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