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