Fundamentals of modern physics

Engineering Physics course materials: by Praveen N Vaidya, SDMCET Dharwad.
Modern Physics
Physics is the oldest branch of science which
deals with the study of matter and various forms
of energy associated with matter and its
applications.
Upto the last decade of 19th
century Classical
physics (Also termed as Classical mechanics)
based on Newton’s laws of motion, low
gravitation and also Maxwell’s theory of
distribution of energy are played major role in
developments of Theories of Physics.
Modern Physics
The series of discoveries and inventions
happened in last decade of 19th century and at the
beginning of 20th
century reveled that, the
classical laws are in good agreement only with
motion of macroscopic matter and with low speed
motion, but unable to explain some observed facts
especially at the motion of microscopic matters
like electrons and sub atomic matters and
behavior matter relativistic motion.
The classical inability of mechanics to
give the explanation for the phenomena like,
Blackbody radiation, photoelectric effect and
Compton effect etc. also proved classical
mechanics is incomplete.
Hence to deal with such observed facts, new
theories are developed in physics namely, Planks
quantum theory, theory of wave particle duality
and Einstein theory of relativity etc. The physics
based on these theories is called as Modern
physics.
The modern Physics defined a microscopic
particle called quantum particle, stream of
quantum particles accelerate with high speed from
source, produce light, which have both particle
and wave nature based on the circumstances.
Hence modern physics also called as Quantum
Mechanics.
Following illustrations proof for are the drawback
of classical laws.
Introduction to Blackbody radiation
spectrum
Blackbody:
A blackbody is an ideal source of electro
magnetic radiation at temperatures higher than the
surrounding and emits the radiation of every
wavelength at any temperature. It is shown in the
following black body radiation curve (Fig)
Spectrum of blackbody radiation:
Lummer and Pringshein studied energy
distribution in the spectrum of blackbody
radiation at various temperatures. The results are
depicted in the graph shown. The observation of
the curves revels following facts:
The black body radiation curve (Fig) shows that
the black body emits the radiation of every
| P a g e 1

wavelength at any temperature. From figure it is
found that curve gets infinitely close to the x-axis
but never touches it. The curve touches at infinite
wavelength.
The intensities of the different wavelengths
emitted by a black body is depending upon its
temperature and this complete set of wave
lengths emitted is called black body spectrum.
For given temperature intensity of all wave
lengths is not same, in the beginning intensity of
radiation increases with wave length, reach
maximum for a given wavelength (λmax) and
decreases after words
Wien’s displacement law:
Total energy emitted by the blackbody increases
with the increase in temperature simultaneously
the wavelength λmax corresponding to the
maximum energy Em emitted by the blackbody
decreases with the increase in temperature. This is
called Wien’s displacement law.
Mathematically, λmax T = Constant.
Explanation of blackbody radiation
spectrum:
Wien’s distribution law:
According to Wien’s distribution law the energy
emitted by the blackbody per unit volume in the
range of wavelength from λ to λ+dλ is given by
Where C1 and C2 are constants and T is absolute
temperature. It was found that Wien’s
distribution law explains energy distribution only
in the shorter wavelength region. Also, the
energy emitted by the blackbody is finite as
temperturetend to infinity, which is contradiction
with Stefan’s fourth power law (E T4
).
Rayleigh-Jeans law:
According to Rayleigh-Jeans law the energy
Where k is Boltzmann constant whose value is
1.38 × 10-23
JK-1
.
Rayleigh-Jeans formula was found to explain
energy distribution only in the longer
wavelength region of the blackbody radiation
spectrum. Further amount of energy in the
shorter wavelength region so that no energy is
available for emission in the longer wavelength
region.
Rayleigh-Jeans law predicts that blackbody
radiates enormous amount of energy in the lower
wave lengths (higher frequencies) like ultraviolet
and x-rays.
But blackbody radiate chiefly in the visible and
infrared region. This discrepancy between
| P a g e 2

experimental results and theoretical predictions
was known as ‘ultraviolet catastrophe’.
Planck’s Radiation Law:
According to Planck’s radiation law whenever
radiation is emitted or absorbed by a blackbody
energy exchange between matter and radiation
takes place only in discrete steps, i.e emission or
absorption of radiation of frequency ν takes place
in steps of integrals multiples of packets of energy
called quanta (hν). Thus Energy emitted or
absorbed by the body is given by
E = nhν
Where h = plank’s constant = 6.254 × 10-34 Js
and n = 1, 2, 3……
According to Planck’s radiation law the energy
------ 1
Where c = velocity of light = 3 × 10 8 m s-1
This equation is suits correctly with the
experimental results for both lower and higher
wave lengths
Planck’s radiation law explains energy
distribution in the spectrum of blackbody
radiation over the wide rang of wavelength, i.e., it
is in agreement with the experimental results.
Further, it can be shown that Planck’s law
reduces to,
Wien’s distribution law in shorter wavelength
region:
At shorter wave length, λT tends to zero hence
kT
hc
kT
hc
ee λλ
⇒−1
There fore equatin (1) becomes
This is given by weins distribution law
Where C1 = 8πhc and C2 = hc/k are constants
Also reduces to Rayleigh-Jeans law in the
longer wavelength region.
At longer wave lengths
−−−+=−
kT
hc
e kT
hc
λ
λ
1
For higher wave lengths higher terms are
negligible
There fore equation (1) becomes
45
88
λ
π
λλ
π kT
kT
hchc
= because kTE
hc
==
λ
This is reyleigh Jean’s law
Photoelectric effect:
The phenomenon of emission of electrons by
the surface of certain materials under the
influence of radiation of suitable frequency is
called photoelectric effect.
The materials which exhibit this property are
called photosensitive while the electrons emitted
are called photoelectrons
According to classical physics -
| P a g e 3

(a) Light is an electromagnetic wave - the
intensity of light is determined by the amplitudes
of these electromagnetic oscillations. When light
falls on an electron bound in an atom, it gains
energy from the oscillating electric field. Larger
the amplitude of oscillations, larger is the energy
gained by the emitted electron - thus energy of the
emitted electrons should depend on the intensity
of the incident light. This is in contrast to what
has been observed in experiment
(b) According to the electromagnetic theory, the
velocity of the emitted electrons should not
depend on the frequency of the incident light.
Whatever may be the frequency of the incident
light, the electron would be emitted if it gets
sufficient time to collect the necessary energy for
emission. So the photoelectric emission is not an
instantaneous effect.
(c) Finally, the incident electromagnetic wave acts
equally on all the electrons of the metal surface.
There is no reason why only some electrons will
be able to collect the necessary energy for
emission from the incident waves. Given
sufficient time, all electrons should be able to
collect the energy necessary for emission. So
there is no reason why the photoelectric current
should depend upon the intensity of the incident
light. However,
In contrast above statements of classical physics,
observed facts are here.
Experimental observations are,
1. Photoelectric effect is an instantaneous process.
i.e., electrons are emitted by the surface as soon as
radiation falls on the surface.
2. If frequency of the incident radiation is less
than certain value, photoelectrons are not emitted
however strong the intensity radiation may be.
The frequency above which photoelectric effect
takes place is called threshold frequency denoted
by ( νo).
3. Kinetic energy of the photoelectrons emitted is
directly proportional to the frequency of the
incident radiation and independent of the intensity
of the radiation.
4. Number of photoelectrons emitted is directly
proportional to the intensity of the intensity of
radiation but independent of the frequency of the
radiation.
5. Photoelectric effect is independent of the
temperature.
6. If the anode plate is made negative w.r.t. to
photo sensitive surface, the photo electron retard
due to repulsion. If the negative potential
increased sufficiently the K.E. energy falls to
zero.
Therefore the sufficient negative potential
applied to anode plate to stop photoelectrons
completely from reaching the anode plate is
called as stopping potential (V).
Here K.E. of photoelectrons = Ve
Where V – negative potential applied
e – charge of electron.
| P a g e 4

Therefore at maximum retarding potenetial
(stopping potential),
V =
e
mv
e
EK
22
.. 2
=
Or
me
p
V
2
2
=
Einstein’s explanation of photoelectric effect
based on quantum theory of radiation:
Free electrons in metal are bound within the
surface of the metal. The energy required to emit
electron from the surface is called Work Function
(Wo). If this energy (Wo)is in the form of light
then Wo = hνo, and νo is threshold frequency
According to Einstein’s quantum theory of light,
light emited from source consists of fast moving
photons, and energy of each photon is E = hν,
where h is Planck’s constant and ν is frequency of
the radiation.
If ν > νo, then each photon interact with each of
the electrons and expel it from the surface and
electron accelerate with a kinetic energy is given
by,
K.E. is directly praportional to frequency of
radiation and photoelectric emission is
spontaneous.
High intensity corresponds to large number of
photons, and they emit large number of
photoelectron, hence Intensity of light is directly
proportional to photoelectric current.
Compton Effect:
Arthur H. Compton observed the scattering of x-
rays from electrons in a carbon target and found
scattered x-rays with a longer wavelength than
those incidents upon the target.
There fore the phenomenon in which the
wavelength of X-rays increases when scattered
from a suitable target is called Compton Effect.
This kind of scattering is also called incoherent
scattering.
Explanation:
Compton explained and modeled the data by
assuming a particle (photon) nature for light (x-
ray) and applying conservation of energy and
conservation of momentum to the collision
between the photon and the electron. The
scattered photon has lower energy and therefore a
longer wavelength according to the Planck
relationship, E = h
The shift of the wavelength increased with
scattering angle according to the Compton
formula
| P a g e 5
Where,
m – mass of electron
p – Momentum of electron
v – velocity of electron

The Compton experiment gave clear and
independent evidence of particle-like behavior.
Compton was awarded the Nobel Prize in 1927
for the "discovery of the effect named after him".
De Broglie’s hypothesis:
We know that the phenomenon such as
interference, diffraction, polirisation etc. can be
explained only with the help of wave theory of
light, while phenomenon such as photoelectric
effect, Compton effect, spectrum of black body
radiation can be explained only with the help of
Quantum theory of radiation. Thus radiation is
assumed to exhibit dual nature. i.e. both particle
and wave nature.
In 1924, Louis de Broglie made a bold hypothesis,
which can be stated as follows
“If radiation which is basically a wave can
exhibit particle nature under certain
circumstances, and since nature likes
symmetry, then entities which exhibit particle
nature ordinarily, should also exhibit wave
nature under suitable circumstances”
Thus according to de Broglie’s hypothesis, there
is wave associated with the moving particle. Such
waves are called Matter waves and wavelength
of the wave associated with the particle is called
de Broglie wavelength.
Expression for de Broglie wavelength (from
analogy):
The radiation consists of stream of particles called
Photons. Energy of each photon is given by
E = hγ
Where h = Plank’s constant and γ is the frequency
of the radiation.
But according to Einstein’s mass energy
relationship E = mc2
de Broglie proposed that same equation is
applicable for matter waves also. Therefore
wavelength of the waves associated with the
moving particle of mass m, moving with the
velocity v is given by
λ is called as de Broglie wavelength.
de- Broglie wavelength of an electron accelerated
by a potential difference of V volts.
| P a g e 6
If θ=90,
h/moc = 0.02426 Å

Consider an electron accelerated by a potential
difference of V volts. Kinetic energy gained by
the electron is given by
Experimental Verification of de-Broglie’s
Hypothesis (Devisson-Germer experiment)
This is the first experiment which confirmed the
wave nature of electrons. The basic experimental
arrangement is shown below along with the nickel
crystal structure. In 1925, Davisson and Germer
were studying electron scattering from various
materials. Their important discovery was made
when nickel was used as the target. Here, the
kinetic energy of electrons can be controlled by
the voltage V.
To conduct experiment Devison and Germer used
un-oxidized nickel crystal, which is done by
heating the crystal in oven almost baked.
The experimental setup:
The experimental setup consists of a filament
called electron gun (E), which emits a stream of
electrons on heated to red hot by the application
of electric field.
The collimated beam of electrons are accelerated
by an energy eV (where V is potential difference
applied to filament) allowed to fall on nickel
target (C).
The target is capable rotation about an axis
perpendicular to the plane of the paper.
A detector called Faraday Chamber detects the
electrons scattered from the nickel target.
The detector is capable of rotation about the same
axis as the target. The detector is fitted with a
galvanometer to measure the current.
Observation:
The energetic electrons emitted from the
electron gun accelerated towards the nickel target
and scatter at an angle φ and detector receives the
scattered electrons at various angles.
The reading is taken for the intensity
against scattering angle for a given energy. The
experiment is repeated for the different energies
of electron so that number of sets of reading
obtained is plotted (fig.).
It is found that in every set, there is a peak
to indicate sudden increase of intensity of
scattered electrons at a particular angle just like
the diffraction pattern of x-rays in a crystal.
| P a g e 7

More over if energy of incident electrons
is increased from smaller value the peak also go
on increases and reach max at particular energy
(eV) and further it go on decreases and this
variation if found periodic (as shown in fig.).
This is analogous to the diffraction pattern.
Conclusion:
The maxima is observed at the energy of electrons
accelerated through a potential 54V, the de-
broglie wavelength for corresponding electron is
given by
meV
h
p
h
2
==λ
m
5410602.1101.92
1063.6
1931
34
×××××
×
=
−−
−
= m10
1066.1 −
×
The wavelength of the diffracted electron beam is
also calculated using Bragg formula for X-
ray diffraction, which is given by
Consider in above set up the electron is diffracted
from the plane of crystal is (111), from this plane
A maximum of the diffraction pattern is observed
at an angle φ = 50o
(in fig.).
Therefore the glancing angle of incidence
(scattering) of this Bragg’s plane is given by


65
2
50180
=
−
=θ
d = inter planer spacing of the nickel crystal =
0.91 x 10-10
m
For m = 1,
Or θλ sin2d= = 2 x 0.91 x 10-10
x Sin 65 o
= 1.65 x10-10
m
The de Broglie formula thus gives a value
of wavelength, for the electron that is in axcellent
agreement with their experimentally measured
wavelength. Therefore Davisson-Germer
experiment not only provides an evidence of
the wave nature of electron but also gives a
striking quantitative confirmation of the de
Broglie formula.
Phase velocity and Group velocity:
Matter in motion will show wave like property,
will not form a single wave because it is
continuous, but matter waves are localized, for
this de-Broglie proposed that, a group of waves of
slightly different wave lengths, superposition of
these number of waves gives a resultant standing
wave just like the resultant wave pattern in beats.
Therefore the superposition of number of waves
of suitable wavelength and amplitude of matter in
motion gives a resultant wave, whose amplitude
varies periodically, called as wave packet. This
wave packet is localized one.
Phase velocity:
The velocity with which an individual wave
travels is called Phase velocity or wave velocity.
It is denoted by vp.
| P a g e 8

Consider the traveling wave equation
)( x
v
tSino
ϖ
ϖψψ −=
Let
v
ϖ
= k, where v – velocity of wave
Or v =
k
ϖ
= Vp ----------------- 1
Where vp is called as wave velocity is also called
as phase velocity
Therefore
k
vp
ω
=
From equation (1)
Vp = k
ϖ
= ν λ
p
E
p
h
h
E
=×=
mv
mc2
=
v
c2
=
Here c – velocity of light and v – particle velocity
It is clear from the above equation that, Phase
velocity is not only greater than the velocity of the
particle but also greater than the velocity of light,
which can never happen. Therefore phase
velocity has no physical meaning in case of
matter waves. Thus a concept of group velocity
was introduced.
Group velocity:
Since phase velocity has no meaning, the concept
of group velocity was introduced as follows.
“Matter wave is regarded as the resultant of
the superposition of large number of
component waves all traveling with different
velocities. The resultant is in the form of a
packet called wave packet or wave group. The
velocity with which this wave group travels is
called group velocity.” The group velocity is
represented by vg.
Expression for group velocity:
Consider two waves A and B having same
amplitude A, but of slightly different wavelengths
and frequencies traveling in the same direction.
The two waves are represented by
When the two waves superimpose upon one
another, the resultant of the two waves is given by
| P a g e 9

The above equation represents a wave traveling
with the velocity ω/k. But amplitude of the wave
is not constant which is given by
The velocity with which variation in the
amplitude is transmitted represents the group
velocity, and is given by
Relation between group velocity and
particle velocity
We know that group velocity is given by
Expression for de-Broglie wavelength
(From Group velocity)
We know that group velocity is given by
If m is the mass, v is the velocity and V is the
potential energy of the moving particle, then its
total energy is given by, …
| P a g e 10

Taking constant as zero we get
That is given by de-Broglie wave equation.
Relation between velocity of light (c), group
velocity (vg) and phase velocity (vp)
2
2
2
22
,
2
2
cvvor
v
c
vtherefore
mv
mc
p
mc
h
mch
E
k
v
particlephase
particle
p
p
=×=
=
====
λ
λ
π
π
ϖ
Following are some of the important
properties of matter waves:
1. Matter waves are waves associated with the
moving particle
2. They are not electromagnetic in nature
3. Wavelength of the matter wave is given by l =
h mv= h/ p
4. The amplitude of the matter wave at the given
point determines the probability of finding the
particle at that point at the given instant of time.
5. The wave nature of matter is explained only by
group velocity, there is no meaning for phase
velocity as its value goes beyond the velocity of
light which can not be possible.
6. The de Broglie concept of matter waves are
suitable for only for microscopic matters like,
electrons, photons and other inter atomic particles
in motion and not suits for macroscopic matters.
Questions:
Example 1.1
A metal has a workfunction of 4.3 V. If the EM
radiation of wavelength 1800Å is incident on
surface calculate the de-Broglie wavelength of
emitted photoelectron.
The corresponding wavelength equals:
Energy of EM radiation
E= J
hc 19
10
834
1010
101800
1031062.6 −
−
−
×=
×
×××
=
λ
K.E. energy of electron
= Energy of EM radiation – work function
= J19
1010 −
× - J19
1089.6 −
× = 3.11x10-19
J
De-Broglie wavelength =
mE
h
p
h
2
==λ
1931
34
1011.3101.92
1062.6
−−
−
××××
×
=
Example 2:
Question: A cricket ball has a mass of 0.150 kg
and is bowled towards a bowler at 40m/s.
Find the de Broglie wavelength of the
cricket ball.
| P a g e 11

We are required to calculate the de Broglie
wavelength of a cricket ball given its mass and
speed. We can do this by using:
The mass of the cricket ball m = 0.150kg
The velocity of the cricket ball
Planck's constant
the de Broglie wavelength
This wavelength is considerably smaller than the
diameter of a proton which is
approximately . Hence the wave-like
properties of this cricket ball are too small to be
observed.
Question: Calculate the de Broglie wavelength
of an electron moving at 40 m·s−1
.
We are required to calculate the de Broglie
wavelength of an electron given its speed. We can
do this by using:
The velocity of the electron
The mass of the electron
Planck's constant
(
Although the electron and cricket ball in the two
previous examples are travelling at the same
velocity the de Broglie wavelength of the electron
is much larger than that of the cricket ball. This is
because the wavelength is inversely proportional
to the mass of the particle.
3. What is the deBroglie wavelength of an
electron with 13.6 eV of kinetic energy? What is
the deBroglie wavelength of an electron with 10
MeV of kinetic energy?
Answer
13.6 eV is much less
than so this is non-
relativistic.
10 MeV is much bigger than for an
electron so it is super-relativistic and we can
use .
Example 1.1
A metal has a workfunction of 4.3 V. If the EM
radiation of wavelength 1800Å is incident on surface
calculate the de-Broglie wavelength of emitted
photoelectron.
The corresponding wavelength equals:
| P a g e 12

Energy of EM radiation
E= J
hc 19
10
834
1010
101800
1031062.6 −
−
−
×=
×
×××
=
λ
K.E. energy of electron
= Energy of EM radiation – work function
= J19
1010 −
× - J19
1089.6 −
× = 3.11x10-19
J
De-Broglie wavelength =
mE
h
p
h
2
==λ
1931
34
1011.3101.92
1062.6
−−
−
××××
×
=
MODERN PHYSICS:
1. Calculate the de Broglie wavelength of a
neutron (mass = 1.67493 x 10¯27
kg) moving at
one five-hundredth of the speed of light
(c/500)
λ = 5.107 x 10¯10
m = 5.107 Å
2. A rain droplet of radius 0.5mm falling under
gravity with terminal velocity of 2m/s,
determine its de-Broglie wavelength (density
of water 1000kg/m2
).
r = 0.5 mm,
v = 2 m/s
λ = 6.33 x 10-28
m.
3. Determine the K.E. in eV, momentum, group
velocity, particle velocity and phase velocity
of an electron of de-Broglie wavelength 2A.
p = h/λ = =3.32x10-24
N-s
Group velocity,
= (3.32 x 10-24
)/
= 0.364 x 107
m/s =particle velocity.
Phase velocity,
vphase = c2
/v
= (3x108
)2
/ (0.364 x 107
)
= 24.72x109
m/s
| P a g e 13

4. Compare the de-Broglie wavelengths of
electron and a proton accelerated by a
potential of 100keV (proton mass = 1.6726 ×
10-27
kilograms).
For proton,
For Electron,
= = = =
2.33x102
therefore, 2.33x102
5. What is the De-Broglie wavelength of the
nitrogen molecule in air at 300K? Atomic mass
of nitrogen is 14.0076amu. (1 atomic mass unit =
1.66053886 × 10-27
kilograms)
=0.391x10-10
m =0.391Å
6. Calculate the De-Broglie wavelength of the
electron of kinetic energy 10eV. (Rest ass energy
of electron is 511 keV and h=6.6x10-34
Js.)
Energy of electron E =10 eV= 16 x10-19
J.
Er = Rest mass energy of electron = 511 KeV
= 511 x103
x1.6 x 10-19
J = 817.6 x10-16
J,
According to Einstein’s mass-Energy relation,
Er = me c2
,  me = Er / c2
= (817.6 x10-16
)/(3 x108
)2
= 9.084 x10-31
kg.
De-Broglie wavelength,
=
= 3.89 x 10-10
m =3.89Å
7. UV ration of wavelength 3200 A focused on a
metal plate. Find the wavelength associated
with the photoelectron emitted, if work
function of metal is 4.2 eV.
Wavelength of light λ = 3200Å,
W = work function of metal = 4.2 eV
Energy of light = E = hc/λ
= (6.64 x 10-34
x 3 x108
)/(3200 x10-10
)
= 6.225 x 10-19
J = (6.225 x 10-
19)/(1.6x10-19
)
= 3.89 eV.
From Photo electric equation,
K.E. of electrons (Ek)
= E – W = 4.2 -3.89 = 0.31 eV
= 0.31 x 1.6 x10-19
J = 0.496 x10-19
J
| P a g e 14
m = 14.0076amu
= 14.0076 x 1.66053886 × 10-27
kg,
T=300K,
(1 amu = 1.66 × 10-27
kg)

De-Broglie wavelength of electron =
= = 2.21 nm
8. Determine the glancing angle of electrons
diffracted from a crystal surface if, a stream
of electron of K.E. 75 eV accelerated towards
that crystal target.
=
=1.42Å.
nλ = 2d sinθ, → sinθ= nλ/2d →
θ = Sin-1
(nλ/2d)
= Sin-1
(1x1.42/2x 1.024) = 43.89 o
= 43o
53’.
11. Find the De-Broglie wavelength of neutron at
1270
C. Given that Boltzmann constant is
1.38x10-23
Jmolecule-1
K-1
. [Ans. 1.264A]
12. The wavelength of the photon is 1.4A. it
collides with the electron. Its wavelength after
the collision is 2A. Calculate the de-Broglie
wavelength of the scattered electron.
13. An alpha particle and proton are accelerated
from the rest through same potential
difference V. Find the ratio of De-Broglie
wavelength associated with them.
14. A particle is moving 3 times as fast as an
electron. The ratio of De-Broglie wavelength
of the particle to that of the electron is
1.83x10-4
. Find the particle mass and identify
the article. Mass of electron is 9.1x10-31
kg.
[Ans. 1.675x10-27
kg, neutron or proton]
15. Calculate the [a] momentum and [b] de-B
wavelength of the electrons accelerated
through a potential difference of 56V. [Ans.
4.02x10-24
kgm/s, 1.64A]
16. An electron and proton have same amount of
kinetic energy. Which of the two will have
greater De-Broglie wavelength?
{Ans. electron]
| P a g e 15

1. ‘E’K.E of particle then, its de Broglie wavelength
is__
2. According to de-Broglie hypothesis matter waves
formed by______
3. De Broglie hypothesis is concerned with________
4. Experiment which confirms the wave nature of
particles is_________.
5. De-Broglie wavelength of an electron accelerated
6. with a 100V _________
7. The relation connecting group velocity and phase
8. velocity is ___________
9. If p momentum of particle, its de-Broglie wave
length is___
10. Particle velocity of a de-Broglie wave is also
called as___
11. If velocity of a particle is 6x105
m/s, it’s de-
Broglie wavelength is______
12. 5. Voltage with which an accelerated electron to
get de-Broglie wavelength 2pF is_______
1. In Davison and Germer experiment, _______ nature
of matter is confirmed by the diffraction of electron.
a) λ > λ’ b) λ < λ’ c) λ = λ’ d) None
2. A small lead ball is allowed to fall freely under
gravity, its de Broglie wavelength
a) Increases b) Decreases
c) Remains same d) None
10. Waves Associated with moving particles are
i) EM waves ii) Mechanical waves
iii) Matter waves iv) None of these
1. Give an illustration for the failure of classical
laws and how the quantum theory overcome
drawback.
2. What is de-Broglie hypothesis? What are
matter waves? What is the significance of
matter waves?
3. Derive an expression for the wavelength of
matter waves, in terms of velocity, energy and
momentum of matter waves.
4. Show the wave nature of a microscopic
particle by experimental proof in Davison and
Germer experiment.
5. What are phase velocity and group velocities?
Write the difference between them.
6. What is particle velocity? Show that particle
velocity of matter is equals to group velocity.
7. Derive an expression for the de-Broglie
wavelength by group velocity concept.
8. Discuss the properties of matter waves.
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Fundamentals of modern physics

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