5-3. E k
= eV o
= p 2
2 m !
2
=
( )
2
2 mc 2
!
2
V o
=
1 e (
( 1240 eV nm )
2
"
5 eV ) ( nm )
2
5-4. !
= h p
= h
2 mE k
=
2 hc
(from Equation 5-2) k
1240 eV nm
(a) For an electron: !
=
"
%
( ) ( .
$
6 eV )( .
$
3 eV ) #
&
(b) For a proton: !
=
$
&
( ) (
(c) For an alpha particle:
1240 eV nm
!
=
$
&
( ) (
#
9 eV )(
#
3 eV ) %
'
1240 eV nm
.
#
6 eV )( .
#
3 eV ) %
'
= # " 4 nm
=
=
#
=
940 V nm
" 4 nm
5-5. !
= / = / 2 mE k
= hc /
$
2 mc 2 ( kT ) (from Equation 5-2)
Mass of N
2
molecule =
!
( / 2 )
=
1240 eV nm
!
=
( ) ( .
%
10 eV ) ( ) (
%
" 5
!
)
4
( 300 K )
2
=
=
.
!
10 / 2 nm
5-11. !
= h p
= h
2 m E k
= 0 25 nm
Squaring and rearranging,
E k
= h 2
2 m p
!
2 n !
= D sin " sin !
=
=
( )
2
2 ( 2 )
!
2
=
(
( 1240 eV nm )
2
6 eV ) ( nm )
2
=
# sin " = n !
/ D = ( )(
" !
= 55 ° nm ) ( nm ) eV

Chapter 5 – The Wavelike Properties of Particles
(Problem 5-5 continued)
 
  
.
 10 eV
1240 eV nm
   
 
5
5-6.
  h
 p h
2 mE k
 hc
2
2 mc E k


 
300 K

1240 eV nm
.
 6 eV
  eV


 nm nm
5-7. (a) If there is a node at each wall, then n
 
/ 2
 
L where n

1, 2, 3,... or
 
2
(b) p
 h /
  hn / 2 E
 p
2 / 2 m
  hn / 2 L

2
/ 2 m
 2 2 h n / 8 mL
2
E n

 
2 hc n
2
8
2 2 mc L
For n = 1: E
1



1240 eV nm
 6 eV
2
  nm

2

For n = 2: E
2

.
eV
 
2 
.
eV eV
5-8. (a)
  
is a nonrelativistic situation, so c
10
2 c




2
2 hc mc E k
  hc mc
2
  mc
2
2 E k

1 2
E k

2 mc
2
   c

2

.
 6
 
2 eV
 eV
(b)
  
is a relativistic for an electron, so c
  h
 mu u c
1
 
/

2
 h mc



 c
1

 
2
 
2




 c


2
 u c
 c

1
  c

2
  u
 h
 m .
.
110

5-12. (a) n !
= D sin
" # D = n !
sin
"
= sin " nhc
2 k
=
( )( eV nm )
( sin .
° ) ( .
#
5 eV ) ( 50 eV )
= nm
(b) sin " = n !
D
=
( nm )
( )( eV nm )
( .
%
5 eV ) ( 100 eV )
!
= sin " 1 ( 0 584 ) = .
°
=
5-17. (a) y = y
1
+ y
2
= m ( 8 0 x m
!
400 / ) + m ( 7 6 x m
!
380 / )
= ( m ) cos
!
&
1
2
( 8 0 x m # 7 6 x m ) #
1
2
( 400 /
# 380 / )
"
'
# cos
!
'
1
2
( 8 0 x m + 7 6 x m ) $
1
2
( 400 /
+ 380 / )
"
(
= m ( 0 2 /
!
10 / ) " ( 7 8 x m
!
390 / )
(b) v
!
= k
=
390 / s m
= 50
(c) v s
=
" !
=
20 / s
= 50
" k m
(d) Successive zeros of the envelope requires that 0 2
"
/
= !
, thus
# x
=
!
5-22. (a)
!
= h p
=
=
5
!
m with
# k
= k
1
" k
2
= h
2 mE k
=
2 hc k
=
(
0 4 m " 1 and
# x
1240 eV nm
=
2
#
!
k
=
.
$
6 eV ) ( 5 eV )
5
!
m .
= nm d sin !
= " / 2 For first minimum (see Figure 5-17).
d
=
!
2 sin
"
= nm
2 sin 5
°
=
3 15 nm slit separation
(b) sin 5 ° = 0 5 cm L where = distance to detector plane L = cm
2 sin 5 °
= cm

5-23. (a) The particle is found with equal probability in any interval in a force-free region.
Therefore, the probability of finding the particle in any interval proportional to
∆ x is
∆ x.
Thus, the probability of finding the sphere exactly in the middle, i.e., with
∆ x = 0 is zero.
(b) The probability of finding the sphere somewhere within 24.9
cm to 25.1
cm is
proportional to ∆ x =0.2
cm . Because there is a force free length L = 48 cm available
to the sphere and the probability of finding it somewhere in L is unity, then the
probability that it will be found in ∆ x = 0.2
cm between 24.9
cm and 25.1
cm
(or any
interval of equal size) is: P x
= ( )( cm ) =
0 00417 .
5-24. Because the particle must be in the box #
0
L
Let u = !
/ ; x = 0 " u = 0 ; dx
L
#
0
A 2 sin 2 ( " ) =
1 x = L " u = !
and dx = ( L /
!
) du , so we have
0
!
( / !
) sin 2 udu = ( / !
)
!
0 sin 2 udu = 1
( L /
!
) A 2
!
)
0 sin 2 udu
= ( L /
!
) A 2
"
' u
2
$ sin 2 u
#
!
4
0
!
A 2
=
2 / L
"
A
= ( 2 / L )
= ( L /
!
) ( !
/ 2 ) =
( ) /
5-25. (a) At x = 0 : Pdx = !
( )
2 dx = Ae 0
2 dx = A dx
(b) At x
= !
: Pdx
=
Ae " !
2 / 4 !
2
2 dx
=
Ae "
2 dx
=
(c) At x
=
2
!
: Pdx
=
Ae " 4 !
2 / 4 !
2
2
=
" 1
2 dx Ae dx
=
(d) The electron will most likely be found at x = 0, where Pdx is largest.

Chapter 5 – The Wavelike Properties of Particles
5-32. Because c
 f

for photon,
 
/

/

/ , so
E
 hc


1240 eV nm
 
3 nm

.
 5 eV and p

E
 c
 8
5 eV
  
7
For electron: p
 h x



15

12 eV s m
  
4
Notice that ∆ p for the electron is 1000 times larger than p for the photon.
5-33. (a) For
48
Ti:

E
 upper state
 
 t

 
14 s



34
J s

13

  
10
MeV

E
 lower state
 
 t

 
12 s



34
J s

13

E
 total
  
E
U
 
E
L
  
10
MeV

E
T
E

(b) For Hα: 
E
U

 
10
MeV
MeV
 

10

8 s



34
J s

19

10

 
and

E
L


E
T
 
 
8 eV also.

8 eV

  
10
MeV

7 eV is the uncertainty in the Hα transition energy of 1.9
eV .
5-34.
 t 1

2
   
1
For the visible spectrum the range of frequencies is

7 5 4 0
 
10
14 
.
 14
Hz
The time duration of a pulse with a frequency uncertainty of
 f is then:
1
2
  f

1
2
 
3 5 10 14 Hz

4 5 10

16 s
 fs
117

5-27. E t h E h /
# t =
"
10 !
7 s (
"
!
34 J s
!
19 )
$ "
!
9 eV
5-35. The size of the object needs to be of the order of the wavelength of the 10 MeV neutron.
!
=
/
=
/ " mu .
" u
E k
= m c 2 ( !
" ) !
" = MeV / 939 MeV
!
= + = =
(
" / 2 ) or u = 0 14 c
Then,
!
=
" h mu
=
&
" hc
( / )
'
=
#
&
( .
)
1240 eV nm
(
%
6 eV ) ( ) $
'
= fm
Nuclei are of this order of size and could be used to show the wave character of
10 MeV neutrons.
5-39. h
2
# !
t
= h
2
!
E
# t =
!
16 eV s
6 eV
=
5-40. (a) For a proton or neutron:
" !
24 s h
2
and !
p m v assuming the particle speed to be non-relativistic.
# v = h
2 #
=
(
"
"
!
27 kg
!
34 J s
)( 10 !
15 m )
= "
7
$ 0 1 c (nonrelativistic)
(b) E k
#
1
2 mv 2
=
(
" !
27 kg )(
2
.
"
7 )
2
= "
!
13 J
=
MeV
(c) Given the proton or neutron velocity in (a), we expect the electron to be relativistic,
in which case, E k
= mc 2 ( !
" )

" p
= h
2
" x
# !
mv
$ !
v
#
2 h
For the relativistic electron we assume v c
!
$ h
2 %
=
(
#
" 31 kg )(
#
" 34 J s
#
8 )( 10 " 15 m )
= 193
" 31 kg )( .
#
8 )
2
( 192 ) = # " 11 J = 98 MeV E k
= mc 2 ( !
" 1 ) =
(
#