Download Physical Chemistry II – Exam 3 Solutions

Chemistry 362
Fall 2015
Dr. Jean M. Standard
November 6, 2015
Name _____________KEY_______________
Physical Chemistry II – Exam 3 Solutions
1.) (14 points) Show that the spherical harmonic function Y11 (θ , φ ) is normalized. This
function is given by
Y11 (θ , φ )
% 3 (1/ 2
= ' €* sinθ e iφ .
& 8π )
To show that the function Y11 (θ , φ ) is normalized, we must show that
€
€
2π π
∫ ∫ Y11*(θ ,φ )Y11(θ ,φ ) sinθ dθ dφ
= 1.
0 0
As shown in the integral above, the requisite integral is actually a double integral over the
two angular coordinates, θ and φ. In addition, the angular portion of the volume element in
€
sin θ dθ dφ , must be included. And finally, when spherical
spherical polar coordinates,
harmonic functions are involved, a complex conjugate (denoted by * above) of one of the
functions must be included (which means change i to –i in the starred function if it is
present).
€
Substituting the expression for the spherical harmonic function Y11 (θ , φ ) into the
normalization integral yields
2π π
∫ ∫ Y11*(θ ,φ )Y11(θ ,φ ) sinθ dθ dφ
€
0 0
1/ 2
& 3 )1/ 2
−iφ & 3 )
sin
θ
e
⋅
sin θ e iφ sin θ dθ dφ
(
+
(
+
∫ ' 8π *
'
*
8
π
0
2π π
=
∫
0
& 3)
= ( +
' 8π *
2π π
∫ ∫ e−iφ e iφ sin3 θ dθ dφ
.
0 0
In the expression above, the normalization constants have been pulled out and written in
−iφ
front of the integral. In addition, the factor of e comes from the complex conjugate of the
€
function. The exponential term may be simplified,
−iφ iφ
0
€e e = e = 1.
2
1.) continued
Substituting, the normalization integral becomes
2π π
∫
0
& 3)
∫ Y11*(θ ,φ )Y11(θ ,φ ) sinθ dθ dφ = (' 8π +*
0
2π π
∫ ∫ sin3 θ dθ dφ
.
0 0
The next step is to break the double integral into separate parts for θ and φ,
€
2π π
∫∫
Y11*
0 0
& 3)
(θ , φ ) Y11 (θ , φ ) sin θ dθ dφ = (' +*
8π
2π
π
∫ sin
3
θ dθ
0
∫ dφ
.
0
The integral over φ may be evaluated immediately,
€
2π
∫ dφ
€
= 2π .
0
Substituting, the normalization integral becomes
2π π
∫∫
0 0
Y11*
€
& 3)
(θ , φ ) Y11 (θ , φ ) sin θ dθ dφ = (' +*
4
π
∫ sin3 θ dθ
.
0
The integral involving θ may be easily evaluated by using tables (#9 on the integral sheet),
€
π
π
& 1
)
sin θ dθ = (− cos θ sin 2 θ + 2 +
∫
' 3
*0
€ 0
& 1
)
& 1
)
= (− cos π sin 2 π + 2 + − ( − cos0 sin 2 0 + 2 +
' 3
*
' 3
*
& 1
)
& 1
)
= (− (-1)(2) + − ( − (1) (2) +
' 3
*
' 3
*
2
2
=
+
3
3
π
4
∫ sin3 θ dθ = 3 .
0
3
(
)
(
)
(
Substituting this into the normalization integral yields
€
2π π
∫ ∫ Y11*(θ ,φ )Y11(θ ,φ ) sinθ dθ dφ
0 0
& 3 )& 4 )
= ( +( +
' 4 *' 3 *
= 1.
Therefore, the spherical harmonic function Y11 (θ , φ ) is normalized.
€
€
)
3
2.) (14 points) An unknown molecule with formula XH exhibits a pure rotational spectrum
in which the J=0 → 1 transition occurs at 15.0262 cm–1 and the J=1 → 2 transition occurs
at 30.0524 cm–1. The equilibrium bond length re of XH is 1.5957 Å. From this
information determine the identity of atom X.
The spacing Δω in wavenumbers between adjacent lines in a rotational spectrum is
Δω = 2Be . Therefore,
€
Δω
30.0524 cm−1 −15.0262cm−1
=
,
2
2
= 7.5131cm−1 .
Be =
€
or Be
Once the rotational constant is calculated, the moment of inertia can be obtained from
€
(
)
6.62618 × 10 −34 Js
h
I =
=
8π 2cBe
8π 2 2.99793 × 1010 cms−1 7.5131cm−1
(
)(
)
I = 3.7259 × 10 −47 kgm2 .
2
From the definition of the moment of inertia, I = µ re , we can solve for the reduced mass µ ,
€
I
µ = 2
€
€ re
=
3.7259 × 10 −47 kgm2
(1.5957 × 10−10 m)
2
µ = 1.4633 × 10 −27 kg .
In units of amu (or g/mol),
€
$
'
1amu
µ = 1.4633 × 10 −27 kg&
)
−27
%1.660565 × 10 kg (
= 0.8812 amu .
The definition of the reduced mass for the XH molecule is
€
µ =
€
m X mH
.
m X + mH
4
2.) continued
Solving for m X ,
mX =
µ mH .
.
mH − µ
Substituting the value calculated for µ and mH = 1.0078amu ,
€
(0.8812amu) (1.0078amu)
mX =
€
€ (1.0078amu − 0.8812amu)
= 7.015 amu .
Therefore, X=Li, and the molecule is LiH.
€
5
3.) (15 points) The wavefunction of the 1s orbital for a hydrogen-like ion has the form
3/2
! 1 $1/2 ! Z $
ψ100 ( r, θ , φ ) = # & # & e−Zr/a0 .
" π % " a0 %
where a0 is a constant. Determine the average value of the potential energy for an
electron in the 1s orbital of the H atom.
€ To determine the average (or expectation) value of the potential energy, we have to use the
expression
∞ π 2π
V
=
∫∫∫
*
ψ100
(r,θ ,φ ) Vˆ ψ100 (r,θ ,φ ) r 2dr sinθ dθ dφ .
0 0 0
Here, a three-dimensional integral is required since the wavefunction is specified to depend
on the r, θ , and φ coordinates.
€
The potential energy operator Vˆ corresponds to the Coulomb interaction between the
€electron
€ and nucleus (given as part of the H atom Hamiltonian operator on the equation
sheet),
€
Ze 2
Vˆ = −
,
4 πε 0 r
with Z=1 for the H atom. Substituting the wavefunction and the potential energy operator
into the expression for the expectation value gives
€
∞ π 2π
V = ∫∫∫
0 0 0
3/2
(! $1/2 ! $3/2
+(
+
2 +(
! $1/2 ! $
*# 1 & # 1 & e−r/a0 - *− e - *# 1 & # 1 & e−r/a0 - r 2 dr sin θ dθ d φ .
*)" π % " a0 %
-, ) 4πε 0 r , *)" π % " a0 %
-,
Pulling out the constants and dividing into separate integrals for r, θ , and φ yields
V
e2
= − 2 3
4 π ε 0 a0
∫ re
−2r / a 0
dr
0
2π
∫
€
0
€
∫ sinθ dθ ∫ dφ
0
The integral over the angle φ is
€
2π
π
∞
dθ = 2π .
€
€0
.
6
3.) continued
The integral over the angle θ is
π
π
∫ sinθ dθ
€
= −cosθ 0
0
= − cos π − ( −cos0)
= 1+ 1
= 2.
Finally, the integral over the angle r is
€
∞
∫0
re−2r / a 0 dr .
From the integral table, the definite integral can be evaluated using
€
∞
∫0
n!
.
a n+1
x n e−ax dx =
Working out the integral,
€
∞
∫0
re−2r / a 0 dr =
1
(2 /a0 )
2
a02
=
.
4
Combining the three integrals, the expectation value of V may be determined,
€
V
2
V
∫ re
2π
π
∞
e2
= − 2 3
4 π ε 0 a0
−2r / a 0
dr
0
∫ sinθ dθ ∫ dφ
0
0
) a2 ,
0
= −
e
+ . (2) (2π )
4 π 2ε 0 a03 * 4 -
= −
e2
.
4 πε 0 a0
Note that this expectation value makes sense, because it corresponds to the Coulomb
potential of interaction between the nucleus and an electron at a distance a0 , which is the
€
average radius of the ground state of the H atom.
€
7
4.) (14 points) The quantized states of the hydrogen-like ion are associated with five
quantum numbers: n, ℓ , mℓ , s, and ms . Give the names of each of these quantum
numbers and describe the range of possible values for each of the quantum numbers.
For the quantum number ℓ , you should also explain its relationship to s, p, and d-type
atomic orbitals.
€
The quantum numbers of the hydrogen-like ion are:
n – principal quantum number
ℓ – orbital angular momentum quantum number
mℓ – magnetic (or azimuthal) quantum number
s – electron spin quantum number
ms – magnetic spin quantum number
The principal quantum number n takes integer values starting with 1:
€
n = 1, 2, 3, …
The orbital angular momentum quantum number ℓ takes integer values starting with 0 and
going up to n–1:
ℓ = 0, 1, 2, … , n–1.
The value of ℓ specifies the atomic orbital type as s, p, d, f, etc. The association is:
ℓ
0
1
2
3
orbital type
s
p
d
f
The magnetic quantum number mℓ ranges from – ℓ to + ℓ in steps of 1:
mℓ = 0, ±1, ±2, ±3, … , ± ℓ .
The electron spin quantum number s takes only one value:
s = 1/2 .
The magnetic spin quantum number ms ranges from –s to +s in steps of 1, which gives two
possible values:
ms = –1/2, +1/2 . €
8
5.) (15 points) The spherical harmonic function for a pz-type orbital, Y10 (θ , φ ) , is given by
% 3 (1/ 2
Y10 (θ , φ ) = ' * cosθ .
& 4π )
€
2
Show by explicit operation that this function is an eigenfunction of the Lˆ operator.
Give the eigenvalue.
€
2
We want to show that Lˆ Y10 (θ , φ ) = CY10 (θ , φ ) , where C is a constant
€ (the eigenvalue). The
2
Lˆ operator has the form
2 )
& 2
ˆL2 = − ! 2 ( ∂ + cot θ ∂ + 1 ∂ + .
2
∂θ
sin 2 θ ∂φ 2 *
' ∂θ
€
€
Substituting,
€
1/ 2
& ∂2
∂
1 ∂ 2 )& 3 )
Lˆ2 Y10 (θ , φ ) = − ! 2 ( 2 + cot θ
+
cos θ .
(
+
+
∂θ
sin 2 θ ∂φ 2 *' 4 π *
' ∂θ
Pulling the constants through and distributing,
€
1/ 2 & 2
2
)
ˆL2 Y (θ , φ ) = − ! 2 &( 3 )+ ( ∂ cos θ + cot θ ∂ cosθ + 1 ∂ cosθ + .
10
' 4 π * ' ∂θ 2
∂θ
sin 2 θ ∂φ 2
*
The individual derivatives may be evaluated,
€
∂2
cosθ = 0 .
∂φ 2
∂2
cosθ = − cosθ .
∂θ 2
∂
cosθ = − sin θ .
∂θ
Substituting,
€
€
ˆ2
L Y10 (θ , φ )
€
2&
1/ 2
2&
1/ 2
3 )
= −! ( +
' 4π *
(−cosθ
3 )
= −! ( +
' 4π *
+ cot θ ⋅ ( −sin θ ) + 0)
&
)
cosθ
⋅ ( −sin θ )+
( −cosθ +
'
*
sin θ
& 3 )1/ 2
= 2! 2 ( + cos θ
' 4π *
Lˆ2 Y10 (θ , φ ) = 2! 2Y10 (θ , φ ) .
2
2
Thus we see that Y10 (θ , φ ) is an eigenfunction of the Lˆ operator with eigenvalue 2! .
€
€
€
€
9
6.) (14 points) The nuclear spin behaves in an analogous way to all the other quantum
mechanical angular momenta, including orbital angular momentum and electron spin.
In particular, the nuclear spin has similar eigenvalue equations, quantum numbers,
commutation
relations, and measurement properties. The nuclear spin vector is
!
denoted I and has magnitude I and z-component Iz .
As a specific example, the 7Li nucleus has a nuclear spin given by I = 3/2 . For the 7Li
€nucleus, determine all the possible angles€that the quantum mechanical nuclear spin
!
vector I can make with the z-axis.
€
Since nuclear spin behaves in the same way as other angular momentm types, the angle θ
€ that the nuclear spin vector makes with the z-axis can be described by the relation,
cosθ =
mI !
Iz
=
I
2
! I(I +1)
=
mI
.
I(I +1)
€
For I = 3/2 , there are 4 possible values of mI , and therefore 4 possible angles. Substituting
I = 3/2 we have,
€
€
cosθ =
€
mI
€ , with mI = −3/2, −1/2, 1/2, and 3/2.
3 3
2 ( 2 +1)
The equation may be solved for the angle θ by taking the inverse cosine. The results are
shown in the table below.
€
mI €
€
–3/2
–1/2
1/2
3/2
€
θ
(degrees)
140.8
105.0
75.0
39.2
10
7.) (14 points) Determine the most probable radius for an electron in the 2p orbital of the
hydrogen atom. The radial wavefunction for the 2p orbital is
3/ 2
2 " Z % " Z r % −Zr / 2a 0
$
' $ ' e
.
3 $# 2a0 '& $# a0 '&
R21 (r) =
2
2
The radial distribution function for this orbital is r [ R21 ( r)] . To find the most probable
location, we need
€ to find where this function is a maximum. To find the maximum of the
radial distribution function, we take
€
d" 2
2%
# r [ R21 ( r)] & = 0 ,
'
dr $
and solve for r. The radial distribution function for the 2p orbital is
€
2
r [ R21 ( r)]
2
3
2
4 " Z % " Zr % −Zr / a 0
= r $
'$ ' e
3 # 2a0 & # a0 &
2
=
4 " Z 5 % 4 −Zr / a 0
.
$
'r e
3 # 8a05 &
Setting Z=1 (for the hydrogen atom) and taking the derivative leads to the expression
€
d" 2
2%
4 ( 1 +/
1 4 −r / a 0 2
#r [ R21 ( r)] & = * 5 - 14r 3 e−r / a 0 −
r e
4 .
'
dr $
3 ) 8a0 , 0
a0
3
Collecting terms yields
€
( 1 +(
d" 2
2%
r + 3 −r / a 0
# r [ R21 ( r)] & = * 5 - * 4 −
.
-re
'
dr $
a0 ,
) 6a0 , )
We now have to set the derivative equal to zero and solve for r to get the most probable
value. Thus, we have
€
" 1 %"
r % 3 −r / a 0
= 0.
$ 5 '$ 4 −
're
a0 &
# 6a0 & #
€
11
7.) continued
Dividing both sides of this equation by
1
e−r / a 0 (which is OK as long as r is not infinity)
6a05
leads to the equation
€
#
r& 3
4
−
%
(r = 0.
a0 '
$
This equation has a root at r=0 (which is a minimum, not a maximum) and at the value of r
that satisfies the equation in parentheses,
€
r
= 0,
a0
or r = 4a0 .
4 −
€