Multiple-Choice Problems
021 qmult 00100 1 1 1 easy memory: molecule defined
1. A stable bound system of electrons and more than one nucleus with some specific recipe for the numbers of each kind of nuclei is:
a) a molecule. b) an atom.
c) a nucleon. d) a fullerene.
e) a baryon.
021 qmult 00100 1 4 1 easy deducto-memory: atoms bound in molecules
2. “Let’s play Jeopardy! For $100, the answer is: Because they are formed by bonding atoms and dissociate into atoms?”
a) What is one good reason for thinking of molecules as bound atoms, Alex? b) What are three bad reasons for thinking of molecules as bound atoms, Alex?
c) What is no reason at all for thinking of molecules as bound atoms? d) What is ambiguous answer, Alex?
e) What is am-Piguous answer, Alex?
021 qmult 00300 2 1 2 easy memory: molecular energy scales
3. Given electron mass m and typical nuclei mass M , the ratio of electronic, vibrational, and rotational energies for a molecule is of order:
a) 1 : 1 : 1.
b) 1 : (m/M )1/2: (m/M ).
c) 1 : (m/M ) : (m/M ). d) 1 : (m/M )1/4: (m/M )1/2.
e) 1 : (m/M )1/4: (m/M )1/3.
021 qmult 00400 1 4 4 easy deducto-memory: Born-Oppenheimer approx. Extra keywords: (Ba-473)
4. “Let’s play Jeopardy! For $100, the answer is: This approximation allows one to treat the nuclei in atoms as though they interacted only with an effective potential constructed from actual potentials and the electronic kinetic energy (i.e., the total electronic energy).”
a) What is the Alpher-Behte-Gamow recipe, Alex?
b) What is the Einstein-Woody-Allen approximation, Alex?
c) What is the linear combination of atomic orbitals method, Alex? d) What is the Born-Oppenheimer approximation, Alex?
e) What is the tight-binding approximation, Alex?
021 qmult 00500 1 4 5 easy deducto-memory: tight-binding theory 131
132 Chapt. 21 Molecules
5. “Let’s play Jeopardy! For $100, the answer is: It posits that overlapping wave functions of bound atoms can be treated to some degree in terms of the orbitals of isolated atoms.”
a) What is the genetic algorithm method, Alex? b) What is the linear variational method, Alex?
c) What is tight-binder theory, Alex? d) What is tight-wadding theory, Alex?
e) What is tight-binding theory, Alex? 021 qmult 00600 1 1 3 easy memory: LCAO
6. In LCAO (linear combinations of atomic orbitals method) one uses atomic orbitals as a non- orthonormalized basis set for constructing:
a) bound states of nucleons. b) bound states of photons.
c) inter-atomic bonding and anti-bonding states. d) intra-atomic stationary states.
e) property-law violating beach-front homes in California.
Full-Answer Problems
021 qfull 00100 2 5 0 moderate thinking: universal sp-coupling parameters Extra keywords: See Ha-95
1. Harrison (Ha-95) presents “universal” sp-bond matrix elements or coupling pararameters:
hs1|H|s2i = Vssσ = − π2 8 h −2 md2 , hpz1|H|pz2i = Vppσ= 3π2 8 h −2 md2 , hs1|H|pz2i = Vspσ= π 2 h −2 md2 , and hpx1|H|px2i = hpy1|H|py2i = Vppπ= −π 2 8 h −2 md2 = Vssσ ,
where 1 and 2 denote two atoms aligned along the z-axis, H is the single-particle Hamiltonian owing to the cores of the two atoms, the |s1i, etc., are single-particle atomic orbitals (radial
parts of some sort times the cubical harmonics for the angular parts) oriented relative to a common set of axes, d is the inter-nuclei separation, and Vspσ has a π, not π2. The s and p
subscripts on the V ’s indicate the atomic orbitals in the coupling and the σ and π indicate the indicates the quantum numbers m2 of L2
z operator of the molecular orbitals that result from
the coupling of the different states: σ is for m2= 0 and π is for m2= 1.
The reason for the complication of using the eigenvalues of the L2
z operator rather than the Lz
operator is that the |pxi and |pyi cubical harmonics are eigenstates of L2z, but not of Lz. Recall
the lowest quantum number spherical harmonics Yℓ,m are
Y0,0= 1 √ 4π , Y1,0= r 3 4πcos(θ) , and Y1,±1= r 3 8πsin(θ)e ±imφ ,
Chapt. 21 Molecules 133 where ℓ is the L2quantum number, m is the L
zquantum number, θ is the angle from the z-axis,
and φ is the azimuthal angle. The cubical harmonics are defined by |si = Y0,0= √1 4π , |pxi = Y1,1+ Y1,−1 √ 2 = r 3 4π x r , |pyi = Y1,1− Y1,−1 i√2 = r 3 4π y r , and |pzi = Y1,0=r 3 4π z r .
a) Verify that polar plots of the “p” cubical harmonics are a touching pair of spheres of radius p3/(4π)/2. (I mean, of course, when you consider the plots as spherical polar coordinate plots.) For |pzi, for example, the spheres touch at the z-axis origin and are aligned with
the z-axis: the upper sphere is “positive” and the lower sphere is “negative”: i.e., the radial position from the origin comes out a negative number: one just plots the magnitude. HINTS: It is sufficient to do the proof for |pzi, since the others are the same mutatis
mutandis. A diagram would help.
b) Interpret the physical significance of the polar plots of the cubical harmonics.
c) Show that |pxi and |pyi are eigenstates of L2z, but not Lz. What other angular momentum
operators are they eigenstates of? HINT: Recall
Lz= h − i ∂ ∂φ .
d) Now we come to the question yours truly wanted to ask before chronic digression set in. Write sp-bond coupling parameters in terms of fiducial values in units eV-˚A2: e.g.,
C d2 A◦
,
where C is a numerical constant (i.e., an actual value) in eV-˚A2 and d
A◦ is mean nuclei
separation in Angstroms. Then evaluate the constants for dA◦ = 3. HINT: Recall
h −2
m = 7.62 eV-˚A
2 ,
where m is the electron mass.
021 qfull 00300 2 3 0 easy math: Li2 with spin Extra keywords: Reference Ha-72
2. Let us consider the single-particle bonding and antibonding states and their energies for Li2.
We assume that the single-particle Hamiltonian of the Li2 molecule is
H = −h− 2 2m ∇ 2+ V (~r ) = −h− 2 2m ∇ 2 + V (~r − ~r1) + V (~r − ~r2) .
where ~r is measured from the midpoint between the nuclei, ~r1 is the position of nucleus 1,
134 Chapt. 21 Molecules
crudest Born-Oppenheimer approximation. The observed equilibrium separation of the nuclei is d = 2.67 ˚A: we take this to be the fixed separation. We will make the LCAO (linear combination of atomic orbitals) approximation.
a) Use the linear variational method to calculate the the bonding and antibonding states and their energies. The basis states are the two atomic orbital 2s states of the atoms: call them |1i and |2i. We assume |1i and |2i are knowns: they have the same energy εs= −5.34 eV.
Assume the basis states are orthogonal: a poor approximation actually, but this problem is intended to be heuristic. Make a reasonable approximation to evaluate the diagonal matrix elements. For off-diagonal or coupling matrix elements use
Vssσ= − π2 8 h −2 md2 = − 9.40 eV-˚A d2 A◦ .
this is one of Harrison’s “universal” sp-bond couplings (Ha-95). Which state is the bonding state and which the antibonding state and why are they so called?
b) What is the component of orbital angular momentum of single-particle states about the inter-nuclear axis? How do you know this? What symbol represents this orbital angular momentum component value for molecules? HINT: Recall that the z-component orbital angular momentum operator is
Lz= h
− i
∂ ∂φ ,
where φ is the azimuthal angle about the z-axis (MEL-23).
c) Now construct six plausible symmetrized two-particle states including spinor from our bonding and antibonding position states: a ground state and the five excited states. So everyone is on the same wavelength let
α = 1 0 and β = 0 1 .
What are the approximate energies of these states? Can we construct any more states from the bonding and antibonding states? We, of course, are assuming that there is no spin operator in the Hamiltonian. NOTE: These states may not be very realistic: this is just an exercise.
021 qfull 03000 2 5 0 moderate thinking: molecular relative coordinates Extra keywords: A misconcieved problem.
3. One usually wishes to separate the center of mass and relative parts of the nuclei part of a molecular wave function. For two nuclei, the situation is a two-body problem and can be treated like hydrogenic systems (Da-334). For the general multiple nuclei case, a different approach is needed that treats all nuclei on the same footing. Let ~ri be nucleus i’s position relative to
an external inertial frame. Let ~r′
i be nucleus i’s position relative to ~R the molecular center of
mass. We make the approximation that the electrons can be neglected in evaluating the center of mass. NOTE: I was fooled into thinking there was neat way of doing this. The cross term doesn’t vanish. But one probably has to construct independent coordinates in a special way for each kind of molecule. But maybe something is salvageable so I’ll leave this around for now.
a) Express ~R and ~r′
i in terms of the positions ~ri.
b) Now express the operators ∂/∂xi and ∂2/∂x2i in terms of operators ∂/∂x′i and ∂/∂X.
For simplicity we only consider the x-components of the various coordinates. The y- and z-components are handled similarly.
Chapt. 21 Molecules 135 c) Now show that
H =X i p2 i 2Mi+ V ({ri}) = p2 cm 2Mtot +X i p′2 i 2M′ i + V ({ri′}) ,
where H is the Hamiltonian of the nuclei in the effective potential V ({ri}) (the curly
brackets mean “set of”) with no external potential present, p′
i are the relative coordinate
momentum operators, and
M′
i =
Mi
[1 − (Mi/Mtot)]2