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