SERWAY HOMEWORK SOLUTIONS

ASSIGNMENT 8

6-3 (a) ( )π

λ = ×

102sin sin 5 10

xA A x so ( )π

λ−= × 10 12

5 10 m , π

λ −= = ××

1010

21.26 10 m

5 10 .

(b) λ

−−

−

×= = = ×

×

3424

106.626 10 Js

5.26 10 kg m s1.26 10 m

hp

(c) −= = ×2

319.11 10 kg2p

K mm

( )( )

−−

−

−

−

×= = ×

× ×

×= =

×

22417

31

17

19

5.26 10 kg m s1.52 10 J

2 9.11 10 kg

1.52 10 J95 eV

1.6 10 J eV

K

K

6-6 ( )

( )( ) ( )( )

ψ

ψ

ψ

ψ

= +

∂= − +

∂

∂= − −

∂

− −− = +

h h

22 2

2

2 2

cos sin

sin cos

cos sin

2 2cos sin

x A kx B kx

kA kx kB kxx

k A kx k B kxx

m mEE U A kx B kx

The Schrödinger equation is satisfied if

( )( )ψ

ψ∂ −

= −∂ h

2

2 22m

E Ux or

( ) ( )( )

2

−− + = +

h

2 2cos sin cos sin

mEk A kx B kx A kx B kx .

Therefore =h

2 2

2k

Em .

6-9 Equation 6.7 is

�� =����ℏ��� =�

���� �� ��

��

=2 2

28n

n hE

mL , so ∆ = − =2

2 1 23

8h

E E EmL

( )( )

( )( )−∆ = =

×

2

26 2 5

1240 eV nm3 6.14 MeV

8 938.28 10 eV 10 nm

cE

c

λ −= = = ×∆ ×

46

1240 eV nm2.02 10 nm

6.14 10 eVhc

E

This is the gamma ray region of the electromagnetic spectrum.

6-16 (a) ( )π

ψ =

sinx

x AL , = 3L Å. Normalization

requires

π

ψ = = =

∫ ∫2

2 22

0 01 sin

2

L L x LAdx A dx

L

so ( )=1 22

AL

( ) ( )ππ πψ φ φ

π π

= = = = − = ∫ ∫ ∫3 3 3 1 2

2 22

0 0 0

2 2 2 3sin sin 0.195 5

6 8

L L xP dx dx d

L L.

(b) π

ψ =

100sin

xA

L, ( )=

1 22A

L

( ) ( )

( )

ππ π πφ φ

π π

π

π π

= = = −

= − = − =

∫ ∫3 100 3

2 2

0 0

1002 2 1 100 1 200sin sin sin

100 50 6 4 3

1 1 2 1 3sin 0.331 9

3 200 3 3 400

L x LP dx d

L L L

(c) Yes: For large quantum numbers the

probability approaches 13.

6-17 (a) The wavefunctions and probability densities are the same as those shown in the two lower curves in Figure 6.16 of the text.

(b) π

ψ = =

∫ ∫3.5 Å 3.5 Å

221

1.5 Å 1.5 Å

2sin

1010 Å

xP dx dx

π

π −

3.5

1.5

1 10sin

5 2 4 5xx

In the above result we used

( )= −∫2 1

sin sin 22 4x

axdx axa .

Therefore,

( ) ( ){ }( ) [ ]

π π π

π π π

π ππ

= − = − − +

= + − = + =

3.5

11.5

1 5 1 5 3.5 5 1.5sin 3.5 sin 1.5 sin

10 5 10 5 51 5 1

2.0 sin 0.3 sin 0.7 2.00 0.0 0.20010 10

xP x

(c)

( ) ( )

( ) ( )[ ] ( )[ ]{ }{ }

ππ π

π π

π π

= = − = −

= + − =

∫3.5 3.53.5

22

1.5 1.51.5

1 1 5 1 5sin sin 0.4 sin 0.4

5 5 5 2 4 10 21

2.0 0.798 sin 0.4 1.5 sin 0.4 3.5 0.35110

x xP dx x x x

(d) Using =2 2

28n h

EmL

we find =1 0.377 eVE and =2 1.51 eVE .

6-32 The probability density for this case is

( )ψ −=22 2

0 0axx C e with ( )π

=1 4

0a

C and ω

=h

ma . For the

calculation of the average position ( )ψ∞

−∞

= ∫2

0x x x dx

we note that the integrand is an odd function, so that the integral over the negative half-axis < 0x exactly cancels that over the positive half-axis (

> 0x ), leaving = 0x . For the calculation of 2x ,

however, the integrand ψ 220x is symmetric,

and the two half-axes contribute equally, giving

( )( )π∞−= =∫

21 2

2 2 2 20 0

0

12 2

4axx C x e dx C

a a .

Substituting for 0C and a gives ω= =

h2 12 2

xa m and

( ) ( )ω∆ = − =

h1 21 22 2

2x x x

m .