in the last lecture we were discussing about the energy transfer and specifically we were
discussing about the non radiative energy transfer process so in the non radiative energy
transfer process what i showed you that d star is present and which is interacting with
a and this energy is transferred from d to a star as a result d is getting deactivated
and a is now in the excited state so if i now tell you the this molecule d star as
the one electron is over here another is over here and this a is present a is in the ground
state then the situation is that these two a and d are actually interacting over a distance
and which is converting this a this d as deactivated and a is now in the excited state and i said
that for a this uh dipole dipole interaction the interaction which is responsible is a
columbic type of interaction and then again i showed you that equation here this k e t
can be written as uh this kappa square divided by tau d r to the power six alright and then
some constant like this way and i said that this kappa squared is where this r equal to
distance between donor and accepter and uh this this should be tau tau d zero and then
kappa squared is my orientation factor
this kappa squared depends on the relative orientation of the donor and the accepter
and if you look at this specific form of this kappa squared let let let me draw this donor
and accepter this dipoles like this way let us say this is the donor dipole oriented in
this fashion and this is this accepter dipole is oriented in this fashion and this is the
distance between donor and the accepter so this is my d this is my a now if you extend
this line the line connecting the d and a and then you measure this angle the angle
between this line and the accepter and in this angle let us say is theta a and this
angle similarly is let us say theta d and if you make a plane with this donor dipole
and this line the line connecting donor and accepter and donor dipole with these if you
make a plane and the accepter dipole and the line connecting the donor accepter with these
if you make another plane like this then the angle between these two plane let us say termed
it as theta t so this kappa squared actually depends on these three theta values with this
form as kappad square cos sin theta t right minus three cos sin theta d into cos sin theta
a and whole square so this is my expression for the kappa squared
ok let me give you three different examples alright so let us say case one let us say
this is my donor dipole and this is my accepter dipole right and the distance between them
is r right so that is my distance but that kappa
squared will not going to depends on that distance it will only depends on that orientation
but total k e t value will obviously depends on the distance and you note here that dependency
of this distance is one over r to the power six little change in r will change the k e
t value in enormously right
ok i will come later i will i will tell all those things in detail so in this case if
i tell you that what is the value of theta a what is the value of theta d and what is
the value of theta t what will be your answer theta a is equal to zero theta d equal to
zero theta t equal to zero so for these values of theta kappa squared will be equal to cos
sin zero minus three cos zero cos zero right so this will be equal to one minus three so
minus two square so is equal to four i will take another case case two this is my accepter
same distance r this is my donor and this is my accepter for this case two i will also
have different values of theta a theta d theta t so theta a is equal to ninety theta d is
equal to ninety and theta t here will be equal to zero so in this square kappa squared will
be equal to one right
ok let me show you another case which is case three in this case let me take my dipoles
this donor and accepter like uh one is like this this is my donor and this is r right
and this is my accepter such way that theta a the theta a is equal to ninety right theta
d is also ninety and the angle between these two is also ninety degree right and this will
going to give me kappa squared equal to cos sin ninety minus three cos sin ninety cos
sin ninety right so this is going to be equal to zero so you see depending on the orientation
of these two dipole mu d and mu a
the value of kappa squared actually varying a lot starting from four to zero and that
is the problem suppose you now you think of a situation you have the donors right and
the how the these donor molecules will orient in solution that will not going to depends
on anything because as we have already seen in the florescence on during our discussion
on florescence that the orientation of these molecules in a solution is random that is
called the isotropic orientation
that means the donor molecules will orient in all possible conditions similarly the accepter
will also orient randomly in all possible orientation that means the theta a theta d
theta t the values of theta a theta d theta t could be anything right so because of this
isotropic orientation of the donor and the accepter for solution but for a rigid
system like a biomolecules ok let i i will talk to the i will talk about this biomolecule
later on but it is for sure that for solution state that the orientations of the donors
and accepters are random and is a isotropic emission so better to have a case four where
i will take this randomizations of the donor and its accepter
so for this case four what i have is the donor and accepter are
oriented randomly this are isotropic right orientation in this case kappa square is retained
as average value right so i will integrate zero to pi cos sin theta t minus three cos
sin theta d cos sin theta a square d theta t d theta d d theta a and normalize it with
zero to theta d theta t theta a and if you now the value of this integral is very simple
which come out comes out to be two by three now you remember our original location this
k e t i had written this as kappa squared by tau d r to the power six then i said this
is a constant right so this is zero
now if i consider this random orientation of this donor and accepter then this this
quantity is also will come inside this constant so i will trite this as a constant a constant
divided by tau d zero r to the power six but remember that this constant has spectral overlap
integral within it it means this is a constant for a definite pair of donor and accepter
so this is a constant for a definite pair of donor and accepter and that constant is
written as r capital r zero to the power six that i can do that is mine my desire right
capital r zero to the power six
and if you know this overlap integral between this donor and accepter and from the expression
of this quantity over here you will be able to calculate the value of r zero and that
value of r zero is fixed for a definite donor and accepter so the final equation will be
k e t is equal to one over tau d zero r zero by r to the power six and that particular
constant r zero is known as forster distance that is known as forster distance
ok now let me define a quantity which is called the efficiency of energy transfer denoted
by capital e right let me define and then i will explain what is that efficiency
of energy transfer is denoted by capital e which is equal to the rate of deactivation
from the excited state to the ground state right the rate constant that is leading to
the energy transfer divided by the total right like similarly for the quantum yield i divided
by k e r by k e r plus k n r for fluorescence quenching what do we had that uh for the quantum
yield that is k e r by k e r plus k n r plus k q into q so in this case the another path
way which is non radiative path way that means energy is taking energy is taking
out by the system this k capital e t right so in this case the efficiency should be k
capital e t divided by the total rate constant right and total rate constant means inverse
of tau d zero that means this k r plus k n r and then that another rate constant which
is responsible for the deactivation is k capital e t rate constant of the energy transfer so
this efficiency of energy transfer can be easily retained as k e t divided by tau d
inverse so in this tau d inverse i am going to have this k r plus k n r right because
you remember tau d tau d equal to one over k r plus k n r right
so i am going to have these all the time this plus k e t right and you also please remember
that this k e t is a function of r small r this is a distance between the donor and accepter
and that distance dependence is very high is r to the power six so i better write here
as a function of r this is a function of r so this is also function of r so efficiency
is also a function of r so then i can just uh simply put the expression of k e t r in
this equation right so if you do that so then this will be tau d so this is again tau d
zero tau d zero inverse into r zero by r to the power six right from here divided by tau
d zero inverse plus tau d zero inverse to r zero by r to the power six right
so if you just simplify this expression what you will going to see is one divided by one
plus r by r zero to the power six you will get such kind of expression right so efficiency
of energy transfer is something like this as i told u earlier that if this is my emission
spectra d in presence of accepter there is a decrease
in the intensity and there is a increase in the intensity of another wavelength right
like this so here the intensity will decrease and here the intensity will increase if you
monitor the fluorescence intensity at this wavelength and fluorescence intensity at this
wavelength then with increase in of this accepter concentration that means the average
distance between the donor and accepter will change you will get a reduced in the fluorescence
intensity and in wavelength the fluorescence intensity will increase right
so that is also in some other words must be corrected with the efficiency of the energy
transfer suppose you have add some molecule a and nothing happen alright the fluorescence
intensity remain same as without the presence of this accepter that mean nothing happen
then the efficiency of energy transfer if i now may say like that then the efficiency
should be zero that is it and let us say for example you8 have added this accepter and
the whole complete quenching of this fluorescence right there is nothing it is like it is like
this zero over here and only this guy is like this it means the efficiency is hundred percent
where the efficiency is equal to one so if i now take this efficiency in a scale between
zero and one so i can tell this efficiency in terms of the fluorescence intensity in
presence of accepter and in in absence of accepter that is it
so if i this efficiency e is typically retained as one minus i d a by i d right so you see
here this efficiency when the i d a that means the intensity of donor in presence of accepter
is same as i d that means there is no change in the emission intensity so then i d a by
i d equal to one one minus one equal to zero that means the energy transfer is completely
inefficient i mean zero nothing now you consider that after addition of this accepter the fluorescence
intensity became zero i mean from a some value it could be hundred from hundred to zero or
thousand to zero but it whatever it is from there it is zero that means this zero by something
is zero one minus zero equal to one so then the efficiency is one that means fully efficient
energy transfer
similarly this efficiency also can be retained in terms of the lifetime in case of energy
transfer this lifetime of this species will also change because of the creation of this
additional pathway what is that additional pathway that i i i i wrote in just few minutes
back as k e t that is my additional pathway that is why this when i have calculated the
efficiency right i have retained this k e t divided by the total rate constant and that
is a rate constant so the fluorescence lifetime will also change right that means i can write
also this efficiency in terms of one minus tau d a divided by tau d so here this tau
d is the lifetime of the donor in absence of this accepter and tau d a is a lifetime
of the donor in presence of accepter
now you you can say that ok if i increase the accepter alright then obviously this will
be different but the reason is not because you have more accepter the reason is that
the average distance between the donor and accepter will decrease we have those one donor
and hundred accepter then from donor to the accepter distance average distance is something
if you have one million then average distance will be much smaller so that i why you are
going to have it because the only thing which is depend here is the distance right and for
a definite donor accepter pair that r capital r zero is a constant
so telling this now i can simply do one thing i can a plot like this where this y axis is
my e and this x axis is my r and now i can comment on that right what will be the value
of e that efficiency as i change my r value right that that that will automatically come
right so what equation i already have i have this equation in involving the efficiency
of the energy transfer as well as the r the distance between the donor and accepter here
here by this way i will going to calculate this efficiency right and then here capital
e equal to one over one plus r by r zero to the power six with this i will plot this right
so let us take a value when r equal to zero that small r equal to zero here is my r when
this small r equal to zero right so it is over here so one plus zero to the power six
and that is uh inverse so this is equal to one so this value will be equal to one
and when r equal to r zero right let us say take a point over here where this value of
r equal to r zero right in that case this r zero by r zero that means one one divided
by two equal to zero point five so here i will get zero point five here is my point
and here is my another point and when r equal to twice r zero then this twice r zero divided
by r zero that is two to the power six that is two to the power six that is sixty four
that one divided by sixty five that will come about zero point zero one five so it is almost
over her almost over here and in between if you plot the value will come what we will
going to see the plot will something like this right so from here i can atleast tell
that when the efficiency of energy transfer is just zero point five that r is equal to
capital r zero which is my first distance right ok so let us finish here and we will
continue our discussion on the next day
thank you very much