WEBVTT
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welcome to the second lecture of the course
entitled basics of fluorescence spectroscopy
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in the last class we have discussed about
some introduction about the fluorescence spectroscopy
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and the brief history of that and then we
discussed about the light matter interaction
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and for us the most important light matter
interaction being through the absorption of
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light by the matter and to discuss that absorption
of light by the matter we discussed the
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lambert beer law and then we bought the concept
of the absorption cross section of the molecule
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and i ended over here that sigma equal to
two point three zero three epsilon l c by
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n l which turns out to be two point three
zero three c by n into epsilon where this
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epsilon as you know is molar absorptivity
or molar extinction coefficient and its unit
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is liter mole inverse centimeter inverse you
can also change it from the centimeter inverse
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to millimeter inverse or length inverse right
so now as you can see here the this c over
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here the unit of the c we generally express
in terms of moles per liter right
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so here is moles per liter and here that n
that n which i have defined as number of molecules
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per c c right so number of
molecules per c c so it means that i have
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to convert these two units accordingly so
this n right in terms of c can be written
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as c into avogadro number divided by thousand
as simply as it is so now if i plug in this
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value of n over here so what i will going
to get is two point three zero three and this
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c is over here and then i will going to have
a thousand factor over here and divided by
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divided by n a coming from here and then c
two epsilon so this c c will cancel out over
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here and i was going to have two point three
zero three in two thousand divided by avogadro
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number into epsilon if you calculate this
this will come as three point eight two into
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ten to the power minus twenty one so this
is going to three point eight two into ten
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to the power minus twenty one to epsilon ok
so what we got is that absorption cross section
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is nothing but that proportionality constant
was used for the lambert beer law and this
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obviously has to be related with the epsilon
value so as epsilon depends on the wavelength
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absorption cross section also has to depends
on the wavelength let us see this the concept
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of absorption cross section here so what we
got we got sigma which is absorption cross
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section is equal to three point eight two
ten to the power minus twenty one into epsilon
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and obviously the unit is important that is
centimeter square because i have expressed
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epsilon in terms of liter mole inverse centimeter
inverse so i got this emitter centimeter inverse
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ok now let me draw a one molecule lets say
i let me draw this h in o four plus this
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molecule so here if i draw this frame of this
molecule looks like this so now the molecular
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cross section molecular cross section is the
whole area of this molecule for example if
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i may now highlight over here so this whole
region is my molecular cross section now if
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we use a light whose wave number is about
thirty four hundred centimeter inverse and
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we know that will going to excite my o h bond
right so vibrational excitation so that particular
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light will going to interact only with this
o h bond so the absorption cross section for
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this thirty four hundred wave number light
is not the whole molecular cross section right
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so in this case for o h bond excitation this
is my absorption cross section for that particular
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frequency of light particular wave number
of light so here obviously i can write that
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this is my molecular cross section and this
one is my absorption
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cross section which is known as sigma so the
if the molecular cross section lets say for
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this particular molecule is ten to the power
minus fifteen centimeter square and this absorption
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cross section for this the small region right
it must be smaller than this and if you somehow
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calculate it this will going to be ten to
the power minus eighteen centimeter square
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it gives me a idea that once write n number
of photons will appear so these n number of
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photons of that particular frequency will
collide with this molecule but only small
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fraction that fraction will determined by
this ten to the power minus fifteen the ten
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to the minus eighteen by ten to the power
minus fifteen so only that fraction of the
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photon will be used to excite the molecules
i mean to excite the o h bond from ground
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vibrational level to the excited state right
so
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depending on the wavelength obviously the
value of this absorption cross section will
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change so later on i will give you that
assignment where you will be able to calculate
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these absorption cross section of a molecule
and you will going to understand the significance
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of absorption cross section and how it is
related to the molar extinction coefficient
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ok so now let us move to another important
phenomena related to the absorption and
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my question is that how much time it will
take for an electronic absorption right so
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let us talk about this electronic for specifically
over here because we are dealing with the
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fluorescence in this particular course
so let me write few few fact that first
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speed of photon is equal to three into ten
to the power eight meter per second right
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i can write this as three into ten to the
power eighteen angstrom per second right and
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let me also assume one thing very simple assumption
that is that size of molecule involved in
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light absorption let us say this is is about
ten angstrom its very valid assumption right
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and now i will going to make another assumption
which
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where i would like to take the size of the
photon right size of the photon is associated
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with the wavelength of the light so light
has said two different kind of property has
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connected by the de broglie hypothesis
so first is light is wave and light is
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photon right so when i say is four hundred
nanometer light that means the wavelength
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of the light is four hundred nanometer but
it also gives the photon right it also behaves
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as a particle so now my assumption here is
that the size of the photon is associated
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with the wavelength of the light so i write
here the wavelength of light associated with
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the size of the photon right it means the
size of a blue photon which is resolving this
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four hundred nanometer is four hundred nanometer
so for example size of blue that means four
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hundred nanometer photon is equal to four
hundred nanometer or lets in the different
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unit is four thousand angstrom right so now
the i can easily calculate because i know
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the speed i know the size so the time taken
for a four hundred nanometer photon to cross
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a point right i can easily calculate so time
taken for a four hundred nanometer photon
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to cross a point is simply given by four thousand
because thats the my size divided by the speed
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right so four thousand angstrom divided by
three into ten to the power eighteen angstrom
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per second right so if i calculate roughly
this will be ten to the power minus fifteen
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second this ten to the power minus fifteen
second means one femtosecond right so this
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is one femtosecond so if you are going to
excite your molecule with this four hundred
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nanometer light so that four hundred nanometer
light will be present to interact with the
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part of this molecule only for one femtosecond
right so if the absorption has to be takes
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place then that absorption of the light has
to be takes place within that time frame right
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and also the velocity of electron in bohr
orbit right we all know that this is ten to
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the power sixteen angstrom per second
so sorry velocity of electron in bohr orbit
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is equal to ten to the power sixteen angstrom
per second right
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but i i have already determined the size of
the molecule is about ten angstrom right so
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electron moves ten angstrom right that is
my size of molecule in ten to the power minus
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fifteen second so what we got here that the
timescale of photon interaction and electron
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motion in the molecules they are same right
so it means that time scale of a photon interaction
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is similar to time scale of electron motion
which is ten to the power minus fifteen second
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or one femtosecond and this is the time scale
for the absorption process ok so now a light
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like electromagnetic radiation consists of
electric field and magnetic field are actually
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coming and interacting with the matter matter
is my molecule for this particular case i
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am taking so now this that is nothing but
the light matter interaction right but what
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i already said in the last class that the
electrometric radiation consists of both electric
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and magnetic field right
so the which are the electric or magnetic
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or both the fields are responsible for this
light matter interaction or one is more responsible
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than the other so that we can easily check
so what i have to look here is the force exerted
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on an electron by the light right so if i
write force exerted on an electron by the
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light right so this f is equal to e e plus
e h v divided by c right so this is this is
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from the physics right so this is the electric
force and this part is my magnetic force
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ok so here e is charge of electron
and this capital e is electric field strength
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h is magnetic field strength
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and this c c speed of light
and v is the velocity of electron
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ok so what we know the value of v and value
of c and with this we can estimate right the
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contribution of these two term so v is equal
to ten to the power eight centimeter per second
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whereas c c is equal to ten to the power ten
right so three into ten to the power ten centimeter
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per second so v by c e h is much much less
than e into e so what we can see is that effect
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of magnetic force is negligible compared
to the effect of electrical
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so for everyday discussion will not going
to include the magnetic field because the
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magnetic field contribution is much smaller
at least for just one percent or less than
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one percent than the electric field right
so we will going to discuss only the electric
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field or we will going to consider only the
electric field when we will discuss the in
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the light matter interaction in the field
of electronic transitions right ok so now
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let me show you the energy levels of the molecules
right i have a molecule i have simply showing
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this molecule as a diatomic molecule now if
i give the energy which corresponds to the
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wavelength of ten to thirty millimeter right
this is in the region of microwave right is
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the microwave region so then this energy can
initiate the molecule to rotate at a different
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speed so this is the rotation of the molecule
so i am going to change the rotational energy
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level right from ground state to the excited
state and other states right so now if i change
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the wavelength from ten to thirty millimeter
to one to ten micrometer which falls in the
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i r region of this electromagnetic rate electromagnetic
spectrum right
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so what is electromagnetic spectrum i am i
will i will come in the next slide so in
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this case this light is responsible to create
the vibrational excitation within this
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molecule however if you tune your light
to hundred u v or visible region like one
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fifty to eight hundred nanometer then this
light will going to excite the molecule
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from ground electronic level to the excited
electronic level so which obviously going
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to change the distribution of the electronic
cloud within this molecule right as we shown
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over here here this is the electronic distribution
once you excited the light the distribution
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changes and it may create the dipole within
this molecule ok so this is the electromagnetic
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spectrum we all know this in this electromagnetic
spectrum as you can see here the left side
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is high energy side over here and you have
this gamma ray and you have this gamma ray
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over here is a high energy
and this side is my low energy
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high low energy means longer wavelength right
low energy means longer wavelength so as you
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can see over here here the length is shorter
and in this case the wavelength is longer
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so we start with the gamma gamma rays where
the frequency so is ten to the power twenty
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and then if we come to ten to the power
sixteen this falls in the u v region and
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then some somewhere between ten to the power
fifteen to ten to the power fourteen we got
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this our visible region so our eye is only
sensitive to this much region eye can only
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detect this light other light by our eye you
cannot see right
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but there are some other instruments some
other devices which can be used to see these
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kind of light like either x ray or infrared
or microwave and so on and so forth so this
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is electromagnetic spectrum and as we can
see here i have many different units over
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here here is written frequency here it is
written a wavelength right so obviously they
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are interconvertible right they are interconvertible
so here this is very important what i want
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to tell you that this unit in spectroscopy
so the characteristics of light can be represented
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not only in frequency not only in wavelength
but also in terms of other parameter and these
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parameters are interconvertible so what is
frequency frequency is number of oscillations
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per unit time and unit is hertz the cycle
per second right and what is wavelength wavelength
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is the distance between adjacent crests of
the wave right so this is my wavelength and
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this frequency and wavelength are interconvertible
over like this way nu equal to c by lambda
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so try to remember this fundamental relationship
nu equal to c by lambda what is the unit of
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nu the unit of nu is second inverse unit of
c is meter per second and this can be represented
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as meter so you see it is second inverse so
easily ok
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so now this another thing which was not there
in that electromagnetic spectrum but we can
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also get this important unit which is defined
as the number of waves per unit distance right
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so it must be related directly inversely proportional
to the wavelength so nu bar this is denoted
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as nu bar and nu bar is one over lambda right
and we also know that the energy carried by
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any electromagnetic radiation that or the
photon is directly proportional to its frequency
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so e equal to h nu so i can write e equal
to h nu or e equal to h c by lambda because
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nu equal to c by lambda or equal to h nu bar
c right so please remember these relationships
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which will be used more frequently at the
later part of this course ok
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now let me show you the different energy levels
i already showed you the rotational level
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right where the small amount of energy is
required to excite from one rotational level
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to the another rotational level vibrational
states well serve showed you and then electronic
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states so in this case right if these are
the two electronic level of this molecule
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each electronic level could have several vibrational
level as shown over here so this is just electronic
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level
and this is each electronic level consists
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of many vibrational levels
and each vibrational level is consist of many
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rotational levels right so when i and and
you see the typical typical energy difference
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between the two levels are like this for the
rotational level these for the vibrational
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level and this is for the electronic level
and at room temperature the energy available
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is this much right so according to boltzmann
distribution right so we have these rotational
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levels maybe populated the higher rotational
level maybe may be populated but the system
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will be in the lowered electronic level as
well as in the lower vibrational level the
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system and the normal condition if i dont
heat up the system normal condition the system
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will be in the lower electronic level over
here not only here but over here i dont talk
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about this part because this energy difference
is too small
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so i will going to have my system over here
so the lowest vibrational level of the ground
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electronic state the lowest vibration level
of the ground electronic states so i will
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going to see my system in this particular
state right ok so now let me let me take a
27:17.710 --> 27:25.010
very simple example of an atom right and to
let me consider that it has two state which
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you are shown over here e i is one state over
here and this e f is another state over here
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now if i give the suitable energy which in
terms of h nu which is equal to e f minus
27:43.010 --> 27:52.550
e i then this system can be promoted from
the e i level to e f level because now you
27:52.550 --> 28:00.370
have given the required amount of energy to
change the energy state of this particular
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atom from e i to e f level right now it
means for atom one electron goes from one
28:12.160 --> 28:18.270
level one orbital to the another orbital so
as shown over here here you see this electron
28:18.270 --> 28:27.090
is moving from one orbital to another orbital
so in this case what i have showed you that
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for the simple atom the meaning the meaning
of these two level means at this position
28:35.610 --> 28:44.840
at this position this is my electronic arrangement
and after excitation these electronic arrangements
28:44.840 --> 28:51.630
refer to this energy state right so please
remember this the these particular electronic
28:51.630 --> 28:58.960
arrangement the electron one is over to here
and one is here these refer to this state
28:58.960 --> 29:13.420
and when the electron moves one is to here
and one is here that refer to this particular
29:13.420 --> 29:21.530
state right ok so let me finish the second
lecture over here and next in the next
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class we will going to discuss about the electronic
excitation in molecules
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thank you