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Unformatted text preview: Chapter 16: The Particle in the Box and the Real World Problem numbers in italics indicate that the solution is included in the Student’s
Solutions Manual Questions on Concepts Q16.1) Why is it necessary to apply a bias voltage between the tip and surface in a scanning tunneling
microscope? The highest occupied energy levels in the tip and sample have the same energy without a bias voltage.
Because tunneling occurs at constant energy, there are no empty levels for the electron to tunnel into.
By applying a bias voltage of the appropriate polarity, the highest ﬁlled states in the tip and sample are
offset, and tunneling can occur from the element that has the higher energy ﬁlled level. Q16.2) The amplitude of the wave on the right side of the barrier in Figure 16.8 is much smaller than
that of the wave incident on the barrier. What happened to the “rest of the wave”? The “rest of the wave” was reﬂected from the barrier, and is traveling away from the barrier. Q16.3) Why isn’t a tunneling current observed in a STM when the tip and the surface are 1 mm apart?
In order for a tunnelng current to be observed, the wave function of the electron tunneling out of the tip
must overlap appreciably with the wave function of an empty accessible state in the surface of the solid
sample. Because the wave functions have signiﬁcant amplitudes than 1 nm from the tip and surface, the
distance between tip and surface must be less than 1 nm for tunneling to occur. Q16.4) Redraw Figure 16.7 for an insulator. For an insulator, the conduction band is completely ﬁlled. Consequently, when the ﬁeld is applied, electrons can not migrate to the other end of the “box” because there are no allowed states in the band 335 Chapter 16/ The Particle in the Box and the Real World Q16.5) Explain how it is possible to create a three dimensional electron conductor that has a continuous
energy spectrum in two dimensions and a discrete energy spectrum in the third dimension. If a very thin (1—1 0 nm) layer of a conductor is evaporated onto the surface of a substrate, it will have a
continuous energy spectrum in the two dimensions parallel to the surface because the lateral dimensions
are on the order of cm. It will have a discrete energy spectrum in the third dimension because the ﬁlm is
only a few atomic layers thick. Q1 6.6) Explain, without using equations, why tunneling is more likely for the particle with E 2 3/4 V0
than for E = l/4V0 in Figure 16.8. Because the decay length, UK, is less for E : 3/4 V0 than for E : 1/4 V0, the exponentially decaying
wave in the barrier has a higher amplitude at the end of the barrier if its energy is near the top of the
barrier. This amplitude determines the amplitude of the traveling wave to the right of the barrier, and
therefore the tunneling probability. Q16.7) What is the advantage of using quantum dots that ﬂuoresce in the near infrared for surgical
applications? The ﬂuorescence can be detected using a near infrared detector while the patient is being illuminated
with visible light to excite the quantum dots and to enable surgeons to work. The important aspect is that
the ﬂuorescence and illumination occur in different parts of the light spectrum Q16.8) The overlap between wave functions can either be constructive or destructive, just as for waves.
Can you distinguish between constructive and destructive overlap for the various energy levels in Figure
16.3? The wave functions for n = l, 3, and 5 on the two wells are in phase and will interfere constructively.
The wave functions for n = 2 and 4 on the two wells are out of phase and will interfere destructively.
Q16.9) Explain how you can use sizequantized quantum dots to create a protein with a barcode that can
be read using light. Different ﬂuorescent tags can be attached to the protein, for example one that emits red light, two that
emit green light, and one that emits blue light. This combination of tags gives a different signal, or bar
code, than a combination in which one tag emits red light, one tag emits green light, and two tags emit
blue light. Q16.10) A STM can also be operated in a mode in which electrons tunnel from the surface into the tip.
Use Figure 16.9 to explain how you would change the experimental setup to reverse the tunneling current. 336 Chapter 16/ The Particle in the Box and the Real World You would simply reverse the polarity of the battery. This would make the highest ﬁlled energy level in the surface higher than that in the tip. Therefore, the tunneling current would ﬂow from the surface to the tip.
Q16.ll) For CdSe quantum dots, the emission wavelength increases from 450. nm to 650. nm as the dot diameter increases from 2 to 8 nm. Calculate the band gap energy for these two particle diameters. The band gap energy is equal to the photon energy. he ‘ 6.626x10‘34 J sx2.998x108 m s"
A 450.x10“" m E: =4.41x10“'9 J The corresponding answer for 650. nm is 3.06 x 10”]. Q16.12) Why is it necessary to functionalize CdSe quantum dots with groups such as organic acids to
make them useful in bioanalytical applications? The role of the functional groups is to make the CdSe quantum dots soluble in aqueous solutions.
Q16.l3) Why must the amplitudes of the ﬁrst derivatives of the energy eigenfunctions in the ﬁnite
depth box and in the adjoining barrier regions have the same value at the boundary? If this were not the case, the second derivative of the wave function would be discontinuous, and the
Schrodinger equation could not be solved in the boundary region. Q16.l4) Why must the amplitudes of the energy eigenfunctions in the ﬁnite depth box and in the
adjoining barrier regions have the same value at the boundary? If this were not the case, the probability of ﬁnding the particle in a region centered at the boundary
would be different depending on which side the boundary is approached from. This is physically
unreasonable. Q16.15) Explain how a quantum dot can absorb light over a range of wavelengths and emit light over a
much smaller range of wavelengths. The minimum photon energy needed to excite a quantum dot is the band gap energy. Higher photon
energies result in an excitation of an electron higher in the band, but the electron rapidly loses energy
without radiating light and decays to an energy level at the band edge. Therefore, light energy can be absorbed over a range of energies, but the quantum dot emits light over a narrow range of energies. Problems
P16.l) In this problem, you will calculate the transmission probability through the barrier illustrated in
Figure 16.8. We ﬁrst go through the mathematics leading to the solution. You will then carry out further calculations.
337 Chapter 16/ The Particle in the Box and the Real World The domain in which the calculation is carried out is divided into three regions for which the potentials are
V (x) = 0 for x S 0 Region I
V(x) 2 V0 for 0 < x < a Region II
V(x) = O for x 2 a Region III The wave functions must have the following form in the three regions if E < V0: (/06) = Aexp[+i 2:171; x]+Bexp[—i 2:215 xJ
ii L 1’ 2 A 6“” + B 62““ Region I
2 — l _
l/f(X)=C6Xp — w x +Dexp + w x
h h
: C e’” + D 6““ Region II
2 2 E
y/(x) =Fexp +1}, ":E x +Gexp 4" m2 x
h“ h
= F em“ + Ge'm‘ Region III Assume that the wave approaches the barrier from the negative x direction. The coefficient B cannot be set equal to zero because Be+"ﬂ2”'E/hi” represents reﬂection from the barrier. However, G can be set equal to zero because there is no wave incident on the barrier from the positive x direction.
a. The wave functions and their derivatives must be continuous at x : 0 and x = a. Show that the coefﬁcients must satisfy the following conditions: A+B=C+D Ce‘k“ +DeW : Fem
A — B = ~l7f(—C + D) —C 6*" + D em = i F em
K 338 Chapter 16/ The Particle in the Box and the Real World 2 , it is useful to manipulate these equations to bi Because the transmission probability is given by IF / A get a relationship between F and A. By adding and subtracting the ﬁrst pair of equations, A and B
can be expressed in terms of C and D. The second pair of equations can be solved individually to give equations for D and C in terms of F. Show that . ka +Ika
zk e+1 +Ke
D=“—3—;K—a“—F
K6
 ka +ika
—zke“ +Ke
C2—2T—F,and
K6 A— (ik—K‘)C+(ik+K)D
2ik c. Substitute these results for C and D in terms of F into _ (ik—K)C+(ik+K)D
2ik A to relate A and F. Show that + tka e
2K ZikA Z [07‘ — K)(*l'k + K)?” + (ik + K)(ik + K)e"“’] F (1. Using the hyperbolic trigonometric functions X —X X e —e e +e
and coshx= 2 2 "X sinh x = and the relationship cosh2 x — sinh2 x=1, show that 2 _ 16(Kk)2 _ 16(Kk)2 +(4(k2 —K2)2 +16(Kk)2)sinh2(Ka)
1 _ 1+ [0:2 + [(2)2 Sinh2(Ka)]/4(Kk)2 F
A e. Plot the transmission probability for an electron as a function of energy for V0 = 1.6 x 10‘19 J and a = 9.0 x 10’10 m up to an energy of 8 x 10“19 J. At what energy is the tunneling probability 0.1? At what energy is the tunneling probability 0.02?
339 Chapter 16/ The Particle in the Box and the Real World f. Plot the transmission probability for an electron of energy 0.50 x 10’“) J as a function of the barrier width for V0 = 1.6 x 10‘19 J between 2 x 10’10 and 8 x 10"10 m. At what barrier width is the transmission probability 0.2? w(x) : Ae "' i l Aem” + 36”“ Region I
Zulu/04) y/(x) = C e7 ’12 2 C ei“+ D e“"" Region II
2mL— 2an y/(x) : Feﬂll; X + Ger! ”2 X : Few” + Ge"""‘ Region [11 At x = 0, the boundary between regions I and 11, set the amplitudes
and the derivatives of the wave functions equal. A+B:C+D Differentiate the wave functions and set x = 0. ikA ~ikB: —K'C + K‘D A—B: 4%(~C+D) Atx = a, the boundary between regions [I and 111, set the amplitudes
and the derivatives of the wave functions equal.
C e’”+ D 6*” : Few".
Differentiate the wave functions and set x 2 a.
a) —KC e‘”’““+ KB 6”" = ikF e””
ik ma 7C e’”+ DE” =—Fe
K 340 Chapter 16/ The Particle in the Box and the Real World b) We begin by eliminating B from the set of equations.
Solving ikA—ik(C+ D—A) = —K‘C+K‘D forA gives :(ik—K)C+(ik+K)D
2ik
Fe/ku ~De+Ka *KG 6 Rewriting Ce’” + De”“ 2 Fe”“1 in the form C: and substituting this result into — KC e“’“’ + K D 6*” = ik F e’l‘ " gives —Ka FeI/(a _De+Ka
_K —__._—_
e je'” +KDe+’“' =ikFe””' 0r ~KFe’k” +2KDem’ = ikFe’ka resulting in an equation relating D and F
ike’k" +Ke’k“ D=——F
2Ke+xa To relate C and F we begin with the two equations C67K0+D ewra :Fe+lka and —KC €7Ka+ KDe+Ka :ikFe+i/ca Solving the second equation for C and using our result for D, we obtain
 'ka ika
zke‘ +Ke Ma . +ika
KDe+Ka_lkFe+lka K[ 2Ke+Ka F16 _lkFe
CZ—T—Z Kewa
 ‘ka +Ika
—zk e“ +Ke
CZ———Ka——F
21a: 0) We now use the expressions for C and D in terms of F to express A in terms of F.
(ik—K)C+(ik+K)D ivin
2ik g g We substitute these expressions for C and D into A = +Ika ‘ ike+lka +Ke+ika
F+(lk+K)‘”—‘—7;Z——“F
2K6  +1ka
(ik—K)—lke +Ke Ka A: 2Ke’ 2ik 341 Chapter 16/ The Particle in the Box and the Real World d) To obtain the ratioﬁ , we proceed as follows. A
. e+tku . . Hm . ‘ 71m
21kA 2 2K [(zk~K)(—zk+1c)e +(lk+K)(lk+K)e 1F
+1ku
2ikA = e [kze’m' +2ikKe+m —K2e+’m —k2e”” +2ikKe'” +Kze‘”]F
2K
Using the deﬁnitions sinhx 2 ex 426” and coshx 2 ex +26%
F ‘ 4ikKe"/k” _ 4ikKeA'k"
A (k2 —K2 )(eﬂm —e"“’)+ 2ikK<e+m +6”) 2(k2 —K2)sinliKa+4ichosh/ca
‘52 : 4ikKe"k” ~4ikKe+'k“
A 2(k2 —K2)sinhKa+4ichosh/(a 2(k2 —K2)sinhKa—4ichoshKa
{F 2 _ l6(kK)2
A 4(k2 —K2)2Slnh2Ka+l6(kK)2COSh2Ka
16(Im)2 4(k2 ~— K2 )2 sinh2 Ka +l6(kK)2 (1+ sinh2 Ka)
In the last step, we have used the relation sinh2 x + cosh2 x = 1
Our ﬁnal result is a rather compact equation. iF—Hiﬁ— l6(Kk)2 _ 1
A A 16(1ck)2 +[4(k2 «2)2 «Irma/013111112 (K a) 1 (k2 +18)2 sinhz (K a)
+ '———__
4 (K k)2
2
e) Plotting 2 as a function of E, we obtain the following graph. By replotting the data over a smaller 2 = 0.11“01~E:1.5><10*19 J and 0.02 for E: 1.1x10“19 J. interval near E = l x 10‘19 J, we ﬁnd that A ‘i $3
on Transmésmm ammaaay
{:3 {:3
332» 0’3 $3
A: Mm we” may” 3x10”
Partieia Emergy (J) 3x13” 342 Chapter 16/ The Particle in the Box and the Real World g m 2% 3 °§ {ma 5 no 5 r: we : .9 : :2; . «g ass; . g : :5; 6.02 é 5w . a {“3494} t) The transmission probability as a function of the barrier width is shown below. The probability is 0.2
for a width ofa = 3.2 X 10“10 m. 33’
4:. Transmission probahiiiw lama
P16.2) Semiconductors can become conductive if their temperature is raised sufﬁciently to populate the (empty) conduction band from the highest ﬁlled levels in the valence band. The ratio of the
populations in the highest level of the conduction band to that of the lowest level in the conduction band
is n conduction __ g conduction e “Ni/k (1'
n valence gvalcnce where AE is the band gap, which is 0.661 eV for Ge and 5.5 eV for diamond. Assume for simplicity that the ratio of the degeneracies is one and that the semiconductor becomes sufﬁciently conductive when ”mm/notion : 1 0 X 104)
n valence 343 Chapter 16/ The Particle in the Box and the Real World
At what temperatures will germanium and diamond become sufﬁciently conductive? Given that the
most stable form of carbon at normal pressures is graphite and that graphite sublimates near 3700 K, could you heat diamond enough to make it conductive and not sublimate it? For germanium, we obtain T * —AE A —0.66leV><1.602x10"9 JeV" _555 K
S, kl [( gVa/cncc j[nmnduclmn ]] 1381X10g23 J Kgl X 111(10X10'6)
n gL'U/ldllc’lll)” l’Il/L’HCL’ For diamond, we obtain
~AE —5.5 evx1.602x10“" J W“1 3
T": :138lx10‘23 JK"><1 (10><10—6):4'6X10 K
k In [ gva/cnce )[Emndm'twn ] ' n '
goona’ucl/on ”valence We predict that diamond would not become conductive before it decomposes.
P16.3) For the zznetwork of ﬂcarotene modeled using the particle in the box, the position—dependent probability density of ﬁnding 1 of the 22 electrons is given by mangrzpuzx] R1(x): The quantum number n in this equation is determined by the energy level of the electron under
consideration. As we saw in Chapter 15, this function is strongly position dependent. The question addressed in this problem is as follows: Would you also expect the total probability density deﬁned by 2 to be strongly position dependent? The sum is over all the electrons in the 7: Bola! (X) : Zn w” (x)
network. z//n(x)l2 using the box length a = 29.0 nm and a. Calculate the total probability density Rom/(x) 2 Zn plot your results as a function of x. Does PW,(x) have the same value near the ends and at the middle of the molecule? b. Determine A130,“,(x)/(P ll W,(x)), where ABUMKX) is the peak—to—peak amplitude omeu,(x) in the interval between 12.0 and 16.0 nm. 0. Compare the result of part (b) with what you would obtain for an electron in the highest occupied energy level. 344 Chapter 16/ The Particle in the Box and the Real World d. What value would you expect for Rm: (x) if the electrons were uniformly distributed over the
molecule? How does this value compare with your result from part (a)? a) We need to take all the electrons into account and do so by summing the probabilities for each electron. Plotting the data for Rom, (x) = Z W" (x),2 gives the ﬁgure shown below. 2C? '5 ‘ it) ‘ 25 x
Rom/(x) varies more strongly near the ends of the chain than at the middle. b) Plotting Ema/(x) over an interval in the middle of the range gives the ﬁgure shown below. APW, (x) Z 0.07x109m’l
<R(),a,(x)> 0.798x109m"1 c) Plotting Pf] 1(x) over an interval in the middle of the range gives the ﬁgure shown below. From this data, we see that = 0.09 {3.1% 0.12 345 Chapter 16/ The Particle in the Box and the Real World A12:.1(x) _ 0.14x109m'l <A17W, (x)> _ 0.07x109m—l
the highest occupied state is approximately 20 times as large as that for the probability density for all
electrons. From this data, we see that = 2 The variation of the probability density for d) There would be 22 electrons uniformly distributed over a length of 29 nm. Therefore the probability
density would be0.76. This is very close to the average value obtained in part (a). P16.4) Calculate the energy levels of the ir—network in hexatriene, C6Hg, using the particle in the box
model. To calculate the box length, assume that the molecule is linear and use the values 135 and 154
pm for C:C and C~C bonds. What is the wavelength of light required to induce a transition from the
ground state to the first excited state? How does this compare with the experimentally observed value of
240 nm? What does the comparison made suggest to you about estimating the length of the ﬂnetwork
by adding bond lengths for this molecule? The length of the box is a = 3 x135pm + 2 x 1 54 pm 2 713 pm. The energy levels are given by 2 2
n h . . .
E" = 8 2 and the transmon 18 between n = 3 and n = 4.
m a c c 8mazc 8x9.11><10“3‘ kg><(7.13><10“'0 m)2><2.998x108ms'1 VIE/h:h(n§—n5): 6.626x10‘34 Jsx(42—32) :239m The calculated wavelength is very good. This result suggests that the effective length of the calculated 7r—network is reasonable. P16.5) Calculate the energy levels of the 7r—network in octatetraene, Cngo, using the particle in the box
model. To calculate the box length, assume that the molecule is linear and use the values 135 and 154
pm for CZC and C~C bonds. What is the wavelength of light required to induce a transition from the
ground state to the ﬁrst excited state? A: The length of the box is a = 4 x 135 pm + 3 x 154 pm 2 1002 pm. The energy levels are given by 2 2
E” = n h 2 and the transition is between n = 4 and n = 5.
8ma
~31 710 2 8 71
,1 c c 8ma2c 8x9.11x10 kgx(10.02x10 m) x2.998x10 ms
—v #E/h—h(n§n12)_ 6.626x10‘34 Jsx(52~42)
=368 nm P16.6) The maximum safe current in a copper wire with a diameter of3.0 mm is about 20 amperes. In a STM, a current of 7.5 x 10‘10 A passes from the tip to the surface in a filament of diameter ~1.5 nm.
Compare the current density in the copper wire with that in the STM. : I ZWNA 2=2.8><10°A/m2 W2 n(1.5x10*~‘ m) The current density in the copper wire is given by J”, 2—3 The current density in the STM is given by AIO
I —I— I — 75X“) A 2=4.2x108A/m2 "W A W2 n(0.75x10"’ m) 346 Chapter 16/ The Particle in the Box and the Real World The current density is greater in the STM by a factor of 150.
P16.7) In this problem, you will solve for the total energy eigenfunctions and eigenvalues for an electron in a ﬁnite depth box. We ﬁrst go through the calculation for the box parameters used in Figure
16.1. You Will then carry out the calculation for a different set of parameters. We describe the potential in this way: V(x) = K) for x S —% Region I
V(x) = 0 for —% < x <% Region 11
V(x) = V0 for x 2% Region 111 The eigenfunctions must have the following form in these three regions: WC) = 2:6Xp[+\/@ x] + B’exp[— MEL—ﬂ x] 6“" + B’ (3" Region I
y/(x)= C sin‘lsz x+Dcos szx
=C sinkx+ Dcoskx Region II
— E 2 — E _
z//(x)=Aexp — MOT—2x +A'exp + Mx
h h‘ _
= Ae "“ +eA’ ”X Region 111 So that the wave functions remain ﬁnite at large positive and negative values of x, A' = 8' =0. An
additional condition must also be satisﬁed. To arrive at physically meaningful solutions for the
eigenfunctions, the wave functions in the separate regions must have the same amplitude and derivatives at the values of x = a/2 and x = —a/2 bounding the regions. This restricts the possible values for the coefﬁcients A, B, C , and D. Show that applying these conditions gives the following equations: 347 Chapter 16/ The Particle in the Box and the Real World 8 {Kw/2) : —C sin k i + D cos k g
2 2 BKeﬁﬁa/Z) 2 C k cos [CELL D ksin kg Ae’KWZ) = Csinkﬂ+ Dcoskﬂ
2 2 —A Ice—KW) = Ckcoskg—Dksinkg These two pairs of equations differ on the right side only by the sign of one term. We can obtain a set of equations that contain fewer coefﬁcients by adding and subtracting each pair of equations to give (A + B) eKW) = 2 Dcos[k%] (A — B) WW” = 2 C sin [k g]
(A + B)K WW) = 2D k sin (Ir—:5]
«(A  B)Ke‘”(“/2) = 2C k cosﬁkgj At this point we notice that by dividing the equations in each pair, the coefﬁcients can be eliminated to sztan[kﬂj or‘l—Zﬂq—(K‘g:£2 = J2mZE tan ‘l2n17E ﬂ and 2 h h h“ 2 —K:kcotan kg or —\/2L(V07;E2=\/2m2E cot \/2m,Eﬂ
2 h” h, h“ 2 Multiplying these equations on both sides by a/2 gives dimensionless parameters and the final equations m(V0 —E)a2 mEar2 ( mEa2
2 = 2 tan 2 and
27? 2h 2h 348 give are Chapter 16/ The Particle in the Box and the Real World m(V0 — E)a2 mEa2 mEa2
2 = 2 cot 2
2ft 2h 2h The allowed energy values E must satisfy these equations. They can be obtained by graphing the two sides of each equation against E. The intersections of the two curves are the allowed energy eigenvalues.
For the parameters in the caption of Figure 16.1, V0: 1.20 x 10‘18 J and a : 1.00 x 10"9 m, the following two graphs are obtained: Maggi 03‘ 31:66 Bum Edncmkm £52,, nubﬂah’m as gammy» Cummings The ﬁve allowed energy levels are at 4.61 X 10“”, 4.09 X 1049, and 1.07 x 10*18 J (top ﬁgure), and 1.84 x 10—19 and 7.13 x 10"19 J (bottom ﬁgure). a. Given these values, calculate k for each energy level. Is the relation It = 2a/ n (for n an integer) that arose from the calculations on the inﬁnitely deep box still valid? Compare the values with the corresponding energy level in the inﬁnitely deep box. Explain why the differences arise. 13. Repeat this calculation for V0— — 5.00 x 10‘19 J and a: 0 900 x 10’9 m. Do you think that there will be
fewer or more bound states than for the problem just worked out? How many allowed energy levels are
there for this well depth and what is the energy corresponding to each level? a) Using the equation i — h a — aJZmE we obtain the following values 349 Chapter 16/ The Particle in the Box and the Real World I: E li/a Illa for inﬁnite box
4.61 x 10"”1 2.29 12.00
1.84X1OTF9J 1.15 1.00
[4 09 x 10 r9‘1 * 0.768 0.667
7.13 x 10 r9J 0.582 0.500
1.07 x 10—” J 0475 0.400 We see that xl/a is always greater for the ﬁnite depth box. This is necessary if the particle is to have a
ﬁnite probability of being outside of the box. Note that if /l/a is less than that for the inﬁnitely deep box,
additional nodes are introduced inside the box. b) Because a ﬁnite depth box has fewer bound states than an inﬁnitely deep box, one should expect that
the number of bound states will decrease as the depth becomes less. Graphing the equations m(V E)a mEa2 mEa2
2h“ 2V 2112 “V 2722
\/m(V a2\/mEa2 \ijaand ~ ——°———)——= cot
2772 271‘ 2722 for VO 2 5.00 x 10‘19 J and a 2 0.900 x 10‘9 m gives {A} in} 1.3 1x10 “W 3313.1} “3‘3 3398“} it} 88.351 “3‘3; 5'34; E1} “lg 8111 350 Chapter 16/ The Particle in the Box and the Real World Eii) saw” 3?? t3 Exist”? Exit?" 4 a to “1‘9 a x in "3513 Em} There are 3 bound states whose energies are 4.76 x 10‘20 J, 1.86 x 10—l9 J, and
3.95 x10‘19 J. Computational Problems Before solving the computational problems, it is recommended that students work through
Tutorials 1— 3 under the Help menus in Spartan Student Edition to gain familiarity with the program. Computational Problem 16.1: Build (a) ethylene, (b) the trans conformation for 1,3 butadiene, and (c) all trans hexatriene and calculate the groundstate (singlet) energy of these molecules using the
B3LYP method with the 6—311+G** basis set. Repeat your calculation for the triplet state, which
corresponds to the excitation of an electron from the highest filled energy level to the lowest unoccupied
energy level. Use a nonplanar input geometry for the triplet states. Compare the energy difference from these calculations to literature values of the maximum in the UV—visible absorption spectrum.
Your solution to this problem should answer the questions in bold type in Steps 5 and 10. Step 1: Create a new ﬁle, and build structures for 1,3—butadiene, 1,3,5hexatriene, 1,3,5,7—octatetraene.
Create each one as a new molecule within the same ﬁle using the “New Molecule” command in the File
menu. Step 2: Go to Setup > Calculations. Set the calculation type to “Equilibrium Geometry” and the method
to B3LYP/631G*. The meaning of the method (Density Functional Theory w/B3LYP functional) and
basis set (631G*) will be explained later in the course, consider this to be a ‘black box’ for now. “Total Charge” should be neutral, and “Global Calculations” should be on. Click “OK.” This will write the
351 Chapter 16/ The Particle in the Box and the Real World parameters for the calculation to the ﬁle, but will NOT start running the calculation. You will submit the
calculation in Step 4.
Step 3: You will now calculate molecular orbitals for butadiene. Use the left and right navigation
buttons at the bottom of the window to select your Butadiene model, then go to Setup —> Surfaces. Click
“Add,” then select “LUMO” from the “Surfaces” menu. (NOT LUMO from the “Properties” menu!).
Click “OK.” Repeat for the HOMO, and HOMO1. Make sure “Apply Globally” is NOT checked.
Step 4: Go to Setup —> Submit. Your calculations should start running. The calculation will most likely
take a few minutes on most current computers.
Step 5: While on butadiene, go to Display —+ Surfaces. A dialog box will appear. Examine the surfaces
one at a time, and comment on the degree of electron delocalization. Is there any evidence that the
system can be modeled as particles in a onedimensional box?
Step 6: Turn off the surfaces, close the dialog box, and then go to Display ——> Properties. Record the
HOMO and LUMO energies in your spreadsheet program of choice. Use the navigation buttons on the
bottom of the screen to advance to the next molecule, and repeat until you have recorded data for all 4
molecules. The method used in this problem is reasonably accurate in calculating HOMO—LUMO gaps.
Quantitatively accurate excited state calculations require advanced methods that are beyond the scope of this course. However, for this exercise, you will be looking for trends in the energy, not exact values, so use of a simpler method is acceptable. Step 7: Calculate the energy gap Eg (in eV) by subtracting the HOMO energy from the LUMO energy.
This is the energy corresponding to the longestwavelength absorption of light by the conjugated molecule. Calculate the wavelength of absorption in another column using equation (1). Be careful about units here!
he
E : ~ 1
,1 ( ) Step 8: Go through each molecule, measure the carboncarbon bond lengths using the “Distance” tool in the tool palette, and add them up to obtain a total conjugation length for each molecule. Record this in your spreadsheet. Step 9: Perform a one—dimensional particle in—a—box calculation for the HOMOLUMO transition
energy for all three of your molecules, using the lengths you measured. Keep track of your units! Example values for 1, 3, 5, 7, 9—decapentaene have been provided to help get you started. The table 352 Chapter 16/ The Particle in the Box and the Real World below does not include wavelengths, however you should calculate wavelengths for all four molecules. If you are stuck here, please reread Section 16.3 of your book. —‘ f Molecule n HOMO L (m) B3LYP B3LYP B3LYP 1D Box Eg
E HOMO E LUMO Eg (eV) (eV)
(6V) (6V)
[ Butadiene [
Hexatriene T
Octatetraene i‘ l
Decapentaene 5 F 1.253x10'9 ~5.116 i —l .796 i332 2.64 Step 10: Generate a table containing the B3LYP and 1D Box predicted energy values, and graph them with # of double bonds on the x—axis and energy in eV on the yaxis. Look at the shape of the curves—— how does the ﬁrst absorption energy change as a function of double bonds? While you may use regression techniques if you would like, a qualitative answer is acceptable. Also, compare your calculated values with the experimental value for butadiene and with some values that are calculated using a higher level theory. What trends do you see? A UV/Visible absorption spectrum for butadiene is on the next page, ﬁnd the strongest peak and use its energy in your comparison. Results for Eg obtained using a higher level theory called time dependent density functional theory (TDDFT) for all four molecules are listed below. B3LYP/631G* TDDF T results for
Butadiene: 5.27 eV
Hexatriene: 4.34 eV Octatetraene: 3.72 eV Decapentaene: 3.27 eV 353 Chapter 16/ The Particle in the Box and the Real World Logarithm epsiton 1,3Butadiene 240 Wavelength (nm)
NEST Chemistry WebBock {http:iiwebbank.nist.govichemtstry) Question from Step 5: 28C.) The HOMO1 is delocalized over the entire molecule, indicating a high degree of electron
delocalization, and consequently, that the system can be modeled as free electrons in a box.
Furthermore, since the molecules are effectively one—dimensional, a 1D box approximation is sufficient. Question from Step 10: Molecule n HOMO L (m) B3LYP B3LYP B3LYP 1D Box Eg
E HOMO (ea E LUMOLEV) Eﬂeu (eV)
Butadiene J 2 4.14E—10 ~6.231 ~0.599 5.63 Tl 10.98 4‘
Hexatriene 3 6.94E10 5.687 —1 . 197 4.49 5.47 i
Octatetraene 4 9.74E—10 —5.333 i551 3.78 3.57 
l Decapentaene 5 l 1.253139 —5.1 16 1.796 I332 2.64 [ The shape of the curve for the B3LYP calculations is very similar to the shape of the curve for the
TDDFT calculations, and it is shifted slightly up the energy axis. Both decay at a decreasing rate, and
the TDDFT result for butadiene is closest to the spectroscopic result. However, none of the methods
provided a particularly good ﬁt to the spectrum—calculation of electronic spectra is difﬁcult—although
use of other advanced techniques, a larger basis set, or different DFT functionals can assist in obtaining a better fit. 354 # {31‘ bands Comﬁarison table {energy in CV} # of bonds PIB B3LYP TDDFT y VVis
2 10.98 5.63 5.27 #3 3 5.47 4.49 4.34 4 3.57 3.78 3.72 F 5 2.64 3.32 3.27 355 ...
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 Fall '10
 Whaley

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