Homework 8 Answer Key
November 5, 2014
1
Q 10.7
Explain why shielding is more effective by electrons in a shell of lower principal quantum
number than by electrons having the same principal quantum number.
Electrons in a shell of lower principal quantum number have a higher probability of being found
between the nucleus and the electrons having the same principal quantum number. Therefore, they are
more effective in shielding.
2
Q 10.16
Show that the Slater determinant formalism automatically incorporates the Pauli exclusion
principle by evaluating the He ground-state wave function of Equation (21.9), giving both
electrons the same quantum numbers.
1
1 1s(1)α(1) 1s(1)α(1)
= √ (1s(1)α(1)1s(2)α(2) − 1s(1)α(1)1s(2)α(2))
ψ(1, 2) = √ 1s(2)α(2)
1s(2)α(2)
2
2
ψ(1, 2) = 0
It is seen that the wavefunction is zero if two electrons have the same quantum numbers.
3
P 10.3
In this problem we represent the spin eigenfunctions and operators as vectors and matrices.
a.
The spin eigenfunctions are often represented as the column vectors
1
0
α=
and β =
0
1
Show that α and β are orthogonal using this representation.
1
0 1 = 1 × 0 + 0 × 1 = 0. Therefore α and β
Using the rules of matrix multiplication, αβ =
0
are orthogonal.
b.
If the spin angular momentum operators are represented by the matrices
h̄ 0 1
h̄ 0 −i
h̄ 1 0
ŝx =
, ŝy =
, ŝz =
2 1 0
2 i 0
2 0 −1
Show that the commutation rule [ŝx , ŝy ] = ih̄ŝz holds.
2 h̄
0
ŝx ŝy − ŝy ŝx =
1
2
2 h̄
i
=
0
2
1
0 −i
0 −i
0 1
−
0
i 0
i 0
1 0
2 h̄
h̄ 1
0
−i 0
1 0
−
= 2i
= ih̄
−i
0 i
0 −1
2
2 0
1
0
= ih̄ŝz
−1
c.
Show that
ŝ2 = ŝ2x + ŝ2y + ŝ2z =
h̄2
4
3
0
0
3
h̄2 0 1
h̄2 0 −i
h̄2 1
0 1
0 −i
+
+
1 0
i 0
4 1 0
4 i 0
4 0
2 2 2 h̄ 1 0
h̄ 1 0
h̄ 3 0
0
+
+
=
1
4 0 1
4 0 1
4 0 3
ŝ2 = ŝ2x + ŝ2y + ŝ2z =
h̄2
=
4
d.
1
0
0
−1
1
0
0
−1
Show that α and β are eigenfunctions of ŝz and ŝ2 . What are the eigenvalues?
h̄ 1 0
h̄ 0
h̄ 1
1 0
0
1
ŝz α =
and ŝz β =
=−
=
0 −1
1
0
2 0
2 0 −1
2 1
The eigenvalues for α and β are
ŝ2 α =
h̄2
4
3
0
0
3
h̄
2
and − h̄2 , respectively.
3h̄2 1
h̄2 3
1
and ŝ2 β =
=
0
0
4
4 0
0
3
3h̄2 0
0
=
1
1
4
2
The eigenvalues are 3h̄4 in each case.
e.
Show that α and β are not eigenfunctions of ŝx and ŝy .
h̄ 0
h̄
h̄ 0 1
h̄ 1
h̄
h̄ 0 1
1
0
=
= β 6= cα,
ŝx β =
=
= α 6= cβ,
ŝx α =
0
1
2 1 0
2 1
2
2 1 0
2 0
2
h̄ 0 −i
h̄ 0
ih̄
h̄ 0 −i
h̄ −i
ih̄
1
0
ŝy α =
=
= β 6= cα, ŝy β =
=
= − α 6= cβ
i
0
0
i
i
0
1
0
2
2
2
2
2
2
4
P 10.15
Why is the magnitude of the electron affinity for a given element smaller than the magnitude of the first ionization energy?
Within Koopmans’ approximation in which it is assumed that the electron distribution in an atom
is unaffected by the addition or removal of an electron, the first ionization energy is the negative of the
energy of the highest occupied atomic orbital and the electron affinity is the negative of the energy of
the lowest unoccupied AO. For atoms with filled subshells, the affinity level is higher in the potential
than the ionization level and consequently the electron affinity is less than the ionization energy.
For atoms with partially filled subshells, this reasoning is inadequate. In this case, it is best to
think of the ionization energy as the difference in the total energy of the atom and the resulting cation
and the electron affinity as the difference in the total energy of the atom and the resulting anion. For
a fixed nuclear charge, the total energy increases and the effective nuclear charge decreases as the number of electrons increases because of electron-electron repulsion and decreased effective nuclear charge.
Therefore, the magnitude of the electron affinity is less than that of the ionization energy.
5
P 10.5
In this problem you will show that the charge density of the filled n = 2, l = 1 subshell is
spherically symmetrical and that therefore L= 0. The angular distribution of the electron
charge is simply the sum of the squares of the magnitude of the angular part of the wave
functions for l = 1 and ml = −1, 0, and 1.
2
a.
Given that the angular part of these wave functions is
Y10 (θ, φ)
1/2
3
8π
1/2
3
8π
1/2
=
Y11 (θ, φ) =
Y1−1 (θ, φ)
3
4π
=
cos θ
sin θeiφ
sin θe−iφ
2 2 2
Write an expression for Y10 (θ, φ) +Y11 (θ, φ) +Y1−1 (θ, φ) .
0
Y1 (θ, φ)2 + Y11 (θ, φ)2 + Y −1 (θ, φ)2 = 3 cos2 θ + 3 sin2 θ + 3 sin2 θ
1
4π
8π
8π
b.
2 2 2
Show that Y10 (θ, φ) +Y11 (θ, φ) +Y1−1 (θ, φ) does not depend on θ and φ.
3
3
3
3
3
2
2
cos θ +
sin θ +
sin2 θ =
(cos2 θ + sin2 θ) =
4π
8π
8π
4π
4π
This is not a function of θ and φ.
c.
Why does this result show that the charge density for the filled n = 2, l = 1 subshell is spherically symmetrical?
If a function is independent of θ and φ, then if has the same value for all θ and φ. This is what we
mean by being spherically symmetrical.
6
Q 11.1
Justify the statement that the Coulomb integral J defined in Equation (22.20) is positive
by explicitly formulating the integral that describes the interaction between two negative
classical charge clouds.
Z Z
e2
1
2
J12 =
[1s(1)]
[2s(2)]2 dτ1 dτ2
−
−
8π0
|→
r1−→
r 2|
6 Z Z
2
1
e2
1
1
r2
−r1 /a0
=√
e
2
−
e−r2 /2a0 dτ1 dτ2
−
−
a0
|→
r1−→
r 2|
32 8π 3 0 a0
The angular coordinates integrate to give a factor 16π 2 , leaving only the integration over r1 and r2 ,
1
, e−r1 /a0 and e−r2 /2a0 are positive as r1 and r2
which can vary between 0 and ∞. The terms →
−
−
r 1 −→
r2
vary over their allowed ranges. The only other factor involving r1 or r2 appears as the square, and is
therefore always positive. Therefore, the integrand is always positive, and we can conclude that J12 must
be positive.
7
Q 11.4
Why is an electronically excited atom more reactive than the same ground-state atom?
It is more reactive because it contains excess energy in the form of electronic energy that can be used
to overcome an activation barrier.
8
P 11.11
How many ways are there to place three electrons into an f subshell? What is the groundstate term for the f 3 configuration, and how many states are associated with this term?
3
See problem P11.36
The first electrons can have any combination of seven ml and 2 ms valuse so that n = 14 and m = 3.
14!
= 364.
The number of states is 3!(14−3)!
Using the method discussed in Example Problem 22.4, MLmax = 6 and MSmax = 3/2. Therefore the
ground-state term is 4 I, which has (2L + 1)(2S + 1) = (2 × 6 + 1)(4) = 52 states.
9
P 11.19
Derive the ground-state term symbols for the following configurations:
a.
d5
b. f 3
c.
p4
(a) MLmax = 0 and MSmax = 5/2. Ground-state term is 6 S.
(b) MLmax = 6 and MSmax = 3/2. Ground-state term is 4 I.
(c) MLmax = 1 and MSmax = 1. Ground-state term is 3 P.
4