WEBVTT
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welcome to the lecture number thirteen so
the last lecture we just have started discussing
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about the solvent effect on fluorescence that
means solvent effect on emission spectra and
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we have taken the solute molecule that is
[re/responsible] responsible for the fluorescence
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right as like like this so this is my solid
molecule right having a dipole moment so i
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am just writing it over denoting it over the
plus and minus sign so the dipole moment would
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be could be mu e or mu g depending on it is
whether it is in the excited state or in the
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ground state and so this solvent molecules
will going to be stabilized this solute and
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because there is there will be a reaction
field we created by the solute molecule to
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the solvent right so that reaction field r
is the electric field in the solvent induced
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by the solute dipole right so this is this
is defined as the electric field electric
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field in the solvent induced by the solute
dipole right and i as i mentioned that the
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day that this r will be a function of the
dipole moment mu the radius over here which
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is also a right and the property of the solvent
some molecule will be polarized more efficiently
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some other not so that will be the property
of solvent so r is written as twice there
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is expression of r twice mu by a cube into
f where where where where f is equal to
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polarizability of the solvent polarizability
of the solvent and a is the radius of the
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cavity that is comparatively to the size of
the molecule right radius of the cavity right
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ok so now this polarizability of the solvent
are two types right one is a electronic polarizability
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another is the orientational polarizability
what is electronic polarizability suppose
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you have a dipole in in are to the solvent
so the electrons of the solvent molecules
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are polarized right so that is the electrical
polarizability and the other one is the orientational
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polarizability that means if you have a charge
pieces like a water let's say take example
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of water so water that o is partially negatively
charged a is especially positively charged
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right so if you have a what positive charge
over here which is solute and what water molecule
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is present like this way because of the influence
of this guy this water will reorient so that
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the final [ar/arrangement] arrangement will
be something like this so that is the reorientation
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of this water so this is called the orientational
polarizability so i can write this f as
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as a two term right so that that will be additive
so first is f e l plus f o r right so this
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f e l can be written as n square minus one
by twice n square plus one right so this is
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the reorientation of electron in the solvent
as it is instantaneous so where n is the
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refractive index of the solvent
ok and this f right that is the total polarizability
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this is a function of the directly constant
so i this can be written as epsilon minus
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one by twice epsilon plus one right so eventually
this f o r is nothing but epsilon minus one
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by twice epsilon plus one minus n square minus
one by twice n square plus one right so
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i can simply write f o r is equal to epsilon
minus one by twice epsilon plus one minus
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n square minus one by twice n square plus
one where this epsilon is the dielectric constant
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of the medium of the solvent this epsilon
is dielectric constant right
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so let's take this granted because i am not
going to show you in detail that how these
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things are coming right however one thing
is clear to me that if the solvent does not
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have a permanent dipole then f o r is going
to be equal to zero right right if the solvent
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does not have a permanent dipole then f o
r going to be equal to zero right and the
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solvation energy of the dipole right so let
me write here solvation energy of the dipole
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in the electric field which electric field
that electric field r this electric field
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how to so this electric field right r can
be written as delta equal to minus mu r right
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this is the solute dipole this mu is my solute
dipole because it has created the electric
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field in the solvent that are the reaction
field so their reaction field going to stabilize
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that solute right so the delta u will be minus
mu into i right ok
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so now let's see that if i have a solute whose
energy is e g vapor in the ground state ok
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let me draw over here so these are the ground
and excited state of the solute right and
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this is in the vapor state g a p vapor state
and now there is no solvent right so if the
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energy over here is e g vapor and this energy
is e e vapor and i also said that this is
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the dipole moment of the ground state and
this dipole moment of the excited state and
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let's take this another condition over here
the condition is mu e is greater than mu g
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so i have i already have imposed this condition
over here let me excited dipole movement is
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more than the ground state dipole moment then
in the vapor state if i if i simply measure
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the absorption i will get this much of energy
transition right so the new absorption
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in vapor right will be equal to mu abs right
there is a maxima let's right max in vapor
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is equal to e e vap minus e g vapor divided
h right e e e e vap minus e g vap divided
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by planck constant ok now if i put this solute
in solvent right then as i showed pictorially
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while discussing the solvent effect on absorption
that this will be stabilized this will also
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be stabilized but the stabilization will be
the different extent right this will stabilize
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less this will stabilize more and will get
lower energy right
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so let's let's see mathematically here right
first ok so here let's say this energy is
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e g isn't it and these energies e e right
so solvation energy in the ground state let
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me write that solvation energy in the ground
state is this much right
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right so is nothing but e g minus e g vapor
this is be equal to minus r into dipole moment
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of the ground state minus r dipole moment
of the ground state because the solute is
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in the ground state right so r into mu g now
r has to component as i showed right here
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r is to mu here r is to mu by a cube into
f f is f e l minus f o r so from here i can
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write r equal to r e l plus r o r which is
equal to two mu by a cube f e l plus two mu
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by a cube f o r so i will going to use this
equation from now and also use a right ok
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so this is going to be equal to minus mu g
r e l minus mu g r o r right so i will just
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simply write the expression of r e l and r
o r right which is nothing but two mu by a
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cube right two mu by a cube and the polarizability
electronic polarizability and the orientational
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polarizability for r o r right so let me write
that minus mu g two mu by a cube is the radius
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of the cavity of your solute a right into
f e l minus mu g two mu by a cube f o r so
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this mu this mu that two mu by a cube this
mu is why which mu this molecule is in the
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ground state so there should be mu g right
so two mu g and this also two mu g right so
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i will just write it as like this way so e
g equal to e g vapor minus two mu g square
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by a cube f e l minus two mu g square by a
cube f o r so i got this value of e g right
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similarly i will going to calculate the value
of e e right so i can i can write easily similarly
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in similar fashion right e e e e minus e e
vapor is equal to minus mu into r but in this
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case that mu is the mu e the excited state
dipole moment of this molecule so mu e into
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r right
so this is equal to minus mu e ar e l minus
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mu e r o r so this is going to equal to minus
mu e twice mu by a cube f e l minus mu e twice
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mu by a cube f o r now the question is that
which mu i should put over here this is electronic
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change electronic change is very fast so that
mu should be mu e however this is the orientational
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change is orientational relaxation so although
although i have excited this molecule from
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ground to excited state but the solvent molecule
which will see right the excited state just
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after the excitation right the though those
molecules will take some time to reorient
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themselves so that means these reaction field
right what we are going to have is very similar
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to the solvent arrangement in the ground
state so this mu will be mu g right right
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so in this case r e l is a function of let
me write clearly over here r e l is a function
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of mu e function of mu e but r o r is a function
of mu g r o r is function of mu g because
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orientation polarization is slow right because
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orientation orientational polarization is
slow right ok so then then the rest of the
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part is simple so what i will going to have
this equal to minus twice mu square by a cube
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f e l twice mu g mu e by a cube f o r
so i can write this as e e equal to e e vapor
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minus twice mu e square by a cube f e l minus
twice mu g mu e by a cube f o r right so now
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if i want to calculate that what is the absorption
energy right similarly i have calculated over
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here you see this lambda look look here look
here mu absorption maximum vapor is like this
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a right so i have calculated so so i i i can
also calculate that mu absorption in the solvent
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right so that then i have to calculate e e
minus e g isn't it e e minus e g so let's
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let's do it over here
e e minus e g is equal to e e vapor and from
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the other equation minus e g vapor right plus
twice mu g square by a cube f e l plus twice
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mu g square by a cube f o r minus twice mu
e square by a cube f e l minus twice mu g
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mu e by a cube f o r right
so this is equal to e e vapor minus e g vapor
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right minus i will take this two f e l by
a cube right two f e l by a cube from here
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and from here then this is equal to mu e square
minus mu g square right so then mu e square
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minus mu g square and then i will also take
minus two f o r by a cube mu g out then this
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will going to be equal to mu e minus mu g
right so look at this term so i have already
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said taken that mu e is greater than
mu g if you remember if mu e is greater than
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mu g then this quantity is a positive quantity
let me change the color organs so this is
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my positive quantity this is my positive quantity
then the whole thing is negative all right
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that means these energy is less than this
energy right because the total thing is negative
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quantity this energy is less than this energy
means there will be a red shift in the absorption
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spectra that means when mu is greater than
mu g there is a red shift that what we have
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seen with with those model system
like while discussing the solvent effect on
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absorption spectra right we have seen but
here what we can see is exactly the same thing
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but here more mathematical treatment i have
shown you over here so when mu is greater
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than mu g then i will see a red shift in the
absorption spectra right ok
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now what i will going to do i will going to
give some time for the relaxation right
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so as i said that once you have created this
excited state e e right the orientational
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polarization is slow that means the solvent
we will take some time to solve it the excited
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state of your solute molecule electronic polarization
is first but the orientational polarization
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is slow right so once you create the excited
state it will take some time the solvent will
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come and reliability relax some come and relax
relax so that extra stabilization of the state
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is because of the solvent reorientation right
which is under the reorientation or orientational
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polarizability of the solvent so now if i
give a enough time then instead of using in
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the here instead of using mu g for the r o
r right for that e e state i will use mu e
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because i have given enough time for that
right so let's see what will happen for that
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so then in this case this is the solution
energy for emission right so salvation energy
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for emission right so in this case e e s
so let's say that this state in a the energy
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of that state is e e s solved it right i just
timed it like that so e e s e e s minus e
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e vapor i can calculate that deadly here the
derivation will be exactly same e only the
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difference is that for r o r i will going
to use mu e not mu g isn't it
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so let me do it quickly over here so this
is equal to minus mu e r this is equal to
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minus mu e into r e l minus mu e into r o
r is equal to minus mu e twice mu e by a cube
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f e l minus mu e twice mu in this case i write
e not g right because i have given time so
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twice mu e by a cube f o r right so this is
equal to minus twice mu e square by a cube
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f e l minus twice mu e square by a cube f
o r right so i can write e e s e e s equal
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to e e vapor minus twice mu e square by a
cube f e l minus twice mu square by a cube
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f o r right
so i got that e e s similarly i will also
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i have to get the e g s right e g s so that
we can also do easily so i can write e g s
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minus e g vapor right e g s minus e g to that
over here so this is another important
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equation here so for that e g s minus e g
vapor this is equal to which state ground
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state right so minus mu g into reduction field
r so minus mu g into r minus mu g r e l minus
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mu g r o r right so this is equal to minus
mu g twice mu g by a cube f e l this means
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that the electron has moved from lumo to homo
right draw a excited state to the ground state
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right so first electronics electronic polarization
is first so i use the mu g however in this
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case you see this mu g r o r that twice mu
by a cube that mu right this is the orientational
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polarizability but the solvent will stay as
that there in the excited state right because
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it takes time orientational [ori/orientational]
orientational relaxation is slow alright so
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in this case i will again use that mu e not
mu g right so this is twice mu e by a cube
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into f o r and i am almost done over here
so this is equal to to minus twice mu g
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square by a cube f e l minus twice mu e mu
g by a cube f o r so i can get this e g s
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equal to e g vapor minus twice mu g square
by cube f e l minus twice mu e mu g by a cube
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f o r so i also got this equation right i
got this e g s that is in the emission after
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emission what is the energy of the ground
state i got this one e e s right before emission
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but after stabilization what is the energy
of the excited state and that's it so with
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this um equation we will continue on the next
lecture about what will be the energy of
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this emission right
thank you very much