5-3.
 hc 
p2
Ek  eVo 

2
2m
2mc 2  2
5-4.

1240eV nm 
1
Vo 
 940V
e 2 5.11  105 eV  0.04nm 2
2
h

p
h

hc

2mEk
2
(from Equation 5-2)
2mc 2 Ek
(a) For an electron:  
(b) For a proton:  
1240eV nm

 2  0.511  106 eV


h

p
h
2m p Ek
 4.5  10 eV 
3
1/ 2
1240eV nm

 2  983.3  106 eV

(c) For an alpha particle:  
5-11.

 4.5  10 eV 
3
1/ 2
 0.0183nm
 4.27  104 nm
1240eV nm

 2  3.728  109 eV

 4.5  10 eV 
3
1/ 2
 2.14  104 nm
 0.25nm
Squaring and rearranging,
 hc  
1240eV nm 
h2

 0.013eV
Ek 
2
2
2
2
2m p 
2 mpc 
2 938  106 eV  0.25nm 
2

2



sin   n / D  1 0.25nm  /  0.304nm 
n  D sin 

sin   0.822
   55
5-12. (a) n  D sin   D 

n
nhc

sin  sin  2mc 2 Ek
11240eV

nm 
 sin 55.6  2 5.11  105 eV

 50eV 
1/ 2
 0.210nm
(b) sin  
11240eV nm 
n

 0.584
D  0.210nm   2 5.11  105 eV 100eV  1 / 2




  sin1  0.584   35.7
5-17. (a) y  y1  y2
 0.002m cos  8.0 x / m  400t / s   0.002m cos  7.6 x / m  380t / s 
1
1

 2  0.002m  cos   8.0 x / m  7.6 x / m    400t / s  380t / s  
2
2

1
1

 cos   8.0 x / m  7.6 x / m    400t / s  380t / s  
2
2

 0.004m cos  0.2 x / m  10t / s   cos  7.8 x / m  390t / s 
(b) v 
(c) vs 

k

390 / s
 50m / s
7.8 / m

20 / s

 50m / s
k 0.4 / m
(d) Successive zeros of the envelope requires that 0.2x / m   , thus

2
x 
 5 m with k  k1  k2  0.4m 1 and x 
 5 m.
k
0.2
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 ∆x is proportional to
∆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 24.9cm to 25.1cm is
proportional to ∆x =0.2cm. Because there is a force free length L = 48cm 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 = 0.2cm between 24.9cm and 25.1cm (or any
interval of equal size) is: Px  1 / 48  0.2cm   0.00417.
Let u   x / L;
 A  L /   sin
2
L
0
0
x  L  u   and dx   L /   du , so we have
x  0  u  0;

L
2
2
 *  dx  1   A sin  x / L  dx  1
5-24. Because the particle must be in the box

2
0
udu  A  L /    sin2 udu  1
2
0


u sin 2u 
  L /   A2  / 2   LA2 / 2  1
 L /   A  sin udu   L /  A  

4 0
2
0
2

2

2

A2  2 / L  A   2 / L 
1/ 2
5-25. (a) At x  0 :
(b) At x   :
(c) At x  2 :
Pdx    0, 0  dx  Ae0 dx  A2 dx
2
2
Pdx  Ae
2
/ 4 2
Pdx  Ae 4
2
2
/ 4 2
2
dx  Ae1 / 4 dx  0.61A2 dx
2
2
dx  Ae 1 dx  0.14 A2 dx
(d) The electron will most likely be found at x = 0, where Pdx is largest.
5-27. E t    E   / t 
5-34.
1.055  1034 J s
 6.6  109 eV
7
19
10 s 1.609  10 J / eV


 t  1  2f t  1
For the visible spectrum the range of frequencies is f   7.5  4.0   1014  3.5  1014 Hz
The time duration of a pulse with a frequency uncertainty of f is then:
t 
1
1

 4.5  1016 s  0.45 fs
14
2f 2  3.5  10 Hz
5-35. The size of the object needs to be of the order of the wavelength of the 10MeV neutron.
  h / p  h /  mu.  and u are found from:
Ek  mn c 2    1 or   1  10 MeV / 939 MeV
  1  10 / 939  1.0106  1 / 1  u 2 / c 2 
1/ 2
or u  0.14c
1240eV nm
h
hc


 9.33 fm
2
 mu  mc  u / c   1.0106  939  106 eV  0.14  


Nuclei are of this order of size and could be used to show the wave character of 10MeV
neutrons.
Then,  


5-36. (a) E  135MeV , the rest energy of the pion.
(b) E t 

2

6.58  1016 eV s

 2.44  1024 s
6
2E 2  135  10 eV
t 
5-40. (a) For a proton or neutron:
xp 
v 

and p  mv assuming the particle speed to be non-relativistic.
2

1.055  1034 J s

 3.16  107 m / s  0.1c (non-relativistic)
27
15
2mx 2 1.67  10 kg 10 m





1.67  1027 kg 3.16  107 m / s
1 2
(b) Ek  mv 
2
2

2
 8.34  1013 J  5.21MeV
(c) Given the proton or neutron velocity in (a), we expect the electron to be relativistic,
in which case, Ek  mc 2    1 and
p 

  mv
2x

v 

2mx
For the relativistic electron we assume v  c

1.055  1034 J s

 193
 
2mcx 2 9.11  1031 kg 3.00  108 m / s 1015 m





Ek  mc 2    1  9.11  1031 kg 3.00  108 m / s
5-41. (a) E 2  p 2 c 2  m 2 c 4
E  hf  

 192   1.58  10
2
p  h/  /k
11
J  98MeV
 2 2   2 k 2 c 2  m 2 c 4
v

k



k
2 k 2c 2  m2c 4
 c 1  m2c 2 / 2 k 2  c
k
d
d

(b) vs 
dk dk
k 2c 2  m2c 4

k
c2k
k 2c 2  m2c 4
2
c2k
c 2 k c 2 p



u


E
(by Equation 2-41)
5-46. (a) E   2 / 2mL2 (Equation 5-28) and E   2 / 2mA2
(b) For electron with A  1010 m :
E
 c 
2
2mc 2 A2


197.3eV
nm 
2

2 0.511  106 eV 101 nm

2
 3.81eV
For electron with A  1cm or A  102 m :

 / 10 nm   3.81 10 eV
1.055  10 J s 

2 100  10 g  10 kg / g  2  10 
E  3.81eV 101
2
(c) E 
2mL2
2
7
2
16
34
3
3
2
2
2
 1.39  1061 J  8.7  1043 eV