CHAPTER – 17
ADVENT OF
MODERN PHYSICS
Qs. What are the basic postulates of
Einstein’s Special theory of relativity. Also give the consequences of the
theory.
EINSTEIN’S
SPECIAL THEORY OF RELATIVITY
Introduction
Einstein examine the motion of objects in frames of references moving relative
to one another. On the basis of his experimental results he proposed a special
theory of relativity in the year 1905. This theory is valid specially for
inertial frames and is to be modified into a general theory for accelerated
frames of reference.
BASIC POSTULATES
The Einstein’s special theory of
relativity is based on two assumptions known as the postulates of special
relativity. The two postulates are states as follows.
First Postulate
The speed of light was regarded as
the universal constant. It means that the speed of light in vacuum is the same
for all observers in uniform transnational motion and is independent of the
motion of the observer and the source.
Second Postulate
According to this postulate the laws
of physics in the frame moving with uniform velocity can be expressed by a
single set of mathematical expression.
This postulate points out if some event takes place in any of the frame and the
frames are moving with uniform velocity the result of the two frames will be
identical. Conversely if the frames are in accelerated motion then the result
will not be identical.
MASS ENERGY RELATION
Einstein proved that energy has
inertia, which is the property of matter and associated with mass. Thus mass is
simply a property attributed to the total energy of the body and only total
energy is required to know total mass of the body. Hence in special theory of
relativity total energy and mass are related by the famous Einstein’s equation.
E = mc(2)
From this relation between mass and
energy it has been predicted that any process that changed the mass by a
detectable amount of energy.
Qs. Write a note on Compton Effect
COMPTON EFFECT
In 1926, Arthur Compton studies this
phenomenon of change in wavelength. On the basis of his experimental results he
proposed a theory based on the idea of photon theory of radiation. Since the
detailed study of phenomenon was made by Compton, the effect is now known as
the Compton’s Effect.
Definition
The phenomenon in which a photon (hv) strike with stationary electron
and after collision both scattered in different direction in such a way v >
v is known as Compton Effect.
Consideration
In order to explain this phenomenon we assume that photon strike with a
stationary electron and after collision both makes an angle θ and ф with
respect to their initial line of motion.
Qs. Write note on Pair Production and
Annihilation of Matter.
PAIR
PRODUCTION
Definition
The phenomenon in which photon collides with heavy nucleus then two
material particles, electron and positron are produced, is called Pair
Production.
Explanation
The positron produced during pair production has been identified to be
identical with an electron in mass and carries an equal positive charge and is
called the anti particle of electron. Since the process of pair production
involves the creation of particle and its anti particle, therefore it is also
known as materialization of energy. This phenomenon is the practical proof of
Einstein’s mass energy equivalence, in which mass and energy of the system
remains constant.
For the production of electron and
positron 1.02 MeV energy is required. I can be calculated by the following
equation
Eo = 2moC(2)
=> Eo = 2 x 9.1 x 10(-33) x (3 x
10(8))2 / 1.6 x 10(-19)
=> Eo = 1.02 x 10(6) cV
=> Eo = 1.02 MeV
If energy of photon is less than 1.02
MeV then this phenomenon cannot produce. If energy of photon is greater than
1.02 MeV then rest of energy is used to accelerate the electron and Positron.
The energy conservation in Pair Production demands.
hv = e+ + e + K.E + K.E+
=> hv = moc² + moc² + K.E + K.E+
=> hv = moc² + K.E + K.E+
ANNIHILATION
Definition
The phenomenon in which electron and positron fuse together to form at least
two photons, is known as Annihilation of matter.
Explanation
Annihilation is the reverse process of pair production. In Pair Annihilation a
particle and one of its anti particle come close enough to be converted
completely into radiation energy of the two photons moving in opposite
direction conserving the total momentum of the creation and annihilation
process. Each photon will have an energy equal to rest mass energy moc of an
electron that is equal to 0.51 MeV.
The energy conservation equation for
the process will be
(mo)e + c² + K.Ee + (mo)e-c² + (K.E_e
= 2hv
Conclusion
The phenomenon of Pair Production and annihilation helps us to conclude that
energy and mass are inter changeable.
Qs. Write note on Uncertainty Principle
UNCERTAINTY
PRINCIPLE
Introduction
In classical physics we can easily determine the momentum and position of
moving body simultaneously with accuracy, that no uncertainties are involved in
it. But for a light particle is found that however refined we make our
instruments there is a fundamental limitation to the accuracy with which the
positron and momentum can be known simultaneously.
This limitation was first expressed by Hersenberg in 1927 and is known as
Uncertainty Principle.
Statement
It is impossible to measure with accuracy both positron and momentum of a
particle simultaneously.
Consideration
Consider a slit of thickness Δy placed near to a screen. Now a particle bean
strikes the slit then after diffraction at very small angle, it reaches at
points A.
Proof
As we know that momentum is a vector quantity, therefore, it can be resolved
into two components. Consider ΔOAB.
tan θ = Perpendicular / Base
=> tan θ P(y) / P(x)
Since θ is very small, therefore
tan θ ≈ θ
=> θ P(y) / P(x)
=> Py = Px θ ——- (I)
From the condition of interference,
mλ = d sin θ
For first maximum,
m = 1
=> λ = Δy sin θ
But,
sin θ ≈ θ
=> λ = Δy θ
=> θ = λ / Δy
Substituting the value of θ in eq (I)
=> P(y) = P(x) λ / Δy
=> P(y)Δy = P(x) λ
From Debroglie’s wave equation
λ = h / P(x)
=> λP(y)Δy = P(x) h / P(x)
=> P(y)Δy = h
Similarly,
P(x)Δx = h
And,
P(z)Δz = h
Conclusion
The uncertainty principle is of no importance in our daily life because plank’s
constant h is very small and so the uncertainties in position and momentum of
even light objects are far too small to be experimentally observed.
Qs. State and explain Debroglie’s
Hypothesis.
DEBROGLIE’S
HYPOTHESIS
Introduction
In 1924, Debroglie proposed an idea called Debroglie’s Hypothesis.
Statement
If light can have particle behaviour then material particles such as electrons
and protons etc can also behave in a wave like manner.
Mathematical
Form
According to Debroglie’s Hypothesis a
particle like electron can possesses a momentum given by
P = mv = h / λ
Where m is the mass of particle. This
relation has related the electron a particle and the wave character of a
frequency. Thus we can write down the wave length associated with the particle
i.e.
λ = h / mv
Conclusion
The Debroglie’s relation was initially developed for the electron but it is
valid for all material objects including particles. However for massive
materials the associated wavelength is too small to be measured.
Qs. Define and Explain Photoelectric Effect
PHOTOELECTRIC
EFFECT
Introduction
In 1887, Hertz discovered the phenomenon of emission of electrons. When ultra
violet light falls on certain metals. On the basis of his experimental results,
he proposed the phenomenon of photoelectric effect.
Definition
The emission of electrons from a solid or liquid surface when it is subjected
to electromagnetic radiation is known as Photo-electric effect.
Experiment
Consider a glass tube in which two electrodes are suspended connected to a
positive and negative terminal of a battery. A milliammeter is connected in
series with the circuit to detect the flow of current.
When ultra violet rays strike the
negative plate, then electrons emit. These electrons are repelled by the
negative (-) plate and attracted by the positive plate. Hence, current flows in
the circuit. The effect is known as Photoelectric effect.
Maximum K.E
of Electrons
The maximum K.E. of electrons can be
achieved by reversing the polarity of the circuit. When ultra violet rays
strike the positive (+) electrode. The kinetic energy possessed by the
electrons can be achieved if it is balanced by the voltage. So we increase the
voltage to such an extent that no electrons emit out. At this stage K.E. is
maximum and can be calculated by
K.E(MAX) = Voe
=> 1/2 mv² = Voe
Where,
m = mass of electron
e = charge of electron
v = velocity of electron
Vo = voltage of circuit
Results
Obtained
The conclusions that were made from
the experiment on Photoelectric effect are
1. Increasing the intensity of source of light increases the number of photoelectrons
but not the velocity with which it leaves the metallic surface.
2. For each substance, there is a certain frequency called the threshold
frequency below which the effect does not occur.
3. The higher the frequency of incident ray, the greater the K.E of electrons.
Photoelectric effect could not be explained on the basis of classical wave
theory, because according to the theory:
There should be no threshold
frequency because by that time electrons might escape from the metallic surface
by absorbing enough energy.
The velocity of photelectrons should depend upon the intensity of the incident
ray rather than the frequency.
Qs. Give Einstein’s explanation of the
photoelectrons effect on the basis of quantum theory of radiation.
EINSTEIN’S
EXPLANATION OF PHOTOELECTRONS EFFECT
Introduction
Albert Einstein was successful in providing an explanation of the photoelectric
effect. He proposed his description on the basis of quantum theory of
radiation.
Explanation
Einstein explained the photoelectric effect on the basis of following
postulates.
1. An electron absorbs neither one
whole photon or it absorbs none.
2. An electron cannot absorb more than one photon.
3. After absorbing a photon, it
acquires energy (hv) equal to photon. The energy is either used up in ejecting
the electrons or it dissipates within the metal surface.
4. An electron may lose some of its
energy before leaving the metal surface and is ejected with a kinetic energy
less than hv.
5. If the energy of the photon is
less than the energy required to overcome the forces then the electron will not
emit.
Mathematical
Expression
The energy of the electron is given
as
Total Energy = Work Function + K.E
=> hv = фo + 1/2 mv²
фo = hvo
=> hv = hvo + 1/2 mv²
=> hv – hvo = 1/2 mv²
=> h(v-vo) = 1/2 mv²
Since, K.E = 1/2 mv² = Voe and v =
c/λ
=> h [c/λ - c/λo] = Voc
=> hc [1/λ - 1/λ] = Voc
The above equation is known as
Einstein’s Photoelectric Equation.
Qs. What is a Photo Cell? Also Write its
Uses.
PHOTO CELL
Construction
The Photocell or photo tube consist of an evacuated glass tube fitted with an
anode and a concave metallic cathode of an appropriate surface.
The material of the cathode can be choosen to respond to the frequency range
over which the photocell operates.
Working
When light of suitable frequency fall on the cathode photoelectrons are emitted
which are attracted by the positive anode and current flows in the external
circuit. The current would cease to flow if the light beam is interrupted.
USES OF PHOTO CELL
1. Simple Photo Cell
A simple photo cell can be used in
any situation where beam of light falling on a cell is interrupted or broken.
Examples are given below.
To count vehicles passing a road or items running on a conveyer belt.
To open door automatically.
To operate burglar alarm etc.
2. Photo Conducting Cell
In this cell internal photoelectric
effect liberates free charge carrier in a material and its electrical
conductivity increases as much as 10,000 times,
Its uses are
For detection and measurement of
infrared radiations where the wavelength is of the order of 10(-6) m.
As relays for switching on artificial
lighting, such as streetlights.
3. Photo Voltaic Cell
Such cells are used as exposure
meters to set the aperture of the camera.
4. Other Uses
Photocells are used for the
production of pictures in television cameras and the sound tracks on motion
pictures. The sound information is stored on the film in the form.
Statement
Radiant energy comes out in discreat amounts or guanta of energy. The energy E
content of each quantum was directly proportional to the frequency v.
Mathematical
Form
E ∞ v
=> E = hv ——– (I)
Where h = Plank’s constant = 6.63 x
10(-34) Js. Since,
c = vλ
=> v = c / λ
Thus equation (I) becomes
E = hc / λ
Where,
c = velocity of light = 3 x 10(8)
m/s.
λ = wavelength of radiation
The energy of ‘n’ photons is given by
E = nλy
Where,
n = 0, 1, 2, 3 ………
