[Donald A. McQuarrie, John D. Simon] Physical Chem(BookZZ.org)
1279 pág.

[Donald A. McQuarrie, John D. Simon] Physical Chem(BookZZ.org)


DisciplinaFísico-química I6.504 materiais97.838 seguidores
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= 1 ifm = n 
mn 
=0 ifm jn 
3-18. Show that the set of functions 
is orthonormal (Problem 3-16). 
P r o b l e m s 
3 - 1 9 . I n g o i n g f r o m E q u a t i o n 3 . 3 4 t o 3 . 3 5 , w e m u l t i p l i e d E q u a t i o n 3 . 3 4 f r o m t h e l e f t b y 1 / f * ( x ) 
a n d t h e n i n t e g r a t e d o v e r a l l v a l u e s o f x t o o b t a i n E q u a t i o n 3 . 3 5 . D o e s i t m a k e a n y d i f f e r e n c e 
w h e t h e r w e m u l t i p l i e d f r o m t h e l e f t o r t h e r i g h t ? 
3 - 2 0 . C a l c u l a t e ( x ) a n d ( x
2
) f o r t h e n = 2 s t a t e o f a p a r t i c l e i n a o n e - d i m e n s i o n a l b o x o f l e n g t h 
a . S h o w t h a t 
a x = . ! ! . _ _ ( 4 ; r
2 
) 1 / 2 
4 J T 3 - 2 
3 - 2 1 . C a l c u l a t e ( p ) a n d ( p
2
) f o r t h e n = 2 s t a t e o f a p a r t i c l e i n a o n e - d i m e n s i o n a l b o x o f 
l e n g t h a . S h o w t h a t 
h 
a P a 
3 - 2 2 . C o n s i d e r a p a r t i c l e o f m a s s m i n a o n e - d i m e n s i o n a l b o x o f l e n g t h a . I t s a v e r a g e e n e r g y 
i s g i v e n b y 
l 
( E ) = - ( p 2 ) 
2 m 
B e c a u s e ( p ) = 0 , ( p
2
) = a
2
, w h e r e a c a n b e c a l l e d t h e u n c e r t a i n t y i n p . U s i n g t h e 
p p 
U n c e r t a i n t y P r i n c i p l e , s h o w t h a t t h e e n e r g y m u s t b e a t l e a s t a s l a r g e a s h
2 
j 8 m a
2 
b e c a u s e 
a x , t h e u n c e r t a i n t y i n x , c a n n o t b e l a r g e r t h a n a . 
3 - 2 3 . D i s c u s s t h e d e g e n e r a c i e s o f t h e f i r s t f e w e n e r g y l e v e l s o f a p a r t i c l e i n a t h r e e - d i m e n s i o n a l 
b o x w h e n a l l t h r e e s i d e s h a v e a d i f f e r e n t l e n g t h . 
3 - 2 4 . S h o w t h a t t h e n o r m a l i z e d w a v e f u n c t i o n f o r a p a r t i c l e i n a t h r e e - d i m e n s i o n a l b o x w i t h 
s i d e s o f l e n g t h a , b , a n d c i s 
( 
8 )
1
1
2 
n n x n n y n ; r z 
1 / f ( x , y , z ) = - s i n _ x _ _ s i n _ Y _ s i n _ z _ 
a b c a b c 
3 - 2 5 . S h o w t h a t ( p ) = 0 f o r t h e g r o u n d s t a t e o f a p a r t i c l e i n a t h r e e - d i m e n s i o n a l b o x w i t h s i d e s 
o f l e n g t h a , b , a n d c . 
3 - 2 6 . W h a t a r e t h e d e g e n e r a c i e s o f t h e f i r s t f o u r e n e r g y l e v e l s f o r a p a r t i c l e i n a t h r e e -
d i m e n s i o n a l b o x w i t h a = b = 1 . 5 c ? 
3 - 2 7 . M a n y p r o t e i n s c o n t a i n m e t a l p o r p h y r i n m o l e c u l e s . T h e g e n e r a l s t r u c t u r e o f t h e p o r p h y r i n 
m o l e c u l e i s 
9 9 
100 Chapter 3 I The Schrodinger Equation and a Particle In a Box 
This molecule is planar and so we can approximate the :n: electrons as being confined inside 
a square. What are the energy levels and degeneracies of a particle in a square of side a? The 
porphyrin molecule has 26 :n: electrons. If we approximate the length of the molecule by 
1000 pm, then what is the predicted lowest energy absorption of the porphyrin molecule? 
(The experimental value is~ 17 000 cm- 1.) 
~ 3-28. The Schrodinger equation for a particle of mass m constrained to move on a circle of 
radius a is 
where I = ma2 is the moment of inertia and e is the angle that describes the position of the 
particle around the ring. Show by direct substitution that the solutions to this equation are 
where n = ±(21 £) 112 jh. Argue that the appropriate boundary condition is lj!(O) = lj!(e + 
2:n:) and use this condition to show that 
n = 0, ±1, ±2, ... 
Show that the normalization constant A is (2rr)- 112\u2022 Discuss how you might use these 
results for a free-electron model of benzene. 
3-29. Set up the problem of a particle in a box with its walls located at -a and +a. Show that 
the energies are equal to those of a box with walls located at 0 and 2a. (These energies may 
be obtained from the results that we derived in the chapter simply by replacing a by 2a.) 
Show, however, that the wave functions are not the same and in this case are given by 
1 n:n:x 1/1 (x) = ~12 sin--
" a 2a 
n even 
1 n:n:x 
= al/2 cos -z;; n odd 
Does it bother you that the wave functions seem to depend upon whether the walls are 
located at ±a or 0 and 2a? Surely the particle "knows" only that it has a region of length 2a 
in which to move and cannot be affected by where you place the origin for the two sets of 
wave functions. What does this tell you? Do you think that any experimentally observable 
properties depend upon where you choose to place the origin of the x-axis? Show that 
axap > h/2, exactly as we obtained in Section 3-8. 
3-30. For a particle moving in a one-dimensional box, the mean value of xis aj2, and the mean 
square deviation is a; = (a2 /12)[1 - (6/rr2n2)]. Show that as n becomes very large, this 
value agrees with the classical value. The classical probability distribution is uniform, 
1 
p(x)dx = -dx 
a 
=0 otherwise 
3-31. This problem shows that the intensity of a wave is proportional to the square of its 
amplitude. Figure 3.7 illustrates the geometry of a vibrating string. Because the velocity at 
u 
Q 
0 
x x + d x 
F I G U R E 3 . 7 
T h e g e o m e t r y o f a v i b r a t i n g s t r i n g . 
a n y p o i n t o f t h e s t r i n g i s a u 1 a t , t h e k i n e t i c e n e r g y o f t h e e n t i r e s t r i n g i s 
1
1 
1 ( a u )
2 
K = - p - d x 
o 2 a t 
w h e r e p i s t h e l i n e a r m a s s d e n s i t y o f t h e s t r i n g . T h e p o t e n t i a l e n e r g y i s f o u n d b y c o n s i d e r i n g 
t h e i n c r e a s e o f l e n g t h o f t h e s m a l l a r c P Q o f l e n g t h d s i n F i g u r e 3 . 7 . T h e s e g m e n t o f t h e 
s t r i n g a l o n g t h a t a r c h a s i n c r e a s e d i t s l e n g t h f r o m d x t o d s . T h e r e f o r e , t h e p o t e n t i a l e n e r g y 
a s s o c i a t e d w i t h t h i s i n c r e a s e i s 
V = [ T ( d s - d x ) 
w h e r e T i s t h e t e n s i o n i n t h e s t r i n g . U s i n g t h e f a c t t h a t ( d s )
2 
= ( d x )
2 
+ ( d u )
2
, s h o w t h a t 
V = 1 T 1 + G : ) - 1 d x 
I I [ 2 ] 1 / 2 ) 
U s i n g t h e f a c t t h a t ( 1 + x )
1
1
2 
~ 1 + ( x / 2 ) f o r s m a l l x , s h o w t h a t 
1 
1
, ( a )
2 
V = - T _ . ! ! . _ d x 
2 0 a x 
f o r s m a l l d i s p l a c e m e n t s . 
T h e t o t a l e n e r g y o f t h e v i b r a t i n g s t r i n g i s t h e s u m o f K a n d V a n d s o 
E = I ! _ _ . ! ! . _ d x + - _ . ! ! . _ d x 
1
' ( a )
2 
T
1
' ( a )
2 
2 0 a t 2 0 a x 
W e s h o w e d i n C h a p t e r 2 ( E q u a t i o n s 2 . 2 3 t h r o u g h 2 . 2 5 ) t h a t t h e n t h n o r m a l m o d e c a n b e 
w r i t t e n i n t h e f o r m 
. n n x 
u ( x , l ) = D c o s ( w t + ¢ ) s m -
n n n n [ 
1 0 1 
~: 
I! 
ii 
~ 102 Chapter 3 I The Schrodinger Equation and a Particle In a Box 
where wn = vnn I l. Using this equation, show that 
and 
Using the fact that v = (TIp) 112 , show that 
Note that the total energy, or intensity, is proportional to the square of the amplitude.