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1. Across what potential difference does an electron have to be accelerated in order to reach
the speed v = 9e7 m/s?
Non-relativistically
V
Relativistically
V
When should you use relativistic calculations?
V
+
Answer:
Required speed of electron v = 9 x 107 m/s
Charge of electron = 1.6 x 10-19 C
Rest mass of the electron m0 = 9.1 x 10-31 kg
+
+
+
Consider an arrangement in which an electron is
placed in between two plates of potential difference
of V volts. Then one of the plates is at negative
potential and the other is at positive as shown in
figure. As electron is a negatively charged particle,
it will get accelerated towards the positive plate. In
the potential zone of V volts, it gains an electric
potential energy and this potential energy then gets
converted into the kinetic energy.
v
–
e–
–
–
–
–
–
–
+
+
+
Non-relativistic kinetic energy = ½ mv2.
= ½ x 9.1 x 10-31 x (9 x 107 )
= ½ x 9.1 x 10-31 x 81 x 1014
= ½ x 9.1 x 10-31 x 81 x 1014
= 368.55 x 10-17 J
Thus Electric Potential Energy = Kinetic energy
i.e
. ch arg e  Potential 
----- (1)
1
mass velocity2
2
(Non-relativistic case)
1 2
mv
2
m0 v 2
V
2e
i.e. e V 
i.e.
i.e
.

i.e.
9.1 10 31  9  10 7
V
2  1.6  10 19
i .e.
V

2
9.1  10 31  81  1014
9.1  81

 10 2  230.34375 volt
19
3.2
3.2  10
Relativistic Case:
When an electron moves with the speed comparable to the speed of light, its rest mass can
not be considered. The mass of the electron will be more than the its rest mass when it
moves at such a high velocity, comparable to the speed of light.
In this case,
The total energy of the electron = Relativistic Kinetic Energy + Rest Mass Energy
 Relativistic Kinetic Energy= Total energy of the electron – Rest Mass Energy
i.e.
EK
= mc2 – m0c2
----- (2)
When an electron moves with the speed comparable to the speed of light, its mass m to its
rest mass m0 relates by the equation
m
m0
1
v2
--- (3)
c2
Using equation (3) in equation (2), we get
EK = mc2 – m0c2
m0c2

 m0c2
2
v
1 2
c






1
 m0c2 
 1
v2


 1 2

c


Substituting the values, we get






1
-31
8 2 
E K  9.1  10 3  10 
 1
2
9  10 7


1



8 2
3

10








1
i.e. E K  9.1  10 -31  9  1016 
 1
81  1014


1



16
9  10










1
i.e. E K  81.9  10 -3116 
 1


-2
 1  9  10



1
i.e. E K  81.9  10 -15 
 1
 1  0.09

 1

i.e. E K  81.9  10 -15 
 1
 0.91 
-15
i.e. E K  81.9  10 1.0482  1
i.e. E K  81.9  0.0482  10 -15 = 3.94758 x 10-15 J
To achieve this much kinetic energy, the required accelerating potential then can be
calculated by using equation (1) as follows.
Electric Potential Energy = Relativistic Kinetic energy
i.e.
i.e.
i.e.
eV
= 3.94758
3.94758  10 -15 3.94758  10 -15
V

 2.4672  10 4 volt
-19
e
1.6  10
V = 24672 volt
Relativistically.
When should you use relativistic calculations?
Relativistic calculations should be used when the speed of a body or particle under
consideration is comparable to that of light.
2. An electron entering Thomson's e/m apparatus (Figure 3.2 and 3.3) has an initial
velocity (in horizontal direction only) of 0.7 107 m/s. Lying around the lab is a
permanent horseshoe magnet of strength 1.3 10-2 T, which you would like to use.
(a) What electric field will you need in order to produce zero deflection of the
electrons as they travel through the apparatus?
V/m
Answer:
Velocity of electrons
v = 0.7 x 107 m/s
Magnetic field induction
B = 1.3 x 10 –2 T
-19
Charge of electron = 1.6 x 10 C
Rest mass of the electron m = 9.1 x 10-31 kg
The deflection of the electrons on the screen will be zero when the forces acting on
the electrons due to electric and magnetic fields will balance each other.
i.e. Electric force
i.e. charge x Electric field
i.e.
eE
i.e.
E
i.e.
E
E
= magnetic force.
= charge x velocity x magnetic field induction
= e vB
= vB
= 0.7 x 107 x 1.3 x 10 –2
= 0.91 x 105 volt/meter = 9.1 x 104 V/m
[ since every quantity is in SI units and therefore electric field must be expressed in SI
unit i.e. in volt/meter or newton/coulomb.]
(b) When the magnetic field is turned off, but the same electric field remains, how large a
deflection will occur if the region of nonzero E and B fields is 2 cm long?
cm
Figure 3.2
Figure 3.3
[Relativistic mass of the electron when it passes through the plates at the speed of
0.7 x 107 m/s
m0
9.1  10 -31
9.1  10 -31
9.1  10 -31
9.1  10 -31
m




2
0.9997
v2
0.49  1014
1  0.0544  10 -2
0.7  10 7
1 2
1

1
2
c
9  1016
3  10 8




i.e. m 
9.1  10 -31
 9.10273  10 -31 kg  m0 ]
0.9997
Here velocity v of the electron is quite less as compared to the speed of light c. Hence the
non-relativistic calculations will also do. In all the calculations then, only rest mass of the
electron is used.
With only electric field and no magnetic field, electron will experience a force only due to
electric field. This electric force eE will produce an acceleration a in the electron in the
upward direction. Electron is a negatively charged particle and hence it will be attracted
towards the positively charged plate, i.e. it gets deflected upward. This acceleration in the
vertical upward direction can be calculated by using Newton’s second law of motion, (i.e.
force equation F=ma) as follows.
Electric Force = Force according to Newton’s Law
i.e.
eE
= ma
eE 1.6  10 19  0.91  10 5 1.6  9.1  10 4  10 19
i.e.
a


 1.6  1016 m / s 2
31
31
m
9.1  10
9.1  10
The electron will experience this acceleration for time t i.e. for the time till it is in the
region of both the plates i.e. in the region of L=2 cm = 2 x 10-2 m.
This time t can be obtained as follows.
Electron passes through this region at a constant initial speed along the length of the tube.
This speed is v = 0.7 x 107 m/s
distance
2  10 2

 2.8571  10 -9 second
Then time required t 
7
speed
0.7  10
vy
+
v0y =0
a
vx
_
When the electron enters the region of the electric field, its vertical velocity v0y
is zero. But the acceleration a in the vertical direction lifts the electron and it starts moving
towards positive plate. As a result, vertical velocity of the electron goes on increasing and
finally, when it exits the region of the electric field, i.e. the region in between the plates, its
vertical velocity becomes vy.
vy
This velocity vy can be obtained as follows.
vy = v0y + at
= 0 + 1.6 x 1016 x 2.8571 x 10-9
= 4.5714 x 107 m/s.
v

vx
Then the deflection () experienced by the electron is decided by the values of vy and vx.
As a result of the vertical component of the velocity, electron deflects along the angle  and
moves with the resultant velocity v.
tan θ 
From the figure,
 vy
θ  tan 1 
 vx
 = 81.29420
vy
vx
 4.5714  10 7

  tan 1 

 0.7  10 7



 4.5714  10 7
  tan 1 

 0.7  10 7



  tan 1 6.5306


3.A photon of wavelength 2.2 nm Compton scatters from an electron at an angle of 90°.
What is the modified wavelength? (Enter your answer correct to 5 significant
figures.)
nm
What is the percentage change,
/ ?
%
Answer:
Wavelength
Angle of deflection
 = 2.2 nm = 2.2 x 10-9 m
 = 900
The modified wavelength ’ relates to the original wavelength  of the photon by the
relation
h
1  cosθ 
λ'λ 
---- (1)
m0c
where
h – Planck’s constant,
h = 6.627 x 10-34 Js
m0 – rest mass of electron
m0 = 9.1 x 10-31 kg
c – speed of light
c = 3 x 108 m/s
 – Angle of scattering
 = 900
Then cos 90 = 0
h
1  0
 λ'λ 
m0c
i.e.
λ' λ 
h
m0c
i.e. Modified wavelength λ'  λ 
h
m0c
 2.2  10 9 
i.e.
6.627  10 34
9.1  10 31  3  10 8
 2.2  10 9  0.24275  10 11
 2.2  10 9  0.0024275  10 9
 2.2024275  10 9
’ = 2.2024 x 10-9 meter = 2.2024 nm
Δλ
 100
λ
λ'-λ
2.202427  10 9 - 2  10 9
0.202427  100

 100 
 100 
9
λ
2
2  10
20.2427

 10.12135 %
2
 %change in the wavelength = 10.12135%
%change =
4. A photon having 27 keV scatters from a free electron at rest. What is the maximum
energy that the electron can obtain?
keV
Answer:
Energy of the incident photon = 27 keV = 27 x 103 eV
= 27  103  1.6  10 19 joule
= 43.2 x 10-16 J
E = 4.32 x 10-15 J
The energy gained by the electron
= Energy of the incident photon – energy of the scattered photon.
= h – h’
= h ( –’)
----- (1)
Thus for the maximum energy gain by the electron,
 –’ must be maximum.
But  is constant

’ must be minimum.
From the relation c = ’’,
the wavelength of scattered photon ’ must be maximum..
Wavelength of the scattered photon ’ is given by
h
1  cosθ 
----- (2)
m0c
Thus wavelength ’ can be maximum only if the term that is to be added to  is maximum
h
1  cosθ  is maximum
i.e
if
m0c
i.e.
if cos = 0
i.e.
if  = 900.
λ'  λ 
Then max. wavelength of the scattered photon is
h
λ max  λ 
----- (3)
m0c
The frequency () of the photon can be obtained from E = h
E 4.32  10 15
i.e.
 
 0.6519  1019  6.519  1018 Hz
34
h 6.627  10
From relation c = , the corresponding wavelength () of the incident photon is
c
3  10 8
λ 
 0.4602  10 10 m
18
ν 6.519  10
Then maximum wavelength of the scattered photon
h
λ max  λ 
m0c
 0.4602  10 10  0.024275  10 10
(the second term taken directly from previous problem calculations)
 max = 0.484475 x 10-10 m
Then the minimum frequency of the scattered radiation
c
3  10 8
ν min 

 6.19227  1018 Hz
λ max 0.484475  10 -10
The maximum energy gained by the electron can be obtained from equation (1).
E = h ( –’)
= h ( –min)
= 6.627 x 10-34 (6.519 x 1018 – 6.19227 x 1018)
= 6.627 x 10-34 x 0.32673 x 10-18
= 2.1652 x 10-16
joule

----------------
2.1652  10 16
19
i.e.
1.6  10
E = 1 . 35327 keV
OR
E = 1353.27 eV
 1.35327  10 3 eV