Class 12 Exam > Class 12 Videos > Application of wave nature - Dual Nature of Radiation & Matter
Application of wave nature - Dual Nature of Radiation & Matter Video Lecture - Class 12
FAQs on Application of wave nature - Dual Nature of Radiation & Matter Video Lecture - Class 12
| 1. What is the dual nature of radiation and matter? |  |
Ans. The dual nature of radiation and matter is a fundamental concept in physics that states that both radiation (such as light) and matter (such as electrons) exhibit characteristics of both particles and waves. This means that they can behave as discrete particles with definite positions and momenta, as well as waves with properties like interference and diffraction.
| 2. How does the wave nature of radiation and matter explain phenomena like interference and diffraction? | |
Ans. The wave nature of radiation and matter explains phenomena like interference and diffraction by considering them as the result of the superposition of waves. When waves interact with each other, they can either reinforce each other (constructive interference) or cancel each other out (destructive interference), depending on their relative phases. This leads to the formation of interference patterns and the bending of waves around obstacles, known as diffraction.
| 3. What is the significance of the dual nature of radiation and matter in quantum mechanics? | |
Ans. The dual nature of radiation and matter is of great significance in quantum mechanics, which is the branch of physics that describes the behavior of particles at the microscopic level. Quantum mechanics relies on the concept of wave-particle duality to explain phenomena such as the behavior of electrons in atoms, the emission and absorption of photons, and the wave functions that describe the probabilities of finding particles in different states.
| 4. How does the dual nature of radiation and matter contribute to the understanding of the photoelectric effect? | |
Ans. The dual nature of radiation and matter helps in understanding the photoelectric effect, which is the phenomenon where electrons are ejected from a material when it is illuminated by light. According to the wave nature of light, the energy of light is spread out over the entire wavefront. However, the photoelectric effect can only be explained by considering the particle nature of light, where light is quantized into discrete packets of energy called photons. These photons transfer their energy to electrons in the material, causing them to be ejected.
| 5. Can you provide an example of an experiment that demonstrates the dual nature of radiation and matter? | |
Ans. One example of an experiment that demonstrates the dual nature of radiation and matter is the double-slit experiment. In this experiment, a beam of light or a stream of electrons is directed towards a barrier with two narrow slits. When the barrier is illuminated, an interference pattern is observed on a screen placed behind it. This interference pattern can only be explained if we consider that light or electrons exhibit wave-like behavior and interfere with themselves as they pass through the slits. This experiment provides strong evidence for the wave-particle duality of radiation and matter.
Text Transcript from Video
hello friends this video on dual nature
of radiation and matter part 11 is
brought to you by exam fee or form no
hope here from exam please make sure
that you have watched all the videos
tell part 10 before going ahead with
part 11 so now this wave nature of
matter also had very important
applications these applications were
mostly in in high study on the higher
fields of study of physics like if you
have something called modern quantum
mechanics I am sure you would have never
started off quantum they have mechanics
at this level but if you study physics
later in your life you will see that
when you go for your higher studies you
will study this very interesting branch
of physics called quantum mechanics
which is very different from your
classical mechanics so far you have
learnt classical mechanics where you
assume that mass of an object is
constant the object you can measure the
position of the object clearly you can
measure the velocity of the object
clearly so it is all it's
straightforward but but classical
mechanics is something which is not
reality it is based on lots of
assumptions we assume that mass is
constant we assume that this is constant
that is constant and so on but when you
actually go into quantum mechanics you
actually see things how things are in
reality how things are in nature so
enemies will not go in into that but
what I am trying to say is this wave
nature of matter this de Broglie
hypothesis Heisenberg's principle they
all set the first stones towards quantum
mechanics electron microscopes they are
a very important application of whatever
we studied so far electron optics is
again another branch which has evolved
from these bases that is which have
evolved from this dual nature of
variation and matter so now that we are
done with our lesson let us go ahead and
try to solve some problems so let us
look at the first problem it says that
the photoelectric cutoff voltage in a
certain experiment is 1.5 volts what is
the maximum kinetic
energy of photoelectrons emitted so in
this case the stopping potential is
given as 1.5 folds right cutoff voltage
is nothing but the stopping potential so
what is cutoff remember and understand
and then do things so what is the
stopping potential at this potential the
photo current becomes 0 so photo current
becomes 0 at this potential so that
means this is the potential at which the
event of electron which is the photo
electron which is moving with the
maximum kinetic energy that means the
maximum kinetic energy of the
photoelectron should become equal to the
energy provided due to this stopping
potential that is e into v naught right
so therefore we can say what is e e is
nothing but 1.6 into 10 to the power
minus 19 to V naught is 1.5 so this
comes out to be 2 point 4 into 10 to the
power minus 19 joules so this would be
the maximum kinetic energy of
photoelectrons emitted because this is
my emitter and this is my collector so
photo electrons are emitted from emitter
right so when we say cutoff voltage that
means there is no photo current that
means there are no electrons reaching
the collector that means the electron
which is having the maximum kinetic
energy that electron is also not able to
reach the collector so the work done by
the stopping potential is equal to the
maximum kinetic energy of the
photoelectron right okay so let us look
at the next problem it says that the
work function of caesium methyl is 2
point 1 4 electron volts so work
function Phi naught is given as 2 point
1 4 electron volts when light of
frequency 6 into 10 to the power 14
Hertz is incident on the metal surface
that means frequency of incident light
is 6 into 10 to the power 14 Hertz photo
emission of electrons take place okay
what is the maximum kinetic energy of
the emitted electrons
so what we will do we will apply
Einstein photoelectric equation so
Einsteins photoelectric equation says
that H nu that is the total energy
supplied to the electron or the total
energy carried by the photon is equal to
work function plus maximum kinetic
energy of the photoelectron right this
is my einstein's equation so from this
we can calculate K max is equal to H nu
minus Phi naught so H is 6 point 6 7
into 10 to the power minus 34 nu is 6
into 10 to the power 14 minus Phi naught
that is 2 point 1 4 electron volt so the
sense this is an electron volt let me
convert this also an electron volt so to
convert into electron volt we will
divide by 1 point 6 into 10 to the power
minus 19 so this gives me zero point 3 4
5 electron volt so this would be the
maximum kinetic energy of the emitted
electrons in the second part it asks us
to calculate the stopping potential so
what is the relation between stopping
potential and kinetic energy
evie naught is equal to K max therefore
stopping potential becomes equal to K
max divided by e so a max is zero point
3 4 5 electron volts so let me convert
this electron volt into joules because
he also will be in Coulomb so then Joule
per Coulomb delegates of zero point 3 4
5 into 1 point 6 into 10 to the power
minus 19 joules divided by electron is 1
point 6 into 10 to the power minus 19
coulombs so this will get cancelled and
will get zero point 3 4 5 Joule per
Coulomb that is zero point 3 4 5 volts
now let us look at the third part it
asks us to calculate the maximum speed
of the emitted photoelectrons now once
we know the maximum kinetic energy of
the emitted photo electron it is the
very easy to calculate the maximum speed
because maximum kinetic energy would be
nothing but half M v-max square right so
we have to calculate V Maps so this will
be 2 into K max divided by M or maximum
velocity is equal to root over 2 into K
max divided by M so this we can write it
as 2 into maximum kinetic energy which
we calculated as zero point 3 4 5
electron volt let me convert it into
twos this divided by M what is n n is
nothing but mass of electron which is 9
point 1 into 10 to the power minus 31
kgs so v-max comes out to be 3 3 2 point
3 kilometers per second
correct okay let us look at the third
problem it says that the energy flux of
sunlight reaching the surface of the
earth is 1.3 into 8 into 10 to the power
3 watt per meter square okay so let me
denote the energy flux as fine which is
given as 1 point 3 8 8 into 10 to the
power 3 watt per meter square right so
what is this energy flux energy flux is
nothing but energy per unit area per
unit time right so this is equal to 1
point 3 8 8 into 10 to the power 3 Jun
per meter square per second because
Joule per second is what right so this
is Joule per meter square second correct
so therefore so this is my
total energy per meter square per second
so we know I know my total energy per
meter square per second so total energy
per meter square per second is equal to
one point three eight eight into 10 to
the power 3 joules right now let us try
to calculate energy of one photon
because I want to know the number of
photons per square meter per second so
let me calculate the energy of one
photon this will be H into nu H is 6
into 6 6 point 6 7 into 10 to the power
minus 34 and what is frequency frequency
we know a relationship between frequency
and wavelength that nu is equal to C by
lambda right and here we are given the
wavelength as 550 nanometers so we can
write it as H into C that is 3 into 10
to the power 8 divided by lambda which
is 550 into 10 to the power minus 9
meters right therefore what would be the
number of photons per meter square per
second so this will be equal to let me
call this as T total energy per meter
square per second divided by energy of
one photon that is H nu so this will be
one point three eight eight into 10 to
the power 3 divided by six point 6 7
into 10 to the power minus 34 into 3
into 10 to the power 8 divided by 5 50
into 10 to the power minus 9 so this
comes out to be three point eight 4 into
10 to the power 21 so you understood the
logic which we followed here if the
question was something like let us
suppose that the total number of
chocolates that was distributed to the
class is 10 total number of chocolates
distributed to the class was 10 you
have to calculate the number of students
in the class provided each student has
got two chocolates that means each
student has got two chocolates so how
will you calculate the number of
students total number of chocolates
divided by number of chocolate with each
student right so that is what you did
here the total energy which was reaching
the surface of the earth was one point
three eight eight into ten to the power
three jewels right and also the energy
of each photon was H nu so you have to
calculate the number of photons which
was equal to the total energy divided by
the energy of each open right thank you
please visit exam comm to watch free
educational videos try free online tests
get the best quality study materials
study from the best tutors and mentors
and much more thank you once again
of radiation and matter part 11 is
brought to you by exam fee or form no
hope here from exam please make sure
that you have watched all the videos
tell part 10 before going ahead with
part 11 so now this wave nature of
matter also had very important
applications these applications were
mostly in in high study on the higher
fields of study of physics like if you
have something called modern quantum
mechanics I am sure you would have never
started off quantum they have mechanics
at this level but if you study physics
later in your life you will see that
when you go for your higher studies you
will study this very interesting branch
of physics called quantum mechanics
which is very different from your
classical mechanics so far you have
learnt classical mechanics where you
assume that mass of an object is
constant the object you can measure the
position of the object clearly you can
measure the velocity of the object
clearly so it is all it's
straightforward but but classical
mechanics is something which is not
reality it is based on lots of
assumptions we assume that mass is
constant we assume that this is constant
that is constant and so on but when you
actually go into quantum mechanics you
actually see things how things are in
reality how things are in nature so
enemies will not go in into that but
what I am trying to say is this wave
nature of matter this de Broglie
hypothesis Heisenberg's principle they
all set the first stones towards quantum
mechanics electron microscopes they are
a very important application of whatever
we studied so far electron optics is
again another branch which has evolved
from these bases that is which have
evolved from this dual nature of
variation and matter so now that we are
done with our lesson let us go ahead and
try to solve some problems so let us
look at the first problem it says that
the photoelectric cutoff voltage in a
certain experiment is 1.5 volts what is
the maximum kinetic
energy of photoelectrons emitted so in
this case the stopping potential is
given as 1.5 folds right cutoff voltage
is nothing but the stopping potential so
what is cutoff remember and understand
and then do things so what is the
stopping potential at this potential the
photo current becomes 0 so photo current
becomes 0 at this potential so that
means this is the potential at which the
event of electron which is the photo
electron which is moving with the
maximum kinetic energy that means the
maximum kinetic energy of the
photoelectron should become equal to the
energy provided due to this stopping
potential that is e into v naught right
so therefore we can say what is e e is
nothing but 1.6 into 10 to the power
minus 19 to V naught is 1.5 so this
comes out to be 2 point 4 into 10 to the
power minus 19 joules so this would be
the maximum kinetic energy of
photoelectrons emitted because this is
my emitter and this is my collector so
photo electrons are emitted from emitter
right so when we say cutoff voltage that
means there is no photo current that
means there are no electrons reaching
the collector that means the electron
which is having the maximum kinetic
energy that electron is also not able to
reach the collector so the work done by
the stopping potential is equal to the
maximum kinetic energy of the
photoelectron right okay so let us look
at the next problem it says that the
work function of caesium methyl is 2
point 1 4 electron volts so work
function Phi naught is given as 2 point
1 4 electron volts when light of
frequency 6 into 10 to the power 14
Hertz is incident on the metal surface
that means frequency of incident light
is 6 into 10 to the power 14 Hertz photo
emission of electrons take place okay
what is the maximum kinetic energy of
the emitted electrons
so what we will do we will apply
Einstein photoelectric equation so
Einsteins photoelectric equation says
that H nu that is the total energy
supplied to the electron or the total
energy carried by the photon is equal to
work function plus maximum kinetic
energy of the photoelectron right this
is my einstein's equation so from this
we can calculate K max is equal to H nu
minus Phi naught so H is 6 point 6 7
into 10 to the power minus 34 nu is 6
into 10 to the power 14 minus Phi naught
that is 2 point 1 4 electron volt so the
sense this is an electron volt let me
convert this also an electron volt so to
convert into electron volt we will
divide by 1 point 6 into 10 to the power
minus 19 so this gives me zero point 3 4
5 electron volt so this would be the
maximum kinetic energy of the emitted
electrons in the second part it asks us
to calculate the stopping potential so
what is the relation between stopping
potential and kinetic energy
evie naught is equal to K max therefore
stopping potential becomes equal to K
max divided by e so a max is zero point
3 4 5 electron volts so let me convert
this electron volt into joules because
he also will be in Coulomb so then Joule
per Coulomb delegates of zero point 3 4
5 into 1 point 6 into 10 to the power
minus 19 joules divided by electron is 1
point 6 into 10 to the power minus 19
coulombs so this will get cancelled and
will get zero point 3 4 5 Joule per
Coulomb that is zero point 3 4 5 volts
now let us look at the third part it
asks us to calculate the maximum speed
of the emitted photoelectrons now once
we know the maximum kinetic energy of
the emitted photo electron it is the
very easy to calculate the maximum speed
because maximum kinetic energy would be
nothing but half M v-max square right so
we have to calculate V Maps so this will
be 2 into K max divided by M or maximum
velocity is equal to root over 2 into K
max divided by M so this we can write it
as 2 into maximum kinetic energy which
we calculated as zero point 3 4 5
electron volt let me convert it into
twos this divided by M what is n n is
nothing but mass of electron which is 9
point 1 into 10 to the power minus 31
kgs so v-max comes out to be 3 3 2 point
3 kilometers per second
correct okay let us look at the third
problem it says that the energy flux of
sunlight reaching the surface of the
earth is 1.3 into 8 into 10 to the power
3 watt per meter square okay so let me
denote the energy flux as fine which is
given as 1 point 3 8 8 into 10 to the
power 3 watt per meter square right so
what is this energy flux energy flux is
nothing but energy per unit area per
unit time right so this is equal to 1
point 3 8 8 into 10 to the power 3 Jun
per meter square per second because
Joule per second is what right so this
is Joule per meter square second correct
so therefore so this is my
total energy per meter square per second
so we know I know my total energy per
meter square per second so total energy
per meter square per second is equal to
one point three eight eight into 10 to
the power 3 joules right now let us try
to calculate energy of one photon
because I want to know the number of
photons per square meter per second so
let me calculate the energy of one
photon this will be H into nu H is 6
into 6 6 point 6 7 into 10 to the power
minus 34 and what is frequency frequency
we know a relationship between frequency
and wavelength that nu is equal to C by
lambda right and here we are given the
wavelength as 550 nanometers so we can
write it as H into C that is 3 into 10
to the power 8 divided by lambda which
is 550 into 10 to the power minus 9
meters right therefore what would be the
number of photons per meter square per
second so this will be equal to let me
call this as T total energy per meter
square per second divided by energy of
one photon that is H nu so this will be
one point three eight eight into 10 to
the power 3 divided by six point 6 7
into 10 to the power minus 34 into 3
into 10 to the power 8 divided by 5 50
into 10 to the power minus 9 so this
comes out to be three point eight 4 into
10 to the power 21 so you understood the
logic which we followed here if the
question was something like let us
suppose that the total number of
chocolates that was distributed to the
class is 10 total number of chocolates
distributed to the class was 10 you
have to calculate the number of students
in the class provided each student has
got two chocolates that means each
student has got two chocolates so how
will you calculate the number of
students total number of chocolates
divided by number of chocolate with each
student right so that is what you did
here the total energy which was reaching
the surface of the earth was one point
three eight eight into ten to the power
three jewels right and also the energy
of each photon was H nu so you have to
calculate the number of photons which
was equal to the total energy divided by
the energy of each open right thank you
please visit exam comm to watch free
educational videos try free online tests
get the best quality study materials
study from the best tutors and mentors
and much more thank you once again
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Nov 26, 2024 Last updated |
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