WEBVTT
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welcome to lecture number thirty five so in
the last lecture we were discussing about
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the excited state proton transfer and today
i would like to show you how to find out the
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rate of this kind of excited state proton
transfer process to start with let me take
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a very basic model of a two state irreversible
proton transfer process
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so let me write it over here i have this h
a star which is converted to a minus star
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because of the removal of proton from it let
me write this rate constant here as k one
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but these you don't forget that this h star
has its own radioactive rate constant right
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so or the deactivation rate constant which
i all together termed as k f over here and
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this one i termed as k f prime so then with
these rate constant it will form the ground
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state of the h a and with these rate constant
it will form the ground state of a minus
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so my whole model for this excited state proton
transfer from h a star to a minus star is
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like this with the three different rate constant
k f k f prime and k one right now if i write
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this rate equations over here i can write
minus d h star d t is equal to the two path
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way right one is k one another is k f right
so i will just write k one plus k f all are
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small k don't confused with the capital and
small capital k here is the small k ok k one
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plus k f into concentration of h star and
i can also write the formation of this a minus
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star right like that way so d a minus star
d t here k one is the formation and the this
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k f prime this one is the decay right
so k one to h a star minus k f prime into
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a minus star right so let me name this as
the equation one and this let's say equation
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two so from equation one what i can write
what i can write is minus l n h a star right
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is equal to k one plus k f to t plus c right
now i put this boundary conditions over
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here so here the boundary condition is at
t equal to zero that means just before the
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starting and just time zero of the starting
of the reaction
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the concentration of h a star is h star zero
so it's a initial concentration so i will
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write this conditions over here at t equal
to zero h a star is equal to h a star zero
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so i can write this equation as h a star is
equal to h a star zero e to the power minus
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k one plus k f into t so you see if you monitor
the h star the concentration of the h star
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will start from initial value and then it
will decay exponentially and the rate constant
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is k one plus k f and the time constant will
be inverse of this right time constant will
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be just inverse of this ok
now let me do one thing now let me take
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from the equation two and from equation two
let me write this as d a minus star d t is
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equal to k one this is my equation two k one
h a star right and this h a star now i will
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replace with this expression right so i will
just write h a star zero e to the power minus
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k one plus k f into t right minus k f prime
to a minus star that's it so i can take
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this k f prime a minus star in the left hand
side and then i will multiply the both side
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with e to the power k f prime into t right
let's do that so d a minus star d t plus k
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f prime a minus star is equal to k one h a
sorry h a star zero e to the power minus k
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one plus k f into t and i am multiplying both
side with e to the power k f prime into t
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so then this whole differentiation i can write
as d d t of a minus star e to the power k
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f prime into t right is equal to k one h a
star zero e to the power minus k one plus
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k f into t into e to the power k f prime into
t right
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so from that equation now i can write a minus
star e to the power k f prime t is equal to
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minus k one h a star zero divided by minus
k f prime plus k one plus k f e to the power
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k f prime minus k one minus k f plus c right
now i need to find out this c i can simply
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first divide both the side with e to the power
k f prime into t and then i will put this
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condition that at t equal to zero a minus
star is equal to zero just after dividing
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both side with this quantity so this this
this this i will divide both side and then
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i will write this one then the equation
what we will get is for c is equal to just
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this one with the positive signs so k one
is h a star zero divided by minus k f prime
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plus k one plus k f and then if i put this
value of c in that equation ultimately the
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equation what we will going to see is a minus
star equal to minus in that quantity k one
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h star divided by minus k f prime plus k one
plus k a f into e to the power minus k one
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plus k f into t then plus exactly the same
quantity will come now k one h a star this
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would be zero divided by minus k f prime
plus k one plus k f e to the power minus k
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f prime into t
earlier what we have seen earlier we have
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seen this h a now h a star how h a star must
decaying with time and we got this formula
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h a star equal to h a star zero e to the power
minus k one plus k f to t so this is the how
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the florescence intensity of h a star is changing
with time this one and this is how the florescence
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intensity of a minus star is changing with
time here we can see for h a star only decay
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only decay because this is a positive right
and this is exponential minus rate constant
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into t one over the rate constant is your
time constant so it will start from initial
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value and will decay
but in this case please note here these two
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are same this this quantity and this quantity
are same now see this is special case right
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for the irreversible to state proton transfer
but most interesting thing is that here you
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have a negative sign and here you have a positive
sign negative signs this exponential means
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the formation so first this a minus star will
form and then a minus star will decay so here
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is my formation here is my this part is my
formation
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and this part is my decay and here is my decay
so that's why people generally don't want
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to tell the florescence decay as in general
we want to say the florescence transient right
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because it may be in some case like these
case it is not a decay it's a formation it's
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a rise right so generally we speak as the
florescence transient right
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so now if i plot this florescence transient
of h a star and a minus star how it will going
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to look like lets plot it quickly to see that
so let's say this y axis is my intensity and
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this x axis is my time axis right and this
is my florescence intensity and let me draw
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this h a with with green colour so here h
a will looks something like this this is just
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a single exponential decay and a single exponential
decay right
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but this a minus will be first there is a
growth right and then there is a decay now
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you see the for the formation this rate constant
and for the decay this rate constant there
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exactly same right so that's why i draw the
formation like from here to till over here
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and then it will start decaying right so this
is for the a minus star and the other one
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is for h h a star right so from your time
resolve measurement either by t c a p c if
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t c a p c is good enough if this process is
slower than hundred picosecond because your
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time resolution t c a p c is about hundred
picosecond but in most of the cases like if
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you do in the water like two napthol in water
this rate of proton transfer right is in the
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order of one picosecond so t c a p c is not
suitable for such kind of experiment what
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you have to do you have to use a up conversion
method because in up conversion the i r f
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is alone two hundred femtosecond one hundred
femtosecond right
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so obviously here the i r f is over here so
this is my i r f right and with this i r f
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it is visible right so this formation is visible
over is my formation ok so as i told you
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this one over this is your time constant
so generally people write this as a time constant
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so we can write the tau one is equal to one
over k one plus k f and tau two is equal to
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one over k f prime and we can write all this
time constant from this experimental data
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now let me just quickly show you that what
happens if this is a reversible two state
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case right so in that case my model is h a
star is forming to a minus star and here is
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k one here is my k two and this is as usual
is k f and this one is k f prime this is making
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h a this is making a minus right
then similarly i should right those equations
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to start with minus d h a star d t is equal
to k one plus k f h a star minus k two this
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is my extra turn which was not present in
case of irreversible a two state proton transfer
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so a minus star and i have d a minus star
d t is equal to k one h a star plus k two
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plus k f prime into a minus star
like similarly what we have done for this
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irreversible case i can also do this for the
reversible case and ultimately this solution
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will be much more difficult than the previous
one and it is out of scope that i will solve
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in this lecture but i would like to show you
the result ultimately if you saw what you
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will get you will get the h star how the h
star is changing with time how the a minus
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star is changing with time those you will
get like in the previous case and those expression
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i i i will show over here
so h star equal to h a star zero divided it
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by m one minus m two to x minus m two e to
the power minus m one t plus h a star zero
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divided by m one minus m two is a complicated
one to m one minus x e to the power minus
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m two into t obviously i have to tell what
is m one m two that i am telling later and
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then a minus star will have the form x minus
m two ok into x minus m one then h a star
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zero divided by m one minus m two and then
k two this is my k two over here i should
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have this exponential part e to the power
here it will come as minus m one t and then
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i will get this minus so the here this minus
sign signified the formation right minus x
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minus m two into x minus m one like this way
h a star zero divided by m one minus m two
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into k two and this is this will be my e to
the power minus m two t in this case x is
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defined as k one plus k f and y is equal to
k two plus k f prime and m one two is defined
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is one and two is equal to half x plus y plus
minus root over x plus y whole square minus
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four into x y minus k one k two ok
so now as you can see here this is this is
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going to very complicated equation and
but but if you monitor the hm decay of
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this h a star or the florescence transients
of a minus star it is possible to evaluate
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the different rate constants like k one k
two k f and k f prime so you see in this case
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the equation is little complicated right
but by monitoring the time resolve emission
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one can get such kind of rate constant
of the proton transfer so we will finish and
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thank you for your kind attention