The Chemistry Maths Book, Second Edition

(Grace) #1

644 Solutions to exercises


Section 14.3


4.f(x,t) 1 = 1 Ae


B(x− 2 t)





6.For separation constant λ


2

;


λ


2

1 = 1 0: f(x,y) 1 = 1 (a 1 + 1 bx)(c 1 + 1 dy)


λ


2

1 > 1 0: f(x,y) 1 = 1 [ae


λx

1 + 1 be


−λx

][c 1 cos 1 λy 1



  • 1 d 1 sin 1 λy]


7.f(x,y) 1 = 1 Ae


(cx−y 2 c)

Section 14.4



  1. (i)E


1,1

1 = 12 ,E


1,2

1 = 1 E


2,1

1 = 15 ,


E


2,2

1 = 18 ,E


1,3

1 = 1 E


3,1

1 = 110 ,


E


2,3

1 = 1 E


3,2

1 = 113 ,


E


1,4

1 = 1 E


4,1

1 = 117 ,E


3,3

1 = 118



  1. (i)


(ii)Degeneracy ifa 1 = 1 b 1 = 1 c:


1 if p,q,r all equal (e.g. E


2,2,2

)


3 if two equal (e.g. E


1,1,2

,E


1,2,1

,E


2,1,1

)


6 if all different (e.g. E


1,2,3

,E


1,3,2

,E


2,3,1

,


E


2,1,3

,E


3,1,2

,E


3,2,1

)


Section 14.5



  1. (i)E


0,1

1 = 1 5.7831,E


±1,1

1 = 1 14.6819,


E


±2,1

1 = 1 26.3744,E


0,2

1 = 1 30.4715,


E


±3,1

1 = 1 40.7070,E


±1,2

1 = 1 49.2186


Section 14.6



  1. (i)ψ


1,0,0

1 = 1 (Z


3

2 π)


122

e


−Zr

(ii)E 1 = 1 −Z


2

22



  1. (i)ψ


2,1,0

1 = 1 (Z


5

232 π)


122

re


−Zr 22

1 cos 1 θ


(ii)E 1 = 1 −Z


2

28



  1. (iii) , n 1 = 1 1, 1 2, 1 3,1=,


l 1 = 1 0, 1 1, 1 2,1=, where the allowed


values ofx


n,l

are the zeros of the


spherical Bessel functionj


l

(x).


(iv)


E


h


ma


10

2

2

8


,

=


ψ


100

1


2


,,

sin )


=


π


π


a


ra


r


(


E


x


ma


nl

nl

,

,

=


22

2

2


A


E


h


m


p


a


q


b


r


c


pqr,,

=++










22

2

2

2

2

2

8


×










×










22


b


qy


bc


rz


c


sin sin


ππ


ψ


pqr

xyz


a


px


a


,,

(, ,) sin=










2 π


fxy Ae


Bx y

(, )


()

=


+

22

Section 14.7











  1. (i)


(ii)


Chapter 15


Section 15.2







  1. (i)


Section 15.3



  1. (i)cos 1 tx 1 = 1 j


0

(t)P


0

(x) 1 − 15 j


2

(t)P


2

(x) 1



  • 19 j


4

(t)P


4

(x) 1 −1-


(ii)sin 1 tx 1 = 13 j


1

(t)P


1

(x) 1 − 17 j


3

(t)P


3

(x) 1



  • 111 j


5

(t)P


5

(x) 1 −1-






Section 15.4



  1. (ii)

  2. (ii)

  3. (ii)

  4. (ii)

  5. (ii)


x


l


2

1


4


4











π

cos



























1


6


2

cos


6 πx


l






ll x


l


22

2

6


41


2


2











π


π


2

cos






1


5


3


  • siin


5 πx


l






















81


3


3


2

33

lx


l


x


l
π

ππ


sin sin


























π


2

222

3


4


1


2


2


3


3


−−+−










cos cosxxxcos






2


1


2


2


3


3


4


4


sin sinxxxxsin sin


−+−+














7


7


−+






cos x






1


2


2


1


3


3


5


5


+−+






π


cos cosxxxcos


V


Q


R


=,=Qqr −


2

0

3

2

22

4


31


πε


(cos )θ


1


4


1


2


5


16


01 2

PP P++


c


aaa


1

135

3


357


= +++














uxy


ya


ba


(,) xa


sinh( )


sinh( )


= sin( )


3


3


3


π


π


π


uxy A n ya n xa


n

n

( , )= sinh( ) sin( )


=


ππ


1


T


x


l


=− 3 Dtl


22

sin exp [ ]


π


π /


yxt


x


ll


(,) sin cos= 3


ππvt

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