Lecture: 1
Modern Physics
Relativity:
Galileo observed that if a ship moved uniformly on the sea, a sailor could not
distinguish between the situations either the ship is in motion and the sea is at
rest or the ship is at rest and the sea is in motion. In Galileo’s view- the only
motion that is measurable is the relative motion between the ship and the sea;
hence the term “Relativity” was manifested first ever.
To consider something relative to other is called relativity.
There are two parts of Einstein’s theory, viz, Special relativity and general
relativity.
Special relativity (1905), treats problems that involve inertial frame of reference.
General relativity (1915), treats problems that involve non inertial frame of
reference.
Frame of Reference:
The co-ordinate system, relative to which any measurement can be done to
specify the state of rest or motion of an object, is known as the frame of
reference.
When we have the origin and the directions in which to measure the distance
from the origin, and a clock to measure the time, we say that we have a frame of
reference or simply a frame.
Inertial Frame:
An inertial frame of reference is one in which Newton’s first law of motion holds
good. In this frame, an object at rest remains at rest and an object in motion
continues to move at constant velocity if no force acts on it. Any frame of
reference that moves at constant velocity relative to an inertial frame is itself an
inertial frame.
Non-inertial Frame:
A reference frame in which a body is accelerated without being acted upon by
external force is called a non-inertial frame of reference. Newton’s laws are not
valid in such a frame of reference.
Failure of Newtonian Mechanics:
For an electron accelerated through a 10 MeV potential difference, a value
reasonable easy to obtain, the speed v equals 0.9988c.
If the energy of 10 MeV is increased by a factor of four (to 40 MeV) experiment
shows that the sped is not doubled to 1.9976 c as we might expect from the
1
Newtonian relation 𝐾𝐸 = 2 𝑚𝑣 2
but remain below c; it increases only from 0.9988 c to 0.9999 c, a chance of
0.11%.
Why Galilean transformation introduced?
Newton’s laws of motion does not tell us whether there is one or many inertial
frames of reference, nor, if there is more than one, how we are to relate the
coordinates of an event as observed from the one inertial reference frame to the
coordinates of the same event as observed in other inertial reference frame.
Galilean Transformation:
The transformation of co-ordinates of a particle from one inertial frame of
reference to another is called the Galilean transformation.
Let us consider we are in an inertial frame
of reference S and the coordinates of an
event that occurs at the time t are (x, y, z).
An observer located in a different inertial
frame S which is moving with respect to
S at constant velocity v, will find that the
same event occurs at the time t and has
the coordinates (x, y, z). Let us suppose
that at t = t = 0 , S and S were coincide.
For convenience, let v is in the +x
direction as shown in the figure.
Galilean transformation
equations
𝑥 ′ = 𝑥 − 𝑢𝑡
𝑦′ = 𝑦
𝑧′ = 𝑧
𝑡′ = 𝑡
Inverse Galilean
Transformation equations
𝑥 = 𝑥 ′ + 𝑢𝑡′
𝑦 = 𝑦′
𝑧 = 𝑧′
𝑡 = 𝑡′
Postulates of special theory of relativity:
1. Principle of equivalence of physical laws: “The laws of physics are the same
in all inertial frame of reference. No preferred inertial system exists”.
2. Principle of constancy of velocity of light: “The velocity of light in free space
(vacuum) has the same value c  2.998×108 m/s in all inertial frames of
reference, independent of the relative velocity of the source and observer”.
Failure of Galilean transformation:
The second postulate calls for the same value of the speed of light c when
determined in S or S. This contradicts with Galilean transformation. If we
measure the speed of light in the x-direction in the S system to be c, however,
in the S system it will be
𝑐′ = 𝑐 − 𝑣
Clearly a different transformation is required if the postulates of special
relativity are to be satisfied.
Concept of Ether:
Huygens' assumed that light wave propagated through the medium named
as ‘Luminiferous Ether’. But problems arose when Ether is attributed to some
strange properties:
1. Since light passes through vacuum, so Ether must be massless and of
zero density and must have perfect transparency to account for its
undetectability.
2. As the velocity of wave propagation depends on the elasticity of the
medium, Ether must be highly elastic. Thus, the whole free space must be
filled up with such an elastic medium to sustain the vibration of light
waves.
A solution of Maxwell’s equation gives the speed of light,
𝑐=
1
= 2.998x108 m/sec
𝜇𝑜 ε 𝑜
This is now well known that light waves are transverse electromagnetic waves
and do not need any medium.
Thank YOU
Lecture: 2
Modern Physics
Concept of Ether:
Huygens' assumed that light wave propagated through the medium named
as ‘Luminiferous Ether’. But problems arose when Ether is attributed to some
strange properties:
1. Since light passes through vacuum, so Ether must be massless and of
zero density and must have perfect transparency to account for its
undetectability.
2. As the velocity of wave propagation depends on the elasticity of the
medium, Ether must be highly elastic. Thus, the whole free space must be
filled up with such an elastic medium to sustain the vibration of light
waves.
A solution of Maxwell’s equation gives the speed of light,
𝑐=
1
= 2.998x108 m/sec
𝜇𝑜 ε 𝑜
This is now well known that light waves are transverse electromagnetic waves
and do not need any medium.
Michelson-Morley Experiment:
A beam of light (parallel wave) from a source S is split by a partially silvered
mirror into two coherent beams.
➢ Beam (1) is transmitted to P and
➢ Beam (2) is reflected from P.
➢ Beam (1) is reflected back to P by the mirror B and
➢ Beam (2) is reflected back to P by the mirror A.
The
interference
will be constructive
or
The returning Beam
(1) is reflected by P
and
Beam
(2)
is
transmitted through P
and they enter into the
telescope
T,
where
they interfere.
destructive
c = speed of light
depending on the
path difference of
the two beams.
Earth Velocity
The mirror P is inclined at 45
to the incident beam directions.
Earth moves with a
velocity v along SP.
Here, D = PB = BP. Time taken by Beam (1) travel from P to B and back to P is
Binomial Expansion
1 + 𝑥 𝑛 = 1 + 𝑛𝑥 if x < < 1
B
The path of Beam (2) travelling from
P to movable mirror A and back is a
cross-stream path through the ether
to return to the mirror p.
Transit time tA is given by
tA = t1 + t2 = t1 + t1 = 2 t1
Now, D2 + V2t12 = c2t12
Or,
t12 (c2 – v2) = D2
Or,
t12 = D2 / (c2 – v2)
𝑡12 =
Or,
𝐷2
𝑣2
𝑐 2 1− 2
𝑐
𝐷
(1 −
𝑐
Or, 𝑡1 =
Therefore,
tA = 2t1 =
2𝐷
𝐶
𝑣 2 −1 𝐷
)2= 𝐶
𝑐2
1+
𝑣2
2𝑐 2
1+
𝑣2
2𝑐 2
The difference in time taken by the two beams of light for a round trip is
If we rotate the apparatus by 90, the time difference will be reversed as A→B
In that case,
Therefore, the change in these two time difference is
Then the corresponding path difference is
For constructive interference we know that,
2 Dv 2
path difference, d = n = 2
c
So the fringe
shift,
In actual experiment, Michelson and Morley used D = 10 m,  = 5000Å and the
earth speed v = 3×104 m/sec. Then fringe shifts
2  10m  (3  104 m / s) 2
n=
= 0.4 fringe
−7
8 2
(5  10 m)  (3  10 )
➢ The arrangement was capable of measuring 1/100th of a fringe shift.
But,
actually no fringe shift was found during this measurement.
➢ The experiment was performed at different seasons and different places with
the same result every time and no fringe shift was detected.
➢ Thus, it was concluded that the relative velocity between the earth and the Ether
is zero and there is no fringe shift at all.
The result of Michelson-Morley experiment was explained by Einstein and he concluded that:
1. There is no ether frame in space, and
2. The speed of light in free space is invariant i.e. it is constant and is
independent of the motion of source, medium or the observer.
Derivation of Lorentz Transformation Equations:
For the constancy of velocity of light, we have to introduce the new
transformation equations which fulfill the following requirements:1. The speed of c must have the same value in every inertial frame of
reference.
2. The transformations must be linear and for low speeds u<<c, they should
approach the Galilean transformations.
3. They should not be based on “absolute time” and “absolute space”.
A reasonable guess about the nature of the correct relationship between x and x is
x = k ( x − ut ) - - - - - - - - - (1)
Here k is a factor that does not depend on either x or t but may be a function of u.
The corresponding equation for x in terms of x and t
x = k ( x + ut ) - - - - - - - - - (2)
The factor k must be the same in both frames of reference since there is no
difference between S and S other than in the sign of u.
Let us consider that a light pulse that starts at the origin of S at t = 0. Since we have
assumed that the origins are coincident at t = t = 0, the pulse also starts at the origin of
S at t = 0. Einstein’s postulates require that the equation for the x-component of the
wave front of the light pulse is
x = ct in frame S
and x  = ct  in frame S.
Substituting ct for x and ct for x
in Eq. (1) and (2), we get
ct = k (ct  + ut ) = k (c + u)t  - - - - (3)
and
ct  = k (ct − ut ) = k (c − u)t
From equation (3),
- - - - (4)
t
c
=
t k (c + u )
Similarly, from equation (4),
t  k (c − u )
=
t
c
and thus, we get
2
c
k (c − u )
c
=
 k2 = 2
k (c + u )
c
c − u2
1
1
k =
k =
2
2
1− u / c
1− u2 / c2
2
Substituting
x = k ( x − ut )
for x in
x = k ( x ' + ut ' ) ,
x = k[k ( x − ut ) + ut ]
= k 2 ( x − ut ) + kut 
or,
x − k 2 ( x − ut )
t =
ku
x−
=
=
=
( x − ut )
(1 − u 2 / c 2 )
 1− u 2 / c2
u
x − u 2 / c 2 x − x + ut
u 1− u2 / c2
ut − u 2 / c 2 x
u 1− u2 / c2
- - - - - - - - - (5)
we get
u
x
2
c
=
1− u2 / c2
t−

u
t− 2 x
u
c
t =
= k (t − 2 x)
c
1− u2 / c2
The Lorentz transformation set is
x = k ( x − ut )
y = y
z = z
ux
t  = k (t − 2 )
c
The Inverse Lorentz transformation set is
x = k ( x + ut )
y = y
z = z
ux 

t = k (t + 2 )
c
- - - - - - - - - (6)
Thank YOU
Lecture: 3
Modern Physics
Derivation of Lorentz Transformation Equations:
For the constancy of velocity of light we have to introduce the new transformation
equations which fulfill the following requirements:1. The speed of c must have the same value in every inertial frame of reference.
2. The transformations must be linear and for low speeds u<<c, they should approach
the Galilean transformations.
3. They should not be based on “absolute time” and “absolute space”.
A reasonable guess about the nature of the correct relationship between x and x is
x = k ( x − ut )
- - - - - - - - - (1)
Here k is a factor that does not depend on either x or t but may be a function of u.
Because the equations must have the same form in both S and S, we need only
change the sign of u (in order to take into account the difference in the direction of
relative motion) to write the corresponding equation for x in terms of x and t:
x = k ( x + ut )
- - - - - - - - - (2)
The factor k must be the same in both frames of reference since there is no difference
between S and S other than in the sign of u.
Let us consider that a light pulse that starts at the origin of S at t = 0. Since we have
assumed that the origins are coincident at t = t = 0, the pulse also starts at the origin of
S at t = 0. Einstein’s postulates require that the equation for the x-component of the
wave front of the light pulse is
x = ct in frame S
and x  = ct  in frame S.
Substituting ct for x and ct for x
in Eq. (1) and (2), we get
ct = k (ct  + ut ) = k (c + u)t  - - - - (3)
and
ct  = k (ct − ut ) = k (c − u)t
From equation (3),
- - - - (4)
t
c
=
t k (c + u )
Similarly from equation (4),
t  k (c − u )
=
t
c
and thus we get
2
c
k (c − u )
c
=
 k2 = 2
k (c + u )
c
c − u2
1
1
k =
k =
2
2
1− u / c
1− u2 / c2
2
Substituting
x = k ( x − ut )
for x in
x = k ( x ' + ut ' ) ,
x = k[k ( x − ut ) + ut ]
= k 2 ( x − ut ) + kut 
or,
x − k 2 ( x − ut )
t =
ku
x−
=
=
=
( x − ut )
(1 − u 2 / c 2 )
 1− u 2 / c2
u
x − u 2 / c 2 x − x + ut
u 1− u2 / c2
ut − u 2 / c 2 x
u 1− u2 / c2
- - - - - - - - - (5)
we get
u
x
2
c
=
1− u2 / c2
t−

u
t− 2 x
u
c
t =
= k (t − 2 x)
c
1− u2 / c2
The Lorentz transformation set is
x = k ( x − ut )
y = y
z = z
ux
t  = k (t − 2 )
c
The Inverse Lorentz transformation set is
x = k ( x + ut )
y = y
z = z
ux 

t = k (t + 2 )
c
- - - - - - - - - (6)
You have to read yourself:
Length Contraction
Time Dilation
Einstien’s Mass And Energy Relation
The relativistic addition of velocities
Let a train move with a velocity v with respect to the ground. If a passenger on
the train, move with a velocity u’ with respect to the train, then according to
classical physics, the passenger's velocity relative to the ground u is just the
vector sum of the two velocities, i.e.
u = u‘ + v
- - - - - - (1)
Let S be the ground frame and S' the frame of the train, whose speed relative
to the ground is v. The passenger's speed in the S' frame is u' and his position
on the train as time goes on can be described by
x’= u’t ‘
- - - - - - (2)
Using Lorentz transformation equations we have,
𝑥′ =
and
𝑡′ =
𝑥 − 𝑣𝑡
1−
𝑣 2 /𝑐 2
𝑣
𝑥
𝑐2
1 − 𝑣 2 /𝑐 2
𝑡−
= 𝑢′ 𝑡 ′
- - - - - - - - - (3)
- - - - - - - - - (4)
[Using equ. (2)]
From equation (3),
𝑥 − 𝑣𝑡 = 𝑢′ 𝑡 ′ 1 − 𝑣 2 /𝑐 2
𝑣𝑥
𝑐2
𝑢′ 𝑣𝑥
′
=𝑢 𝑡− 2
𝑐
= 𝑢′ 𝑡 −
𝑜𝑟,
𝑢′ 𝑣𝑥
𝑥 + 2 = 𝑢′ + 𝑣 𝑡
𝑐
(𝑢′ + 𝑣)
𝑜𝑟,
𝑥=
×𝑡
𝑢′ 𝑣
1+ 2
𝑐
[ Using equation (4) ]
- - - - - - - - - (5)
If the passenger's speed relative to the ground be u, then his ground location as
time goes on is given by
x = ut
- - - - - - - - - (6)
Comparing eqn. (5) with eqn. (6), we obtain
(𝑢′ + 𝑣)
𝑢=
𝑢′ 𝑣
1+ 2
𝑐
This is the relativistic addition of velocities.
Case-1: If u' and v are very small compared to c, then the second term in the
denominator of is negligible compared to one (1).
Then above equation reduces to the classical result, i.e., u = u + v.
Case-2: If, on the other hand, u' = c i.e., our passenger on the train is a light pulse,
then
(𝑐 + 𝑣) (𝑐 + 𝑣) (𝑐 + 𝑣)
𝑢=
𝑐𝑣 =
𝑣 = (𝑐 + 𝑣) = 𝑐
1+ 2
1+𝑐
𝑐
𝑐
Example: Two electrons leave a radioactive sample in opposite directions, each
having a speed of 0.67c with respect to the sample. What is the speed of one
electron relative to the other?
Solution:
Here v x = 0.67 c and u = 0.67c
The speed of one electron relative to other
v x + u
0.67c + 0.67c
1.34c
vx =
=
=
= 0.92c
u
0.67c
1.45
1 + 2 v x 1 + 2  0.67c
c
c
This is means that speed of one electron relative to the other is less than c.
Relativity of mass
Consider two frames of reference S and S'. S' is moving with a constant velocity
v relative to S, in the positive x-direction.
S'
x
'
Two exactly similar elastic balls A and B in the frame S'
Same mass, moving in opposite direction with equal velocity u and –u.
Alter collision they coalesce into one.
According to the law of conservation of momentum
mu + m (-u) = momentum of coalesced mass = 0
Thus, the coalesced mass must be at rest in S' frame.
Consider the collisions with respect to the frame S. Let u1 and u2 be the
velocities of the balls as observed from S.
Then,
and
v is the relative velocity of coalesced mass, u1 is the velocity of m1 (A) and
u2 is the velocity of m2 (B) with respect to frame S.
Since the total momentum of the balls must be conserved, we have
m1u1 + m2u2 = (m1 + m2)v
From equations (i), (ii) and (iii), we get,
(iii)
Thank YOU
Online Lecture – 4
Modern Physics
Relativity of Simultaneity:
Two events are said to be simultaneous if they occur at the same time. According to the
relativity of simultaneity, if two observers are in relative motion, they will not agree as
to whether two events are simultaneous. If one observer finds them to be simultaneous,
the other generally will not.
The simultaneity of the two events is not an absolute concept and depends on the frame
of reference. In fact, each observer is correct in his own frame of reference.
Two events that appear simultaneous to an observer A will not be simultaneous to
an observer B if B is moving with respect to A.
Let us consider an example to clarify the above statement:
Figure: Two tennis ball ejected at a same time with same velocity in a moving train as
observed (a) by an observer on the train itself and (b) by an observer standing on the ground.
Let us consider two frames of reference S and S. The frame reference S is moving
with velocity u relative to the frame of reference S along +ve direction of x-axis and
the two events occur simultaneously in S. Since the events are simultaneous in frame
S, therefore we have t1 = t2.
If t1 and t 2 are the corresponding times of the same two events with respect to
system S, then we have from Lorentz transformation equations:ux2
ux1

t
=
k
(
t
−
)
and

t1 = k (t1 − 2 )
2
2
2
c
c

ux 2
ux1
)
−
k
(
t
−
)
1
2
2
c
c
u ( x1 − x 2 )
=k
since t1 = t 2
2
c
t 2 − t1 = k (t 2 −
Thus, if the events are simultaneous in frame S, t1 must be equal to t 2 or t 2 − t1
must be equal to zero, but it is not so because x1 is not equal to x2. Therefore, the
same two events are not simultaneous in frame S.
Relativistic kinetic energy (KE) to Classical KE
We know
and
m0 c 2
E = mc 2 =
1− v2 / c2


1

K = m c − m0 c =
− m 0 c = m0 c 
− 1
2
2
2
2
1− v / c
 1− v / c

2
2
m0 c 2
2
2
Using the binomial approximation (1 + x) n  1 + nx with x  1 ,
For low speeds,
v / c  1


1
K = m0 c 2 
− 1
2
2
 1− v / c


1 v2
2

= m0 c 1 +
−
1
2
2
c


1
= m0 v 2
2
we get
At low speed, the
relativistic
expression
for the K.E. of a moving
body reduces to the
(9)
classical one.
Mass-less particles: We know,
and relativistic momentum
total energy
p=
m00vu
E=
m0 c 2
1− v / c
2
2
,
1− v2 / c2
(Case-1) When m0 = 0 and v < c, i.e., E = p = 0 i.e., A mass less particle with a
speed less than that of light can have neither energy nor momentum.
(Case-2) When mo = 0 and v = c, E = 0/0 and p = 0/0 which are indeterminate: E
and p can have any values. The above equations are consistent with the existence
of mass-less particles that possess energy and momentum provided that they
travel with the speed of light.
we can write
Subtracting
p 2c 2
2
2
2
m0 c 4
2
E =
1− v2 / c2
and
from E2 , we get,
m0 v 2
p =
,
1− v2 / c2
2
or,
m0 v 2 c 2
p c =
1− v2 / c2
2
2
2
2
m0 c 4
m0 v 2 c 2
2
2 2
E −p c =
−
2
2
1− v / c
1− v2 / c2
c2
v2
2 2
= m0 c  2
− 2
2
2
c
−
v
c
−
v

 2
c

2
2


c
−
v
4

= m0 c  2
2 
c −v 
2
= m0 c 4
2

(12)
E 2 = m0 c 4 + p 2 c 2
2
Therefore, for all particles, we have
E = mo c 4 + p 2 c 2 =
2
Eo + p 2 c 2
2
According to this formula, if a particle exists m0 = 0, then the relationship between
its energy and momentum is given by
mass - less particles
𝑜𝑟,
E = pc
𝑝=
𝐸 ℎυ
=
𝑐
𝑐
Theory of Light
There are different theories on the nature of the light. The important theories are as
follows:
1. Newton's corpuscular particle theory:
Isaac Newton stated in his Hypothesis of Light on 1672 that light was composed of
corpuscles (tiny, light and elastic particles) which were emitted in all directions from
a source.
2. Huygens, Wave Theory:
Christiaan Huygens worked out a mathematical wave theory of light in 1679 and
published it in his treatise on light in 1690. He proposed that light was emitted in all
directions as a series of waves in a medium called the Luminiferous ether.
3. Maxwell's electromagnetic theory:
Maxwell concluded that light was a form of electromagnetic radiation, he first stated
this result in 1862.
4. Planck's Quantum theory:
In 1900 Max Planck, attempting to explain black-body radiation, suggested that light
energy is a combination of "quanta“.
5. Dual nature of light:
It is established that light has both the nature of wave and particles.
Planck's Quantum theory:
Postulates of Planck's quantum theory:
1) Energy is not emitting continuously but discontinuously in the form of small
packets of energy known as quanta of energy (photon).
2) The energy of each quantum is directly proportional to the frequency of
radiation i.e., E directly proportional to ʋ, where ʋ is frequency?
Photo-electric effect:
W Smith around 1885 found
that resistance of a metallic
wire varies when sun ray is
incident on it. W Smith was
a telegraph operator.
When electromagnetic radiation is incident
on a metal surface then electrons are emitted
from that surface, this phenomenon is known
as Photoelectric effect. The emitted electrons
are known as photoelectric electron and the
current flow due to the emission of these
electrons is known as photoelectric current.
Characteristics (laws) of photoelectric
emission:
1. Emission commences at the instant
the surface starts to be irradiated.
2. For any incident radiation of a
constant frequency ʋ, and if the
accelerating potential V is constant,
then the number of electron emitted
per second, i. e., photoelectric current
is proportional to the intensity of the
radiation.
3. Emission occurs only if the
frequency of the incident radiation is
above some minimum frequency
called the threshold frequency.
The value of threshold frequency is
different for different metal.
With the increase in frequency of the
incident radiation the energy of the
emitted electron increases.
Characteristics (laws) of photoelectric
emission:
4. If both the intensity and frequency of the
incident radiation are kept constant, as the
accelerating potential is reduced from zero
to negative values the photoelectric current
does not immediately drops to zero. This
proves that electrons are emitted with
some definite velocity with sufficient
enough to give kinetic energy to the
electron so as to overcome the retarding
electric field between the electrodes.
When the negative potential is made at
large enough as Vs then the current is
reduced to zero. Vs is known as the
stopping potential.
Vs is independent of the intensity of the
incident radiation.
5. Stopping potential varies linearly
with the frequency.
Thank YOU
Lecture: 5
Modern Physics
Failure of wave theory of light to explain Photoelectric effect:
(i) According to wave theory
Energy ∝ (Intensity)2
or, E ∝ I2
➢ Light of low frequency with sufficient intensity is much
effective than light of higher frequency with low
intensity. But it is contradictory to the experimental
result.
(ii) It is equally difficult to explain that there is no time lag
between the arrival of the radiation on the metal surface
and the emission of photoelectrons.
• Consider violet light incident on a sodium
surface.
• Detectable photoelectric current will result
when 10-6 watt of electromagnetic energy is
absorbed by 1 m2 of the surface.
1m2
1m
1m
1m
Fig. 1: Sodium cube
Now there are about 1019 atoms in a layer of sodium one atom thick and 1m2 in area.
Consider light absorbed by 10 uppermost layers of sodium atom.
Therefore, 10-6 watt is distributed by 1020 atoms.
Therefore, energy received per atom is 10-26 watt
= 10-26 J s-1
= 6.25 x 10-8 eV s-1
So it should be taken 1.6 x 107 sec to receive 1eV per atom.
= 185 days
If 2 eVis the work function of that metal, then almost 1 Year is required to emit
electrons from that surface. But, practically, it is an instantaneous event.
Einstein Photoelectric equation:
Energy of incident Photon = Energy required for getting the electron free from the
atom of metal surface + kinetic energy of the electron
or, hʋ = hʋo + ½ mv2
Determination of Planck's Constant:
The stopping potential for a particular metal were determined for different
frequencies. A graph is plotted with ʋ along x-axis and Vs along y-axis. The resulting
graph is a straight line.
Light-matter interaction:
The photoelectric effect is the emission of
electrons when electromagnetic radiation, such as
light, hits a material. Electrons emitted in this
manner are called photoelectrons.
Thomson scattering is the elastic scattering of
electromagnetic radiation by a free charged
particle. It is the low-energy limit of Compton
scattering; the particle's kinetic energy and photon
frequency do not change as a result of the
scattering. This limit is valid as long as the photon
energy is much smaller than the mass-energy of the
particle/electron: hʋ << mc2, i. e., the wavelength of
the light is much greater than the Compton
wavelength of the particle.
Compton scattering, discovered by Arthur Holly Compton, is the scattering of a
photon by a charged particle, usually an electron. It results in a decrease in energy
(increase in wavelength) of the photon. Part of the energy of the photon is transferred to
the recoiling electron.
Pair production is the creation of a subatomic particle and its antiparticle
from a photon. Examples include creating an electron and a positron, a muon
and an antimuon, or a proton and an antiproton etc. Pair production often
refers specifically to a photon creating an electron–positron pair near a
nucleus. For pair production to occur, the incoming energy of the photon
must be above a threshold of at least the total rest mass energy of the two
particles, and the situation must conserve both energy and momentum.
However, all other quantum numbers (angular momentum, electric charge,
lepton number) are also conserved. Thus the created particles shall have
opposite values of each other. For instance, if one particle has electric charge
of +1 the other must have electric charge of −1.
The probability of pair production in photon–matter interactions increases
with photon energy and also increases approximately as the square of atomic
number of the nearby atom.
The energy of a photon can be converted into an electron–positron pair:
The photon's energy is converted to particle mass in accordance with
Einstein’s equation,
E = m ⋅ c2
Photodisintegration is a nuclear process in which an atomic nucleus absorbs
a high-energy gamma ray, enters an excited state, and immediately decays by
emitting a subatomic particle (such as, the neutron and proton, etc.). The
incoming gamma ray effectively knocks one or more neutrons, protons, or an
alpha particle out of the nucleus. The reactions are called (γ,n), (γ,p), and
(γ,α).
Photofission is a process in which a nucleus, after absorbing a gamma ray,
undergoes nuclear fission and splits into two or more fragments.
Applications of photo-electric effect
Photo-electric cell:
Two types
(1) Vacuum tube
(2) Gas filled tube
C => Semi-cylindrical plate coated with
potassium or caesium known as emitter.
A => A wire of platinum or nickel fixed
along the axis of the cylinder known as
collector.
When photons are incident on the emitter
then photoelectric current is produced in
the circuit is detected by the
galvanometer.
If the tube is vacuum then current is very weak.
On the other hand, if the tube is filled by gas (usually Argon or Helium) then
photo-electron bombarded the gas molecules. In this way gas molecules are
ionized. As a result a large current is flow through the circuit.
Applications of the photoelectric cell
The photoelectric cell is used to automatic turn on and off the street bulbs.
The photoelectric cell is used to automatic traffic control.
The photoelectric cell is used in exposure meter. The exposure meter is used
along with a camera to know the correct time of exposure for having a good
photograph.
The photoelectric cell is used in lux-meter. It is used to determine the
intensity of light.
The photoelectric cell is used in a burglar alarm. This device is kept near to
door to protect form a thief.
Photomultiplier tube:
A photomultiplier tube, useful for light detection of very weak signals, is a
photo-emissive device in which the absorption of a photon results in the
emission of an electron.
Photomultipliers
are
typically constructed with
an evacuated glass tube,
containing a photocathode,
several dynodes, and an
anode. Incident photons
strike the photocathode
material. Electrons are
ejected from the surface
are directed by the focusing
electrode
toward
the
electron multiplier, where
electrons are multiplied by
the process of secondary
emission.
The electron multiplier consists of a number of electrodes called dynodes.
Each dynode is held at a more positive potential, by ≈100 Volts, than the
preceding one. A primary electron leaves the photocathode with the
energy of the incoming photon. A small group of primary electrons is
created by the arrival of a group of initial photons. The primary electrons
move toward the first dynode because they are accelerated by the positive
electric field. They each arrive with ≈100 eV kinetic energy. Upon striking
the first dynode, more low energy electrons are emitted, and these
electrons are in turn accelerated toward the second dynode. The geometry
of the dynode chain is such that a cascade occurs with an exponentiallyincreasing number of electrons being produced at each stage.
For example, if at each stage an average of 6 new electrons are produced
for each incoming electron, and if there are 11 dynode stages, then at the
last stage one expects for each primary electron about 611 ≈ 3x108
electrons. This last stage is called the anode. This large number of electrons
reaching the anode results in a sharp current pulse that is easily detectable.
Photo-voltaic cell:
In the case of photo-electric cell an external potential is required.
But the photo-voltaic cell is a self-generating cell where the electrons ejected by
the light, themselves produce a potential difference between two plates, which
causes the current to flow through the external circuit, even though no external
battery is applied.
The most commonly used photovoltaic cell are:
iron-selenium cells
In this, selenium layer is placed on
an iron base.
An extremely thin semi-transparent layer of
silver is formed on the selenium to act as a
front electrode. A contact ring on the silver
layer acts as one electrode and the iron base
acts as the other electrode.
Solar cell construction and working principle:
Solar cell is a device that convert the solar energy into the electrical energy. The
basic principle behind the function of solar cell is based on photo-voltaic cell.
The electricity supplied
by the solar cell is DC.
Which is same like
provided by a battery
but a little bit different
in the sense the
battery is providing
constant voltage.
Construction of solar cell
Mainly solar cell is constructed using the crystalline silicon that consists of a ntype semiconductor. This is the upper layer also known as emitter layer. The
second layer is p-type semiconductor layer known as base layer. Both the layer
are sandwiched and hence there is formation of p-n junction between them.
The surface is coated with anti-reflecting coating to avoid the loss of incident
light energy due to reflection.
Working of solar cell
As soon as the solar cell is exposed to sunlight, the solar energy is absorbed by semiconductor materials. Due to this absorbed energy the phenomena of photo-voltaic
occurs and electrons are liberated and produce the external DC current. The DC current
is converted into 220 volt AC current using and inverter for different applications.
An atom of silicon has 14 electrons, the outer shell has four electrons. Therefore a
silicon atom will always look for ways to feel up its last shell and to do this it will share
electron with four nearby atoms.
Now we use phosphorus (with 5 electrons in its outer shell). Therefore when it
combines with silicon, one electron remains free. When energy is added to pure silicon
it can cause few electron to break free of their bonds and leave their atoms. These are
called free carriers, which move randomly around the crystalline lattice looking for
holes to fall into and carrying an electrical current.
The other part of solar cell is doped with the element boron (with 3 electrons in its
outer shell) to become p-type silicon. Now when these two type of silicone interact, an
electric field form at the junction which prevents more electrons to move to p-site.
When photon hits solar cell, its energy break apart electron-hole pairs. Each photon
with enough energy will normally free exactly one electron, resulting in a free hole as
well. If this happens close enough to the electric field, this causes disruption of
electrical neutrality, and if we provide an external current path, electron will flow. This
electron flow provides the current.
Electron-hole formation
Types of solar cell and efficiency levels
Solar panel/solar array and Solar module
You have to read yourself
Automatic bell alarm circuit?
Math problems?
Compton Effect:
When a photon of energy hʋ strikes a free electron, then energy is communicate
between them. In this case the incident photon loss its energy. Consequently it
acquire a lower frequency or higher wavelength than that of incident photon.
These change in wavelength or frequency is known as Compton effect.
Generally, the electron
considered to be at rest
and the photon is
usually considered to
be an energetic photon
such as an X-ray
photon.
Compton Theory:
According to the law of conservation
of energy,
ℎ𝜗 + 𝑚𝑜 𝑐 2 = h𝜗 ′ + 𝑚𝑐 2
-------- (1)
According to the law of conservation of momentum,
Along the direction of incident photon – (x-axis),
ℎ𝜗
ℎ𝜗 ′
-------- (2)
+0=
𝑐𝑜𝑠𝜃 + 𝑚𝑣 𝑐𝑜𝑠∅
𝑐
𝑐
Along the direction perpendicular to the incident photon – (y-axis),
0+0=
ℎ𝜗′
𝑐
sin𝜃 − 𝑚𝑣 𝑠𝑖𝑛∅
-------- (3)
From equations (2) and (3)
𝑚𝑣𝑐 𝑐𝑜𝑠∅ = ℎ𝜗 − ℎ𝜗 ′ 𝑐𝑜𝑠𝜃
-------- (4)
𝑚𝑣𝑐 𝑠𝑖𝑛∅ = ℎ𝜗 ′ 𝑠𝑖𝑛𝜃
-------- (5)
4
2
+ 5
2
𝑚2 𝑣 2 𝑐 2 = ℎ2 𝜗 2 + ℎ2 𝜗 ′2 𝑐𝑜𝑠 2 𝜃 − 2ℎ2 𝜗𝜗 ′ 𝑐𝑜𝑠𝜃 + ℎ2 𝜗 ′2 𝑠𝑖𝑛2 𝜃
or,
𝑚2 𝑣 2 𝑐 2 = ℎ2 𝜗 2 + ℎ2 𝜗 ′2 − 2ℎ2 𝜗𝜗 ′ 𝑐𝑜𝑠𝜃
-------- (6)
From equ. (1)
𝑚𝑐 2 = ℎ𝜗 − ℎ𝜗 ′ + 𝑚0 𝑐 2
or, 𝑚2 𝑐 4 = ℎ2 𝜗 2 + ℎ2 𝜗 ′2 + 𝑚02 𝑐 4 − 2ℎ2 𝜗𝜗 ′ − 2ℎ𝜗 ′ 𝑚0 𝑐 2 + 2ℎ𝜗𝑚0 𝑐 2 -------- (7)
(7)-(6)
𝑚2 𝑐 2 𝑐 2 − 𝑣 2 = 𝑚02 𝑐 4 − 2ℎ2 𝜗𝜗 ′ − 2ℎ𝜗 ′ 𝑚0 𝑐 2 + 2ℎ𝜗𝑚0 𝑐 2 + 2ℎ2 𝜗𝜗 ′ 𝑐𝑜𝑠𝜃
We know,
𝑚=
𝑚0
𝑣2
1− 2
𝑐
2 2
𝑚
𝑐
0
𝑚2 = 2
𝑐 − 𝑣2
𝑚2 𝑐 2 − 𝑣 2 = 𝑚02 𝑐 2
𝑚2 𝑐 2 𝑐 2 − 𝑣 2 = 𝑚02 𝑐 4
---- (8)
From equ. (8)
𝑚02 𝑐 4 = 𝑚02 𝑐 4 − 2ℎ2 𝜗𝜗 ′ 1 − 𝑐𝑜𝑠𝜃 + 2ℎ𝑚0 𝑐 2 (𝜗 − 𝜗 ′ )
or,
2ℎ𝑚0 𝑐 2 𝜗 − 𝜗 ′ = 2ℎ2 𝜗𝜗 ′ (1 − 𝑐𝑜𝑠𝜃)
or,
𝜗 − 𝜗′
ℎ
=
(1 − 𝑐𝑜𝑠𝜃)
𝜗𝜗 ′
𝑚0 𝑐 2
or,
1
1
ℎ
−
=
(1 − 𝑐𝑜𝑠𝜃)
𝜗′ 𝜗
𝑚0 𝑐 2
or,
𝑐
𝑐
ℎ
−
=
(1 − 𝑐𝑜𝑠𝜃)
𝜗′ 𝜗
𝑚0 𝑐
or,
λ′
or,
ℎ
−λ=
(1 − 𝑐𝑜𝑠𝜃)
𝑚0 𝑐
Δλ =
ℎ
(1 − 𝑐𝑜𝑠𝜃)
𝑚0 𝑐
Δλ =
2ℎ
𝑚0 𝑐
𝑠𝑖𝑛2
𝜃
2
Thank YOU
Lecture: 6
Modern Physics
Derive an expression for
(a) Kinetic energy of recoil electron and
(b) Direction of recoil electron
Math problem (Compton effect)
Wave Particle duality/de-Broglie hypothesis:
In 1924, de-Broglie suggested that matter, like radiation, has dual nature, i., e.,
matter which is made up of discrete particles (atoms, protons, electrons etc)
might exhibit wave like properties.
de-Broglie wavelength:
Let the quantity that undergoes periodic changes giving rise to matter waves be Ψ.
Let its value at a point xo, yo, zo at any instant of time to be given by,
Ψ = Ψ𝑜 sin 2π𝜗𝑜 𝑡𝑜
-------- (1)
Where Ψo is the amplitude and 𝜗o be the frequency as observed at rest with
respect to the particle.
By applying the Lorentz transformation equation, we can write,
𝑣𝑥
𝑐2
𝑣2
1− 2
𝑐
𝑡−
𝑡𝑜 =
-------- (2)
From equations (1) and (2) we get,
𝑣𝑥
)
𝑐2
𝑣2
1− 2
𝑐
(𝑡 −
Ψ = Ψ𝑜 sin 2π𝜗𝑜
-------- (3)
This equation may be compared with,
𝑥
𝑦 = 𝐴 sin 2π 𝜗(𝑡 − )
𝑢
-------- (4)
By comparing equations (3) and (4) we get,
𝑐2
𝑢=
𝑣
and
𝜗=
𝜗𝑜
𝑣2
1− 2
𝑐
-------- (5)
From Einstein’s mass-energy relation,
ℎ𝜗𝑜 = 𝑚𝑜
𝑐2
=>
𝑚𝑜 𝑐 2
𝜗𝑜 =
ℎ
-------- (6)
From equations (5) and (6) we get,
𝑚𝑜 𝑐 2
𝑚𝑐 2
𝜗=
=
2
ℎ
𝑣
ℎ 1− 2
𝑐
Therefore, wavelength of matter wave,
λ=
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑢
=
𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 𝜗
Using equations (5) and (7) we get,
𝑐2
𝑐 2ℎ
ℎ
𝑣
λ=
=
=
𝑚𝑐 2 𝑣𝑚𝑐 2 𝑚𝑣
ℎ
λ=
ℎ
𝑚𝑣
-------- (7)
Maths:
de-Broglie wavelength
Rutherford's atom
model:
The main points of
Rutherford's atom
model are# An atom consists of a
positively charged
nucleus, which is
surrounded by
electrons moving
around it as planet
revolve around the sun
.
# Electrons and the
nucleus are held
together by coulombic
force of attraction.
Gold atom
Drawbacks/Limitations of Rutherford’s Atom
Model:
Rutherford’s model explains the structure of atom in
very simple way, but it suffers from the following
drawbacks:
# Rutherford's atom model could not explain the
stability of the atom. According to Rutherford's model
electrons revolve around the nucleus as planets revolve
around the sun according to classical theory of
mechanics, during uniform revolution, any body
accelerates. An accelerating charged particle must emits
radiation and lose energy. As a result, the electron will
follow a spiral path, and ultimately fall into the nucleus .
In such cases the atom would be highly unstable
and would collapse. However, this is not consistent with
real-world observations as the atoms are stable.
# The Rutherford Model of atom does not say anything
about the arrangement of electrons in an atom.
Bohr’s Atom Model:
In 1913 Bohr proposed an atom model to resolve the problems of
the Rutherford's atom model. Bohr proposed that classical
mechanics break down into the atom model. Bohr applied quantum
mechanics in Rutherford's atom model. Bohr’s proposal are:
1. The angular momentum of the electrons are whole number
ℎ
ℎ
multiple of , 𝑖. 𝑒. , 𝑛 . Thus, the angular momentum is quantized.
2𝜋
2𝜋
This means that electrons can have only certain radii called energy
levels.
2. The electrons of an atom are orbiting along some particular
orbits. During rotation electrons never emit any radiation (energy).
3. An electron can jump from one orbit of energy E1 to another orbit
of energy E2. In doing so electron either emits or absorbs an
electromagnetic radiation of frequency ϑ, so that E1  E2 = hϑ.
• The Bohr model explain why the spectra are not continuous because
the electron are in only certain discrete energy level or state.
3
ℎ
2𝜋
2
ℎ
2𝜋
ℎ
2𝜋
An electron revolves around the nucleus, in the n
th orbit of radius rn then,
Limitation of Bohr’s atom model:
Bohr was successful in introducing the idea of
quantum energy state, but it has some
limitations:
1. Bohr’s model could not explain the spectra of
atoms containing more than one electron. His
theory can only explain the hydrogen spectrum
or ions containing one electron, eg., He+, Li2+.
2. In the presence of a magnetic field, each
spectral line get split up into closely spaced line
(Zeeman effect) could not be explained by Bohr’s
model. Similarly, the splitting of spectral line
under the effect of applied electric field (Stark
effect), could not be explained by Bohr’s model.
de-Broglie hypothesis on atom Model:
de-Broglie suggested that,
1. The motion of electron can be represented by standing matter
wave.
2. Only certain wavelength can exist, i. e., its energy can be only
certain discrete values (the motion of electron is quantized).
According to de-Broglie,
2𝜋𝑟𝑛 = 𝑛λ
ℎ
We know matter wavelength, λ =
where, n = 1, 2, 3, - - -
𝑚𝑣
ℎ
𝑚𝑣
ℎ
𝑛
2𝜋
or, 2𝜋𝑟𝑛 = 𝑛
or, m𝑣𝑟𝑛 =
 𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑀𝑜𝑚𝑒𝑛𝑡𝑢𝑚 = 𝑛
ℎ
2𝜋
Which satisfy the Bohr postulates.
Fig.1: Standing wave fit to a circular
Bohr orbit. In this particular diagram
three wavelength are fit to the orbit.
Corresponding to the n = 3 energy
state of the Bohr theory.
Pauli's exclusion principle
In 1925, Wolfgang Pauli
The Pauli exclusion principle says that no two identical fermions can simultaneously
occupy the same quantum state.
All fermions, such as protons, neutrons, Quarks (up and down), leptons (electrons,
electron neutrinos, muons, muon neutrinos, taus, and tau neutrinos), etc., obey
Fermi-Dirac statistics; this includes obeying the Pauli’s exclusion principle. their spin is
always some odd whole-number multiple of one-half.
The Pauli exclusion principle does not apply to bosons: these are particles that obey
Bose-Einstein statistics; they all have integer values of spin. Photons, gluons, gravitons,
and the W, Z and Higgs bosons, etc., are all bosons.
Quantum numbers of electron should be considered as address of each electron
within an atom, each address has four components.
Pauli’s Exclusion principle states that no two electrons in an atom can have the same
four quantum numbers. If two electrons occupy the same orbital, they must have
different spins. Pauli’s Exclusion Principle limits the numbers of electrons that a shell
or a subshell may contain.
The four quantum numbers
Principal Quantum Number n
This quantum number describes the electron shell or energy level of an atom.
The value of the principal quantum number can be any integer with a positive value
that is equal to or greater than one.
A larger value of the principal quantum number implies a greater distance between
the electron and the nucleus (which, in turn, implies a greater atomic size).
The value n=1 denotes the innermost electron shell of an atom, which corresponds to
the lowest energy state or the ground state of an electron. Thus, the principal
quantum number, n, cannot have a negative value or be equal to zero.
Angular momentum quantum number ℓ
The angular or orbital quantum number, describes the sub-shell and gives the
magnitude of the orbital angular momentum.
A value of the azimuthal quantum number can indicate either an s, p, d, or f subshell
which vary in shapes. This value depends on the value of the principal quantum
number, i.e., the value of the azimuthal quantum number ranges between 0 and (n-1).
ℓ = 0 is called an s orbital, ℓ = 1 is p orbital, ℓ = 2 is d orbital, and ℓ = 3 is f orbital.
The value of ℓ ranges from 0 to n − 1 because the first p orbital (ℓ = 1) appears in the
second electron shell (n = 2), the first d orbital (ℓ = 2)appears in the third shell (n = 3),
and so on.
This quantum number specifies the shape of an atomic orbital and strongly influences
chemical bonds and bond angles.
Magnetic Quantum Number mℓ
mℓ is associated with the orbital orientation.
mℓ describes energy levels available within a sub-shell
mℓ yields projection of orbital angular momentum along a specified axis.
Value of mℓ ranges from −ℓ to +ℓ, with integer steps between them.
The s sub-shell (ℓ=0) contains one orbital, and therefore the mℓ of an electron in an s
sub-shell will always be 0.
p sub-shell (ℓ=1) contains three orbitals, so mℓ of an electron in a p sub-shell will be
−1, 0, or 1.
The d sub-shell (ℓ=2) contains five orbitals, with mℓ values of −2, −1, 0, 1, and 2.
Spin Quantum Number ms
Spin quantum number describes the intrinsic angular momentum of the electron
within that orbital and gives the projection of the spin angular momentum (s) along
the specified axis.
Value of ms ranges from −s to +s.
An electron has spin s = ½, consequently ms will be ±½, corresponding to spin and
opposite spin
Each electron in any individual orbital must have different spins because of the Pauli
Exclusion Principle. Thus, an orbital can never contain more than two electrons.
The electron spin quantum number is independent of the values of n, ℓ, and mℓ . The
value of this number gives insight into the direction in which the electron is spinning
and is denoted by the symbol ms.
.
Thank YOU
Lecture: 7
Modern Physics
In 1932 Chadwick discover the neutron
Atomic Mass Unit
One atomic mass unit (1 u) is equal to one twelfth of the mass
of the most abundant form of the carbon (12) atom.
Atomic mass unit: 1 u = 1.6606 x 10-27 kg = 931.502 MeV/c2.
TABLE: Masses in Kilograms, Unified Atomic Mass Units, and 𝑴𝒆𝑽/𝒄𝟐
Mass
Objects
Kg
U
𝑴𝒆𝑽/𝒄𝟐
Electron
9.1094 x 10-31
0.00054858
Proton
1.6726 x 10-27
1.007276
938.27
Neutron
1.6749 x 10-27
1.008665
939.57
1
1𝐻
1.6735 x 10-27
1.007825
938.78
atom
0.511
Properties of Nucleus
• Static nuclear properties (Time-independent):
Nuclear radius/size, mass, density, charge, spin angular
momentum, magnetic dipole moment, electric quadrupole
moment.
• Dynamic properties (Time-dependent):
Self-induced (Radioactive decay), Forced (Nuclear reaction).
Nuclear Size:
Rutherford’s experiment of α-scattering indicates
Atom
Nucleus
Radius:
10-10m
10-14 to 10-15m
The empirical formula of nuclear radius,
R = Ro A1/3
where, Ro = 1.3 x 10-15m
A => Mass number
Nuclear Mass:
Mass of the nucleus = Z mp + (A-Z) mn .
It is about 99.97% of the mass of atom
Nuclear Density:
Nuclear density is unbelievably high since is extremely small size and
large mass.
Nuclear mass
The nuclear density, ρ = -----------------------Nuclear volume
The mass of 2cm3 of nucleus is 3.7 x 1011 kg
Mass of all people of the world = 790,0000000 x 45 kg = 3.56 x 1011kg
Nuclear Charge:
The charge of the nucleus is due to the charge of the proton
which is Ze.
Spin angular momentum:
Both the proton and neutron like electron, have an intrinsic spin.
The spin angular momentum is given by,
𝐿𝑠 =
𝐿𝑠 =
𝑙 𝑙+1
3 ℎ
2 2𝜋
ℎ
2𝜋
where, the quantum number 𝑙, is
commonly called the spin, is equal to 1/2.
Nuclear magnetic dipole moment:
The nuclear magnetic moment is the magnetic moment of an atomic
nucleus and arises from the spin of the protons and neutrons.
The nuclear magnetic moment varies from isotope to isotope of an
element. For a nucleus of which the numbers of protons and of
neutrons are both even in its ground state (i.e. lowest energy state),
the nuclear spin and magnetic moment are both always zero. In cases
with odd numbers of either or both protons and neutrons, the
nucleus often has nonzero spin and magnetic moment.
According to Dirac’s, the magnetic dipole moment of a spinning
proton is given by,
ℎ
𝜇=
𝑒2𝜋
2𝑚𝑝
where, mp is the mass of a proton.
Nuclear quadrupole moment:
In general, a nucleus is considered to be spherical in shape. Electrical quadrupole
moment is the measure of deviation of nucleus from spherical symmetry. Electrical
quadrupole moment is denoted by symbol “Q“ and is given by
2
𝑄 = 5 𝑧(𝑏2 − 𝑎2) ………. (i)
Where z is atomic number and b is radius of nucleus along symmetric axis and a is
radius of nucleus along perpendicular to symmetric axis.
Now TWO cases arises
1. For nuclei with mass number A < 20, the number of protons
and number of neutrons in nucleus almost similar, so
charge distribution inside the nucleus will be uniform and
shape of nucleus will be a perfect SPHERE. Here b = a and
from equation (i), Q = 0.
2. For nuclei with greater mass number A, number of protons
and neutrons inside the nucleus are unequal, so charge
distribution is non-uniform inside the nucleus .
Now there are following 2 possibilities
Now there are following 2 possibilities
(a) If charge distribution is more along symmetric axis, then in this
situation b > a and from equation (i), Q > 0 and Q = +ve. So, there
will be some deviation in the nucleus from spherical symmetry and
nucleus will be PROLATE SPHEROID as shown in figure.
(b) If charge distribution is more along perpendicular to
symmetric axis, then in this situation b < a and from equation (i),
Q < 0 and Q = -ve. So, there will be some deviation in the nucleus
from spherical symmetry and nucleus will be OBLATE SPHEROID
as shown in figure.
Electrical quadrupole moment (Q) is measured in BARN.
1 BARN = 10-28 m2
Max. Value of Q is found to be for Lutetium (Lu, Z=71, A=176) which
is + 7 BARNS.
Min. Value of Q is found to be for Antimony (Sb, Z=51, A=123)
which is - 1.2 BARNS.
As range of values of Q is small, so we can say that deviation in
nucleus from spherical symmetry is small.
So, our assumption that nucleus is spherical is justified.
Radioactive decay:
Radioactivity is the property exhibited by certain types of matter
of emitting energy and subatomic particles spontaneously. The
emissions of the most common forms of spontaneous
radioactive decay are the alpha (α) particle, the beta (β) particle,
the gamma (γ) ray.
Radionuclide is an atom that has an unstable nucleus (i.e., an
excess of energy, characterized by an excess of protons or
neutrons). It will decay (i.e., loss its excess energy) by emitting
particles (alpha and beta decay) or photons (gamma rays or Xrays) to reach a stable configuration.
Nuclear reaction:
A process in which the nucleus of an atom is changed by being
split apart or joined with the nucleus of another atom
Mass defect:
The total mass of a nucleus is always less than the sum of the
masses of its separate protons and neutrons.
∆𝑚 = 𝑍𝑚𝑝 + 𝐴 − 𝑍 𝑚𝑛 − 𝑀𝑛𝑢𝑐𝑙𝑒𝑢𝑠
Where has the mass gone?
Binding Energy:
The minimum energy required to break up a nucleus into its
constituent nucleons is known as binding energy.
B. E. = Mass defect x c2
= ∆𝑚 ∗ c2
or,
𝑜𝑟,
𝐵. 𝐸. 𝐽 = ∆𝑚 𝑘𝑔 × 𝑐 2 𝑚𝑠 −1
2
𝐵. 𝐸. 𝑀𝑒𝑉 = ∆𝑚 𝑎𝑚𝑢 × 931
Binding Energy per nucleon:
The binding energy per nucleon for any nucleus is found by dividing its
total binding energy by the total number of nucleons it contains.
𝑇𝑜𝑡𝑎𝑙 𝑏𝑖𝑛𝑑𝑖𝑛𝑔 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑎 𝑛𝑢𝑐𝑙𝑒𝑢𝑠
𝐵𝑖𝑛𝑑𝑖𝑛𝑔 𝐸𝑛𝑒𝑟𝑔𝑦 𝑝𝑒𝑟 𝑛𝑢𝑐𝑙𝑒𝑜𝑛 =
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑛𝑢𝑐𝑙𝑒𝑜𝑛𝑠
Binding Energy per nucleon:
Nuclear force:
Nucleus consist of protons and neutrons
How does the positive protons and neutral neutrons are kept
together in the nucleus?
In 1935 Yukawa a Japanese scientist proposed a meson field
exists between the protons and neutrons.
The particle responsible for nuclear force: π (pi) mesons.
π (pi) mesons is 272 times heavier than electron
Positive π+ mesons are associated with the force between: p and n
Negative π- mesons are associated with the force between: p and p
Neutral πo mesons are associated with the force between: n and n
Nuclear force:
Nucleus consist of protons and neurons
How does the positive protons and neutral neutrons are kept
together in the nucleus?
• Strong nuclear force is an attractive force that attracts
among all nucleons (protons and neutrons alike).
• Protons attract each other via strong nuclear force at the
same time they repel each other via electrostatic force.
• Strong nuclear force > electrostatic force.
• Neutrons (electrically neutral) only attract other neutrons or
protons via strong nuclear force.
• Strong nuclear force is a short-range force. It acts only over
a very short distance.
• It is very strong between two nucleons if they are < 10-15m apart.
• It is zero if they are separated by a distance > 10-15 m apart.
• If the nuclide contents too fever or too many neutrons relative to
the number of protons, the binding of nucleus reduce (nuclide
unstable).
• Nuclear force is charge independent. It is same for p-p, p-n and
n-n interaction.
• The energy needed to separate a nucleon from a nucleus is about
8MeV in contrast a few eV is required to separate an electron
from an atom.
• Nuclear force is limited in range (saturation property). As a
result, each nucleon interacts with only a limited number of
nucleons around it.
Thank YOU
Lecture: 8
Modern Physics
Nuclear chain reactions
Over Christmas vacation in 1938, nuclear chemist Otto Hann observed when
Uranium-235 is struck by a thermal neutron it split up and produce neutrons
and enormous amount of energy.
Physicist Leo Szilard made an important realization: if fission emits neutrons,
then neutrons from the fission of one nucleus could cause the fission of
another nucleus.
Nuclear chain reactions are series
of nuclear fissions, which continued
by a neutron produced in a
preceding fission. For example, on
average 21/2 neutrons are released
by the fission of each uranium-235
nucleus that absorbs a neutron of
moderate/thermal energy.
Thermal neutron:
A thermal neutron is a free neutron with a kinetic energy of
about 0.025 eV (about 4.0×10−21 J, hence a speed of 2.19
km/s), at a temperature of about 290 K (17 °C or 62 °F),
Source: Internet
Multiplication factor, k:
The condition for a chain reaction is usually expressed in terms of a multiplication
factor, k, which is defined as the ratio of the number of fissions produced in one step
(neutron generation) in the chain to the number of fissions in the preceding
generation.
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑛𝑒𝑢𝑡𝑟𝑜𝑛𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝑖𝑛 𝑎𝑛𝑦 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛
𝑘=
𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑛𝑒𝑢𝑡𝑟𝑜𝑛 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑟𝑒𝑐𝑒𝑑𝑖𝑛𝑔 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛
If k is less than unity, a chain reaction cannot be sustained.
If k = 1, a steady-state chain reaction can be maintained; and
if k is greater than 1, the number of fissions increases at each step, resulting in a
divergent chain reaction.
Although two to three neutrons are produced for every fission, not all of these
neutrons are available for continuing the fission reaction. If the conditions are such
that the neutrons are lost at a faster rate than they are formed by fission, the chain
reaction will not be self-sustaining. At the point where the chain reaction can
become self-sustaining, this is referred to as critical mass.
Critical mass:
In nuclear physics, critical mass is the minimum amount of a given fissile
material necessary to achieve a self-sustaining chain reaction. It depends on
several factors, including the kind of fissile material used, its density and
purity, temperature, the composition and geometry of the surrounding.
The critical mass of a bare sphere of uranium-235 at normal density is
approximately 47 kg (104 pounds); for plutonium-239, critical mass is
approximately 10 kg (22 pounds).
By surrounding the fissionable material with a suitable neutron reflector, the
loss of neutrons can reduced and the critical mass can also be reduced.
By using a neutron reflector, only about 11 pounds (5 kilograms) of nearly
pure plutonium 239 and about 33 pounds (15 kilograms) uranium 235 is
needed to achieve critical mass.
Neutron reflector:
A layer of material immediately surrounding a fissionable material that reflects
back the neutrons into the core. Without reflector many neutrons that would
otherwise escape. The returned neutrons can then cause more fissions and
improve the neutron economy of the chain reaction.
A reflector made of a light material like graphite or beryllium will also serve as
a neutron moderator reducing neutron kinetic energy, while a heavy material
like lead will have less effect on neutron velocity.
Critical size:
The actual size of fissionable material which allows the escape of neutrons to
such an way that at least one neutron is positively left behind per fission
reaction is known as critical size. The critical size must at least include enough
fissionable material to reach critical mass. If the size of the fissionable material
is less than a certain minimum, too many fission neutrons escape through its
surface and the chain reaction is not sustained.
In order to minimize the losses through the surface, the area of the surface
‘S’ (through which fission neutrons escape) must be made as small as
possible in comparison to the volume ‘V’ (in which fission take place). Larger
the size of the fissionable material, smaller the surface to volume (S/V) ratio
becomes.
Surface area of a sphere, S = 4πr2
4
Volume of a sphere, V = πr3
3
Therefore, S/V = 3/r = 6/d, i. e., inversely proportional to size of the material
If, d = 0.2 cm ; S/V = 30
Again, if, d = 20 cm (100 times larger); S/V = 0.3 (100 times smaller)
Bare sphere critical masses and size
U-233 => 15 kg, 11 cm of diameter
U-235 => 52 kg, 17 cm of diameter
Pu-239 => 10 kg, 9.9 cm of diameter
Nuclear Fission:
The process by which a heavy nucleus is split up into two fragments of
roughly equal mass nuclei is known as nuclear fission.
Source: Internet
+ Energy
+Energy
Nuclear Fission:
+ Energy
+Energy
+ Energy
+ Energy
1.0087amu + 235.0457amu = 140.9177amu + 91.8859amu + 3 x 1.0087amu + Energy
Mass Difference = 236.0544amu – 235.8297amu = 0.2247amu
Therefore released energy = 210MeV
Nuclear Fusion:
The process by which two small nuclei fused together to form a relatively
big nucleus is known as nuclear fusion.
Energy
2.0147amu + 2.0147amu → 4.0039amu + Energy
Mass Difference = 4.0294amu –4.0039amu = 0.0255amu
Therefore released energy = 24MeV
Second World War
Germany, Italy, Japan, Rassia . . . . .
1 September, 1939, Germany attacked the Poland
3 September, 1939, England and France announced war against Germany
Germany, Japan, Italy, Romania….
USA, Rassia, England, France, Canada….
7Dec, 1941, Japan attacked PEARL HARBOR, then USA joined
German surrendered: 8 May 1945
Hiroshima: 6 Aug 1945
Nagasaki: 9 Aug 1945
Japan surrendered 14 Aug 1945
Germany
Heisenberg
1939 U split
1943 Lab Destroyed
1945 May 3, Heisenberg arrested
and migrated to England
USA
J. Robert Oppenheimer
Einstein Letter to Franklin D. Roosevelt
1942 Manhattan Project
Within 1945, July; 3 atom bomb
At 5:30 a.m. on July 16, 1945, Trinity Test: Gadget.
Total Expenditure: 2 Billion Dollar
On August 6, 1945, at about 8:15am local time, the US aircraft Enola Gay
dropped an untested uranium-235 gun-assembly bomb nicknamed "Little Boy"
over Hiroshima. Hiroshima was immediately flattened. The resulting explosion
killed 70,000 people instantly; by December 1945, the death toll had risen to
140,000. Within next few years about more than 200,000 people died. The
radius of total destruction was reportedly 1.6km.
Source: Internet
On August 9, 1945, at about 10:58am local time, Boeing B-29 dropped a
plutonium-239 implosion type bomb nicknamed “Fat Man" over Nagasaki.
Source: Internet
Little Boy: A Gun-Type Bomb
The Little boy was 3.0 m long, 0.71 m in diameter, and weighed 4,400 kg (64Kg
of U) and blast yield was 15 KTon of TNT (63 x 1012 J). The Little Boy designed
as a gun-type, that fired one mass of uranium 235 at another mass of uranium
235, thus creating a supercritical mass. Once the two pieces of uranium are
brought together, the initiator introduces a burst of neutrons and the chain
reaction begins, continuing until the uranium become finished.
Fat Man: Implosion-Type Bomb
The device was 3.3 m long, 1.5 m in diameter, and weighed 4,600 kg (6.4 Kg of
Pu) and blast yield was 21 KTon of TNT (88 x 1012 J). “Fat Man” was an
implosion-type device using plutonium 239. A subcritical sphere of plutonium
was placed in the center of a hollow sphere of high explosive. Thirty-two pairs
of detonators located on the surface of the high explosive were fired
simultaneously to produce a powerful inward pressure on the core, squeezing
it and increasing its density, resulting in a supercritical condition and a nuclear
initiation.
The first was
the "Gadget“.
Known
as
Trinity test on
July 16, 1945.
Source: Internet
Tsutomu Yamaguchi was a Japanese marine engineer and a survivor of
both the Hiroshima and Nagasaki atomic bombings during World War II.
Nuclear Power Reactor
Source: Internet
Source: Internet
Nuclear Power Reactor
A nuclear reactor produces and controls the release of energy from splitting
the atoms of certain elements. In a nuclear power reactor, the energy
released is used as heat to make steam to generate electricity. The principles
for using nuclear power to produce electricity are the same for most types of
electric power plant. The following are the stages/components in the working
of a power reactor.
Enrichment
The percentage of U in the naturally occurring uranium is only about 0.7%. In
a nuclear reactor, the percentage of U-235 is increased by a process called
enrichment. This is usually achieved either by gaseous diffusion or by a
centrifuge. In the diffusion process, natural uranium, which is first converted
to gaseous uranium hexafluoride (UF) is passed through a series of semiporous membranes which permit passage of the gas that contains the lighter
U-235 but not the component containing the heavier isotope U-238. The
technique is continued/repeated until achieve the desired concentration of
about 4 to 5% of U-235. In the centrifuge technique the natural UF is spun at
high speed which separates the heavier and the lighter components as they
are subjected to different centrifugal forces.
Fuel
The fuel (Uranium U235, U238, Thorium Th232, Plutonium Pu239, Pu240, Pu241 etc.)
are used in running the reactor. Usually pellets of uranium oxide (UO2) are
arranged in tubes to form fuel rods. The rods are arranged into fuel assemblies
in the reactor core. In a 1000 MW class Pressurized Water Reactor (PWR)
there might be 51,000 fuel rods with over 18 million pellets. The mass of the
fissile material required to sustain chain reaction is said to be critical mass. If a
reactor cannot sustain chain reaction, it is called sub-critical. On the other
hand if the chain reaction proceeds uncontrolled, as in an atom bomb, the
reactor is called supercritical.
Control rod
These are made with
neutron-absorbing material
such as cadmium or boron,
and
are inserted or
withdrawn from the core to
control the rate of reaction,
or to halt it to make sure
that the reactor does not
become supercritical.
In nuclear industry cadmium is commonly used as a thermal neutron
absorber due to very high neutron absorption cross-section
of 113Cd. There is a cadmium cut-off energy for the absorption. Only
neutrons of kinetic energy below the cadmium cut-off energy (~0.5
eV) are strongly absorbed by 113Cd. Therefore, cadmium is widely used
to absorb thermal neutrons in nuclear reactor.
In general, the absorption cross-section is an effective area that
quantifies the likelihood of certain interaction between an
incident object and a target object.
Coolant
A variety of substances, including light water, heavy water, air, carbon dioxide,
helium, liquid sodium etc., have been used as coolants. Such substances are,
in general, good conductors of heat, and they serve to carry the thermal
energy from the fuel to the steam-generating equipment of the nuclear
power plant.
Moderator
In many cases, the same substance functions as both coolant and moderator,
as in the case of light and heavy water. Graphite and beryllium are also used as
moderator. The moderator slows the fast (high-energy) neutrons emitted
during fission to energies at which they are more likely to induce fission. In
doing so, the moderator helps initiate and sustain a fission chain reaction.
Neutrons reflector
A reflector is a region of unfueled material surrounding the core. Its function is
to scatter neutrons that leak from the core, thereby returning some of them
back into the core. The reactor allows for a smaller core size. In addition,
reflectors “smooth out” the power density by utilizing neutrons that would
otherwise leak out through fissioning within fuel material located near the
core’s outer region.
Two types of reactor
Boiling Water Reactor (BWR) In a boiling water reactor, water in pipes
circulates inside the core. The water gets heated, generates steam which is
then used to drive turbines. As the water enters the core, there is a
possibility of its becoming radioactive. Further, in case of a rapture of the
pipe due to extreme heat, it could lead to accidents.
Pressurized Water Reactor (PWR) In these reactors water is extracted in two
steps. The primary coolant, as in the case of BWR circulates inside the core.
However, the water circulates under great pressure so that it does not
become steam. The heat is transferred through a heat exchanger to a
secondary coolant, which may be used to drive a turbine. As the secondary
coolant does not enter the core, it does not become radioactive.
Source: Internet
Thank YOU
Lecture: 9
Modern Physics
The fusion of nuclei can take place in several different reaction sequences, the
most common of which, the proton-proton cycle as suggested by Hans Bethe.
A neutrino is a subatomic particle has no electrical
charge and a very small mass, which might even be
zero.
p-p cycle
1. 1H1 + 1H1  1H2 + e+ + neutrino
2. 1H2 + 1H1  2He3 + Gamma
3. 2He3 + 2He3  2He4 + 1H1 +1H1+ 24.7MeV
= 4 x 10-12 J
Which is a better option to generate power, nuclear fission or fusion?
Fusion is much better than fission in a number of ways:
Firstly, nuclear fusion requires less fuel than fission.
Secondly, fusion is carried out by using deuterium (an isotope of hydrogen) as
fuel, which is quite abundant in nature. In contrast, the fuel necessary for
fission (uranium, plutonium or thorium) is very hard to get, and very
expensive!
Thirdly, unlike fission, nuclear fusion does not produce any radioactive waste;
it only produces helium atoms as a byproduct, which we can actually use to
our benefit in various ways.
Fourthly, Since fusion doesn’t produce chain reactions it should be easy to
control.
So, if fusion is so great, and better than fission in so many respects, why aren’t
we using fusion to generate power already?
Difficulties against using nuclear fusion
When hydrogen atoms fuse, the nuclei must come together. However, the protons in
each nucleus will tend to repel each other because they have the same charge.
High temperature:
The high temperature gives the hydrogen atoms enough energy to overcome the
electrical repulsion between the protons.
Fusion requires temperatures about 100 million Kelvin (approximately six times hotter
than the sun's core).
At these temperatures, hydrogen is a plasma, not a gas. Plasma is a high-energy state
of matter in which all the electrons are stripped from atoms and move freely about.
The sun achieves these temperatures by its large mass and the force of gravity
compressing this mass in the core.
High pressure and density:
High pressure squeezes the hydrogen atoms together. They must be within 1x10-15 m
of each other to fuse.
The sun uses its mass and the force of gravity to squeeze hydrogen atoms together in
its core. This high density ensure that collisions among protons are frequent.
There are two ways to achieve the temperatures and pressures necessary
for hydrogen fusion to take place:
Magnetic confinement uses magnetic and electric fields to heat and
squeeze the hydrogen plasma. The ITER (the International Thermonuclear
Experimental Reactor) project in France is using this method.
Inertial confinement uses laser beams or ion beams to heat and squeeze
the hydrogen plasma. Scientists are studying this experimental approach at
the National Ignition Facility of Lawrence Livermore Laboratory in the
United States.
Let's look at magnetic confinement first. Here's how it would work:
Microwaves, electricity and neutral particle (they do not leave tracks of
ionized particles or curve in magnetic fields) beams from accelerators heat
a stream of hydrogen gas. This heating turns the gas into plasma. This
plasma gets squeezed by super-conducting magnets, thereby allowing
fusion to occur. The most efficient shape for the magnetically confined
plasma is a donut shape (toroid) called TOKAMAK.
Let's take a closer look at the ITER fusion reactor to see how magnetic
confinement works.
ITER ("The Way" in Latin) is one of the most ambitious energy projects in
the world today.
In southern France, 35 nations (The ITER Members China, the European
Union, India, Japan, Korea, Russia and the United States) are collaborating
to build the world's largest tokamak.
ITER project
Inside Tokamak (Vacuum chamber)
Construction/The main parts of the tokamak reactor are:
Vacuum vessel - holds the plasma and keeps the reaction chamber in a
vacuum
Neutral beam injector (ion cyclotron system) - injects particle beams from
the accelerator into the fusion fuel to help the fuel to critical temperature to
make plasma.
Magnetic field coils (poloidal, toroidal) - super-conducting magnets that
confine shape and contain the plasma using magnetic fields.
Central solenoid/Transformers - supply electricity to the magnetic field coils.
Cryostat (cooling equipment) - cool the magnets, provide a supercool
vacuum environment.
Blanket modules - made of lithium; absorb heat and absorb high-energy
neutrons from the fusion reaction to transfer their energy to heat water.
Divertors - exhaust the helium products of the fusion reaction.
Basic working process
1. The fusion reactor will heat a stream of deuterium and tritium fuel to form
high-temperature plasma. It will squeeze the plasma so that fusion can take
place.
2. The lithium blankets outside the plasma reaction chamber will absorb highenergy neutrons from the fusion reaction also get heated.
3. The heat will be transferred by a water-cooling loop to a heat exchanger to
make steam.
4. The steam will drive electrical turbines to produce electricity.
5. The steam will be condensed back into water to absorb more heat from the
reactor in the heat exchanger.
Hydrogen Bomb
On November 1, 1952, the United States detonated a 10.4-megaton hydrogen device in
the Pacific on the Marshall Islands. The test, code-named "Mike," was the first
successful test.
Those who had witnessed
atomic tests were stunned
by the blast. The cloud,
when it had reached its
furthest extent, was about
100 miles wide and 25
miles high. The fireball
reached about 3.25 mile
in diameter. Elugelab, the
island on which Mike’s
detonation took place,
was indeed vaporized,
leaving a crater 2km in
diameter and 50m deep.
The MIKE Test (Operation Ivy)
Fusion Reactors: Inertial Confinement
The National Ignition Facility (NIF) at Lawrence Livermore Laboratory, USA is
experimenting with using laser beams to induce fusion. In the NIF device,
about 192 laser beams will focus on single point in a 10-meter-diameter
target chamber called a hohlraum. A hohlraum is a cavity whose walls are in
radiative equilibrium (Radiative equilibrium is the condition where the total
thermal radiation leaving an object is equal to the total thermal radiation
entering it).
At the focal point inside the target chamber, there will be a pea-sized pellet
of deuterium-tritium encased in a small, plastic cylinder. The power from
the lasers (1.8 million joules) will heat the cylinder which convert the pellet
into plasma and compress (rocket like blowoff) it until fusion occurs. The
fusion reaction will be short-lived, about one-millionth of a second, but will
yield about 50 times more energy than is needed to initiate the fusion
reaction.
Like the magnetic-confinement fusion reactor, the heat from inertialconfinement fusion will be passed to a heat exchanger to make steam for
producing electricity.
Flibe is a molten salt made from a
mixture of lithium fluoride (LiF) and
beryllium fluoride (BeF2).
Hyperspectral imagery (HSI) has
several image bands captured at
distinct
wavelengths
in
the
electromagnetic spectrum.
Figure: Laser Inertial Confinement Fusion Reactor
conceived at Lawrence Livermore Laboratory.
During the final part, the fuel
core reaches around 1000 times
of its initial density (about 20
times the density of lead) and
about 107 to 108 0 C.
Nuclear models
The aim of nuclear models is to understand how neutrons and protons
form a nucleus.
The liquid drop model
The liquid drop model of the nucleus, proposed by Bohr and derived by
Von Weizsacker in 1935, was one of the earliest phenomenological
successes constructed to account for the binding energy of a nucleus.
They marked few similarities between a nucleus and a liquid drop, these
are follows:
(i) In the stable state, a nucleus is supposed to be spherical in shape, just
as a liquid drop is spherical due to the symmetrical surface tension forces.
(ii) Just as the force of surface tension acts on the surface of the liquid
drop, there is a potential barrier acts at the surface of the nucleus.
(iii) The density of a liquid drop is independent of its volume. Similarly, the
density of the nucleus is also almost independent of its volume.
(iv) The intermolecular forces in a liquid are short range forces. The
molecules in a liquid drop interact only with their immediate neighbors.
Similarly, the nucleons in the nucleus also interact only with their
immediate neighbors. That means, the nuclear forces are also sort rang
forces. This leads to the saturation of nuclear forces and a constant
binding energy per nucleon.
(V) When the temperature of the liquid is raised, the molecules from a
liquid drop evaporate due to their increased energy of thermal agitation.
Similarly, when a nucleus is bombarded with nuclear projectiles it gains
energy and a compound nucleus is formed. But the compound nucleus
emits nuclear radiation almost immediately.
(vi) When a small drop of liquid allow to oscillate, it breaks up into two
similar drops of equal size. Similarly, in nuclear fission the nucleus breaks
up into two nuclei of almost equal mass numbers.
(vii) Two small liquid drop make a large drop by combining together.
Similarly two deuterium produce helium by fusion process.
Semi-empirical mass formula
The liquid drop model can be used to obtain an expression for the binding
energy of the nucleus. We shall start by assuming that the energy associated
with each nucleon-nucleon bond has some value U.
Volume energy: Since each bond energy is shared by two nucleons, the
binding energy associated with each nucleon 1/2U. When an assembly of
spheres of the same size is packed together in the smallest possible volume,
each interior sphere has 12 other spheres in contact with it. Hence each
interior nucleon in a nucleus has a binding energy of (12) (1/2U) or 6U. if all
the A nucleons in a nucleus could be considered to be in the interior of the
nucleus, the total binding energy of the nucleus would be
Ev = (6U)(A) = 6 AU = a1 A …………… (1) (Volume energy)
The energy Ev is called volume energy of a nucleus and is proportional it A.
Thank YOU
Lecture: 10
Modern Physics
Semi-empirical mass formula
The liquid drop model can be used to obtain an expression for the binding
energy of the nucleus. We shall start by assuming that the energy associated
with each nucleon-nucleon bond has some value U.
Volume energy: Since each bond energy is shared by two nucleons, the
binding energy associated with each nucleon 1/2U. When an assembly of
spheres of the same size is packed together in the smallest possible volume,
each interior sphere has 12 other spheres in contact with it. Hence each
interior nucleon in a nucleus has a binding energy of (12) (1/2U) or 6U. if all
the A nucleons in a nucleus could be considered to be in the interior of the
nucleus, the total binding energy of the nucleus would be
Ev = (6U)(A) = 6 AU = a1 A …………… (1) (Volume energy)
The energy Ev is called volume energy of a nucleus and is proportional it A.
Surface energy: Actually not all the nucleons are within the interior the
nucleus. Some nucleons are on the surface of the nucleus and therefore
have fewer than 12 neighbors. The number of such nucleons depends on
the surface area of the nucleus. A nucleus of radius R has an area of
4πR2 = 4π𝑅02 𝐴2/3
Hence the number of nucleons with fewer than the maximum number of
bonds is proportional to A2/3. Thus the total binding energy is reduced by
Es = -a2A2/3 …………..(2) (Surface energy)
It is most significant for the lighter nuclei since a greater fraction of their
nucleons are on the surface. Because natural system always tend to evolve
toward configurations of minimum potential energy, nuclei tend toward
configurations of maximum binding energy. Hence a nucleus should exhibit
the same surface-tension effects as a liquid drop. Therefore, in the
absence of external forces it should be spherical, since a sphere has the
least surface area for a given volume.
Coulomb energy:
The electrostatic force of repulsion between two protons (a pair of protons) in
a nucleus also decreases the binding energy of the nucleus. The potential
energy of a pair of protons separated by a distance r is given by
𝑒2
𝑉=−
4𝜋𝜀0 𝑟
Since there are Z(Z-1)/2 pairs of protons in a nucleus, coulomb energy,
𝑍(𝑍 − 1)
𝑍 𝑍 − 1 𝑒2 1
𝐸𝑐 =
.𝑉 = −
2
8𝜋𝜀0
𝑟
where
1
𝑟 𝑎𝑣
is the value of
1
𝑟
averaged over all proton pairs. If the protons are
uniformly distributed throughout a nucleus of radius R,
and hence to
1
𝐴1/3
𝑎𝑣
1
is
𝑟 𝑎𝑣
proportional to
1
𝑅
, so that
𝑬𝒄 = −𝒂𝟑
𝒁(𝒁−𝟏)
………… (3)
𝑨𝟏/𝟑
(Coulomb energy)
The Coulomb energy is negative because it arises from an effect that opposes nuclear
stability.
The total binding energy of the nucleus according to liquid drop model ought
to be the sum of its volume, surface and coulomb energies:
Eb = Ev + Es + Ec = a1A – a2A2/3 - 𝒂𝟑
𝒁(𝒁−𝟏)
𝑨𝟏/𝟑
The binding energy per nucleon is therefore,
𝑬𝒃
𝑨
=
𝑩𝑬
𝑨
= 𝒂𝟏 −
𝒂𝟐
𝑨𝟏/𝟑
− 𝒂𝟑
𝒁(𝒁−𝟏)
…
𝑨𝟒/𝟑
… … … (4)
All terms of eqn. 4 is
plotted in Fig. 1 versus A.
Since the theoretical curve
is fit so well with the Semiempirical mass formula
means that the analogy
between a nucleus and a
liquid drop has some
validity at least.
Fig. 1
The Shell Model
The shell model assumes that the energy structure (energy levels) of the
nucleons in a nucleus is similar to that of electrons in an atom. According to
this model, the protons and neutrons are grouped in shells in the nucleus in a
manner similar to grouping of electrons in various shells outside the nucleus.
The shells are regarded as "filled“ when they contain a specific number of
protons or neutrons or both. It should be remembered that the case is not so
simple as it appears. In the extranuclear shells only one type of particles (i.e.,
electrons) is to be arranged in different shells and Pauli's exclusion principle is
applied. In the case of the nucleus there are two types of particles, protons
and neutrons and the shell arrangement is only empirical and based upon the
study of the stability and interactions of the nuclides which are known.
Nuclei that have 2, 8, 20, 28, 50, 82 and 126 neutrons or protons are more
abundant than other nuclei of similar mass number, suggesting that their
structures are more stable. Many evidence also points up the significance in
nuclear structure of the numbers 2, 8, 20, 28, 50, 82 and 126. These numbers
known as magic numbers. The main points in favour of this inference are:
(i) The inert gases with-closed electron shells exhibit a high degree of
chemical stability. Similarly, nuclides whose nuclei contain a magic number of
nucleons exhibit more than average stability.
(ii) Nuclides of isotopes of elements having an isotopic abundance greater
than 60% contain magic numbers.
(iii) The stable end product of the three main radioactive series (viz, the
uranium, actinium and thorium series) is 82Pb208 which contains 82 protons
and 126 neutrons. Thus, 82Pb208 is the most stable isotope. This shows that
the number 82 and 126 indicate stability.
(iv) Tin (50Sn) has ten stable isotopes, while (20Ca) has six stable isotopes. So,
elements with Z = 50, 20 are usually more stable.
(v) 8O17, 36Kr87 and 54Xe137 isotopes spontaneously emit neutrons. For these
isotopes, the number of neutrons are 9, 51 and 83 respectively which can be
written as 8+1, 50+1 and 82+1. If this loosely bound neutron is interpreted as
a valency neutron, then the neutron numbers 8, 50 and 82 represent greater
stability than other neutron numbers.
𝑬𝒃
𝑨
Asymmetry energy:
=
𝑩𝑬
𝑨
𝒂
𝟐
= 𝒂𝟏 − 𝑨𝟏/𝟑
− 𝒂𝟑
𝒁(𝒁−𝟏)
…
𝑨𝟒/𝟑
… … … (4)
The binding energy formula as given by eqn. 4 can be improved by two effects that
do not fit the simple liquid drop model, but which can be readily explained in
terms of a model that take into account nuclear energy levels (shell model). One of
these effects occur when the neutrons in a nucleus outnumber the protons or vice
versa, which means that higher energy levels have to be occupied than would be
the case if N and Z are equal.
Let the uppermost neutron and proton energy
levels as shown in Fig. 1 which have the same
energy spacing 𝜺. In order to produce a
neutron excess, say, N-Z = 8 without changing
1
A, 2(N- Z) = 4 protons converted to neutrons in
an original nucleus in which N = Z. The new
neutrons would occupy levels higher in energy
by 2 𝜀 = 4 (𝜀 /2). Each shifted proton (new
neutron) raised in energy by
1
(𝑁
2
𝜀
− 𝑍) 2 .
Fig. 1
The total raised in energy,
ΔE = number of new neutron x increased energy by each neutron
=
1
(𝑁
2
− 𝑍)
1
(𝑁
2
𝜀
𝜀
− 𝑍) 2 = 8 𝑁 − 𝑍
2
The same formula will hold if Z> N, since(N-Z)2 is always positive.
Now, N= A-Z, therefore (N-Z)2 = (A-2Z)2, and
𝜀
∆𝐸 = 𝐴 − 2𝑍 2
8
The greater the number of nucleons in a nucleus, the smaller is the energy level
spacing 𝜀, with 𝜀 proportional to 1/A.
The departure from N=Z introduces an asymmetry N-Z, which results in a decrease in
binding energy.
This means that the asymmetry energy Ea due to the difference between N and Z can
be expressed as
𝐸𝑎 = −∆𝐸 =
𝐴−2𝑍 2
−𝑎4
…………………(4)
𝐴
(Asymmetry energy)
The asymmetry energy is negative because it reduces the binding energy of the
nucleus.
Pairing energy:
The final correction term arises from the tendency of protons and neutrons to occur
in pairs. Even-even nuclei are the most stable and hence have higher binding
energies than would otherwise be expected. Thus even-even nuclei like 2He4, 6C12,
16
8O appear as peaks on the empirical curve of binding energy per nucleon. On the
other hand, odd-odd nuclei have both unpaired protons and neutrons and have
relatively low binding energies. The pairing energy Ep, is positive for even-even
nuclei, 0 for odd-even and even-odd nuclei, and negative for odd-odd nuclei and
seems to vary with A as A-3/4 . Hence
1
𝐸𝑝 = ±, 0 𝑎5 𝐴3/4 … … … … …(5) (Pairing energy)
Total Binding Energy
The final expression for the binding energy of a nucleus of atomic number Z and
mass number A was first obtained by C. F. Von Weizsacker in 1935 and is known as
Weizsacker’s semi-empirical mass formula. This Is given by
𝐸𝑏 = 𝑎1 𝐴 −
2
𝑎2 𝐴3
− 𝑎3
𝑍 𝑍−1
1
𝐴3
− 𝑎4
𝐴 − 2𝑍
𝐴
The best values of the constants, expressed in MeV, are
a1 = 15.76, a2 = 17.81, a3 = 0.711, a4 = 23.702 and a5 = 34
2
(±, 0)𝑎5
1
𝐴3/4
Radioactivity
• Properties of ∝ - particles, 𝜷- particles and 𝜸- rays
• Radioactive transformation
• Law governing radioactive disintegration/ Decay law
𝒅𝑵
∝𝑵
𝒅𝒕
= 𝑵𝟎 𝒆−λ𝒕
−
𝑵
• Half life period
𝟎. 𝟔𝟗𝟑
𝑻𝟏/𝟐 =
λ
• Average life period (or Mean life) of a radioelement
𝟏
𝑻𝒂 =
λ
Average life period (or Mean life) of a radioelement
Since the radioactive atoms are constantly disintegrating one after another, the life of
every atom is different. The atoms which disintegrate earlier have very short life
whereas others which disintegrate at the end have a very long time. Thus, the
possible time of existence of a radioactive atom may vary from zero to infinity.
The average life period (or mean life ) of a radioactive clement is defined as the ratio
of the sum of lifetime of all the radioactive atoms to the total number of such atoms
in it.
Let N0 be the total number of radioactive atoms in the beginning at t= 0.
Let N be the number of atoms at that element after a time t.
Now let the dN atoms disintegrate between t and (t + dt). Since dt is very small, we
may assume without serious error that all these atoms have lived for a time t. Hence
the total lifetime of these dN atoms
= (-dN)t.
The possible life of any atom varies from 0 to ∝, the total life period of all atoms is
given by
∝
=
−𝑑𝑁 𝑡
0
Therefore, Average life period,
𝑆𝑢𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑓𝑒 𝑝𝑒𝑟𝑖𝑜𝑑 𝑜𝑓 𝑎𝑙𝑙 𝑡ℎ𝑒 𝑎𝑡𝑜𝑚𝑠
𝑇𝑎 =
𝑇𝑜𝑡𝑎𝑙 𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑎𝑡𝑜𝑚𝑠
∝
0
𝑢𝑣𝑑𝑥 = 𝑢
𝑣𝑑𝑥 −
𝑑𝑢
𝑑𝑥
−𝑑𝑁 𝑡
=
𝑁0
∝
λ 𝑁0 𝑒 −λ𝑡 𝑡𝑑𝑡
0
=
𝑁0
𝑣𝑑𝑥 𝑑𝑥
∝
−𝑑𝑁
∝𝑁
𝑑𝑡
𝑑𝑁 = −λ 𝑁 𝑑𝑡
= −λ𝑁0 𝑒 −λ𝑡 𝑑𝑡
𝑡𝑒 −λ𝑡 𝑑𝑡
=λ
0
= λ 𝑡.
𝑒 −λ𝑡
−λ
−
1.
𝑒 −λ𝑡
−λ
Therefore, Average life period, 𝑇𝑎 =
1
λ
𝑑𝑡
0
1 −λ𝑡 ∝
−λ𝑡
= −𝑡𝑒
− 𝑒
λ
0
=− 0+0 − 0+
∝
1
λ
Thank YOU