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University of Washington
Department of Chemistry
Chemistry 453
Winter Quarter 2012
Homework Assignment 2
Due at 5 p.m. on 1/18/12 at the Catalyst drop box.
Recommended Problems: 12.12, 12.15, 14.7, 14.12
12.12)
−1
Etotal (T ) ∞ 8π hν 3 hν / kT
h
e
dν
=∫
− 1) dν ⇒ x = hν / kT ⇒ dx =
(
3
V
c
kT
0
4
−1
Etotal (T ) ⎛ kT ⎞ 8π h ∞ 3 x
⎛ kT ⎞ 8π h π
∴
=⎜
⎟ 3 ∫ x ( e − 1) dx = ⎜
⎟ 3
V
⎝ h ⎠ c 0
⎝ h ⎠ c 15
12.15)
4
14.7)
4
14.12)
This homework is worth a total of 10 points.
1) Einstein’s crystal refers to a lattice of atoms each of which oscillates in 3 dimensions.
U
3hν
Planck’s theory says the average energy per atomic oscillator is E =
= hν / kBT
,
NA e
−1
where the factor of three accounts for the fact that each atom vibrates in the x, y, and z
directions.
⎛ ∂U ⎞
a) The heat capacity is defined as: CV = ⎜
⎟ . Obtain an expression for the heat
⎝ ∂T ⎠V
capacity CV of Einstein’s crystal using Planck’s theory. Hint: You will need to use
the chain rule.
Solution
−1
3 N hν
∂ hν / kBT
⎛ ∂U ⎞
U = hν / kBAT
e
⇒ CV = ⎜
− 1)
(
⎟ = 3N A hν
e
−1
∂T
⎝ ∂T ⎠V
: = −3 N A hν ( e hν / kBT − 1)
−2
−2 ⎛
∂ hν / kBT
hν hν / kBT ⎞
= −3 N A hν ( e hν / kBT − 1) ⎜ −
e
e
(
)
⎟
2
∂T
⎝ k BT
⎠
2
−2
⎛ hν ⎞ hν / kBT hν / kBT
= 3N AkB ⎜
− 1)
e
(
⎟ e
⎝ k BT ⎠
b) The Law of DuLong and Petit states that at very high temperature, the heat
capacity of an atomic crystal approaches, 3R, i.e. CV=3R. Using your result from
part a for the heat capacity of Einstein’s crystal, prove the Law of Dulong and
Petit. Hint: Evaluate the heat capacity expression in the limit that kBT>>hν. Use
the fact that e x ≈ 1 + x for x<<1.
Solution:
2
2
−2
⎛ hν ⎞ hν / kBT hν / kBT
⎛ hν ⎞ ⎛
hν ⎞ ⎛ hν ⎞
− 1) ≈ 3N A k B ⎜
CV = 3N A k B ⎜
e
(
⎟ e
⎟ ⎜1 +
⎟⎜
⎟
⎝ k BT ⎠
⎝ k B T ⎠ ⎝ k B T ⎠ ⎝ k BT ⎠
⎛
hν ⎞
= 3N Ak B ⎜1 +
⎟ ≈ 3 N A k B = 3R
⎝ k BT ⎠
−2
c) What value does the heat capacity of Einstein’s crystal approach as T approaches
0K? Explain any assumption you make in getting your answer.
Solution: For hν >> k BT … e hν / kBT − 1 ≈ e hν / kBT . Then
2
2
−2
⎛ hν ⎞ hν / kBT hν / kBT
⎛ hν ⎞ hν / kBT −2 hν / kBT
− 1) ≈ 3 N A k B ⎜
CV = 3N A k B ⎜
e
e
(
⎟ e
⎟ e
⎝ k BT ⎠
⎝ k BT ⎠
2
⎛ hν ⎞ − hν / kBT
= 3N Ak B ⎜
≈0
⎟ e
⎝ k BT ⎠
2
⎛ hν ⎞
…because e
decays faster than ⎜
⎟ grows as T approaches 0. You can prove
⎝ k BT ⎠
this using L’Hospital’s rule.
− hν / k BT
2)
This problem explores when you do and do not have to quantize motions.
1) Assume argon atoms translate in one dimension.
a) What is the AVERAGE translational kinetic energy per argon atom? Assume
T=1000K
Solution: The average energy per argon atom for translatino in one dimension
−23
−1
k T (1.38 ×10 JK ) (1000 K )
is E = B =
= 6.9 ×10−21 J
2
2
b) Assume you can model the energy of translation of an argon atom using the
particle in a one dimensional box model. Assume the box is 1 m in length. For
what value of the quantum number n is the particle in the box energy equal to the
average energy calculated in part a? FYI: the atomic weight of argon is
0.018kgmol-1.
Solution: m =
0.018kgmol −1
= 3 ×10−26 kg
23
−1
6.02 × 10 mol
n 2 h 2 k BT
=
= 6.9 × 10−21 J
8ma 2
2
n=
=
8ma 2 ( 6.9 × 10−21 J )
h
2
=
2a
2m × 6.9 × 10−21
h
2m
6 × 10−26 kg × 6.9 × 10−21 J
−34
6.62 ×10 Js
= ( 3.02 × 1033 mJ −1s −1 )( 41.4 × 10−47 kg 2 m 2 s −2 )
1/2
= ( 3.02 × 1033 kg −1m −1s1 )( 2.03 ×1023 kgms −1 ) ≈ 6.13 × 1010
c) Based on your answers in parts a and b, how important are quantum effects in the
translation of argon at T=1000K?
Answer: There are 61.3 billion energy levels between the ground state energy (n=1)
and the average energy. So the energy level separation ∆E is very much smaller than
kBT and quantum effects are unimportant.
d) Suppose the box within which argon translations is 0.01 nm in length. Calculate
the n for the particle in the box energy that is equal to kBT/2.
2a
n=
2m × 6.9 × 10−21
h
2 ×10−11 m
=
6 × 10−26 kg × 6.9 ×10−21 J
6.62 ×10−34 Js
= ( 3.02 × 1022 mJ −1s −1 )( 41.4 ×10−47 kg 2 m 2 s −2 )
1/2
= ( 3.02 × 1022 kg −1m −1s1 )( 2.03 × 10−23 kgms −1 ) ≈ .613
e) Based on your answer in part d, how do quantum effects vary with the size of the
box a? Do quantum effect increase or decrease with box size?
Answer: As the box size decreases and approaches atomic dimensions, the energies
and energy level splittings for particle in the box increase and approach kBT. When
∆E approaches kBT, quantum effects become important.
f) Based on your answers to parts a-e, discuss why the vibrational heat capacity is
almost zero at T=100K, but the translational heat capacity is 3R/2. Hint: atomic
vibrations have amplitudes of about 0.01 Angstroms.
Answer
For vibrations the amplitude of motion is so small that ∆E>>kBT at low T and heat is
not absorbed. So the heat capacity for vibration is zero. For translations the amplitude
of motion is large…so ∆E<<kBT, heat is absorbed at low T and the heat capacity is
3R/2.
3) β-carotene, a precursor of retinal, a visual pigment found in the retina of the eye,
has the formula given below, right . β-carotene contains 11 conjugated double bonds
which contribute 22 π
electrons that are
delocalized along the length of the molecular chain. Suppose we can model the
energies of these electrons with the particle-in-the box model.
a) Assuming each particle-in-the-box energy level is occupied by at most two
electrons, what is the quantum number n associated with the highest
occupied energy state?
Solution: 22 electrons occupy 11 energy levels so n=11 is the highest
occupied state.
b) When the electrons in the highest occupied state adsorb energy they make
a transition to the lowest unoccupied energy state. Calculate the energy
change ∆E that occurs when the electrons make a transition in β-carotene
from the highest occupied energy state to the lowest unoccupied energy
state. If β-carotene absorbs radiation with a wavelength of λ=480 nm,
calculate the energy change ∆E
Solution:
∆E = hν =
hc
λ
( 6.623 ×10
=
−34
Js )( 3 × 108 ms −1 )
−9
480 × 10 m
= 4.14 ×10−19 J
c) Using you result from parts a and b, calculate the length of b-carotene in
units of nm.
6.623 × 10−34 Js ) ( 23)
(
h2
−19
2
2
∆E = 4.14 × 10 J =
(12 − 11 ) = 8 9.1×10−31 kg L2
8mL2
( )(
)
2
∴L =
2
( 6.623 ×10
(8) ( 9.1×10
−31
−34
Js ) ( 23)
2
kg )( 4.14 × 10
−19
J)
= 3.35 ×10−18 m 2
L = 3.35 × 10−18 m 2 = 1.83 ×10−9 m = 1.83nm
c) Using you results from parts a-c, calculate the average momentum p
and the average momentum squared p 2 for an electron in the highest
occupied energy state of β-carotene.
Solution:
n=11. So…
2
2 nπ
⎛ nπ x ⎞ ⎛ d ⎞ ⎛ nπ x ⎞
⎛ nπ x ⎞
⎛ nπ x ⎞
sin ⎜
sin ⎜
⎟⎜
⎟ sin ⎜
⎟ dx =
⎟ cos ⎜
⎟ dx
∫
∫
L 0 ⎝ L ⎠ ⎝ i dx ⎠ ⎝ L ⎠
Li L 0
⎝ L ⎠
⎝ L ⎠
L
p =
L
L
2 nπ L
⎛ nπ x ⎞
2
2
sin 2 ⎜
=
⎟ = ( sin nπ − sin 0 ) = 0
L i L 2nπ
L
iL
⎝
⎠0
−34
2
n 2 h 2 n 2 h 2 11 ( 6.623 × 10 Js )
= 2mEn = 2m
=
=
= 3.96 ×10−48 kg 2 m 2 s −2
2
2
2
−
9
8mL
4L
( 4 ) (1.83 ×10 m )
2
p
4)
2
A particle in a two dimensional square box has energies E ( nx , n y ) =
and wave functions ψ ( x, y ) =
h2
n 2 + n y2 )
2 ( x
8mL
2
⎛ n π x ⎞ ⎛ n yπ y ⎞
sin ⎜ x ⎟ sin ⎜
⎟ . Porphyrin rings occur, for
L
⎝ L ⎠ ⎝ L ⎠
example, in myoglobin and chlorophyll. The 26 π electrons of a porphyrin ring can be
treated as particles in a thin square box. Assume the sides of the porphyrin “box” have
side of length L=1nm.
a) Order the 13 occupied energy levels from the lowest energy to the highest energy.
Designate each energy level by its quantum numbers (nx,ny). Note any
degeneracies.
Solution:
(1,1)<(1,2)=(2,1)<(2,2)<(3,1)=(1,3)<(3,2)=(2,3)<(4,1)=(1,4)<(3,3)<(4,2)=(2,4)
b) Calculate the energy change De that results when a porphyrin absorbs a photon
and promotes and electron from the highest occupied energy state to the lowest
unoccupied energy state.
Solution:
The highest occupied energy level is (4,2)=(2,4). The electron is promoted to (4,3)=(3,4).
6.623 ×10−34 Js )
(
h2
2
2
2
2
n + n y , f − nx , i − n y , i ) =
∆E =
42 + 32 − 42 − 22 ) = 3.01× 10−19 J
2 (
2 ( x, f
−
−
31
9
8mL
(8) ( 9.1×10 kg )(10 m )
2
c) Calculate the frequency and wavelength of radiation absorbed by this porphyrin
ring. Give your answer for wavelength in units of nm.
Solution:
∆E
3.01×10−19 J
∆E = hν ⇒ ν =
=
= 4.54 × 1014 s −1
−34
h
6.623 × 10 Js
8
−1
c
3 ×10 ms
= 661× 10−9 m = 661nm
λ= =
ν 4.54 ×1014 s −1