welcome back to the lecture number thirty one today we will going to discuss a defined
topic which is called the energy transfer as i told you briefly during our discussion
on fluorescence quenching like this this is the broad topic of fluorescence quenching
is the quenching of the fluorescence intensity by the presence of some other molecule which
we termed as the quencher molecule and the reason for the fluorescence quenching are
of many fault for example the electron transfer is one of them so when a molecules is being
excited to the excited state and the fluorescence comes from the excited state of the molecule
now if you have one (Refer Time: 01:00 electron diffusion molecules or electron molecule present
nearby to this fluoropore which is in the excited state that the other molecule can
donate one electron to the homo of that fluoropore which makes the fluoropore to be
to have a filled homo and the fluorescence will be deactivated immediately or in other
case it can take the excited electron from this fluoropore when the other molecule is
electron deficient right similarly this energy transfer is another case of the fluorescence
quenching right i can i can describe this energy transfer as one of the case of the
fluorescence quenching for example like if you add right some kind of molecule then the
emission intensity will decrease of the original fluoropore that excess of energy could be
transferred to the other molecule and that may create a new emission band of the other
molecule let me write point by point
so what i was trying to tell you is that you have a molecule i name this molecule as
d and it is these molecule is in the excited state so it is d star so in absence of any
other thing these molecule will going to give you emission or fluorescence alright and you
can record the fluorescence intensity as a function of wavelength the emission wavelength
which is nothing but the fluorescence spectrum or emission spectrum of these particular molecule
now consider you have another molecule present in the solution that molecule i write as a
and the result is that these molecule is being deactivated right and simultaneously these
other molecules is getting excited now here this has a different emission maxima
let us say this emission lambda emission maxima or d but this molecule will have a different
emission maxima lambda emission maxima of a right so i can have so as a result what
you will see you will see the fluorescence now now if i if i now may plot right this
is my fluorescence intensity i and then this is my wavelength ambda
so initially when there was no a molecule what you would have seen is that this is your
fluorescence intensity and this corresponding maxima is your lambda emission maxima of d
as you are adding this a molecule then these fluorescence intensity will decrease but a
new fluorescence will appear like this way if you add more a then this fluorescence will
further decrease and and then this new fluorescence intensity will be much more and if you had
further it will be something like this
so this new emission maxima is because of the a not because of the d so this is lambda
emission maxima of a so now if you look at this wavelength right you will see that fluorescence
is quenching but in this particular case here at this particular wavelength the fluorescence
is actually increasing also since it is special type of thing right so what just looking at
i can simply tell that here this molecule this molecule means if this d star the energy
from d star is getting transferred to a so that as a result that d is getting deactivated
not d is in the ground state and a is now in the excited state so by denoting it to
star that means this d star is my energy donor and this a is my energy accepter
so simply we write this as donor and this as donor and this as accepter now question
is that how the energy is getting transferred right so i can think of a simple path way
by which the energy is transferred let me write this as d star is getting converted
to the ground state of the donor and plus the emitted photon was the h mu right so i
can write d two d plus h mu i can simply write that and then i can write these h mu is somehow
being absorbed by this a molecule to make this a from the ground state to the excited
state
so i have first radiative emission of this photon this photon and this photon is being
absorbed again by this accepter a so this this is a two step process and this is known
as radiative energy transfer so i can write this as case one radiative energy transfer
and this is a two step process now just looking at this mechanism as as i have written
here it is evident that if the donor quantum yield is high right that means right the non
radiative rate constant is much higher than the non radiative rate constant so this process
must depend on the quantum yield of the donor again if you look at this you see that this
emitted photon has to be absorbed by this a and that clearly tells me if i have more
a present nearby the d then this process right then these process is are more feasible right
so it also directly proportional to the concentration of a molecules present in the solution if
you again look at this diagram automatically you understand that the emission wavelength
of this donor must be very similar to the absorption of this accepter right so i can
write these three different conditions over here
so the condition number one is the fluorescence quantum yield of donor two i can write
the concentration a and three i can write that overlap of emission
say like these are the three condition if the fluorescence quantum yield is more than
obviously the emitted photons will be more and then concentration of a is more then the
possibility of the emitted photon will be absorbed this is also more and then the third
is the overlap of emission of d with the absorption of a right if these three thing satisfy then
efficiency of this energy transfer will increase right let me this in the point number and
point number two is easy to understand but let me elaborate this point number three
for example let us say this is the absorption spectrum now let me draw first let us say
the absorption spectrum of the donor so i will just write here abs of donor depending
on i mean suppose you are exciting at the maxima so here is your lambda excitation
alright then you will get the corresponding emission spectrum of this let us say this
is your emission spectrum of the donor alright you see i have drawn as a mirror image just
because the i just assumed that here the in the excited state there is no structure change
in this molecule so mirror image rules holds good in this case
now if i use another color and draw the absorption spectrum of the accepter you see this is the
absorption spectrum of the accepter now you see the photon which are emitted at this wavelength
this wavelength this wavelength cannot be absorbed by this accepter right cannot be
absorbed by this accepter one of the condition is the delta e has to be matched with the
incoming photon h mu right that should match otherwise it is not possible so this is my
absorption spectrum of the accepter so it is not possible right however if the absorption
spectrum is not like this but but it is like this this is absorption spectrum of the accepter
in this case what you can see that the photons which are emitted at these wavelengths starting
from here to here the absorption of this a is actually matching there is a certain probability
that these photons will be absorbed obviously the absorption cross section at the different
wavelengths will be different so the probability of the different h mu will be different but
there is some probability there is a probability of the absorption
so what we cans tell that these common area if this common area is large then the emitted
photon has a good probability for the absorption by this accepter a if this shaded region is
small then the probability that the emitted photon will be absorbed by this accepter will
be less and this shaded region is known as spectral
overlap integral and denoted as j mu bar so this j mu mu bar i mean just is a inverse
of the wavelength alright and they say usually the unit we referred is as a centimeter
inverse and the expression for this that should contempt the absorption spectrum the
emission spectrum emission spectrum of the donor and the absorption spectrum of the accepter
right and this is generally evaluated as integration zero to infinity right florescence spectrum
or emission spectrum of donor obviously this is a plotted as a function of wavenumber you
know you can i already discussed that we can change the units right that it change from
the wavelength to wavenumber to frequency whatever you want right so i am just using
here the wavenumber into absorption spectrum absorption spectrum and epsilon the more and
less coefficient is same it is just divided by the concentration into path length so this
is just a epsilon this is same thing epsilon of the accepter which is plotted against nu
bar d nu bar ok so this is my overlap integral
so here this number three this three point is telling me if this overlap integral j mu
bar is more then this is more favorable the energy transfer is more favorable like considering
all this thing if you now try to find out the fraction let me write the fraction
of emitted photon by d star right and consequently absorbed by a so that can be evaluated and
that is generally expressed as a small a and then i should have all these things over here
right in the expression i am not i am just writing the expression but all these this
one this one this one everything should be present in this formula alright
so this is this is equal to one over phi d zero where this phi d zero is the quantum
yield of the donor in absence of i should write somewhere i will write it so one of
our phi d is zero c a this is concentration and this is that path length right of the
sample and then integration zero to infinity if d nu bar epsilon a nu bar d nu bar where
phi d zero is fluorescence quantum yield of the donor in of absence of the accepter and
c a is your molar concentration of a and a is the thickness of the sample
alright now i will tell you the another possibility that other possibility is that
you have excited the donor but before the donor actually emitted the photon the energy
is transferred to the accepter a alright so in this case this is the radiative energy
transfer as i told you and the other possibility is that non radiative energy transfer and
this radiative energy transfer is a one step process right so let me show you that so the
other option is d star was there and in presence of a this d be[comes] d star become d and
a become a star and this is fall under this category of non radiative energy transfer
which is a one step process now obviously i have to discuss about that what are the
variables on which these non radiative energy transfer depends
so if you look at here obviously it depends on the lifetime of this donor because if the
excited state will remain there for a longer time then only such kind of interaction is
possible for example if you just excite and immediately it will come back then there is
no time for the interaction with this molecule what kind of interaction that is also important
that i am coming after few minutes but what you understand is that the rate of non radiative
process must be low so that the molecule will stay for longer at the excited state then
only the interaction is more favorable alright if it is immediately comeback then interaction
is not that favorable right so it depends on the quantum yield because that is also
depends on the radiative non radiative path and it depends on the lifetime this is the
same thing alright in absence of the accepter right it als[o]
now here it is not like that that photon is emitted right and that photon is absorbed
by this a so the concentration that how many are present that is not important the important
parameter in this case is that because they have to interact these d and a these two have
to interact with each other so the important parameter in these case is how close are they
if they are far apart then obviously any kind of interaction i i will discuss what type
of interaction it is but if the distance are far apart then those interactions are not
visible as understandable ok if they are close by then obviously those interactions are visible
right
so let me let me produce by telling you that both these donor and this accepter
right they are dipoles right in rare case they are not dipoles right i mean that they
have some permanent dipole movement that same dipole movement i was talking about when we
were discussing the on the fluorescence like a it has a dipole movement of mu d mu d in
the excited state mu d in the ground state they may be same may be different but in most
of the cases they have this dipole so in this case the d has a dipole or d is a dipole and
this a is also a dipole so there is a possibility of dipole dipole interaction if i now describe
this situation this d star and a is as my initial state and d plus a star as my final
state this transformation form the initial to final must be accompanied with some and
that is the interaction between this donor and the accepter right so if i write the wave
function of this initial state as psi i which is psi d star psi a and the wave function
of this final state is psi f i i am just writing as a product of this wave function of d and
a so in this case this is psi d psi a star then i can write that particular transition
right that means i two f right or i can write this probability probability of i to f so
this is written as psi i that is my psi f right
now question is that what is this h as i said that they are dipole and nobody stop us to
tell that is a columbic type of interaction so if i take this h as a columbic type of
interaction let me write that so here it is my h is columbic interaction and in this
case i will simply write this as h c alright columbic type of interaction so this rate
constant of this energy transfer k e t will be given by psi d star psi a h c psi d psi
a star square right so this will be rate constant of this energy transfer energy transfer
now the interaction energy between the two dipole right now let me talk about this
this columbic interaction so the interaction energy between two dipoles here i said that
one is mu d another is mu a one is for the donor another is for the accepter right so
that is proportional to mu a mu d by r cube and one factor if theta which actually tells
me about the dilutive orientation of this two dipole right
so it will such kind of form that means this k e t that rate of rate constant of this energy
transfer right i should write as psi d star psi a mu a mu d by r cube f theta psi d psi
a star so here you see this r cube term is over here and here is a square term so whatever
it is whatever the expression is i what i will get is this is proportional to r to the
power six whatever the other term is present over here i will get that this k e t is proportional
to one over r to the power six that means here you can see that the distance between
the donor and accepter is really really one of the important parameter to decide about
this fate of the energy transfer from d star to a to d to a star right
we will going to see obviously the full expression and as i told you this also should
depends on this f theta that is also important that is the relative orientation relative
orientation of the two dipoles right
now if i if i write the complete expression right it will look something like this this
kappa square by i will define this term tau d to r to the power six phi d right and then
there is some constant which i am not writing over here and then integration zero to infinity
if t mu bar f epsilon a mu bar d mu bar you see as i have commented it depends on phi
d it depends on tau d it depends on the distance it depends on this overlap integral obviously
it has to it depends on the overlap integral because the energy should match otherwise
this is not possible so it depends on the overlap integral and it also depends on this
factor kappa square where this kappa square known as the orientation factor orientation
factor
now considering that obviously this tau d i should write as tau d zero because this
is the lifetime of the donor in absence of the accepter right and obviously i should
also write this phi d as phi d zero because this is a fluorescence of the donor in the
absence of the accepter so these whole equation right can be converted in a small equation
if you consider the fact that here this is a constant term this tau d zero phi d zero
is a constant because it is a property of the donor this overlap integral is a constant
for a particular pair if i define this is my d this is my a then overlap integral has
to be constant for a particular solvent right
so all these things are constant right only this kappa square is not constant because
it depends on the relative orientation of the donor is donor is like this accepter is
like this kappa is defined donor and accepter are like that kappa will is different like
this so so this kappa square right so i can convert this equation as kappa square and
obviously r is not a constant because the distance between donor accepter can change
can vary by r to the power six i deliberately also take this tau d zero over here right
and rest of the things are constant ok let me stop here and we will continue our discussion
on next day
thank you very much