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Chemistry 362
Dr. Jean M. Standard
Problem Set 10 Solutions
1.
!
Determine the possible angles that the quantum mechanical electron spin angular momentum vector S
can make with the z-axis.
The angle θ that the spin angular momentum vector makes with the z-axis is described by the relation
€
€
For an electron, we have s =
1
2
2
! s(s + 1)
.
and ms = ± 12 . Substituting, the equation becomes
€
€
ms!
Sz
=
S
cos θ =
cos θ =
€
± 12 !
! 2 12 ( 12 + 1)
or cos θ = ±
€
,
1
.
3
!
!
The two angles that we get are θ = 125.3 and θ = 54.7 .
€
€
2.
€
2
Determine the result of measuring S and Sz for an electron with a wavefunction corresponding to the
"spin up" state.
The wavefunction for the "spin
€ up"€state of an electron is denoted α . The eigenvalue equations for the "spin
up" state are
3 2
Sˆ 2α = ! 2 s( s + 1€) α =
! α
4
1
Sˆzα = mℓ ! α =
!α .
2
2
2
Since the "spin up" wavefunction is an eigenfunction of the operators Sˆ and Sˆz , measurement of S and Sz
€
3
1
2
will yield the eigenvalues, with the result ! 2 for S and ! for Sz .
4
2
€
€
€
€
€
€
€
€
2
3.
Give the explicit form of the Hamiltonian operator (in atomic units) for the helium atom. Your
expression should not include any summations (expand them out).
The general form of the Hamiltonian operator (in atomic units) for an atom is
1
Hˆ = −
2
N
N
ˆ2
∇
i
∑
−
i=1
∑
i=1
Z
+
ri
N
N
∑∑
i=1 j =i+1
1
,
rij
where the "del squared" operator is defined as
€
2
2
2
ˆ2 = ∂ + ∂ + ∂ .
∇
i
2
2
∂x i ∂y i ∂z i2
For He, there are two electrons (N=2) and the atomic number Z=2, so the operator becomes
€
1
Hˆ = −
2
2
∑
2
ˆ2
∇
i
i=1
−
∑
i=1
2
+
ri
2
2
∑∑
i=1 j =i+1
1
.
rij
Writing out each term in the operator explicitly, we have
€
1 ˆ2
Hˆ = − ∇
1
2
−
1 ˆ2
2
2
1
∇2 −
−
+
.
2
r1
r2
r12
The first two terms are the kinetic energy operators for electrons 1 and 2, while the third and fourth terms are
the electron-nuclear attractions,
and the last term is the electron-electron repulsion.
€
3
4.
Give the explicit form of the Hamiltonian operator (in atomic units) for the beryllium atom. Your
expression should not include any summations. Classify the various terms that are included in the
Hamiltonian operator.
The general form of the Hamiltonian operator (in atomic units) for an atom is
1
Hˆ = −
2
N
N
ˆ2
∇
i
∑
−
i=1
∑
i=1
Z
+
ri
N
N
∑∑
i=1 j =i+1
1
,
rij
where the "del squared" operator is defined as
€
2
2
2
ˆ2 = ∂ + ∂ + ∂ .
∇
i
2
2
∂x i ∂y i ∂z i2
For Be, there are four electrons (N=4) and the atomic number Z=4, so the operator becomes
€
1
Hˆ = −
2
4
∑
i=1
4
ˆ2
∇
i
−
∑
i=1
4
+
ri
4
4
∑∑
i=1 j =i+1
1
.
rij
Writing out each term in the operator explicitly, we have
€
1 ˆ2
1 ˆ2
1 ˆ2
1 ˆ2
4
4
4
4
ˆ
H = − ∇
∇2 −
∇3 −
∇4 −
−
−
−
1 −
2
2
2
2
r1
r2
r3
r4
1
1
1
1
1
1
+
+
+
+
+
+
.
r12
r13
r14
r23
r24
r34
The first four terms are the kinetic energy operators for electrons 1, 2, 3, and 4, while terms 5-8 are the electronnuclear attractions,
and the last six terms (on the second line) correspond to the electron-electron repulsions.
€
4
5.
2
1
For a Li atom with electron configuration 1s 2s , list the possible sets of quantum numbers for electrons
1, 2, and 3. Make sure to include all five quantum numbers, n, ℓ , mℓ , s, and ms for each electron. How
many different sets of quantum numbers are possible?
Electrons 1 and 2 occupy the same 1s orbital, so we must €
apply
€ the Pauli€Principle to these electrons. In
particular, we must make sure that they have different sets of quantum numbers. The table below lists the
possible quantum numbers for the two 1s electrons.
Quantum
number
n
ℓ
mℓ
s
ms
Electron
1
1
0
0
1/2
1/2
Electron
2
1
0
0
1/2
–1/2
€
€
Note that all the quantum numbers are required to be the same (because we are specifying that the particles are
electrons and they are
€ in the 1s orbital) except the ms quantum numbers, which have to be different for the two
electrons in order to satisfy the Pauli Principle.
Electron 3 is the only electron in the 2s orbital,
so we do not have to worry about the Pauli Principle for that
€
electron. There are two possibilities for the quantum numbers of electron 3:
Quantum
number
n
ℓ
mℓ
s
ms
Option
1
2
0
0
1/2
1/2
Option
2
2
0
0
1/2
–1/2
€
€
This electron can be either spin up or spin down. Therefore, there are two possible sets of quantum numbers for
the electrons in Li. €
These are listed in the table below. The two sets correspond to the same choice for the two
1s electrons (the only option) and either option 1 or option 2 for the 2s electron.
Quantum
number
n
ℓ
mℓ
€
€
€
s
ms
Electron
1
1
0
0
1/2
1/2
Set 1
Electron
2
1
0
0
1/2
–1/2
Electron
3
2
0
0
1/2
1/2
5
5.
Continued
Quantum
number
n
ℓ
mℓ
€
€
€
s
ms
Electron
1
1
0
0
1/2
1/2
Set 2
Electron
2
1
0
0
1/2
–1/2
Electron
3
2
0
0
1/2
–1/2