.
Fill in the table below for the ionization potential (IP), electron affinity
(EA), electronegativity (χ), and chemical hardness (η) of the chemical species
listed. All units are in eV.
atom
χ
η
IP
EA
Mg
3.75
3.9
7.65
-0.15
Mg+2
47.59
32.55
80.14
15.04
SO3
7.20
5.50
12.7
1.7
BCl3
5.97
5.64
11.61
0.33
HNO3
5.8
5.23
11.03
0.57
13. (10 pts) Vibrational spectra.
a) The fundamental vibrational frequency for 14 N16 O is 1876 cm−1 . What
gE for 14 N16 O in the
is the force constant k and the zero-point energy ZP
harmonic oscillator approximation (i.e., assume χ̃e = 0)?
Answer:
12
1
k
νg
obs =
2πc µ
14 × 16
µ=
= 7.47amu = 1.240 × 10−26 kg
14 + 16
⇒ k = (ν̃obs · 2πc)2 · µ
= (1876 × 100 × 2π × 3 × 108 )2 × 1.240 × 10−26
= 1549kg · s−2
gE = 1 ν0 = 938cm−1 = 1.86 × 10−20 J
ZP
2
b) The first overtone in the 14 N16 O spectrum occurs at 3724 cm−1 (because the vibrations are not exactly harmonic). What are values of ν̃e , χ̃e ,
gE?
and ZP
Answer:
ν̃0→1 = ν˜e − 2χ˜e ν˜e = 1876
ν̃0→2 = 2ν˜e − 6χ˜e ν˜e = 3724
⇒ ν̃e = 3 · 1876 − 3724 = 1904cm−1
3724
χ˜e ν̃e = ·1876 −
= 14
2
⇒ χ̃e = 7.353 × 10−3
gE = 1 ν̃e − 1 χ̃e ν̃e = 948.5cm−1
ZP
2
4
14. (10 pts) Rotational spectra.
molecule
B̃e
(cm−1 )
α̃e
(cm−1 )
D̃
(cm−1 )
H19 F
20.9557
0.7980
2.15×10−3
Calculate B̃v , Re (v), and ν̃obs for the J → J + 1 rotational transition
(including centrifugal distortion and vibration-rotation interaction) of H19 F
for:
a) v = 0; J = 0. Answer:
1
B̃0 = B̃e − α̃e (0 + ) = 20.5567cm−1
2
ν̃obs = F (1) − F (0) = 2B̃v − 4D̃ = 41.11048cm−1
h
B̃v = 2
8π cI
h
⇒I=
8π 2 cB̃v
6.63 × 10−34
= 1.363 × 10−47
= 2
8 × 20.5567 × 100
8π
×
3
×
10
s
⇒ Re =
I
= 9.29 × 10−11 m = 92.9pm
µ
b) v = 2; J = 3. Answer: Same procedure as in a)
B̃v = 18.9607cm−1
ν̃obs = 151.1352cm−1
Re = 9.678 × 10−11 m = 96.78pm
15. (5 pts) Vibrational-rotational spectra.
molecule
B̃e
(cm−1 )
α̃e
(cm−1 )
ν̃
(cm−1 )
H79 Br
8.4649
0.2333
2558.54
Calculate B̃0 , B̃1 , and the first 3 lines of the R branch (ν̃R ) of the
vibrational-rotational spectrum of H79 Br (no centrifugal distortion).
Answer:
1
= 8.3483cm−1
2
2
B̃1 = B̃e − α̃e · = 8.1150cm−1
3
ν̃R = ν̃ + 2B̃1 + (3B̃1 − B̃0 )J + (B̃1 − B̃0 )J 2
= 2574.77cm−1 f orJ = 0
= 2590.53cm−1 f orJ = 1
= 2605.83cm−1 f orJ = 2
B̃0 = B̃e − α̃e ·
16. (15 pts) Electronic spectra.
gE, T̃e and ν̃0,0 in the figure below.
a) Label D̃0 , D̃e , ZP
see Figure 13.7 in your text
b) The frequencies of the first few vibronic transitions to an excited electronic state of BeO are as follows:
transition
0→2
0→3
0→4
0→5
ν̃obs / cm−1
12569.95
13648.43
14710.85
15757.50
A quadratic equation can be fit to this data with the result that
2
ν̃obs = −8.0v 0 + 1118.2v 0 + 10365
gE 0 , and ν̃0,0 ?
What are the values of ν̃e0 , χ̃0e ν̃e0 , ZP
ν̃obs = ν̃0,0 + ν̃e0 v 0 − χ̃0e ν̃e0 v 0 (v 0 + 1)
= ν̃obs = ν̃0,0 + (ν̃e0 − χ̃0e ν̃e0 )v 0 − χ̃0e ν̃e0 v 02
Equate like terms with the fit...
= ν̃obs = ν̃0,0 + (ν̃e0 − χ̃0e ν̃e0 )v 0 − χ̃0e ν̃e0 v 02 = 10365 + 1118.2v 0 − 8.0v 0
2
ν̃0,0 = 10365 cm−1
χ̃0e ν̃e0 = 8.0 cm−1
ν̃e0 = 1126.2 cm−1
1
1
ZP˜ E = (1126.2 cm−1 ) − (8.0 cm−1 ) = 561.1 cm−1
2
4
c) What is the condition that must be satisfied for an allowed transition
between two states 1 and 2, with wavefunctions Ψ1 and Ψ2 , respectively?
Non-zero transition dipole:
µψ2 →ψ1 =
Z
ψ1∗ µ̂ ψ2 dτ 6= 0
(10 pts) Extra credit.
State the 2 fundamental principles of density-functional theory.
Answer:
I. The ground-state electronic energy is uniquely determined by electron density. Thus, energy is a functional of electron density.
E0 = E[ρ]
II.Energy variational principle for wave functions applies also to the density:
E0 ≤ E[ρ̃]
where ρ̃ is a trial density.