Resolução exercicios ímpares do capitulo 9 do livro Fisicoquímica - Atkins, 8ed (em inglês)

Prévia do material em texto

9 
Quantum theory: 
techniques and 
applications 
Answers to discussion questions 
09.1 In quantum mechanics, particles are said to have wave characteristics. The fact of the existence of the 
particle then requires that the wavelengths of the waves representing it be such that the wave does not 
experience destructive interference upon reflection by a barrier or in its motion around a closed loop. 
This requirement restricts the wavelength to values A = 21n x L , where L is the length of the path and 
n is a positive integer. Then using the relations A = hi p and E = p2/2m, the energy is quantized at 
E = /12h2 18mL 2. This derivation applies specifically to the particle in a box, the derivation is similar 
for the particle on a ring; the same principles apply (see Section 9.6). 
09.3 The lowest energy level possible for a confined quantum mechanical system is the zero-point energy, 
and zero-point energy is not zero energy. The system must have at least that minimum amount of 
energy even at absolute zero. The physical reason is that, if the particle is confined, its position is not 
completely uncertain, and therefore its momentum, and hence its kinetic energy, cannot be exactly zero. 
The particle in a box , the harmonic oscillator, the particle on a ring or on a sphere. the hydrogen atom, 
and many other systems we will encounter, all have zero-point energy. 
09.5 Fermions are particles with half-integral spin, 1/2, 3/2, 5/2, ... , whereas bosons have integral spin, 0, 
I , 2, .... All fundamental particles that make up matter have spin 1/2 and are ferrnions , but composite 
particles can be either fermion s or bosons. 
E9.1(b) 
Ferrnions: electrons, protons, neutrons, 3He, .... 
Bosons: photons, deuterons. 
Solutions to exercises 
/12/72 
E = --2 [9.4a] 
8meL 
/72 (6.626 x 10- 34 J s)2 _ 
.,----,-----'-..,......,;:-:-------'---,,-----,;- = 2.678 X 10- 20 J 
8(9.109 X 10- 31 kg) x (1.50 x 1O- 9 m)2 
The conversion factors required are 
leV = 1.602 x 10- 19 J ; I cm- I = 1.986 x 10- 23 J ; leV = 96.485 kJ mol - I 
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS 177 
h2 20 (a) E3 - EI = (9 - 1)--2 = 8(2.678 x lO- J) 
8meL 
= 12.14 x 10- 19 J 1 = 11.34eV I = 11.08 x 104 cm- 1 1= I 129kJmol-1 I 
h2 (b) E7 - E6 = (49 - 36)--2 = 13(2.678 X 10-20 J) 
8meL 
= 13.48 x 10- 19 J I = 12.17eV I = 11.75 x 104 cm-1 1= 12lOkJmol-1 1 
E9.2(b) The probability is 
E9.3(b) 
E9.4(b) 
P = J 1fr*1fr dx = ~ J sin2 C~X) dx ~ 2~X sin2 C~X) 
where t.x = 0.02L and the function is evaluated at x = 0.66 L. 
(a) For n = I 2(0.02L) ~ P = L sin2(0.66JT) = ~ 
(b) For n = 2 2(0.02L) ~ P = L sin2[2(0.66JT)] = ~ 
The expectation value is 
(P) = J 1fr*p1fr dx 
but first we need p1fr 
A • d (2) 1/2. nJT x . (2) 1/2 nJT nJT x p1fr = -In dx L sm ( L ) = -In L L cos ( L ) 
A -2innJT rL (nJT X) (nJTX) Inl 
so 1jJ) = ---u- Jo sin L cos L dx = L2.J 
for all n. So for n = 2 
The zero-point energy is the ground-state energy, that is, with nx = ny = nz = I: 
(n2 + n2 + n2)h2 3h2 
E = x Y 2 z [9.12b with equal lengths] = --2 
8mL 8mL 
Set this equal to the rest energy me2 and solve for L: 
3h2 
me2 =--
8mL2 
soL = (~)1 /2 ~ = (~)1 /2 AC 
8 me 8 
where AC is the Compton wavelength of a particle of mass m. 
E9.5(b) 
E9.6(b) 
E9.7(b) 
178 STUDENT'S SOLUTIONS MANUAL 
( 2)1 /2 (5JTX) 1/15 = L sin L 
M · d' " P() d dP(x) 0 aXlma an rrumma III x correspon to -- = 
dx 
d d1/l 2 (5JT X) (5JTX) (IOJTX) dx P (x) ex dx ex sin L cos L ex sin ~ 
sin e = 0 when e = 0, JT , 2JT , . . . , n' JT (n' = 0, 1, 2, . .. ) 
10JTX , 
--=nJT 
L 
n'L 
for 11' < 10 so x = -
- 10 
[2 sin ex cos ex = sin 2ex] 
x = 0, x = L are minima. Maxima and minima alternate, so maxima correspond to 
11' = 1, 3, 5, 7, 9 x =1 ~ I, 1 ~~ I, [TI, 1 ~~ I, 1 ~~ 1 
The energy levels are 
where £1 combines all constants besides quantum numbers. The minimum value for all the quantum 
numbers is 1, so the lowest energy is 
The question asks about an energy 14/3 times this amount, namely 14£1 . This energy level can be 
obtained by any combination of allowed quantum numbers such that 
The degeneracy, then, is []], corresponding to (nl , n2 , 113) = (1 , 2,3), (I , 3, 2), (2, 1, 3), (2, 3, I), 
(3 , 1, 2), or (3, 2, I). 
£ = ~kT is the average translational energy of a gaseous molecule (see Chapter 17). 
3 (n2 + n2 + n2)h2 n2h2 £ = -kT = 1 2 3 
2 8mL2 8mL2 
£ = G) x (1.381 x 10- 23 JK- 1) x (300K) = 6.214 x 10- 21 J 
8mL2 
n2 = --£ h2 
E9.8(b) 
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS 179 
(6.626 X 10- 34 J S)2 = 1.180 x 10- 42 J 
( 
0.02802 kg mol-I ) 2 (8) x x 100m 
6.022 x 1023 mol-I 
- 21 2 6.214 x 10- J - 21. 
n = 42 = 5.265 x 10 , n= !7.26 x 101O! 1.180 x 10- J 
( 
h 2 ) ( h2 ) 14.52 x 1O IO h2 
t:;.E = (2n + I ) x --2 = [(2) x (7.26 x 1010) + 1] x --2 = 2 8mL 8mL 8mL 
= (14.52 x 1010) x (1.180 x 10-42 J) = ! 1.71 X 10-31 J! 
The de Broglie wavelength is obtained from 
h h 
A = - = - [8.12] 
p mv 
The velocity is obtained from 
EK = !mv2 = ~kT = 6.214 x 10- 21 J 
6.214 X 10- 21 J -
v2 = = 2.671 X 105 m2 s- 2; v = 517 m S- I 
(
1) (0.02802 kg mol-I ) 
2" x 6.022 x 1023 mol- I 
6.626 x 10- 34 J s 
A= =2.75 x 10- 11 m=127.5pml (4.65 x 10-26 kg) x (517 m S- I) 
The conclusion to be drawn from all of these calculations is that the translational motion of the nitrogen 
molecule can be described classically. The energy of the molecule is essentially continuous, 
t:;.E 
E «< 1. 
The zero-point energy is 
1 1 (k)1 /2 1 ( 285 Nm- I )1 /2 Eo = -!'u.v = -Ii - = -(\.0546 X 10-34 Js) x 26 
22m 2 5.16 x 10- kg 
= !3.92 x 10- 21 J ! 
E9.9(b) The difference in adjacent energy levels is 
( k) 1/ 2 t:;.E = Ev+ 1 - Ev = !'u.v [9.26] = Ii;;; [9 .25] 
m(t:;.E)2 
so k = --::--1i2 
(2 .88 X 10- 25 kg) x (3 . 17 X 10-21 J)2 ! _I ! 
( 1.0546 X 10-34 J s)2 = 260 N m 
180 STUDENT'S SOLUTIONS MANUAL 
E9.10(b) The difference in adjacent energy levels, which is equal to the energy of the photon, is 
and 
,,(_k )1 /2 ~E = Iiw = hv so " 
m 
A = h; (~y/2 = 2Jre(T) 1/2 
he 
A 
8 I ((l5.9949U) x (1.66 x 1O-27 kg U- I»)1 /2 
= 27T(2.998 x 10 ms- ) x I 544Nm-
A = 1.32 x 10-5 m = 113.2 I1m 'l 
E9.11 (b) The difference in adjacent energy levels, which is equal to the energy of the photon, is 
E9.12(b) 
( k) 1/2 ~E = Iiw = hv so n -;;; he A 
he (k ) 1/2 (m) 1/2 
and A = - - = 27Te -
n 111 k 
Doubling the mass, then, increases the wavelength by a factor of 21 / 2. So taking the result from Exercise 
9 .1O(b), the new wavelength is 
A = 21 /2 (1 3.2 11m) = 118.7 11m I 
~E = Iiw = hv 
(a) ~E = hv = (6.626 X 10-34 1 Hz-I ) x (33 x 103 Hz) = 12.2 x 10- 29 1 I 
(b) ~E = Iiw = n(~) 1/2 
meff [ 
I 1 1 ] 
- = - + - with ml = m2 
meff m, m2 
For a two-particle oscillator meff , replaces 111 in the expression for w. (See Chapter 13 for a more complete 
discussion of the vibration of a diatomic molecule.) 
(
2k)I /2 ( (2) x ( ll77Nm-l) )1 /2 ~E = n -;;;- = (1.055 X 10-34 1 s) x (16.00) x (1.6605 x 10-27 kg) 
=13.14 x 10- 20 11 
E9.13(b) The first excited-state wavefunction has the form 
1{1 = 2NIyexp (-!l ) 
where NI is a collection of constants and y == x(mw/n) 1/2. To see if it satisfies Schrodinger's equation, 
we see what happens when we apply the energy operator to this function 
. n2 d2 1{1 1 2 2 
H1{I=---+-111W X 1{1 
2m dx2 2 
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS 181 
We need derivatives of l/f 
dl/f = dl/f dy = (nut» 1/2 (2NI) X (l _ i) x exp (- ~i) 
dx dy dx n 2 
and :~ = ~:~ (:r = (m;) x (2NI) x (-3y + y3) X exp ( -~i) = C:W ) x (i - 3)l/f 
, n2 (mw) 2 1 2 2 So Hl/f = -- x - x (y - 3)1/f + -mw x l/f 
2m n 2 
1 2 1 2 3 
= -"2 nw x (y - 3) x l/f + "2 nwy l/f = "2nwl/f 
Thus, l/f is a solution of theSchrOdinger equation with energy eigenvalue 
E9.14(b) The harmonic oscillator wavefunctions have the form 
( 
1 ) x ( n2 ) 1/4 
1/fv(x ) = NvHv(y) exp -"2i with y = -;;; and a = mk [9 .28] 
The exponential function approaches zero only as x approaches ±oo, so the nodes of the wavefunction 
are the nodes of the Hermite polynomials. 
Hs (y ) = 32i - 160y3 + 120y = 0 [Table 9.1] = 8y(4l- 20i + l5) 
So one solution is y = 0, which leads to x = O. The other factor can be made into a quadratic equation 
by letting z = i 
4z2 - 20z + 15 = 0 
-b ± .Jb2 - 4ac 20 ± .J202 - 4 x 4 x 15 5 ± .JTO 
so z = 
2a 2 x 4 2 
Evaluating the result numerically yields z = 0.92 or 4 .08, so y = ±0.96 or ±2.02. Therefore 
x=10,±0.96a, or ±2.02a I. 
COMMENT. Numerical values could also be obtained graphically by plotting H5(y)' 
E9.1S(b) The zero-point energy is 
Eo = ~nw = ~n(~) 1/ 2 
2 2 meff 
For a homonuclear diatomic molecule, the effective mass is half the mass of an atom, so 
Eo = - ( 1.0546 X 10- 34 J s) x I (2293.8Nm- 1 )1 /2 
2 ~(l4.0031 u) x (1.66054 X 10-27 kgu- I) 
Eo = 12.3421 X 10- 20 J 1 
182 STUDENT'S SOLUTIONS MANUAL 
E9.16(b) Orthogonality requires that 
f 1/1,;,1/1" dr = 0 
if m f=. n. 
Performing the integration 
If m f=. /1 , then 
f 1/1* 1/1 dr = N 2 ei(II_lIIl</>1 27T = N 2 (I - 1) = 0 III" i(n - m) 0 i(/1 - m) 
Therefore, they are orthogonal. 
E9.17(b) The magnitude of angular momentum is 
Possible projections onto an arbitrary axis are 
where m, = 0 or ± I or ±2. So possible projections include 
I 0, ± 1.0546 x 10-34 J sand ±2.1109 x 10-34 J s I 
E9.18(b) The cones are constructed as described in Section 9.7(d) and Figure 9.40(b) of the text; their edges are 
of length {6(6 + J)} 1/ 2 = 6.48 and their projections are mj = +6, +5 , ... , -6. See Figure 9.1 (a). 
The vectors follow, in units of n. From the highest-pointing to the lowest-pointing vectors (Figure 9. J (b », 
the values of m, are 6, 5, 4, 3, 2, 1,0, -I , - 2, - 3, - 4, -5, and -6. 
~=====! In = + 6 
~-->..-----"~:z.. + 5 
"'"->-----~z + 4 
~~------~~ +3 ~~--------~~~.+2 S~===:::::::====:::20+ I 
~~------------~ - 1 ~------------~~ -2 
- 3 
-r------~ - 4 
- 5 
Figure 9.1(a) 
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS 183 
Figure 9.1(b) 
Solutions to problems 
Solutions to numerical problems 
P9.1 n
2h2 3h2 
E = 8mL2 ' E2 - EI = 8mL2' 
We take m(02) = (32.000) x (1.6605 X 10-27 kg), and find 
(3) x (6.626 x 10- 34 J s)2 1 39 1 E2 - E 1 = = 1.24 x 10- J. (8) x (32.00) x (1.6605 x 10- 27 kg) x (5.0 X 10-2 m)2 
n2h2 1 
We set E = 8mL2 = 2kT and solve for n. 
From above, h2 18mL 2 = (E2 - E I) 13 = 4.13 x 10- 40 J; then 
n
2 x (4. 13 x 10-40 J) = G) x (1.381 x 10-23 JK- 1) x (300K) = 2.07 x 10-21 J. 
(
2.07 X lO- 2I J)1 /2 9 
We find n = 40 = 12.2 x 10 I· 4.13 x 10- J . . 
At this level, 
h2 h2 h2 
E" - E,,_I = (n2 - (n - 1)2 ) x --2 = (2n - I) x --2 ~ (2n) x --2 
8mL 8mL 8mL 
= (4.4 x 109 ) x (4. 13 x 1O-4o J) ~ 11.8 x lO-30 J 1 [or 1.1I!Jmol- I]. 
P9.3 m
2Ji2 m2Ji2 
E = -' -[9.38a] = -' - [l = mr2] . 21 2mr2 
Eo = 0 [m, = 0]. 
Ji2 (1.055 x 10-34 J s)2 1 1 
EI = -- = = 130 xlO- 22 J 2mr2 (2) x (1.008) x (1.6605 x 10-27 kg) x (160 X 10- 12 m)2 .. 
The minimum angular momentum is 1 ±Ji I. 
184 STUDENT'S SOLUTIONS MANUAL 
P9.S (a) Treat the small step in the potential energy function as a perturbation in the energy operator: 
H(I ) = {O for 0 :'S x :'S (l / 2)(L - a) and (l / 2)(L + a) :'S x :'S L 
£ for (l / 2)(L - a) :'S x :'S (l / 2)(L + a) . 
The first-order correction to the ground-state energy, EI, is 
Io
L j(l /2)(L+a) (2) 1/ 2 Jtx (2) 1/ 2 Jtx E~I ) = o/iO)*H(i ) o/~O) dx = - sin (-) £ - sin (-) dx, 
o (I /2)(L- a) L L L L 
( I ) 2 £ j(i /2)(L+a) . 2 (JtX) £ ( (JtX) . (JtX)) li / 2(L+ a) EI = - Sin - dx = - Jtx - L cos - Sin - , 
L (I /2)(L- a) L LJt L L i/2(L-a) 
E (I) = sa _ ~ cos (Jt(L + a») sin (Jt(L + a») + ~ cos (Jt(L - a») sin (Jt(L - a)). 
i L Jt 2L 2L Jt 2L 2L 
This expression can be simplified considerably with a few trigonometric identities. The product of 
sine and cosine is related to the sine of twice the angle: 
cos Sin = - Sill = - Sin Jt ± - , ( Jt(L±a») . (Jt(L±a») 1. (Jt(L±a») 1. ( Jta) 2L 2L 2 L 2 L 
and the sine of a sum can be written in a particularly simple form since one of the terms in the sum 
is Jt : 
sin ( Jt ± 7) = sin Jt cos (~a) ± cos Jt sin (~a) = 'f sin (~a) . 
Thus E = - + - Sin - . (I) sa £. (Jta) 
I L Jt L 
(b) If a = L/ IO, the first-order correction to the ground-state energy is 
(I ) £ £ . (Jt) I I EI = 10 + ;- Sin 10 = 0.1984 £. 
P9.7 The second-order correction to the ground-state energy, EI, is 
2h2 ( I ) _ (0) _ . nJtx _ n 
where H - mgx'o/lI - Sin --, and En - --2' 
L 8mL 
The denominator in the sum is 
E(O) _ E(O) = _h_2 _ _ _ n2_h_2 = _( 1_-_n--;:2 );-h_2 
i n 8mL2 8mL2 8mL2 
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS 185 
The integral in the sum is 
1/I (O)* H ( I ) 1/1 (0) dx = -- x sinaxsinbx dx 
lo
L 2mg loL 
o /I I L 0 ' 
where a = nnl L and b = n i L. 
The integral formulas given with the problem allow this integral to be expressed as 
2mg d loL. d (cos (a - b)x cos(a + b)X) IL 
- --- cos axsmbxdx = -mg- - --'---
L da 0 da 2(a - b) 2(a + b) 0 
_ 2mg (-x Sin(a-b)X cos(a-b)x xsin(a + b)x cos(a+b)x)I L 
- - - + + ------:0-
- L 2(a - b) 2(a - b)2 2(a + b) 2(a + b)2 0 . 
The arguments of the trigonometric functions at the upper limit are: 
(a - b)L = (n - I)n and (a + b)L = (n + I)n. 
Therefore, the sine terms vanish. Similarly, the cosines are ± 1 depending on whether the argument is 
an even or odd mUltiple of n ; they simplify to (_I)Il+ I . At the lower limit, the sines are still zero, and 
the cosines are all I. The integral, evaluated at its limits with Tt l L factors pulled out from the as and bs 
in its denominator, becomes 
mgL (-1)11+ 1 - I _ ( -1)11+ 1 - I) = mgL[(-I)n+1 - I] (n + 1)2 - (n _ 1)2) 
n 2 (n - 1)2 (n + 1)2 n 2 (n _ 1)2(n + 1)2 ' 
4mgL[(-I)n+1 - l]n 
n 2(n2 - 1)2 
The second-order correction, then, is 
Note that the terms with odd n vanish. Therefore, the sum can be rewritten, changing n to 2k, as 
The sum converges rapidly to 4.121 x 10- 3 , as can easily be verified numerically; in fact, to three 
significant figures , terms after the first do not affect the sum. So the second-order correction is 
£ (2) _ 1 -
186 STUDENT'S SOLUTIONS MANUAL 
The first-order correction to the ground-state wavefunction is also a sum: 
./. (I ) = '""" C ./. (0) 
'PO ~ no/n ' 
II 
4mgL[(-I )II+ 1 - l]n 
rL ./. (0)0 HI ./.(0) dx 2 2 2 
where c = _JO 'I'll '1'1 = _ Jt (n - I ) 
II £,\0) _ £~O) «n2 _ I )h2 j 8mL2) 
Once again, the odd n terms vanish. 
32m2gL 3[(-l)n + I]n 
Jt 2h2(n2 - 1)3 
How does the first-order correction alter the wavefunction? Recall that the perturbation raises the potential 
energy near the top of the box (near L) much more than near the bottom (near x = 0); therefore, we 
expect the probability of finding the particle near the bottom to be enhanced compared with that of 
finding it near the top. Because the zero-order ground-state wavefunction is positive throughout the 
interior of the box, we thus expect the wavefunction itself to be raised near the bottom of the box and 
lowered near the top. In fact, the correction terms do just this. First, note that the basis wavefunctions 
with odd n are symmetric with respect to the center of the box; therefore, they would have the same 
effect near the top of the box as near the bottom. The coefficients of these terms are zero: they do not 
contribute to the correction. The even-n basis functions all start positive near x = 0 and end negative 
near x = L; therefore, such terms must be multiplied by positive coefficients (as the result provides) to 
enhance thewavefunction near the bottom and dimini sh it near the top. 
Solutions to theoretical problems 
PS.9 The text defines the transmission probability and expresses it as the ratio of IA11 2j 1A12, where the 
coefficients A and AI are introduced in eqns 9.14 and 9.17. Eqns 9.18 and 9.19 list four equations for 
the six unknown coefficients of the full wavefunction. Once we realize that we can set BI to zero, these 
equations in five unknowns are: 
(a) A + B = C + D, 
(b) Cel<L + De- I<L = A/eikL , 
(c) ikA - ikB = KC - KD, 
(d) KCel([ - KDe- I<L = ikA /eikL . 
We need AI in terms of A alone, which means we must eliminate B, C, and D. Notice that B appears 
only in eqns (a) and (c). Solving these equations for B and setting the results equal to each other yields 
KC KD 
B=C+D-A=A- ik + ik · 
Solve this equation for C: 
2A+D(~ -I) 
C= lk 
K 
ik + I 
2Aik + D(K - ik) 
K +ik 
Now note that the desired AI appears only in (b) and (d). Solve these for AI and set them equal: 
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS 187 
Solve the resulting equation for C, and set it equal to the previously obtained expression for C: 
(K + ik )De - 2KL 
K - ik 
Solve this resulting equation for D in terms of A: 
(K + ik)2e- 2KL - (K - ik)2 2Aik 
-'---------'--------,------'---D=--(K - ik ) (K + ik) K + ik ' 
2Aik(K - ik) 
D = (K + ik )2e-2KL _ (K - ik)2 . so 
2Aik + D (K - ik ) 
K + ik 
Substituting this expression back into an expression for C yields 
Substituting for C and D in the expression for A' yields 
The transmission coefficient is 
The denominator is worth expanding separately in several steps. It is 
(K + ik )2(K - ik)2e - 2KL - (K - ik )4 - (K + ik )4 + (K - ik )2(K + ik )2e2KL 
= (K2 + k2)2(e2KL + e-2KL ) _ (K2 _ 2iKk _ e)2 _ (K2 + 2iKk _ k2)2 
= (K4 + 2K2k2 + k4) (e2KL + e-2KL ) _ (2K4 _ l 2K 2k2 + 2k2). 
If the 12K2k2 term were -4K2k2 instead, we could collect terms still further (completing the square), 
but of course we must also account for the difference between those quantities, making the denominator 
So the coefficient is 
We are almost there. To get to eqn 9.20a, we invert the expression 
188 STUDENT'S SOLUTIONS MANUAL 
Finally, we try to express (K2 + k2 ) / k2 K2 in terms of a ratio of energies, c = £ / V. Egns 9.14 and 9.16 
define k and K. The factors involving, 2, n, and the mass cancel leaving K ex (V - £)1 /2 and k ex £1 / 2, 
so 
V2 [£ + (V - £)f 
£(V - £) £(V - £) c (l - c)' 
which makes the transmission coefficient 
P9.11 We assume that the barrier begins at x = 0 and that the barrier extends in the positive x direction. 
P9.13 
(a) 
(b) 
Question. Is N a normalization constant? 
p2 ,n d 
withT=- andp=--. 
2m i dx 
T=-~~= - n2 d2 =-~nw~ [x =ay, a 2 =mnw] 2m dx2 2ma2 dyZ 2 dyZ' 
which implies that 
We then use 1jr = NHe- i /2 , and obtain 
From Table 9.1 
H~ - 2yH~ = -2vHv 
i Hv = y GHv+I + VHv- I) = ~ (~Hv+2 + (v + l)Hv) + v (~Hv + (v - 1)Hv- 2) 
= ~Hv+2 + v(v - I)Hv-2 + (v + D HI'. 
d
2
1jr _ [ 1 ( I ) ] _ y2/2 Hence, --2 - N -Hv+2 + v(v - I)Hv-2 - v + - HI' e . dy 4 2 
P9.1S 
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS 189 
Therefore, 
(a) 
(T) = N2 (-~IUv) i: Hv [~Hv+2 + v(v - I)Hv-2 - (v + D Hv] e-i dx 
[dx = ady) 
= aN2(-11Uv) [0 + 0 - (v + 1)n l /22V v!] 
[i:oo HvHv'e-y2 dy = 0 if v' "1 v, Comment 9.2] 
(x) = faL (~)' /2 sinC~X)x(~)' /2 sinC~X)dx 
= (~) faL xsin2 axdx [a = n:] 
= (~) X (X2 _ xsin2ax _ cos2ax)IL = (~) x (L2) 
L 4 4a 8a2 0 L 4 
L 
= - [by symmetry also]. 
2 
[ 
L 2 ( I) L 2 ] 1/ 2 ( I 1) 1/ 2 
ox = "3 I - 6n2n2 -"4 = L T2 -~ 
(P) = 0 [by symmetry, also see Exercise 9.2(a)), 
(p2) = n2h2/4L2 [from E = p2/2m, also Exercise 9.2(a)]. 
o = (n2h2)1 /2 = \ nh \' 
p 4L2 2L 
nh ( I I) 1/ 2 nh ( I) 1/ 2 Ii 
opox = 2L x L 12 - 2n2n2 = 2v'3 I - 24n2n2 > "2 ' 
(b) (x) = a 2 i: 1jr2ydy [x = ay] = 0 [by symmetry,y is an odd function), 
( 2) 2 ( I 2) 2 x = k "2kx = k (V ) 
190 STUDENT'S SOLUTIONS MANUAL 
since 2 (T ) = b (V) [9.35, (T ) == EK] = 2 (V) [V = axb = ~kx2, b = 2] 
or (V) = (T) = ~ (v +~) h.w [Problem 9.13] . 
(X2 ) = (v+ D x (¥) = (v+ D x (w~) = (v+ D x (~:)' /2 [9.32]. 
ox = [(v + D ~ r /2 
(P) = 0 [by symmetry, or by noting that the integrand is an odd function of x]. 
(p2) = 2m (T ) = (2m) x G) x (v + D x h.w [Problem 9.13]. 
COMMENT. Both results show a consistency with the uncertainty principle in the form !!..p!!..q :::: ~ as given 
in Section 8.6, eqn 8.36a. 2 
P9.17 Use the first two terms of the Taylor series expansion of cosine: 
V = Vo(l - cos3¢) ~ Vo (I _ I + (3~)2 ) = 9~0 ¢2 . 
The SchrOdinger equation becomes 
This has the form of the harmonic-oscillator wavefunction (eqn 9.24). The difference in adjacent energy 
levels is: 
EI - Eo = h.w [9.26] where w = (9~0 ) 1/2 [adapting 9.25] . 
If the displacements are sufficiently large, the potential energy does not rise as rapidly with the angle as 
would a harmonic potential. Each successive energy level would become lower than that of a harmonic 
oscillator, so the energy levels will become progressively closer together. 
Question. The next term in the Taylor series for the potential energy is - (27gVo) ¢ 4. Treat this as 
a perturbation to the harmonic oscillator wavefunction and compute the first-order correction to the 
energy. 
P9.19 
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS 191 
e2 I 
V = - -- . - [10.4 with Z = I] = axb with b = -I [x -+ r] 
4Jt t:Q r 
Since 2 (T) = b (V) [9.35 , (T ) == Ed 
2 (T) = - (V ) . 
Therefore, I (T) = - ~ (V ) I· 
P9.21 The elliptical ring to which the particle is confined is defined by the set of all points that obey a certain 
equation. In Cartesian coordinates, that equation is 
as you may remember from analytical geometry. An ellipse is similar to a circle, and an appropriate 
change of variable can transform the ellipse of this problem into a circle. That change of variable is most 
conveniently described in terms of new Cartesian coordinates (X, Y) where 
X = x and Y = ay/ b. 
In this new coordinate system, the equation for the ellipse becomes 
which we recognize as the equation of a circle of radius a centered at the origin of our (X, Y) system. 
The text found the eigenfunctions and eigenvalues for a particle on a circular ring by transforming from 
Cartesian coordinates to plane polar coordinates. Consider plane polar coordinates (R, cP) related in the 
usual way to (X , Y): 
X = R cos cP and Y = R sin cP. 
In this coordinate system, we can simply quote the results obtained in the text. The energy levels are 
m2n 
E = dl [9.38a] 
where the moment of inertia is the mass of the particle times the radius of the circular ring 
1 = ma2 . 
The eigenfunctions are 
eim/<I> 
l/f = --'-/2 [9.38b]. (2Jt) 
It is customary to express results in terms of the original coordinate system, so express cP in terms first 
of X and Y, and then substitute the original coordinates: 
Y 
- = tan cP 
X 
so cP = tan-' ~ = tan - ' ay . 
X bx 
192 STUDENT'S SOLUTIONS MANUAL 
P9.23 The SchrOdinger equation is 
1i2 
- -\121{1 = £1{1 [9.48 with V = 0], 
2m 
2 I a2(r1{l) I 2 
\1 1{1 = - --2- + -ZA 1{1 [Table 8,1 ] , 
r ar r 
Since r = constant, the first term is eliminated and the Schrodinger equation may be rewritten 
1i2 2 1i2 2 2 2 2/£1{I 
---A 1{1 = £1{1 or - -A 1{1 = £1{1 [l = mr ] or A 1{1 = ---~~ II ~ 
2 I a2 I a , a 
where A = ---- + - ,-- sm8-, 
sin28 a¢2 sm 8 a8 a8 
Now use the specified 1{1 = YI,III/ from Table 9,3, and see if they satisfy this equation, 
(a) Because Yo,o is a constant, all derivatives with respect to angles are zero, so A 2 Yo,O = ~ implying 
that £ = ~ and angular momentum = ~ [from (l(l + I)} 1/21ij, 
(b) 2 1 a2Y2_ 1 1 a , l1aY2-1 ' n i¢ A Y2 I=-----'-+---smu--'- where Y2 ,_ I=Ncos8smue-, 
,- sin2 8 a¢2 sin 8 a8 a8 
(c) 
I a ay~ - I 1 a, '¢ 2 ' 2 
-- - sin 8---' - = -- - sm 8Ne-' (cos 8 - sm 8) 
sin 8 a8 a8 sin 8 a8 
N -i¢ ) 
= -~- (sin 8( -4cos8 sin 8) + cos 8(cos2 8 - sin2 8) 
sm8 
=Ne-'¢ -6cos8sin8+ - ,-, ( COS8) 
sm8 
a2Y2,_1 -Ncos8sin8e- i¢ -Ncos8e-i¢ 
sin2 8 ~ sin2 8 sin 8 
so A2Y2,_1 = Ne-i¢(-6cos8 sin8 ) = -6 Y2 ,-1 = -2(2 + I) Y2 ,-1 [i,e, I = 2] 
and hence 
2/£ ~ 
- 6Y2,-1 = -/i2Y2,-I , implyingthatCTI-I-
and the angular momentum is {2(2 + I)} 1/21i = 161 /21i \' 
2 1 a2Y3,3 1 a , naY3,3 3 3¢ A Y33 = -- - -- + -- - sm u -- where Y3 ,3 = N sin 8e • , 
, sin2 8 a¢2 sin 8 a8 a8 
aY33 ' 2 3'''' 
--' = 3Nsm 8cos8e 'Y', 
a8 
P9.2S 
P9.27 
QUANTUM THEORY: TECHNIQUES AND APPLICATI ONS 193 
1 a a Y3 3 I a 3 3 '" 
-- - sin e --' = -- - 3N sin e cos ee ''+' 
sin e ae ae sin e ae 
a2 Y3 ,3 
sin2 e a</!2 
so A2Y3,3 = -12N sin3 e e3ir/> = -1 2Y3,3 = -3(3 + I)Y3,3 [i .e. l = 3] 
and hence 
21E ~ 
-
12Y3.3 = -/i2Y3.3 , implying that ~-I-
and the angular momentum is (3(3 + I)} 1/2 II. = 12.)311.1. 
From the diagram in Fig. 9.2, cose = mt!(l(l + 1)}1 / 2 and hence e = arccos m, (l(l+I ) }1 /2 
II 
Figure 9.2 
1 I 
For an ct electron, m, = +2' s = 2 and (with m, -+ m,., 1 -+ s) 
The minimum angle occurs for m, = I: 
lim emin = lim arccos ( l 1/2 ) = lim arccos~ = arccos 1 = [2]. 
I--> 00 I--> 00 (l (l + I ) } I--> 00 l 
j k 
i=r XP= x y Z [see any book treating the vector product of vectors] 
Px py pz 
P9.29 
194 STUDENT'S SOLUTIONS MANUAL 
Therefore, 
Iy = (zPx - xpz) = ~ (z~ - x~) , 
I ax az 
Iz = (ipy - YPx) = ~ (x~ -y~) . 
I ay ax 
A Ii a A A AA AA 
We have used Px = i ax ' etc. The commutator of Ix and Iy is (lxly - 1), /.,), We note that the operations 
always imply operation on a function . We form 
AA 2(a a)( a a) Ixl,! = - Ii y - - z- z- -x- f 
. az ay ax az 
and AA 2( a a)( a a) /,Ix! = -Ii z- - x- y - - z- f 
.I ax az az ay 
Since multiplication and differentiation are each commutative, the results of the operation Ixl)' and Iylx 
differ only in one term. For lyl.J , x(af lay) replaces y(af lax). Hence, the commutator of the operations, 
(Ixl)' -1)'lx) is _ 1i2 (y~ -x~) orl -~Iz I. ax ay I 
COMMENT. We also would find 
~A ~ ~ ~A ~ A ~ A ~ A 
We are to show that [I , IJ = [I; + l" + Iz , Izl = [Lx ' Izl + [I)', Izl + [Iz , Izl = 0 
The three commutators are: 
A2 A /2 A A /2 A3 A3 [lz , lzl = lz Iz - Izlz = lz - lz = 0, 
A2 A /2 A A /2 /2 A A A A A A A A /2 [lx, lzl = lxlz -l:'x = '-.Iz -l.tlzlx + lxlzlx - Iz'-. 
= Ix(lxlz - Izlx) + (l.Jz - Izlx)lx = Ix [lx, Izl + [lx, Izl/x 
= Ix(-ilily) + (-ilil)' )lx = - ilia.l)' + I)x) [9.S6aJ, 
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS 195 
12 ' 12' , 12 12' , , , , , , , 12 [iy ,lzl = ly Iz - Izly = Iy Iz - lyizly + Iyll ly - Il ly 
= ly(/ylZ -Illy) + (Iyll -IZly)ly = Iy[ly, III + [I.", IzJly 
= Ly(inLx) + (inLx)ly = in(Lylx + lx/y) [9.56al. 
Therefore, [/2, IzI = -in(LxLy + Lvlx) + in(lx/y + LyLx) + 0 = O. 
, , , , ,,' '2 
We may also conclude that [12, Ix I = 0 and [12, Iyl = 0 because Ix, Iy, and Iz occur symmetrically in I . 
Solutions to applications 
P9.31 (a) The energy levels are given by 
and we are looking for the energy difference between n = 6 and n = 7: 
Since there are 12 atoms on the conjugated backbone, the length of the box is 11 times the bond 
length, 
L = 11(140 x 10- 12 m) = 1.54 x 10-9 m, 
_ (6.626 x 10- 34 J s)2(49 - 36) = 1330 10- 19 J 1 
so f.,E - 8(9.11 x 10- 31 kg)(1.54 x 10-9 m)2· x. 
(b) The relationship between energy and frequency is 
f.,E =hv so f.,E 3.30 x 10-
19 J 1 1 
v = - = = 4.95 X 10- 14 s-I. 
h 6.626 x 10- 34 J s 
(c) Look at the terms in the energy expression that change with the number of conjugated atoms, N. The 
energy (and frequency) are inversely proportional to L 2 and directly proportional to (n + 1)2 - n2 = 
2n + I , where n is the quantum number of the highest occupied state. Since n is proportional to 
N (equal to N / 2) and L is approximately proportional to N (strictly to N - 1), the energy and 
frequency are approximately proportional to N- 1• So the absorption spectrum of a linear polyene 
shifts to 1 lower !trequency as the number of conjugated atoms 1 increases I. 
P9.33 In effect, we are looking for the vibrational frequency of an 0 atom bound, with a force constant equal 
to that of free CO, to an infinitely massive and immobile protein complex. The angular frequency is 
where m is the mass of the 0 atom, 
m = (I6.0u)(1.66 x 10- 27 kgu - I) = 2.66 x 1O- 26 kg, 
196 STUDENT'S SOLUTIONS MANUAL 
and k is the same force constant as in Problem 9.2, namely, 1902 N m- I: 
( 
1902Nm-1 )1 /2 
W = = 12.68 X 1014 s-I I. 2.66 x 10 26 kg 
P9.35 The angular momentum states are defined by the quantum number m[ = 0, ± I, ±2, etc. By rearranging 
eqn 9.42, we see that the energy of state m[ is 
and the angular momentum is 
(a) If there are 22 electrons, two in each of the lowest II states, then the highest occupied state is 
m[ = ±5, so 
lz = ±5h = ±5 x (1.055 x 10-341 s) = 15.275 x 10- 34 1 s I 
25h2 
andE±5 = --. 21 
The moment of inertia of an electron on a ring of radius 440 pm is 
1 = mr2 = 9.11 x 10- 31 kg x (440 x 10-12 m)2 = 1.76 x 10-49 kg m2. 
= 25 x (1.055 x 10- 341 s)2 = 17.89 x 10- 19 1 I. 
Hence E±5 2 x (1.76 x 10 49 kg m2) 
(b) The lowest unoccupied energy level is m[ = ±6, which has energy 
_ 36 x (1.055 x 10-34 1 s)2 _ 1 14 10- 18 1 E±6 - 2 - . x . 
2 x (1.76 x 10-49 kg m ) 
Radiation that would induce a transition between these levels must have a frequency such that 
_ _ /""E _ (11.4 -7.89) x 10- 19 J -152 1014H I hv - /""E so v - - - 34 -. x z . h 6.626 x 10- J s 
This corresponds to a wavelength of about 570 nm, a wave of visible light. 
P9.37 The Coulombic force is 
F--~ qlq2 _ ~ 
- dr 4rrfor - 4rrf or2' 
For two electrons 2.0 nm apart, the force is 
(1.60 x 10- 19 C)2 I I F- 92=5.8xlO- II N . 
- 4rr x (8.854 x 10 I2C2l I m I) x (2.0 x 10 m) 
QUANTUM THEORY: TECHNIQUES AND APPLICATIONS 197 
P9.39 (a) In the sphere, the Schrodinger equation is 
li2 (a 2 2 a I 2) 
-- - + -- + -A 1/1" = E1/1" [9.5 Ia] 2m ar2 r ar r2 
where A 2 is an operator that contains derivatives with respect to () and 4> only. 
Let 1/1"(r ,(}, 4» = X(r)Y«(},4» . 
Substituting into the Schrodinger equation gives 
-- Y-+--+-A Y =EXY. li
2 (a2X 2yaX x 2 ) 
2m ar2 r ar r2 
Divide both sides by XY: 
The first two terms in parentheses depend only on r , but the last one depends on both r and angles; 
however, multiplying both sides of the equation by r2 will effect the desired separation: 
Put all of the terms involving angles on the right-hand side and the terms involving distance on the 
left: 
Note that the right side depends only on () and 4>, while the left side depends on r. The only way 
that the two sides can be equal to each other for all r, (), and 4> is if they are both equal to a constant. 
Call that constant -(li21(l + 1))/2m (with 1 as yet undefined) and we have, from the right side of 
the equation, 
li2 2 li21(l + I ) 
- A Y = - - ...,-----'-
2mY 2m 
so A2y = -1(1 + I )Y. 
From the left side of the equation, we have 
_~ (r2 a2x + 2r ax ) _ Er2 = _ li21(1 + I). 
2m X ar2 X ar 2m 
After multiplying both sides by X / r2 and rearranging, we get the desired radial equation 
_~ ( a2X + ~ ax ) + li21(l + I ) X = EX. 
2m ar2 r ar 2mr2 
Thus the assumption that the wavefunction can be written as a product of functions is a valid one, 
for we can find separate differential equations for the assumed factors. That is what it means for a 
partial differential equation to be separable. 
198 STUDENT'S SOLUTIONS MANUAL 
(b) The radial equation with l = 0 can be rearranged to read 
a2x 2 ax 2mEX 
-+------
ar2 r ar - IL2· 
Form thefollowing derivatives of the proposed solution: 
ax = (2rcR)-1 /2 [coS(nrcr/R) (nrc) _ Sin(nrcr/R)] 
ar r R r2 
d a2x _ 2 -1 / 2 [ sin(nrcr/R) (nrc)2 2cos(nrcr/R) (nrc) 2Sin(nrcr/R)] an - - ( rcR) - - - - + ---;:---
ar2 r R r2 R r3 . 
Substituting into the left side of the rearranged radial equation yields 
(2rcR) - 1/2 [_ sin(nrcr/R) (nrc)2 _ 2cos(nrcr/R) (nrc) + 2Sin(nrcr/R)] 
r R r2 R r3 
+ (2rcR)-1 /2 [2COS(nrcr/ R) (nrc) _ 2Sin(nrcr/R)] 
r2 R r3 
_1 /2sin(nrcr/R) (nrc)2 (nrc)2 
= -(2rcR) - = - - x. 
r R R 
Acting on the proposed solution by taking the prescribed derivatives yields the function back 
mUltiplied by a constant, so the proposed solution is in fact a solution. 
(c) Comparing this result to the right side of the rearranged radial equation gives an equation for the 
energy 
E = (nrc) 2 If-. = n2rc2 (.!!...)2 = n2h2 . 
R 2m 2mR2 2rc 8mR2 
	Part 2 - Structurenull
	9 - Quantum Theory: Techniques and Applicationsnull