SPH 202 MODERN PHYSICS

1.0 Relativity
1.1 Introduction to relativity*
1.2 Galilean relativity*
1.3 Time dilation*
1.4 Length contraction*
1.5 Relativistic mechanics*
1.6 Einstein’s mass energy relation*
2.0 Wave particle duality
2.1 Black body radiation
2.2 Photoelectric effect *
2.3 Gravitational red shift
2.4 Spectral lines
2.5 de Broglie waves
2.6 Uncertainty principle

3.0 Atomic physics

3.1 Thomson atomic model

3.2 Rutherford atomic model

3.3 Bohr model of the atom

4.0 Nuclear Physics

4.1 Nuclear structure

4.2 Nuclear stability

4.3 Radioactivity

Introduction to relativity 0786164094

0721334140
Introduction to special theory of relativity . Relativistic mechanics. Einstein’s mass-energy
relation. Black body radiation. Photoelectric effect. Gravitational red shift Spectral lines.
Bohr model of the atom. Quantization of angular momentum. de Broglie waves.
Heisenberg’s uncertainty principle. Standing wave model of the atom. The Davisson-
Germer experiment. Quantized atomic levels. Concepts of nuclear physics. Nuclear
models, the binding energy curve. Fission and fusion. Radioactive decay.
Most intelligent people have funny stories Isaac Newton (1643-1727) died a virgin at 84
years. Albert Einstein his brain was so special that the pathologist on call, Thomas Harvey,
stole it…….where is Einstein brain? Albert Einstein did not want to be worshiped or his body
studied, he therefore left instructions to be cremated immediately.
Albert Einstein (1879-1955) who was a genius died in April 1955 at 01:15 hr, and was
cremated later that day in New Jersey. But the following day, his son Hans Albert learned
that the body in the coffin had not been intact. The pathologist who conducted the autopsy,
Dr Thomas Harvey, had gone further than simply identifying the cause of death-a burst aorta.
He had sawed open Einstein’s cranium and removed the brain….

Introduction to special theory of relativity

In 1905, at the age of only 26, Albert Einstein postulated his theory of special relativity. This theory came to
solve problems where Newtonian mechanics had failed. Newtonian mechanics was formulated to describe
motion of objects that move with speed much less than that of light.
Newtonian mechanics however fails when applied to particles whose speeds approach that of light c =
299,792,458 or 2.997 x 108 m/s
Consider you are travelling in a jet airliner from Nairobi to New York where your sister is waiting to receive
you. Your wrist watch on your hand record the passage of time as experienced by yourself. Amazingly, did you
know, the duration of trip as measured by your earth bound sister in new York would be slightly longer. How
can high speed travel affect something as regular as the ticking of a clock?
The Michelson Interferometer
The interferometer invented by American Physicist A. A Michelson (1852-1932) splits a light beam into two
parts to form an interference pattern. The device can be used to measure wavelengths or other lengths with great
precision.

Mirror M, inclined at 45 ℃ to the incident light beam splits the beam into two. One beam is
transmitted to mirror M 2 and the other is reflected towards mirror M 1. After reflection from
M 1 and M 2 the two rays eventually recombine at M to produce interference pattern which can
be viewed through the telescope. The glass plate P, equal in thickness to mirror m, ensures
that the two returning rays travel the same thickness of glass. Mirror M 1 if adjusted, causes
series of bright and dark fringes due to interference.

The Michelson –Morley Experiment

The most famous experiment designed to detect small changes in the speed of light was first
performed in 1881 by Michelson and later repeated under different conditions by Michelson
and Edward Morley (1838-1923)
The experiment was designed to determine the velocity of the earth relative to that of
hypothetical ether. Arm 2 was aligned along the direction of the earth’s motion through
space. The earth moving through the ether at speed v is equivalent to the ether flowing past
the earth in opposite direction with speed v.
This ether wind blowing in the direction opposite the direction of earth’s motion should cause
the speed of light measured in the earth frame to be c-v as the light approaches mirror M 2
and c+v after reflection.
The two beams reflected from M 1 and M 2 recombine and an interference pattern consisting
of alternating dark and bright fringes is formed.
Suppose the length of each arm was L, the time of travel to the light hand side is
L
c−v
distance
(Side work time= )
speed
And the time of travel to the left is
L
c+ v
Total time for round trip becomes
2 −1
L L 2 Lc 2 Lc v
t 1= + c−v = 2 2 = c (1− 2 )
c+ v c +v c
Now consider the beam travelling along arm 1. The speed of the beam relative
1
to the earth’s is (c 2−v 2) 2 see the figure below and recall we learnt relative
speeds in Sph 100: Mechanics 1
 1
2 2 2
a=(c −v )

The time of travel for each half of the trip is

L
1
2 2
(c ¿ ¿ 2−v ) ¿
Total time of travel for round trip is
2L 2l v
2 −1
t 2= 1 = (1− 2 ) 2
2
(c −v )
2 2 c c
Thus time difference between horizontal and vertical beams ∆ t becomes

[ ]
2 −1 2 −1
2L v v
∆ t=t 1−t 2= (1− 2 ) −( 1− 2 ) 2
c c c
2
v
Because 2
≪¿ 1 we can simplify using binomial expansion after dropping all
c
terms higher than second order
(1−x ) ≈ 1−nx for x << 1
n

2
v
In our case, x= 2 , and we find that
c
2
Lv
∆ t=t 1−t 2 ≈ 3
c
Question 3
a) Calculate the relativistic momentum of a proton moving at a speed of v
=0.86c in both
i. kg .m/ s
ii. eV
Relativistic Mechanics

1. Relativistic Momentum

In classical mechanics, we write momentum as p=mv 1

mv
But relativistic momentum becomes p=γmv =
√ 1−
v2
c
2
2

If v << c, equation 2 reduce to equation 1

Where m is rest mass or proper mass of an object, its mass when measured at
rest.

Question 3

a) Calculate the relativistic momentum of a proton moving at a speed of v

=0.86c in both

i. kg .m/ s(joules)
ii. eV

Check if m c 2=938 MeV

1.67x10−27 x(3.0x10pow8)squared = 938x1000000x1.602x10−19

2. Relativistic second law

 dv dp
In classical mechanics, second law is written as F=ma=m = 3
dt dt

dv dp d
Relativistic second law becomes F=γma=γm =γ = (γmv) 4
dt dt dt

When v<< c equation 4 reduce to equation 3

Example

Find acceleration of a particle of mass m and velocity v when it is acted upon by a constant
force F, where F is parallel to v

dv
Since a=
dt

(√ )
d v
d m
F= (γmv) = dt v2 5
dt 1−
c2

Recall derivative of a product

dγ dv
F=mv + γm 6
dt dt
2 −1 /2
d v
F=mv [(1− 2 ) ]+ γma 7
dt c

Use the chain rule to evaluate γ

dy/du =dy/du*du/dv
2
v −3
Let u=1− 2 du/d=-1/2U 2
c
−3

dy/dv= -1/2U
2
( −2c v )( dvdt )
2

[ ( ) ( ) dt )]+ γma
2 −3
F=mv
−1
2
v
1− 2
c
2
( −2 v
dv
c
2 8

Revise implicit differentiation in maths

Factor out common factor γma

 [ ]
2
v
2
c
F=γma +1 9
v2
1− 2
c

( ) ( ) ( )
2 −3 2 −1 2 −1
v 2 v 2 v
Note sidework 1− 2 = 1− 2 . 1− 2
c c c

Find a common denominator in 9

[ ]
v2 v2
2
+1− 2
c c
F=γma 10
v2
1− 2
c

( ) ( )( ) ( )
2 3 2 1 2 1 2 3
v v v 3 v
Note 1− 2 2 = 1− 2 2 . 1− 2 hence γ = 1− 2 2

c c c c

Equation 10 becomes

[ ]
1
F=γma
v2
1−
c2
3
F=γ ma

F
a= 3
mγ

( )
2 3/ 2
F v
a= 1− 2 acceleration of the particle
m c

Mass and Energy

(The most famous relationship Einstein obtained E o=m c2 )

The work done W on an object by a constant force F through a distance s moving with a
velocity v is W = Fs. If no other force work and the particle starts from rest, all the work
done on it become kinetic energy
s

K.E = ∫ Fds 1
0
In non relativistic physics the formula for kinetic energy is

1 2
K . E= m v 2
2

To the correct relativistic formula for kinetic energy, we start from relativistic form of
Newton second law
s mv v
dγmv
K . E=∫ ds=∫ vdγmv=∫ vd ¿ ¿ 3
0 dt 0 0

v
K . E=∫ vd ¿ ¿ 4
0

Integrating equation 4 by parts ∫ xdy =xy−∫ ydx

v
Let y be and x be v hence dx is dv
√¿ ¿ ¿
v
mvv v dv
K . E= −m∫
√ 1−v 2
/c
2
0 √ 1−v 2 /c2
Research on steps missed here
v
v dv
Side work m∫
0 √ 1−v 2 /c 2
v
2
du −2 v −2 vdv 2 du
Let u=1− 2 therefore = 2 or du= or vdv=−c
c dx c c
2
2
2 1
−c du

√
v 2 −1 2 2
v
−c du mc u v2
m∫
v v
v dv 2 2
−c u du =-
m∫ =m ∫ 1 2 = 1 =
m ∫ 2 2∗1 =−mc 2
1−
0 √ 1−v /c
2 2 2
o
0
2 u2 0
c
u 2
2

2
K . E=
mv
√ 1−v /c
2 2
+ [ mc √ 1−v /c ] v
2 2 2

0 () 5

2
mv
+ mc √ 1−v /c −m c
2 2 2 2
K . E= 6
√ 1−v /c
2 2
 ( )
2
2 v
2
m c 1− 2
mv c 2
K . E= + −m c 7
√ 1−v
( )
2 2 2 1 /2
/c v
1− 2
c

( )
2
v
1− 2
c
Note 7 above √ 1−v 2 2
/c =¿
( )
1/ 2
v2
1− 2
c

m v2 m c 2−m v 2
K . E= + −m c2
√ 1−v /c 1− v
( )
2 1 /2
2 2
8
c2

2 2 2
m v + mc −mv
K . E= -m c 2 9
√ 1−v 2 2
/c
2
mc 2
K . E= −m c 10
√ 1−v /c
2 2

2
K . E=γm c −mc =(γ −1)m c
2 2
11

Equation 11 says that, the kinetic energy of a particle is equal to the difference
between γm c 2∧¿ m c 2

Equation 11 may be written as

2
γm c =mc + K . E
2
12

Hence

Total Energy 2
E=γm c =m c + K . E
2
13

If we interpret γm c 2 as total energy E of a particle we see that, when it is at rest

and K.E is zero, it possesses the energy m c 2. m c 2 is called rest energy E o of a
particle whose mass is m.
E=EO + K . E

Rest energy EO =m c2
If the object is moving, its total energy is
 2
mc
2
E=γm c =¿
√1−v 2 /c 2

Example 1.6
A stationary body explodes into two fragments each of mass 2.5 kg that move apart at speeds
of 0.6c relative to the original body. Find the mass of the original body.
Solution

The rest energy of the original body must equal the sum of the total
energies of the fragments. Hence EO =m c2 rest energy

When the object starts moving then rest energy is converted to

the total energy of the body
2
EO =γm c
2 2
m1 c m2 c
2
γm c = +
√1−v 2 /c 2 √1−v 2 /c 2
Eo m1 m2 2.5 kg 2.5 kg
m= = + = +
c
2
√ 1−v 2
/c
2
√1−v 2
/c
2
√1−¿ ¿ ¿ √1−¿ ¿ ¿
2 x(2.5 kg) 5 5
m= = 2 = =6.24 Kg
√ 1−¿ ¿ ¿ √1−0.6 0.64
ENERGY AND MOMENTUM
How they fit together in relativity
Total energy and momentum are conserved in an isolated system, and the rest energy of
a particle is invariant. Hence these quantities are in some sense more fundamental than
velocity or kinetic energy, which are neither. Let us look into how the total energy, rest
energy, and momentum of a particle are related.
We begin with Equation for total energy,
2
mc
2
E=γm c =¿ and squaring it to give
√1−v 2 /c 2
2 4
2 m c
E= 2 2
1−v /c

From momentum equation

mv m v c
2 2 2

√ we can manipulate to get p c = 2

p=γmv = v2 2 2
v
1− 2
1− 2
c c

Subtract p c
2 2
from E2
 2 2 2
m v c
2 4
m c
E −p c =
2
2 2 2
2 2
−¿ v
1− 2
= m2 c 4 ¿¿
1−v / c
c

2 2
E − p c =( mc )
2 2 2

2 2
E = (m c ) + p c
2 2 2
Energy and momentum
The relationship between E and p for a particle with m=0
E=pc massless particle

Electronvolts
In atomic physics the usual unit of energy is the electronvolt (eV), where 1 eV is the energy
gained by an electron accelerated through a potential difference of 1 volt.
1eV = 1.602 x 10−19 joules
Example 1.8
An electron (m = 0.511 MeV/c 2) and a photon (m = 0) both have momenta of 2.000 MeV/c.
Find the total energy of each.
2 2
E = (m c ) + p c
2 2 2

√ 2 2
E= ( mc ) + p c =
2 2
√ ((0.511 MeV /c )c ) +(2.0 MeV /c ) c
2 2 2 2 2
= 2.064 MeV

E=pc = (2.000 MeV /c) x c = 2.000MeV

Wave Particle duality

Classical physics, treats particles and waves as separate components. The mechanics of
particles and the optics of waves are traditionally independent disciplines, each with its own
chain of experiments and principles based on their results.

We regard electrons as particles because they possess charge and mass and behave according
to the laws of particle mechanics in such familiar devices as television picture tubes. We shall
see, however, that it is just as correct to interpret a moving electron as a wave manifestation
as it is to interpret it as a particle manifestation. We regard electromagnetic waves as waves
because under suitable circumstances they exhibit diffraction, interference, and polarization.
Similarly, we shall see that under other circumstances electromagnetic waves behave as
though they consist of streams of particles.
ELECTROMAGNETIC WAVES
In 1864 the British physicist James Clerk Maxwell made the remarkable suggestion that
accelerated electric charges generate linked electric and magnetic disturbances that can travel
indefinitely through space. If the charges oscillate periodically, the disturbances are waves
whose electric and magnetic components are perpendicular to each other and to the direction
of propagation, as in Fig. 2.1.

Maxwell was able to show that the speed c of electromagnetic waves in free space is given by

During Maxwell’s lifetime the notion of em waves remained without direct experimental
support. Finally, in 1888, the German physicist Heinrich Hertz showed that em waves indeed
exist and behave exactly as Maxwell had predicted.

Hertz determined the wavelength and speed of the waves he generated, showed that they have
both electric and magnetic components, and found that they could be reflected, refracted, and
diffracted. Light is not the only example of an em wave. Although all such waves have the
same fundamental nature, many features of their interaction with matter depend upon their
frequencies. Light waves, which are em waves the eye responds to, span only a brief
frequency interval, from about 4.3 x 1014 Hz for red light to about 7.5 x 1014 Hz for violet
light. Figure 2.2 shows the em wave spectrum from the low frequencies used in radio
communication to the high frequencies found in x-rays and gamma rays.
BLACKBODY RADIATION

Following Hertz’s experiments, the question of the fundamental nature of light seemed clear:
light consisted of em waves that obeyed Maxwell’s theory. This certainty lasted only a dozen
years. The first sign that something was seriously amiss came from attempts to understand
the origin of the radiation emitted by bodies of matter.

We are all familiar with the glow of a hot piece of metal, which gives off visible light whose
color varies with the temperature of the metal, going from red to yellow to white as it
becomes hotter and hotter. In fact, other frequencies to which our eyes do not respond are
present as well. An object need not be so hot that it is luminous for it to be radiating em
energy; all objects radiate such energy continuously whatever their temperatures, though
which frequencies predominate depends on the temperature. At room temperature most of the
radiation is in the infrared part of the spectrum and hence is invisible.

It is convenient to consider as an ideal body one that absorbs all radiation incident upon it,
regardless of frequency. Such a body is called a blackbody.

One of the problems faced by scientists in 19th century was how to explain the spectra of
electromagnetic radiation emitted by hot objects. At everyday temperatures this radiation is
almost all in the infra region (IR) and thus not visible to our eyes. At a temperature of 1000 K
an object begins to emit an appreciable amount of radiation in the long wavelength end of the
visible spectrum, observed as reddish glow.

Still higher temperatures cause radiation to shift to even shorter wavelengths and the color to
change to yellow orange. Above 2000 k an object glows yellowish white like the filament of
a light bulb and gives off appreciable amounts of all the visible colors (wavelengths) but with
different percentages, Figure 3.3 shows intensity versus wavelength curves for the thermal
radiation from a blackbody at different temperatures. Why does the blackbody spectrum
have the shape shown?

A blackbody radiates more when it is hot than when it is cold, and the spectrum of a hot
blackbody has its peak at a higher frequency than the peak in the spectrum of a cooler one.
We recall the behavior of an iron bar as it is heated to progressively higher temperatures: at
first it glows dull red, then bright orange-red, and eventually it becomes “white hot.” The
spectrum of blackbody radiation is shown in Fig. 2.6 for two temperatures.
The Ultraviolet Catastrophe
Why does the blackbody spectrum have the shape shown in Fig. 2.6? This problem was
examined at the end of the nineteenth century by Lord Rayleigh and James Jeans. Lord
Rayleigh and James Jeans developed the following formula in a quest to explain the spectrum
emitted by a black body.
o

The failure of classical physics led Max Plank to the discovery that radiation is emitted in
quanta whose energy is hv where v- frequency. But where does this radiation originate?

Planks formula fits the observed data very well as shown in figure 3.16
Explanation of black body radiation

 Can only be explained by Plank theory of quantization. Which states that energy is
absorbed or emitted as discrete packets.
 Atoms at low energies exist at ground state, when heated they jump to a higher energy
level absorbing a photon of energy of a given colour, on increased heating the
electron will jump to even higher energy level and emit a photon of light of different
colour

Planck’s quantum theory

According to Planck’s quantum theory,

1. Different atoms and molecules can emit or absorb energy in discrete quantities only.
The smallest amount of energy that can be emitted or absorbed in the form of
electromagnetic radiation is known as quantum.
 2. The energy of the radiation absorbed or emitted is directly proportional to the
frequency of the radiation.

Meanwhile, the energy of radiation is expressed in terms of frequency as,

E=hν

Where,

E = Energy of the radiation

h = Planck’s constant (6.626×10–34 J.s)

ν= Frequency of radiation

Interestingly, Planck has also concluded that these were only an aspect of the processes of
absorption and emission of radiation. They had nothing to do with the physical reality of the
radiation itself. Later in the year 1905, famous German physicist, Albert Einstein also
reinterpreted Planck’s theory to further explain the photoelectric effect. He was of the opinion
that if some source of light was focused on certain materials, they can eject electrons from the
material. Basically, Planck’s work led Einstein in determining that light exists in discrete
quanta of energy, or photons

3.2 PHOTOELECTRIC EFFECT

The energies of electrons liberated by light depend on the frequency

of the light
Electrons are usually emitted when light of sufficient frequency falls on a metal surface. This
phenomenon is known as the photoelectric effect and the emitted electrons are called
photoelectrons.
It is one of the ironies of history that the same work to demonstrate that light consists of em
waves also gave the first hint that this was not the whole story.
Figure 2.9 shows how the photoelectric effect was studied. An evacuated tube contains two
electrodes connected to a source of variable voltage, with the metal plate whose surface is
irradiated as the anode.
Some of the photoelectrons that emerge from this surface have enough energy to reach the
cathode despite its negative polarity, and they constitute the measured current.
The slower photoelectrons are repelled before they get to the cathode. When the voltage is
increased to a certain value V0, of the order of several volts, no more photoelectrons arrive,
as indicated by the current dropping to zero. This extinction voltage corresponds to the
maximum photoelectron kinetic energy.

Quantum Theory of Light (wave theory of light)

When Planck’s derivation of his formula appeared, Einstein was one of the first— perhaps
the first—to understand just how radical the postulate of energy quantization was.

In 1905, Einstein realized that energy in light is not spread out but exists as discrete packets
called qunta or photons. Each photon of light of frequency v have energy

E=v h 1
Since v=c / λ 2
We get
ch
E= 3
λ
We often speak of photons as if they were particles and as concentrated bundles of energy
they have particle like properties. Photons travel with speed of light and so they must obey
the relativistic relationship p=E /c (since E = m c 2 = (mc)c = pc )

From equation 1 above we get

h
p =
λ
Photons carry linear momentum as well as energy and thus they share characteristic of
particles. Because photons travel with speed of light, they must be of zero energy. Otherwise
its energy and momentum would be infinite.

Suppose v o is the minimum frequency of light below which no photoelectrons are emitted.
The energy represented by equation 1 becomes

is called the work function of the metal. The greater the work function of a material the
greater the energy required for an electron to leave its surface. Some examples of
workfunctions for different metals are illustrated on the table below

To pull an electron from a metal surface generally takes about half as much energy as that
needed to pull an electron from a free atom of that metal
Example
Ultraviolet light of wavelength 350 nm is directed at a potassium surface. Find the maximum
KE of the photoelectrons.

Assignment
Explain what is meant by gravitational red shift of spectral lines

DE BROGLIE WAVES
A moving body behaves in certain ways as though it has a wave nature

Louis de Broglie (1892–1987), His doctoral thesis in 1924 contained the proposal that
moving bodies have wave like properties that complement their particle properties: