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