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

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```a l o g y w i t h E q u a t i o n 1 . 2 7 , a n d s h o w t h a t t h e p a r t i c l e e x p e r i e n c e s a n a c c e l e r -
a t i o n g i v e n b y
t . v t . e
a = l i m - = v l i m - = v w
L ' > t - - > 0 L ' . t L ' > t - - > 0 L ' . t
( 1 . 3 0 )
2 9
30 Chapter 1 I The Dawn of the Quantum Theory
Thus we see that the particle experiences an acceleration and requires an inward force equal
to ma = m vw = m v2 / r to keep it moving in its circular orbit.
1-42. Planck's distribution (Equation 1.2) law gives the radiant energy density of electromag-
netic radiation emitted between v and v + dv. Integrate the Planck distribution over all
frequencies to obtain the total energy emitted. What is its temperature dependence? Do you
know whose law this is? You will need to use the integral
1oo x3 dx = n4 0 ex - 1 15
1-43. Can you derive the temperature dependence of the result in Problem 1-42 without
evaluating the integral?
1-44. Ionizing a hydrogen atom in its electronic ground state requires 2.179 X w-lS J of
energy. The sun's surface has a temperature of ~ 6000 K and is composed, in part,
of atomic hydrogen. Is the hydrogen present as H(g) or H+(g)? What is the tempera-
ture required so that the maximum wavelength of the emission of a blackbody ionizes
atomic hydrogen? In what region of the electromagnetic spectrum is this wavelength
found?
M A T H C H A P T E R A
C O M P L E X N U M B E R S
T h r o u g h o u t p h y s i c a l c h e m i s t r y , w e f r e q u e n t l y u s e c o m p l e x n u m b e r s . I n t h i s m a t h -
c h a p t e r , w e r e v i e w s o m e o f t h e p r o p e r t i e s o f c o m p l e x n u m b e r s . R e c a l l t h a t c o m p l e x
n u m b e r s i n v o l v e t h e i m a g i n a r y u n i t , i , w h i c h i s d e f i n e d t o b e t h e s q u a r e r o o t o f - 1 :
i = v C l
( A . 1 )
o r
i
2
= - 1
( A . 2 )
C o m p l e x n u m b e r s a r i s e n a t u r a l l y w h e n s o l v i n g c e r t a i n q u a d r a t i c e q u a t i o n s . F o r e x a m -
p l e , t h e t w o s o l u t i o n s t o
z
2
- 2 z + 5 = 0
a r e g i v e n b y
z = 1 ± R
o r
z = 1 ± 2 i
/
~ . ;
. - D - - &quot; ' ' &quot; <
L .
~--'
' \
\ . ' /
&quot; ' ' ,
\ . ;
I '
~ s--~1
' \ -
: : : : L - - '
w h e r e 1 i s s a i d t o b e t h e r e a l p a r t a n d ± 2 t h e i m a g i n a r y p a r t o f t h e c o m p l e x n u m b e r z .
G e n e r a l l y , w e w r i t e a c o m p l e x n u m b e r a s
Z = X + i y
( A . 3 )
w i t h
x = R e ( z )
y = I m ( z )
( A . 4 )
3 1
32
I
Math Chapter A I C 0 M P L E X N U M B E R S
We add or subtract complex numbers by adding or subtracting their real and
imaginary parts separately. For example, if z1 = 2 + 3i and z2 = 1 - 4i, then
z1 - z2 = (2- 1) + (3 + 4)i = 1 + 7i
Furthermore, we can write
2z1 + 3z2 = 2(2 + 3i) + 3(1 - 4i) = 4 + 6i + 3 - 12i = 7 - 6i
To multiply complex numbers together, we simply multiply the two quantities as
binomials and use the fact that i 2 = -1. For example,
(2- i)( -3 + 2i) = -6 + 3i + 4i- 2i 2
= -4 + 7i
To divide complex numbers, it is convenient to introduce the complex conjugate
of z, which we denote by z* and form by replacing i by - i. For example, if z = x + i y,
then z* = x - iy. Note that a complex number multiplied by its complex conjugate is
a real quantity:
zz* = (x + iy)(x- iy) = x 2 - i 2 / = x 2 + l (A.5)
The square root of zz* is called the magnitude or the absolute value of z, and is denoted
by lzl.
Consider now the quotient of two complex numbers
2+i
z = 1 + 2i
This ratio can be written in the form x + iy if we multiply both the numerator and the
denominator by 1 - 2i, the complex conjugate of the denominator:
EXAMPLE A-1
Show that
z = 2 + i ( 1 - 2i) = 4- 3i = ~ - ~i
1 + 2i 1 - 2i 5 5 5
_ 1 X iy z --------
- x2 + y2 xz + l
SOLUTION: _ 1 1 1 1 (x-iy) x-iy
z = ~ = X + iy = X + iy X - iy = x 2 + y2
X iy
x2 + y2- xz + l
Because complex numbers consist of two parts, a real part and an imaginary part,
we can represent a complex number by a point in a two-dimensional coordinate system
M a t h C h a p t e r A I C 0 M P L E X N U M B E R S
w h e r e t h e r e a l p a r t i s p l o t t e d a l o n g t h e h o r i z o n t a l ( x ) a x e s a n d t h e i m a g i n a r y p a r t i s
p l o t t e d a l o n g t h e v e r t i c a l ( y ) a x i s , a s i n F i g u r e A . l . T h e p l a n e o f s u c h a f i g u r e i s c a l l e d
t h e c o m p l e x p l a n e . I f w e d r a w a v e c t o r r f r o m t h e o r i g i n o f t h i s f i g u r e t o t h e p o i n t
z = ( x , y ) , t h e n t h e l e n g t h o f t h e v e c t o r , r = ( x
2
+ l )
1 1 2
, i s l z l , t h e m a g n i t u d e o r t h e
a b s o l u t e v a l u e o f z . T h e a n g l e e t h a t t h e v e c t o r r m a k e s w i t h t h e x - a x i s i s t h e p h a s e
a n g l e o f z .
E X A M P L E A - 2
G i v e n z = 1 + i , d e t e r m i n e t h e m a g n i t u d e , l z l , a n d t h e p h a s e a n g l e , 8 , o f z .
S 0 L U T I 0 N : T h e m a g n i t u d e o f z i s g i v e n b y t h e s q u a r e r o o t o f
z z * = ( 1 + i ) ( l - i ) = 2
o r l z l = 2
1
/
2
\u2022 F i g u r e A . 1 s h o w s t h a t t h e t a n g e n t o f t h e p h a s e a n g l e i s g i v e n b y
tan8=~=1
X
o r ( ) = 4 5 ° , o r n / 4 r a d i a n s . ( R e c a l l t h a t 1 r a d i a n = 1 8 0 ° j n , o r l a = n / 1 8 0 r a d i a n . )
W e c a n a l w a y s e x p r e s s z = x + i y i n t e r m s o f r a n d e b y u s i n g E u l e r ' s f o r m u l a
e i e = c o s e + i s i n e ( A . 6 )
w h i c h i s d e r i v e d i n P r o b l e m A - 1 0 . R e f e r r i n g t o F i g u r e A . l , w e s e e t h a t
a n d s o
I m ( z )
r
e
X = r C O S e a n d y = r s i n e
Z = X + i y = r C O S e + i r s i n e
= r ( c o s e + i s i n e ) = r e ; e
( A . 7 )
\u2022 ( x , y )
R e ( z )
F I G U R E A . 1
R e p r e s e n t a t i o n o f a c o m p l e x n u m b e r
z = x + i y a s a p o i n t i n a t w o - d i m e n s i o n a l
c o o r d i n a t e s y s t e m . T h e p l a n e o f t h i s f i g u r e
i s c a l l e d t h e c o m p l e x p l a n e .
3 3
34
I
Math Chapter A I C 0 M P L EX N U M B E R S
where
and
y
tan()=-
X
(A.8)
(A.9)
Equation A.7, the polar representation of z, is often more convenient to use than
Equation A.3, the Cartesian representation of z.
Note that
(A.lO)
and that
(A.ll)
or r = (zz*) 112 \u2022 Also note that z = e;e is a unit vector in the complex plane because
r 2 = (e;8 )(e-i8 ) = 1. The following example proves this result in another way.
EXAMPLE A-3
Show that e-ie = cos e - i sine and use this result and the polar representation of z
to show that le;e I = 1.
S 0 L UTI 0 N: To prove that e-ie = cos 8 - i sin 8, we use Equation A.6 and the fact
that cos e is an even function of e [cos( -8) = cos 8] and that sine is an odd function
of 8 [sin( -8) = -sin 8]. Therefore,
Furthermore,
e-ie = cose + i sin(-8) = cose- i sine
le;el = [(cose + i sin8)(cose- i sin8)] 112
= (cos2 8 + sin2 8) 112 = I
P r o b l e m s
A - 1 . F i n d t h e r e a l```