CBSE Class 11 > Class 11 Videos > Electromagnetic waves : numericals - Structure of Atom
Electromagnetic waves numericals - Structure of Atom Video Lecture - Class
Text Transcript from Video
hello friends this video instruction of
atom part 17 is brought to you by exam
fear calm
no more fear from exam before watching
this video please make sure that you
have watched part 1 to part 16 the
question says when the electromagnetic
radiation of wavelength 300 nanometer
falls on a serving of sodium this is my
sodium surface let's suppose I have my
electromagnetic waves of 300 nanometer
falling in this electrons comes out with
a kinetic energy of 1 point 6 8 into 10
to power 5 Joule per mole please note
this Joule per mole is talking about per
mole what is the energy required to
remove electron from sodium so what does
this mean is this wavelength of energy
was given this level of energy was used
to was used to remove electron plus give
kinetic energy so I can say that energy
from radiation was divided into energy
to remove electron plus kinetic energy
of the electron has right you have some
energy energy of radiation that was used
to kick out one electron that is two
energy to remove electron plus the
kinetic energy of its electron has got
correct so our we have to find minimum
energy to remove electrons we have to
find this part energy of radiation we
can easily find because we are given the
wavelength of the radiation correct so
let's find yi of radiation is nothing
but since we have wavelengths we use SC
by lambda this formula this is nothing
but 6 point 6 to 6 into 10 to power
minus 34 joule-seconds into C's speed of
light 3 10 to power 8 meter per second
divided by wavelength that is 300
nanometer
I will convert this into meter 303 bar
into 10 to power 9 meter so if you solve
this you get answer as 6 point 6 to 6
into 10 to the power minus 19 Joule this
is for 1 molecule this is this is energy
of radiation one molecule but the kinda
gauge is given per mole so let's convert
this into per mole so let's write energy
of radiation or one mole for bumble it
will be will have this six point zero 2
3 into 10 to the power 23 this is per
mole into will multiply with this energy
we have got right six point six to 16 to
the power minus 19
right because one mole has this many
molecules so what you'll get is 3 point
9 9 into 10 to the power 5 Joule per
mole so Joule an energy in Joule per
mole for this radiation is 3 point 9 9
correct so if you put this rather in
this equation 1 what you get is 3 point
9 9 into 10 to the power 5 Joule per
mole is equal to energy to remove
electron plus kinetic energy that is 1
point 6 8 into 10 to the power 5 Joule
per so if you solve this equation what
you will get energy to remove electron
will be nothing but 3 point 9 9 minus 1
point 6 8 that is 2 point 3 1 into 10 to
power 5 Joule well so that answer I have
the minimum energy required to remove
electron from sodium is 2 point 3 1 into
10 to the power 5 Joule per mole now the
question is what is the maximum
wavelength that can cause the photo to
mean so in that case we will use this
much energy itself because see any extra
energy we use that converts into kinetic
energy so we will as
that there is no guiding energy to the
electron we just want a wavelength that
will eject the electron off so we'll use
this one right so this is Joule per mole
I can find per molecule also very easily
so this Joule per mole I can write
energy to remove per molecule got more
molecule will be two point three one
into 10 to power 5 Joule per mole
divided by six point zero 2 3 into 10 to
the power 23 bonds so what I will get is
3 point 8 4 into 10 to the power minus
19 joules so per molecule this is the
energy required and this will come from
here you'll see vom na SC my lambda
because you want to find the wavelength
which can give this energy right so
we'll use the same formula na we'll put
we'll put H is going to applying
constant C is 300 parade so we'll solve
for this and we'll solve we get lambda
as 5 1 7 into 10 to the power minus 9
meter and that is nothing but 5 1 7
nanometre and there is math correct easy
one see what I have done is in this case
I found the wavelength of radiation was
given I found a energy per molecule then
I found energy per mole because
everything was born and then with that I
was given the kinetic energy of the
electrons i found the minimum energy
required to remove my electron from the
sodium metal per mole since it was per
mole i can write into per molecule
because i wanted to find a wavelength I
got this per molecule this many joules
and then I use this is nothing but HC by
lambda right the energy is nothing but
IC by lambda or molecule then I have
used this formula and I got the lambdas
5 and 7 nanometre it's a common example
the threshold frequency of a metal is
7.0 into
for first second calculate the kinetic
energy of the electron emission when the
radiation of frequency of this hits the
metal it's pretty easy question again
see there is a metal whose V naught is
given that is seven point zero into 10
to power 14 per second and we define the
kinetic energy of the electron this is
Ray light the electron comes out to find
the kinetic energy of this electron when
the frequency of this radiation is
nothing but 10 to power 15 per se or
nothing but 10 into 10 to power 14
because the power is 1414 Here I am
breaking 14 this becomes 10 in a bar
these become tender the power looking
person so we know that kinetic energy is
nothing but half into maassluis square
and this is nothing but H mu minus mu
naught where mu is the frequency of the
radiation this time mu naught is the
frequency of the metal let's put the
value H is what 6.6 2 3 into 10 to the
power 20 minus minus 34 joule-seconds
what is the value of mu mu is 10 into 10
to the power 14 per second and you not 7
into 10 to the power looking for say you
solve this you get something like this 1
point 9 8 8 into 10 to the power 90 this
is the kinetic energy of the electron
easy question let us see one more
question
the VRT Station of Delhi broadcast the
frequency of this kilohertz we had to
find the wavelength of electromagnetic
radiation emitted and need to find which
part of the electromagnetic spectrum it
blocks the frequency is given 1 3 6 8
kilo Hertz that is
very first we have to find the
wavelength lambda is C by V we know that
C speed of life 10 to the power 8 a 2
meter per second by V is 1 3 6 8 to 10
to power 3 husk we saw this you get two
one nine point three meters that is
minus so it comes with radio waves this
let's take for example the wavelength of
a physical visible spectrum extend from
400 nanometer to 750 nanometer express
this wavelength and frequency easy only
select this V lambda one and this will
have got two so let's find B when V one
is going to C by lambda 1 C is 3 into 10
to the power 8 meter per second lambda 1
has to be in meter or on a diameter is
nothing but 4 into power minus 9 meter
you solve this you get 7 point 5 into 10
to the 14 first similarly frequency 2 is
nothing but C by lambda 2 C is 3 into 10
to the power 8 meter per second lambda 2
is 750 nanometers is nothing but 750 to
10 over 9 meter you solve this you get 4
into 10 to the power 14 Hertz so the
frequency raises from this to this and
there is man's easy one you need to find
the wave number and frequency of a
yellow radiation having wavelength of
this wave number is nothing but 1 by
lambda so let's put the value 1 by 500
Ichiro Armstrong 1 angstrom is 10 to
power minus 10 meter one Armstrong is
Hinkley power minus 10 meter this is
what I get you solve this you get 1
point 7 2 4 into 10 to the power 6 and
there's my answer
find the energy of each of the photons
which corresponds the light of frequency
this hurts and wavelength off so if you
note quotient talks about two different
photons in one Photon
has their frequency of this the other
photons have a wavelength of this so
formula we know if you have the
frequency the formula is easier to H nu
you want to find the energy of photon
this become H is Planck's constant 6.6 2
3 into 10 power minus 34 joule-seconds
into of frequencies 3 into 10 to the
power 15 Hertz the pin Hans is per
second so if you solve this this becomes
1 point 9 8 8 into 10 to power minus 18
G this is first case second case photon
has a wavelength of some value so we
will use the formula HC by lamda same
thing H is flying constant 6.6 to the
minus 34 joule-seconds C is speed of
light 3 into 10 to power 8 meter per
second and lambda is here point form 5
ounce over this point 5 into 10 to power
minus M it
please note everything s if you see I am
converting into SI unit and this gives 3
point 9 8 into 10 to the power minus 15
June and that is master correct so here
you define wavelength frequency and
drain them of the light whose period as
this much so we know that frequency is
nothing but 1 by period let's find the
frequency first 1 by 10 to power minus
10 seconds that is nothing but 5 into 10
to the power 8 per second or Hertz this
is my frequency frequency is done like
wave length is C by mu frequency C is
speed of light 10 to the power 8 meter
per second and frequencies we have
formed 5 into 10 to power 8 pulse so you
get here 6 into so you go to the benign
here so here you get 9 here you get 6
into 10 to the power minus 2
meter this is my fave length we talk
about wave number that is nothing but 1
by wavelength that is 1 by 6 into 10 to
power minus 2 that is 100 by 6 that is
50 that is my per meter and that is my
baby thank you visit exam Viacom 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 thanks and serene
atom part 17 is brought to you by exam
fear calm
no more fear from exam before watching
this video please make sure that you
have watched part 1 to part 16 the
question says when the electromagnetic
radiation of wavelength 300 nanometer
falls on a serving of sodium this is my
sodium surface let's suppose I have my
electromagnetic waves of 300 nanometer
falling in this electrons comes out with
a kinetic energy of 1 point 6 8 into 10
to power 5 Joule per mole please note
this Joule per mole is talking about per
mole what is the energy required to
remove electron from sodium so what does
this mean is this wavelength of energy
was given this level of energy was used
to was used to remove electron plus give
kinetic energy so I can say that energy
from radiation was divided into energy
to remove electron plus kinetic energy
of the electron has right you have some
energy energy of radiation that was used
to kick out one electron that is two
energy to remove electron plus the
kinetic energy of its electron has got
correct so our we have to find minimum
energy to remove electrons we have to
find this part energy of radiation we
can easily find because we are given the
wavelength of the radiation correct so
let's find yi of radiation is nothing
but since we have wavelengths we use SC
by lambda this formula this is nothing
but 6 point 6 to 6 into 10 to power
minus 34 joule-seconds into C's speed of
light 3 10 to power 8 meter per second
divided by wavelength that is 300
nanometer
I will convert this into meter 303 bar
into 10 to power 9 meter so if you solve
this you get answer as 6 point 6 to 6
into 10 to the power minus 19 Joule this
is for 1 molecule this is this is energy
of radiation one molecule but the kinda
gauge is given per mole so let's convert
this into per mole so let's write energy
of radiation or one mole for bumble it
will be will have this six point zero 2
3 into 10 to the power 23 this is per
mole into will multiply with this energy
we have got right six point six to 16 to
the power minus 19
right because one mole has this many
molecules so what you'll get is 3 point
9 9 into 10 to the power 5 Joule per
mole so Joule an energy in Joule per
mole for this radiation is 3 point 9 9
correct so if you put this rather in
this equation 1 what you get is 3 point
9 9 into 10 to the power 5 Joule per
mole is equal to energy to remove
electron plus kinetic energy that is 1
point 6 8 into 10 to the power 5 Joule
per so if you solve this equation what
you will get energy to remove electron
will be nothing but 3 point 9 9 minus 1
point 6 8 that is 2 point 3 1 into 10 to
power 5 Joule well so that answer I have
the minimum energy required to remove
electron from sodium is 2 point 3 1 into
10 to the power 5 Joule per mole now the
question is what is the maximum
wavelength that can cause the photo to
mean so in that case we will use this
much energy itself because see any extra
energy we use that converts into kinetic
energy so we will as
that there is no guiding energy to the
electron we just want a wavelength that
will eject the electron off so we'll use
this one right so this is Joule per mole
I can find per molecule also very easily
so this Joule per mole I can write
energy to remove per molecule got more
molecule will be two point three one
into 10 to power 5 Joule per mole
divided by six point zero 2 3 into 10 to
the power 23 bonds so what I will get is
3 point 8 4 into 10 to the power minus
19 joules so per molecule this is the
energy required and this will come from
here you'll see vom na SC my lambda
because you want to find the wavelength
which can give this energy right so
we'll use the same formula na we'll put
we'll put H is going to applying
constant C is 300 parade so we'll solve
for this and we'll solve we get lambda
as 5 1 7 into 10 to the power minus 9
meter and that is nothing but 5 1 7
nanometre and there is math correct easy
one see what I have done is in this case
I found the wavelength of radiation was
given I found a energy per molecule then
I found energy per mole because
everything was born and then with that I
was given the kinetic energy of the
electrons i found the minimum energy
required to remove my electron from the
sodium metal per mole since it was per
mole i can write into per molecule
because i wanted to find a wavelength I
got this per molecule this many joules
and then I use this is nothing but HC by
lambda right the energy is nothing but
IC by lambda or molecule then I have
used this formula and I got the lambdas
5 and 7 nanometre it's a common example
the threshold frequency of a metal is
7.0 into
for first second calculate the kinetic
energy of the electron emission when the
radiation of frequency of this hits the
metal it's pretty easy question again
see there is a metal whose V naught is
given that is seven point zero into 10
to power 14 per second and we define the
kinetic energy of the electron this is
Ray light the electron comes out to find
the kinetic energy of this electron when
the frequency of this radiation is
nothing but 10 to power 15 per se or
nothing but 10 into 10 to power 14
because the power is 1414 Here I am
breaking 14 this becomes 10 in a bar
these become tender the power looking
person so we know that kinetic energy is
nothing but half into maassluis square
and this is nothing but H mu minus mu
naught where mu is the frequency of the
radiation this time mu naught is the
frequency of the metal let's put the
value H is what 6.6 2 3 into 10 to the
power 20 minus minus 34 joule-seconds
what is the value of mu mu is 10 into 10
to the power 14 per second and you not 7
into 10 to the power looking for say you
solve this you get something like this 1
point 9 8 8 into 10 to the power 90 this
is the kinetic energy of the electron
easy question let us see one more
question
the VRT Station of Delhi broadcast the
frequency of this kilohertz we had to
find the wavelength of electromagnetic
radiation emitted and need to find which
part of the electromagnetic spectrum it
blocks the frequency is given 1 3 6 8
kilo Hertz that is
very first we have to find the
wavelength lambda is C by V we know that
C speed of life 10 to the power 8 a 2
meter per second by V is 1 3 6 8 to 10
to power 3 husk we saw this you get two
one nine point three meters that is
minus so it comes with radio waves this
let's take for example the wavelength of
a physical visible spectrum extend from
400 nanometer to 750 nanometer express
this wavelength and frequency easy only
select this V lambda one and this will
have got two so let's find B when V one
is going to C by lambda 1 C is 3 into 10
to the power 8 meter per second lambda 1
has to be in meter or on a diameter is
nothing but 4 into power minus 9 meter
you solve this you get 7 point 5 into 10
to the 14 first similarly frequency 2 is
nothing but C by lambda 2 C is 3 into 10
to the power 8 meter per second lambda 2
is 750 nanometers is nothing but 750 to
10 over 9 meter you solve this you get 4
into 10 to the power 14 Hertz so the
frequency raises from this to this and
there is man's easy one you need to find
the wave number and frequency of a
yellow radiation having wavelength of
this wave number is nothing but 1 by
lambda so let's put the value 1 by 500
Ichiro Armstrong 1 angstrom is 10 to
power minus 10 meter one Armstrong is
Hinkley power minus 10 meter this is
what I get you solve this you get 1
point 7 2 4 into 10 to the power 6 and
there's my answer
find the energy of each of the photons
which corresponds the light of frequency
this hurts and wavelength off so if you
note quotient talks about two different
photons in one Photon
has their frequency of this the other
photons have a wavelength of this so
formula we know if you have the
frequency the formula is easier to H nu
you want to find the energy of photon
this become H is Planck's constant 6.6 2
3 into 10 power minus 34 joule-seconds
into of frequencies 3 into 10 to the
power 15 Hertz the pin Hans is per
second so if you solve this this becomes
1 point 9 8 8 into 10 to power minus 18
G this is first case second case photon
has a wavelength of some value so we
will use the formula HC by lamda same
thing H is flying constant 6.6 to the
minus 34 joule-seconds C is speed of
light 3 into 10 to power 8 meter per
second and lambda is here point form 5
ounce over this point 5 into 10 to power
minus M it
please note everything s if you see I am
converting into SI unit and this gives 3
point 9 8 into 10 to the power minus 15
June and that is master correct so here
you define wavelength frequency and
drain them of the light whose period as
this much so we know that frequency is
nothing but 1 by period let's find the
frequency first 1 by 10 to power minus
10 seconds that is nothing but 5 into 10
to the power 8 per second or Hertz this
is my frequency frequency is done like
wave length is C by mu frequency C is
speed of light 10 to the power 8 meter
per second and frequencies we have
formed 5 into 10 to power 8 pulse so you
get here 6 into so you go to the benign
here so here you get 9 here you get 6
into 10 to the power minus 2
meter this is my fave length we talk
about wave number that is nothing but 1
by wavelength that is 1 by 6 into 10 to
power minus 2 that is 100 by 6 that is
50 that is my per meter and that is my
baby thank you visit exam Viacom 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 thanks and serene
About this Video
Apr 26, 2026 Last updated
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