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Theory of barrierless electronic relaxation in solution. Delta function sink models
Volume 204, number 5,6
CHEMICAL PHYSICS LETTERS
26 March 1993
Theory of barrierless electronic relaxation in solution. Delta function sink models NaliniChalcravarti and K.L. Sebastian l Departmentof AppliedChemistry,Cochin Universityof Science and Technology,Cochin682022, India Received 17 August 1992; in final form 14 January 1993
We give a general method for Ending the exact solution for the problem of electronic relaxation in solution, modelled by a particle undergoing diffusive motion in a potential in the presence of a delta function sink. The diffusive motion is described by the Smoluchowski equation and the sink could be a delta function of arbitrary position and strength. The solution requires the knowledge of the Laplace transform of the Green function for the motion in the absence of the sink, We use the method to find the solutions for two different models. These are: (a) the particle in a box and (b) the particle on a ring model. Explicit expre& sions are obtained for the rate constants for these two models. We also analyze the short term behaviour of the excited state decay rate and arrive at analytical expressions which are of general validity.
1. Introduction Relaxation of an excited electronic state of a molecule which is in a polar solvent has attracted attention both from experimental [ l-31 and theoretical points of view (see the review of Bagchi and Fleming [ 41 for a recent survey). A molecule immersed in a liquid is put on an excited state potential energy surface (PES) by the absorption of radiation. The molecule executes a walk on the PES which may be taken to be random as it is immersed in the solvent. As it moves, it may undergo non-radiative decay from certain regions of the surface. It also undergoes radiative decay from anywhere on the surface. From the theoretical point of view, the problem is to calculate the probability that the molecule will still be in the excited state after a time t. We denote the probability that the molecule would survive on the excited PES by P,(t). It is common to assume the motion on the excited PES to be one-dimensional and diffusive, the relevant co-ordinate being denoted by x. In our discussion we refer to x as the position of a particle and to the de-excitation of the molecule as the absorption (annihilation) of the particle. It is also usual to assume that the motion on the PES is overdamped. Thus, the probability P(x, t ), that the particle may be found at x at the time t obeys a modified Smoluchowski equation [ 4,5 ] y
=[sp--kJ(x)-%]P(x,
t) .
(1)
In the above, Y=A
AL + 2 dV(x) ax2 ax
dx
.
(2)
V(x) is a potential causing the drift of the particle, and is determined by the shape of the excited PES. S(x) is a position-dependent sink function, taken to be normalized (i.e. j’ra S(x)&= 1), for convenience. ko is the ’ Present address: Abteilung Theorie, Fritz-Haber-Institut der Max-Planck-Gesellschaft, Faradayweg 4-6, W- 1000 Berlin 33, Germany.
496
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rate of non-radiative decay and k, is the rate of radiative decay. We have taken k, to be independent of position. A is the diffusion coefficient. It is related to the friction coefficient
am 0 =syx,
-
at
t)
(3)
)
subject to P(0, t) =O! This equation can be solved quite easily by the method of images and from the solution one can obtain P,(t), k, and kp Cases where the sink is not at the origin and is not of infinite strength are interesting and have been studied (see refs. [ 7, lo] 9. However, the investigation was numerical and was found to converge slowly in the limit of large b. ’ In the following, we make use of a general procedure for finding the exact solution of the problem with a delta function sink to fmd solutions for different probl&s. As the general procedure has already been reported [ 111 and made use of for solving the problem of delta function sink in the presence of a parabolic potential, we give only a minimum amount of details of the method in section 2. In sections 3 and 4 we use it to solve two models - a particle in a box and a particle on a ring - the sink being of the delta function type in both cases.
2. Exact results for the d&a function sink We find it convenient to define the Laplace transform 9(x, s) of P(x, t) by 03 g(x,s)=
s 0
dtP(x, t) exp(
-a).
(4)
Laplace transformation of eq. ( 1) gives [s-2’+k,,S(x)+k,]9’(x,s)=Po(x).
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As k, is a constant, the solution of this equation may be expressed in terms of the Green function g(x, s 1x0), which obeys [s-Y+@Q)]
b(x,s]x())=6(x-x0).
(6)
‘3(x, slxo) describes motion in the case where there is no radiative decay. In terms of this, the solution of eq.
(5) is
(7) Using the operator notations of quantum mechanics, we can write
~~~,~I~~~=~~I~~-~+ko~)-‘lx~>,
(8)
where Ix> (1x0) ) denotes the position eigenket with an eigenvalue x (x0). Now we use the operator identity (s-Y+f?os)-i=(s-_!?)-I
-(s-2)-‘k&s-_!?ysks)-’
(9)
to obtain ~(x,s~xo)=(x~(s-Y)-‘~xo>-(x~(s-~)-’~s(s-u+ks)-’~x~)
(10)
Inserting the resolution of identity 1=JZ, dy ]y) (y] in between the two inverses in the second term of the above equation, we arrive at the Lippman-Schwinger-type equation
~(~~~I~o)=~(~,~~~,)-~o
(11)
f dy q(x,sly)S(y)B(y,slxo). -co
F&,(x, sl x0) is defined by %(x, q&l) =
Ix0 >
(12)
and corresponds to propagation of the particle placed initially at x0, inthe absence of any sink. Note that it is the Laplace transform of G,(x, t]xo) which is the probability that a particle starting at x0 may be found at x at the time t, given that there is no decay. It’obeys the diffusion equation Go(x, tlx,)=b(x-x0)
.
(13)
The above equation has no sink term in it. As there is no sink, there is no absorption of the particle. Therefore, J?, dx Go(x, t 1x0)= 1. From this one concludes that co I -0)
b %b(&SIX)=~IS,
(14)
a result that we make use of in the following. If S(y) =6(y-x,),
~~x,~lxo~=4~~,~l~o~-ko~b(~,~l~.~~~~~,~l~o~
+
then eq. ( 11) becomes
(15)
We now solve eq. ( 15) to find
%G,~l~o)=
~%(~,,~I~o~~~+ko~~~a~~l~*)l-l~
This, when substituted back into eq. ( 15 ) gives
(16)
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Volume 204, number 5.6
(17)
~~~,~l~o~=%~~,~l~~~-ko%b(~,~I~s~%~~,,~l~o~~~+ko%~~,,~l~,~l-’ .
Using the above equation in (7) gives us 9 (x, S) explicitly, which we do not bother to write down, as our interest is only in the survival probability PJ t) =J’??, dx P(x, t). It is possible to calculate the Laplace transform 3(s) ofP,(t) directly. Ye(s) isrelatedto P(x,s) by sP,(s)=JoD,dx~P(x,s). Therefore,fromeqs. (7), (14) and (17), we get
s;o=(~-~l+~%~x..s+k.lx.~l-~~o
J a0 4(x,,s+k,,xo)P~(x~))/(sf~~) --o
The average and long time rate constants may be found from g(s). of the pole of $(s), closest to the origin. From eq. ( 18), we get k,_‘=(l-4[1+%$(x.,~lx,)l-’
1
ho
%(i,klxoU’ohd)/k.
*
(18)
Thus, kc ’ = PO(O) and k,=negative
(19)
--m Obviously, kI is dependent on the initial probability distribution P,(x). On the other hand, kL= - (pole of ~[l+ko~o(xSP st kl x,) 1(s t h)}- ’ ), closest to the origin, on the negative s axis, and is independent of the initial distribution. The expressions that we have obtained for ge(s), kI and kL are quite general and are valid for any V(x). However, their utility is limited by the fact that in order to make use of them one must know %(x, s]xo), which is somewhat difficult to determine. It is possible to find q(x, $1~~) only in a few limited cases. We report the solution in two such cases in the following. The problem of parabolic potential has been solved using the same procedure [ 111.
3. The particle in a box with delta function sink 3.1. The Green function S’dx, $1xd In the following, we give results for the case where V(x) =0 if the particle is between x=0 and x=a, and Y(x) ZOOotherwise. In this case, the Green function satisfies
(s-& )
%(x,s~xo)=s(x-x,)
)
(20)
if 0 < x < a. As the particle is reflected at x = 0 and at x= a, the above equation has to be solved subject to d % (x, slxo)/dx=O at these two points. The solution is easy and one obtains %(x, slxo)=JYx, slxo)ls
(21)
with F(x,slx,,)=cosh[m(
a-x,)]cosh(mx,)m/sinh(am)
(22)
where x, =min(x, x0) and x, =max(x, x0). 3.2. The rate constants
To get a qualitative idea of the behaviour of the rate constants, we imagine the initial distribution, PO(x), to be sharply peaked at x0, and to be well represented by 6(x-x0). Then, we obtain 499
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CHEMICAL PHYSICS LETTERS
W(& k 1x0)
ki’=kF’ I- ~+kiJF(X,,k,IX,)
26 March 1993
(23)
Further, k,=-[valueofsforwhichs+L~,,F(x,,s~x,)=O]+k,.
(24)
We note that k1 is dependent on the initial position x0 and k, in a rather complex fashion. On the other hand, kL is independent of x0 and depends on k, linearly. In the following we let h-0, in which limit eqs. (23) and (24) simplify. Note that this is the limit studied earlier [ 71. By analysing this limit, we arrive at conclusions which we expect to be valid even when k, is finite. As /c+ 0, F(x,, k, 1x0 ) and F( x,, k, 1x, ) + 1/a so that w( x,, k,lxo)/[k+@(x,, L+lx,)]+l. Hence one can write
(25) We now take xo
On carrying out the differentiation, we obtain ki’ EU/~ •I (Xf -X6)/U.
(27)
If, on the other hand, xo>xS, we obtain ki’ =LZ/~ -t [ (u-&)~-
(cz-xO)~]/U.
(28)
In the limit &co, the particle would be absorbed with certainty if it reaches x,. This is referred to as the pinhole sink. In this limit, the first term onthe right-hand sides of eqs. (27) and (28) vanishes. Then eq. (27) for ki I reduces to that of Bagchi’s for the staircase model [ 71, wherein the reactive motion occurs on a flat surface with the reflecting barrier at the origin and the absorbing barrier at x=x,. For &/A CK1, one gets k,x&/a, independent of the viscosity. If k&A>> 1, then kI is proportional to A and hence inversely to the viscosity. 4 The long term rate constant kL is determined by the value of s, which satisfies s+ w( x,, s 1x,) = 0. Ifs obeys this equation, then k,= --s. Defining o by u2A/a2=kL, we write this equation as w=cos[w( 1-x,/a)
] cos(wx,/a)/c,a/Asin(w)
.
(29)
As 0
4. Particle on a ring with delta function sink In the following, we consider a problem quite similar to the above. However, we now imagine that the particle is on a ring and its position is specified by the angle co-ordinate 0. This is appropriate for non-radiative transitions occurring from an excited state in which rotational motion is important. Further, in the b-tlimit, this reduces to the Oster-Nishijima model. 500
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4.1. The Green function $,(9, SI 19&
The Green function obeys (30)
For this problem, we denote the diffusion coefftcient by the symbol D. We now solve this equation, taking the origin to be at the initial position of the particle, and B varying from 0 to 2x. The solution of the above equation is ?&(0,s]O)=Bcosh[~(8-x)]+Csinh[m(B-x)].
(31)
The ring topology implies that ?$(O, s] 0) = +$,(2x, s]O), from which we find C=O. The value of B may be obtained from the equation satisfied by the derivative of go (0, s ]0), which is (32) Thus we get B=1/[2@sinh(&@x)],
so that
4(~,40)=~(~,4O)/~,
(33)
where P(B,slO)=JslDcosh[JslD(B-A)]/2sinh(JslDx).
(34)
If $ is not at the origin, ~(e,~ieo)=li(e,~~eo)i~
(35)
and F(B,s]B,)=mcosh m(0) = f3-$+2nrr,
{m
[m(e)-~]}/2sinh(~x).
(36)
where n is a positive or negative integer, chosen such that Od m(6) 62x.
4.2. The rate constants As in the earlier case, we imagine the initial distribution, Po( e), to be sharply peaked at do, and to be well represented by S( 8- 8,). Then we obtain (37) As +O, the expression within the brackets in the equation above goes to 0. Hence one can write, just as in the earlier case, (38)
Then we get ki’ =m(O,) [2x-m(8,)
]/20+2rr/k,,
.
(39)
In the limit &+oo, the particle would be absorbed with certainty on reaching 0, (the pinhole sink condition). 501
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In this limit, the second term on the right-hand side of the above equation vanishes. The long term rate constant kL is determined by the value of s which satisfies
2,,/%sinh(xm)+k,cosh(nm)=O.
(40)
If s obeys this equation, then kL= -s. Defining w2= k,/D, we find w tan(orc)=kJ2D.
(41)
If ko* D, then kLz k,,/2Dx and is independent of viscosity while in the opposite limit of b > D, kLz {D and is inversely proportional to viscosity.
5. The
shortterm
behaviour
of P.(t)
In the earlier investigations, analytic expressions for the survival probability were obtained only in the limit Bagchi [ 7, lo] has investigated the problem numerically for motion on a parabolic potential. This problem was solved analytically by Sebastian [ 111. Here we address the question: what would be the experimental manifestations of a finite value of k+,?For specificity, we take the particle on a ring model as the example. In this case, the long time rate constant is determined by eq. (41). An approximate solution of this equation, valid in the large /c,,limit is k,-m.
kL=tD-D2/b+LO(l/k;).
(42)
If one tried to fit the long time experimental data assuming the pinhole model, the experimentally determined diffusion coefficient D, would be related to the actual one by D.=D-40*/h and w?uld clearly be smaller than the actual one. At short times, the decay is dominated by diffusion. If this D, is used to predict the short time behaviour, one would obtain lower decay rates than found experimentally. This could be a reason for the disagreement between the parabolic pinhole model and experiment found by Ben-Amotz and Harris [ 21. It is interesting to ask: Can one derive explicit expressions for the short time behaviour of P,( t)? The answer is: for sufficiently small t, it is always possible. For this, consider the particle on a ring problem. It is quite easy to carry out a similar investigation for the box problem too, and the conclusions are similar. From eqs. ( 18), (21) and (22), we find (43)
In the above, we have put k=O. Defining the probability of finding the system in the ground electronic state PB(t) = 1-P,(f), and its Laplace transform s(s), we find, for the particle on a ring problem, (44)
In the ring topology the particle can follow two paths to arrive at the sink. Here we consider the situation where only the shorter one is important. This means that the other path is at least two times longer and that we consider only t-x ti/D. In terms of the Laplace variable s, this implies that sn2/Dx- 1. If in addition, s% kg/D (i.e. t-x D/k:), then the second term in the denominator of (44) may be neglected and the sinh and cash functions approximated by the larger of the exponential terms (here we throw out the longer path! ). Then,
~(~)=koexp[-dB,~JslDl/~, where 502
4(8,)=~-~~(8,)-X~.
(45) From this, we find
CHEMICALPHYSICSLETTERS
Volume 204, number 5,6
Fig. 1. Plotsof (l/D) (P,(f)/df)
againstDt. (-) the exact results for k,,/D=O.ZS and b/0=25,
26 March 1993
anti-(+,.) are respectively.
(-a-) is for kJO=O.25, using the approximation of eq. (47), (---) is the result using eq. (49). Notethat this does not
while
depend on k,,. All the plots are for 19,=x/2 and &= 0.
Dt
(46)
@e(t) @JO
-dt=-=
On the other hand, if s*c k$D, so that t sDfk& ps(r)=
(47)
dt
then we get
exp[-JslD444)l/s,
(48)
so that
(49) This is the expression appropriate for the pinhole sink limit. Note that for sufficiently low t( t<
Acknowledgement The work of NC was supported by the Council of Scientific and Industrial Research (India).
References [ 1 ] D. Bendmotx and C.B. Harris, J. Chem. Phys. 136( 1987) 4856. [ 21 D. Ben-Amotz and C.B. Harris, J. Chem. Phys. 86 ( 1987) 5433.
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[ 31 D. Ben-Amotz, R. Jeanloz and C.B. Harris, J. Chem. Phys. 86 (1987) 6119. [4] B. Bagchi and G.R. Fleming, J. Phys. Chem. 94 (1990) 9. [ 51B. Bagchi, G.R. Fleming and D.W. Oxtoby, J. Chem. Phys. 78 (1983) 7375. [6] H. Sumi and R.A. Marcus, J. Chem. Phys. 84 (1986) 4894. [7] B. Bagchi, J. Chem. Phys. 87 (1987) 5393. [8] G.OsterandN. Nishijima, J.Am. Chem. See. 78 (1956) 1581. [9] K. Schulten, Z. Schulten and A. Szabo, Physica A 100 ( 1980) 599. [lo] B. Bagchi, Chem. Phys. Letters 138 (1987) 315. [ 111K.L. Sebastian, Phys. Rev. A 46 (1992) 1732.
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Theory of barrierless electronic relaxation in solution. Delta function sink models NaliniChalcravarti and K.L. Sebastian l Departmentof AppliedChemistry,Cochin Universityof Science and Technology,Cochin682022, India Received 17 August 1992; in final form 14 January 1993
We give a general method for Ending the exact solution for the problem of electronic relaxation in solution, modelled by a particle undergoing diffusive motion in a potential in the presence of a delta function sink. The diffusive motion is described by the Smoluchowski equation and the sink could be a delta function of arbitrary position and strength. The solution requires the knowledge of the Laplace transform of the Green function for the motion in the absence of the sink, We use the method to find the solutions for two different models. These are: (a) the particle in a box and (b) the particle on a ring model. Explicit expre& sions are obtained for the rate constants for these two models. We also analyze the short term behaviour of the excited state decay rate and arrive at analytical expressions which are of general validity.
1. Introduction Relaxation of an excited electronic state of a molecule which is in a polar solvent has attracted attention both from experimental [ l-31 and theoretical points of view (see the review of Bagchi and Fleming [ 41 for a recent survey). A molecule immersed in a liquid is put on an excited state potential energy surface (PES) by the absorption of radiation. The molecule executes a walk on the PES which may be taken to be random as it is immersed in the solvent. As it moves, it may undergo non-radiative decay from certain regions of the surface. It also undergoes radiative decay from anywhere on the surface. From the theoretical point of view, the problem is to calculate the probability that the molecule will still be in the excited state after a time t. We denote the probability that the molecule would survive on the excited PES by P,(t). It is common to assume the motion on the excited PES to be one-dimensional and diffusive, the relevant co-ordinate being denoted by x. In our discussion we refer to x as the position of a particle and to the de-excitation of the molecule as the absorption (annihilation) of the particle. It is also usual to assume that the motion on the PES is overdamped. Thus, the probability P(x, t ), that the particle may be found at x at the time t obeys a modified Smoluchowski equation [ 4,5 ] y
=[sp--kJ(x)-%]P(x,
t) .
(1)
In the above, Y=A
AL + 2 dV(x) ax2 ax
dx
.
(2)
V(x) is a potential causing the drift of the particle, and is determined by the shape of the excited PES. S(x) is a position-dependent sink function, taken to be normalized (i.e. j’ra S(x)&= 1), for convenience. ko is the ’ Present address: Abteilung Theorie, Fritz-Haber-Institut der Max-Planck-Gesellschaft, Faradayweg 4-6, W- 1000 Berlin 33, Germany.
496
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26 March 1993
rate of non-radiative decay and k, is the rate of radiative decay. We have taken k, to be independent of position. A is the diffusion coefficient. It is related to the friction coefficient
am 0 =syx,
-
at
t)
(3)
)
subject to P(0, t) =O! This equation can be solved quite easily by the method of images and from the solution one can obtain P,(t), k, and kp Cases where the sink is not at the origin and is not of infinite strength are interesting and have been studied (see refs. [ 7, lo] 9. However, the investigation was numerical and was found to converge slowly in the limit of large b. ’ In the following, we make use of a general procedure for finding the exact solution of the problem with a delta function sink to fmd solutions for different probl&s. As the general procedure has already been reported [ 111 and made use of for solving the problem of delta function sink in the presence of a parabolic potential, we give only a minimum amount of details of the method in section 2. In sections 3 and 4 we use it to solve two models - a particle in a box and a particle on a ring - the sink being of the delta function type in both cases.
2. Exact results for the d&a function sink We find it convenient to define the Laplace transform 9(x, s) of P(x, t) by 03 g(x,s)=
s 0
dtP(x, t) exp(
-a).
(4)
Laplace transformation of eq. ( 1) gives [s-2’+k,,S(x)+k,]9’(x,s)=Po(x).
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As k, is a constant, the solution of this equation may be expressed in terms of the Green function g(x, s 1x0), which obeys [s-Y+@Q)]
b(x,s]x())=6(x-x0).
(6)
‘3(x, slxo) describes motion in the case where there is no radiative decay. In terms of this, the solution of eq.
(5) is
(7) Using the operator notations of quantum mechanics, we can write
~~~,~I~~~=~~I~~-~+ko~)-‘lx~>,
(8)
where Ix> (1x0) ) denotes the position eigenket with an eigenvalue x (x0). Now we use the operator identity (s-Y+f?os)-i=(s-_!?)-I
-(s-2)-‘k&s-_!?ysks)-’
(9)
to obtain ~(x,s~xo)=(x~(s-Y)-‘~xo>-(x~(s-~)-’~s(s-u+ks)-’~x~)
(10)
Inserting the resolution of identity 1=JZ, dy ]y) (y] in between the two inverses in the second term of the above equation, we arrive at the Lippman-Schwinger-type equation
~(~~~I~o)=~(~,~~~,)-~o
(11)
f dy q(x,sly)S(y)B(y,slxo). -co
F&,(x, sl x0) is defined by %(x, q&l) =
Ix0 >
(12)
and corresponds to propagation of the particle placed initially at x0, inthe absence of any sink. Note that it is the Laplace transform of G,(x, t]xo) which is the probability that a particle starting at x0 may be found at x at the time t, given that there is no decay. It’obeys the diffusion equation Go(x, tlx,)=b(x-x0)
.
(13)
The above equation has no sink term in it. As there is no sink, there is no absorption of the particle. Therefore, J?, dx Go(x, t 1x0)= 1. From this one concludes that co I -0)
b %b(&SIX)=~IS,
(14)
a result that we make use of in the following. If S(y) =6(y-x,),
~~x,~lxo~=4~~,~l~o~-ko~b(~,~l~.~~~~~,~l~o~
+
then eq. ( 11) becomes
(15)
We now solve eq. ( 15) to find
%G,~l~o)=
~%(~,,~I~o~~~+ko~~~a~~l~*)l-l~
This, when substituted back into eq. ( 15 ) gives
(16)
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(17)
~~~,~l~o~=%~~,~l~~~-ko%b(~,~I~s~%~~,,~l~o~~~+ko%~~,,~l~,~l-’ .
Using the above equation in (7) gives us 9 (x, S) explicitly, which we do not bother to write down, as our interest is only in the survival probability PJ t) =J’??, dx P(x, t). It is possible to calculate the Laplace transform 3(s) ofP,(t) directly. Ye(s) isrelatedto P(x,s) by sP,(s)=JoD,dx~P(x,s). Therefore,fromeqs. (7), (14) and (17), we get
s;o=(~-~l+~%~x..s+k.lx.~l-~~o
J a0 4(x,,s+k,,xo)P~(x~))/(sf~~) --o
The average and long time rate constants may be found from g(s). of the pole of $(s), closest to the origin. From eq. ( 18), we get k,_‘=(l-4[1+%$(x.,~lx,)l-’
1
ho
%(i,klxoU’ohd)/k.
*
(18)
Thus, kc ’ = PO(O) and k,=negative
(19)
--m Obviously, kI is dependent on the initial probability distribution P,(x). On the other hand, kL= - (pole of ~[l+ko~o(xSP st kl x,) 1(s t h)}- ’ ), closest to the origin, on the negative s axis, and is independent of the initial distribution. The expressions that we have obtained for ge(s), kI and kL are quite general and are valid for any V(x). However, their utility is limited by the fact that in order to make use of them one must know %(x, s]xo), which is somewhat difficult to determine. It is possible to find q(x, $1~~) only in a few limited cases. We report the solution in two such cases in the following. The problem of parabolic potential has been solved using the same procedure [ 111.
3. The particle in a box with delta function sink 3.1. The Green function S’dx, $1xd In the following, we give results for the case where V(x) =0 if the particle is between x=0 and x=a, and Y(x) ZOOotherwise. In this case, the Green function satisfies
(s-& )
%(x,s~xo)=s(x-x,)
)
(20)
if 0 < x < a. As the particle is reflected at x = 0 and at x= a, the above equation has to be solved subject to d % (x, slxo)/dx=O at these two points. The solution is easy and one obtains %(x, slxo)=JYx, slxo)ls
(21)
with F(x,slx,,)=cosh[m(
a-x,)]cosh(mx,)m/sinh(am)
(22)
where x, =min(x, x0) and x, =max(x, x0). 3.2. The rate constants
To get a qualitative idea of the behaviour of the rate constants, we imagine the initial distribution, PO(x), to be sharply peaked at x0, and to be well represented by 6(x-x0). Then, we obtain 499
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W(& k 1x0)
ki’=kF’ I- ~+kiJF(X,,k,IX,)
26 March 1993
(23)
Further, k,=-[valueofsforwhichs+L~,,F(x,,s~x,)=O]+k,.
(24)
We note that k1 is dependent on the initial position x0 and k, in a rather complex fashion. On the other hand, kL is independent of x0 and depends on k, linearly. In the following we let h-0, in which limit eqs. (23) and (24) simplify. Note that this is the limit studied earlier [ 71. By analysing this limit, we arrive at conclusions which we expect to be valid even when k, is finite. As /c+ 0, F(x,, k, 1x0 ) and F( x,, k, 1x, ) + 1/a so that w( x,, k,lxo)/[k+@(x,, L+lx,)]+l. Hence one can write
(25) We now take xo
On carrying out the differentiation, we obtain ki’ EU/~ •I (Xf -X6)/U.
(27)
If, on the other hand, xo>xS, we obtain ki’ =LZ/~ -t [ (u-&)~-
(cz-xO)~]/U.
(28)
In the limit &co, the particle would be absorbed with certainty if it reaches x,. This is referred to as the pinhole sink. In this limit, the first term onthe right-hand sides of eqs. (27) and (28) vanishes. Then eq. (27) for ki I reduces to that of Bagchi’s for the staircase model [ 71, wherein the reactive motion occurs on a flat surface with the reflecting barrier at the origin and the absorbing barrier at x=x,. For &/A CK1, one gets k,x&/a, independent of the viscosity. If k&A>> 1, then kI is proportional to A and hence inversely to the viscosity. 4 The long term rate constant kL is determined by the value of s, which satisfies s+ w( x,, s 1x,) = 0. Ifs obeys this equation, then k,= --s. Defining o by u2A/a2=kL, we write this equation as w=cos[w( 1-x,/a)
] cos(wx,/a)/c,a/Asin(w)
.
(29)
As 0
4. Particle on a ring with delta function sink In the following, we consider a problem quite similar to the above. However, we now imagine that the particle is on a ring and its position is specified by the angle co-ordinate 0. This is appropriate for non-radiative transitions occurring from an excited state in which rotational motion is important. Further, in the b-tlimit, this reduces to the Oster-Nishijima model. 500
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4.1. The Green function $,(9, SI 19&
The Green function obeys (30)
For this problem, we denote the diffusion coefftcient by the symbol D. We now solve this equation, taking the origin to be at the initial position of the particle, and B varying from 0 to 2x. The solution of the above equation is ?&(0,s]O)=Bcosh[~(8-x)]+Csinh[m(B-x)].
(31)
The ring topology implies that ?$(O, s] 0) = +$,(2x, s]O), from which we find C=O. The value of B may be obtained from the equation satisfied by the derivative of go (0, s ]0), which is (32) Thus we get B=1/[2@sinh(&@x)],
so that
4(~,40)=~(~,4O)/~,
(33)
where P(B,slO)=JslDcosh[JslD(B-A)]/2sinh(JslDx).
(34)
If $ is not at the origin, ~(e,~ieo)=li(e,~~eo)i~
(35)
and F(B,s]B,)=mcosh m(0) = f3-$+2nrr,
{m
[m(e)-~]}/2sinh(~x).
(36)
where n is a positive or negative integer, chosen such that Od m(6) 62x.
4.2. The rate constants As in the earlier case, we imagine the initial distribution, Po( e), to be sharply peaked at do, and to be well represented by S( 8- 8,). Then we obtain (37) As +O, the expression within the brackets in the equation above goes to 0. Hence one can write, just as in the earlier case, (38)
Then we get ki’ =m(O,) [2x-m(8,)
]/20+2rr/k,,
.
(39)
In the limit &+oo, the particle would be absorbed with certainty on reaching 0, (the pinhole sink condition). 501
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In this limit, the second term on the right-hand side of the above equation vanishes. The long term rate constant kL is determined by the value of s which satisfies
2,,/%sinh(xm)+k,cosh(nm)=O.
(40)
If s obeys this equation, then kL= -s. Defining w2= k,/D, we find w tan(orc)=kJ2D.
(41)
If ko* D, then kLz k,,/2Dx and is independent of viscosity while in the opposite limit of b > D, kLz {D and is inversely proportional to viscosity.
5. The
shortterm
behaviour
of P.(t)
In the earlier investigations, analytic expressions for the survival probability were obtained only in the limit Bagchi [ 7, lo] has investigated the problem numerically for motion on a parabolic potential. This problem was solved analytically by Sebastian [ 111. Here we address the question: what would be the experimental manifestations of a finite value of k+,?For specificity, we take the particle on a ring model as the example. In this case, the long time rate constant is determined by eq. (41). An approximate solution of this equation, valid in the large /c,,limit is k,-m.
kL=tD-D2/b+LO(l/k;).
(42)
If one tried to fit the long time experimental data assuming the pinhole model, the experimentally determined diffusion coefficient D, would be related to the actual one by D.=D-40*/h and w?uld clearly be smaller than the actual one. At short times, the decay is dominated by diffusion. If this D, is used to predict the short time behaviour, one would obtain lower decay rates than found experimentally. This could be a reason for the disagreement between the parabolic pinhole model and experiment found by Ben-Amotz and Harris [ 21. It is interesting to ask: Can one derive explicit expressions for the short time behaviour of P,( t)? The answer is: for sufficiently small t, it is always possible. For this, consider the particle on a ring problem. It is quite easy to carry out a similar investigation for the box problem too, and the conclusions are similar. From eqs. ( 18), (21) and (22), we find (43)
In the above, we have put k=O. Defining the probability of finding the system in the ground electronic state PB(t) = 1-P,(f), and its Laplace transform s(s), we find, for the particle on a ring problem, (44)
In the ring topology the particle can follow two paths to arrive at the sink. Here we consider the situation where only the shorter one is important. This means that the other path is at least two times longer and that we consider only t-x ti/D. In terms of the Laplace variable s, this implies that sn2/Dx- 1. If in addition, s% kg/D (i.e. t-x D/k:), then the second term in the denominator of (44) may be neglected and the sinh and cash functions approximated by the larger of the exponential terms (here we throw out the longer path! ). Then,
~(~)=koexp[-dB,~JslDl/~, where 502
4(8,)=~-~~(8,)-X~.
(45) From this, we find
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Fig. 1. Plotsof (l/D) (P,(f)/df)
againstDt. (-) the exact results for k,,/D=O.ZS and b/0=25,
26 March 1993
anti-(+,.) are respectively.
(-a-) is for kJO=O.25, using the approximation of eq. (47), (---) is the result using eq. (49). Notethat this does not
while
depend on k,,. All the plots are for 19,=x/2 and &= 0.
Dt
(46)
@e(t) @JO
-dt=-=
On the other hand, if s*c k$D, so that t sDfk& ps(r)=
(47)
dt
then we get
exp[-JslD444)l/s,
(48)
so that
(49) This is the expression appropriate for the pinhole sink limit. Note that for sufficiently low t( t<
Acknowledgement The work of NC was supported by the Council of Scientific and Industrial Research (India).
References [ 1 ] D. Bendmotx and C.B. Harris, J. Chem. Phys. 136( 1987) 4856. [ 21 D. Ben-Amotz and C.B. Harris, J. Chem. Phys. 86 ( 1987) 5433.
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[ 31 D. Ben-Amotz, R. Jeanloz and C.B. Harris, J. Chem. Phys. 86 (1987) 6119. [4] B. Bagchi and G.R. Fleming, J. Phys. Chem. 94 (1990) 9. [ 51B. Bagchi, G.R. Fleming and D.W. Oxtoby, J. Chem. Phys. 78 (1983) 7375. [6] H. Sumi and R.A. Marcus, J. Chem. Phys. 84 (1986) 4894. [7] B. Bagchi, J. Chem. Phys. 87 (1987) 5393. [8] G.OsterandN. Nishijima, J.Am. Chem. See. 78 (1956) 1581. [9] K. Schulten, Z. Schulten and A. Szabo, Physica A 100 ( 1980) 599. [lo] B. Bagchi, Chem. Phys. Letters 138 (1987) 315. [ 111K.L. Sebastian, Phys. Rev. A 46 (1992) 1732.
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