Extension worksheet – Topic 6 - Cambridge Resources for the IB

Cambridge Physics for the IB Diploma
Answers to Coursebook questions – Chapter H4
1
a
E  m0 c2  9.11031  9 1016  8.19 1014 J  5.1105 eV  0.51 MeV .
b
i
Using classical mechanics: the work done is the change in kinetic energy
and so
1 2
2qV
2 1.6 1019 1.00 109
mv  qV  v 

 1.9 1010 m s 1
2
m
9.11031
which is greater than the speed of light.
ii
2
Using relativity: the work done is 1.00 109 eV  1.00 103 MeV and this is
the change in kinetic energy, i.e.
1000
EK  (  1)  0.511  1.00 103    1 
   1958 .
0.511
This implies that the speed of the electron is essentially the speed of
1
v2
1
v
1
 1 2 
  1
 0.99999987 .
light: 1958 
2
c 1958
c
19582
v2
1 2
c
EK  (  1)m0 c 2  10m0c 2    11 .
v2
1
v
1
 1  2  2   1  2  0.996 .
Hence, 11 
2
c 11
c
11
v
1 2
c
1
3
ET  5m0c 2  5  938  4690 MeV .
From E 2  (m0c2 ) 2  p 2c2 , pc  E 2  (m0c2 )2  46902  9382  4595 MeV .
Hence, p  4.6 103 MeV c1 .
1
4
The gamma factor is  
5
p  685 MeV c1 
6
The total energy is E  (m0c2 )2  p2c2  9382  5002  1063 MeV .
 2.29 .
1  0.92
Hence, ET   m0 c2  2.29  0.511  1.17  1.2 MeV .
685 106 1.6 1019
 3.7 1019 N s .
8
3.0 10
Hence, EK  1063  938  125 MeV .
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7
This question is a careless remnant of the old syllabus in which the mass was assumed
to change with speed and m   m0 . In other words, ET  mc2   m0 c2 .
m
ET 10.0 109 1.6 1019

 1.8 1026 kg .
c2
9 1016
8
Both are essentially the speed of light, since the total energy is so much greater than the
rest energy. A calculator would give v  c , a computer gives the answers in the
textbook (see page 815 in Physics for the IB Diploma).
9
The work done is 2.0 MeV and this is the change in kinetic energy. Since the electron
was accelerated from rest this is also the final kinetic energy.
Hence EK  (  1)m0c2  (  1)  0.511 MeV  2.0 MeV ,
leading to   1  3.91    4.91 .
v2
1
v2
1
 1 2 
 2  1
 0.9585
Therefore, 4.91 
2
2
c
4.91
c
4.912
v
1 2
c
v
and so  0.9585  0.979  0.98 .
c
1
10
The gamma factor is  
1
 3.57 .
1  0.962
Hence the kinetic energy is EK  (  1)m0 c 2  2.57  938  2.4 103 MeV . The change
in kinetic energy is the work done which in turn equals qV . Hence V  2.4 103 MV .
1
11
The gamma factor is  
12
The gamma factors in each case are
 7.1 .
1  0.992
Hence the kinetic energy is EK  (  1)m0c 2  6.1 938  5.7 103 MeV .
a

b

c

1
1  0.502
1
1  0.902
1
1  0.992
 1.155
 2.294
 7.089
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So in each case the kinetic energy of the electron would be
a
EK  (  1)m0c 2  0.155  0.511  0.079 MeV requiring an accelerating
voltage of 0.079 MV .
b
EK  (  1)m0 c 2  1.294  0.511  0.661 MeV requiring an accelerating voltage of
0.661 MV .
c
EK  (  1)m0c 2  6.089  0.511  3.11 MeV requiring an accelerating voltage of
3.11 MV .
13
a
The work done is
W  Fd  eEd  1.6 1019  5.0 106 103  8.0 1010 J  5.0 109 eV  5.0 103 MeV
Therefore EK  5.0 103 MeV (or 5.0 GeV).
b
The total energy is ET  (938  5.0 103 )  5.938 103 MeV .
Hence  
ET
5.938 103

 6.33 .
m0c2
938
Hence the speed is 6.33 
1
1
14
a
v2
1
v
1

  1
 0.987 .
2
2
c
6.33
c
6.332
From p   m0v and E   m0 c2 we get by dividing side by side to get rid of the
gamma factor:
b
v2
c2
 1
pc2
p v

 2 . Hence, v 
E
E c
pc2
m02c4  p 2c2
as required.
For the electron with momentum 1.00 MeV c1 ,
m02c4  p 2c2  0.5112  1.002  1.123 .
For the proton,
m02c4  p2c2  9382  1.002  938 .
Hence the ratio of the speeds is
c
ve
938

 835 .
vp 1.123
For a momentum of 1.00 GeV c1 , for the electron
m02c4  p2c2  0.5112  (103 )2  103 and for the proton
m02c4  p2c2  9382  (103 )2  1371 so that
d
ve 1371

 1.37 .
vp 1000
As the momentum increases we may neglect the rest energy in which case both
speeds tend to become the speed of light and so the ratio approaches 1.
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15
a
The momentum of the fragments must be zero since the original
momentum before the breakup was zero. The gamma factor is
1
1


 1.90
2
v
1  0.852
1 2
c
and so the momentum is 1.90 125 MeV c2  0.85c  202 MeV c1 . This is also
the momentum of the other fragment. Hence the total energy of each of the
fragments is E1  1252  2022  238 MeV and E2  2502  2022  321 MeV .
b
The total energy of the system therefore is 238  321  559 MeV . This is the rest
energy of the particle that broke up and so its rest mass is 559 MeV c2 .
16
17
a
The total momentum of the electron–positron pair is zero. If only one photon is
produced it will have momentum violating the law of momentum conservation.
b
Again because of momentum conservation.
c
The total energy of the electron is E  mc2  EK  0.51  2.0  2.51 MeV .
The positron has the same energy.
The total energy is then ET  2  2.51  5.0 MeV .
The photons must have the same energy because they move in opposite directions
with the same momentum (magnitude) and hence the same wavelength. So each
has an energy of about 2.5 MeV .
a
Initially the electron has zero momentum.
h
hc
The photon has energy
and momentum
in the x-direction.


After the interaction, the electron moves with some kinetic energy and the photon
h
hc
has energy
and momentum
in the direction shown in the diagram (see


page 676 in Physics for the IB Diploma).
Conservation of momentum implies that
h h
 cos   px
 
h
0  sin   p y

From these two equations we deduce that
p 2  px2  p y2
h h
h
 (  cos  ) 2  ( sin  ) 2
 

2
h
2h
h
 ( )2 
cos   ( ) 2

 

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Conservation of energy implies

 m0 c 2 
hc
 (m0 c 2 ) 2  p 2c 2 ,

hc
 m0c 2 ) 2 .
 
Hence, using the expression for momentum above,
hc
2h 2 c 2
hc
hc hc
(m0c2 ) 2  ( ) 2 
cos   ( ) 2  (   m0c 2 ) 2 . Expanding all,

 

 
and so (m0c 2 ) 2  p 2c 2  (
hc
hc

2h 2 c 2
hc
hc
hc
2h 2 c 2
hc
hc
cos   ( ) 2  ( ) 2  ( ) 2  ( m0c 2 ) 2 
 2 m0c 2  2 m0c 2

 



 


2 2
2 2
2h c
2h c
hc
hc

cos   
 2 m0 c 2  2 m0 c 2
 
 


h
   
(1  cos  )
m0 c
(m0 c 2 ) 2  (
hc
)2 
b
Some of the energy of the incoming photon has been used to provide the kinetic
energy of the electron and so the scattered photon has reduced energy.
hc
Since the energy of a photon is E 
the wavelength has increased.

c
Applying the formula just derived,
   

h
(1  cos  )
m0 c
6.63 1034
 1
 1  
31
8
9.110  3.0  10  2 
 1.211012 m
Hence    3.00 1012  1.211012  4.211012 m .
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18
The initial momentum of the photon is
p
h

6.63  1034
5.0  1012
 1.326  1022 N s

Because
   
h
(1  cos  )
m0 c
6.63 1034
 1  (1) 
9.11031  3.0 108
 4.857 1012 m

The wavelength of the scattered photon is therefore    9.857 1012 m .
The momentum of the scattered photon has magnitude
p
h

6.63 1034
9.857  1012
 0.673 1022 N s

And so the change in the momentum of the photon is
p  1.326 1022  (0.673 1022 )
 2.0 1022 N s
This means that the electron moves off with this momentum. The total energy of the
electron is therefore
ET  (9.11031  9 1016 ) 2  (2.0 1022  3 108 ) 2  1.015 1013 J  0.6345 MeV .
Hence the kinetic energy is EK  0.6345  0.511  0.124 MeV .
0.6345
 1.242 .
0.511
1
v2
1
v
1
 1 2 
  1
 0.593 .
Hence, 1.242 
2
2
c 1.242
c
1.2422
v
1 2
c
The gamma factor is  
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19
The gamma factor is  
1
1
v2
c2

1
1  0.802

5
3
5
and so the total energy of the pion is ET  135  225 MeV . The momentum of the
3
2
2
pion is: 225  (135)  ( pc)  pc  2252  1352  180 MeV  p  180 MeV c 1 .
Conservation of energy and momentum gives:
225  hf A  hf B
180 
hf A hf B

c
c
To simplify things set c = 1, which is all right since we are going to take a ratio and
units will not be important. Then
225  hf A  hf B
180  hf A  hf B
Adding, f A 
405
45
f
405
, subtracting, f B 
. The ratio is therefore A 
9.
2h
2h
fB
45
20
The momentum of the two bodies is zero and so the particle they form is produced at
1
1
5

 and so the momentum of each
rest. The gamma factor is  
v2
1  0.802 3
1 2
c
5
particle is  3.0  0.80  3 108  1.2 109 N s .
3
Hence the total energy of each of the particles is
E  (3.0  9.0 1016 ) 2  (1.2 109  3 108 ) 2  4.5 1017 J .
The rest energy of the particle that is formed and so its rest mass is therefore
9.0 1017
2  4.5 1017  9.0 1017 J and hence the rest mass is
 10 kg .
9.0 1016
21
a
p   m0v
b
E   m0 c 2
c
Eliminating the gamma factor from a and b we get
Hence v 
d
p v
 .
E c2
pc 2
as required.
E
For a massless particle, E  0  p 2c2  pc , hence v 
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pc2 pc2

 c.
E
pc
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