SSEVEN DERIVATIONS OF THE LAMBERT-BEER LAW
Hern´an R. S´anchez
Centro de Qu´ımica Inorg´anica. CONICET La Plata - UNLP, Argentina. (cid:0) [email protected]
Abstract
Seven derivations of the Lambert-Beer law are proposed in this paper. They were designed to be simple and intuitive. Most ofthem are suitable for the classroom. Readers can also benefit from them by looking at the phenomenon from different perspectives,which gives valuable resources when explaining it in class.
Electromagnetic radiation beams are attenuated by passingthrough an absorbent material. The Lambert-Beer (L-B) lawdescribes how this attenuation depends on the concentrationof the absorbent particles and on the optical path, providedthat certain conditions are met . This work has two principalaims: firstly to provide simple yet rigorous derivations of theL-B law useful to be taught in the classroom. Secondly, tobroaden our current understanding of this law by approachingit from different viewpoints.Many derivations of the L-B law have been proposed .From an abstract point of view, the derivations at some pointstate a relationship between internal transmittance ( T ) andconcentration ( c ) or optical path ( b ) satisfied only by theexponential function. For this, different approaches can befollowed. In §2.1 a very brief and simple derivation of theL-B law is proposed. What makes this derivation accessibleis that the relationship employed is the exponential identity a x + y = a x a y which is simple and known from introductorycourses.Berberan-Santos and Daniels proposed proofs closelyconnected to gas kinetic theory. These are pretty rigorousand provided a clear picture of the phenomenon. A derivationon the same lines is proposed in §2.2. It differs from the onesjust mentioned in that it is mathematically much simpler.In the above derivations a photon is absorbed whenever itfinds an absorbing particle. This is not true under the alterna-tive picture employed in the derivation proposed in §2.3.The above derivations have a strong probabilistic approachinvolving spatial variables. The latter are related to the widthof solution layers or the position of absorbing particles. Thederivations proposed in §§2.4 and 2.5 maintain the probabilis-tic approach but move the focus to the time that a photonremains unabsorbed. That of §2.4 uses basic elements of sur-vival analysis and chemical kinetics, while the one from §2.5models the phenomenon as a Poisson process .The derivations based in calculus often take the form ∂ T ( x , y ) ∂ x = k y yT ( x , y ) where ( x , y ) represents ( b , c ) or ( c , b ) and k y is a proportional- ity constant. Commonly, radiant power ( P ) or light intensityare used in place of internal transmittance but the relationabove can be obtained by a simple change of variables. Bothalternatives, that is ( x , y ) = ( b , c ) and ( x , y ) = ( c , b ) , can becombined in one system of two equations which also leads tothe L-B law .According to Bare, the derivations found in most under-graduate texts require the use of calculus concepts, and thesestart by considering that the absorption in a layer of infinites-imal width is proportional to that width, for example in thefollowing form: dP = k c cPdb . A rigorous derivation requiresstaring from self-evident premises. He argued that it is notobvious what absorption properties an infinitesimally thinfilm should have because we have no physical experiencewith such a film. Bare stressed that the linear relationship isvalid only for infinitesimal widths , and implicitly criticizedthe approach of Lykos who assumed this for thin but finitelayers. Berberan-Santos implied that it is not clear what thewidth of those layers is and figured it to be similar to that ofa molecule . What is clear is that it is not obvious to studentshow to interpret something on which experts disagree. In§2.6 a calculus-based derivation is provided. It reveals andhighlight the probabilistic nature of the process and does notbegin by making assumptions about the properties of thinlayers.Finally, a derivation based on the continuity equation isgiven in §2.7. It could be considered as a straightforwardcalculus-based derivation free from the issues mentionedabove. The transmittance is defined as the ratio of the transmittedradiant power to that incident on the sample . Among ofthe many mechanisms of energy loss only that of absorptionis considered here. Specifically, absorption due to processesinvolving a single photon, which are the most common for rea-sonably low radiant powers. We will consider that the sampleis homogeneous and isotropic and that the incident radiationis monochromatic, collimated and normal to the surface ofthe sample. In such case, as electromagnetic radiation can be1 a r X i v : . [ phy s i c s . c l a ss - ph ] A ug hought as a stream of photons, transmittance equals the frac-tion of non-absorbed photons. This corresponds to the proba-bility that a randomly chosen photon will pass through thesample. Thus, this probability equals the transmittance andthe problem may be stated as finding how the former dependson the path length and the concentration of the absorbingparticles, lets call this function P( c , b ) . These considerationswill be made in the most of the proposed derivations. The process of a photon traveling through the sample oflength b + b can be subdivided into two processes, onecorresponding to the photon traveling in the first section(of length b ) and the other to the photon traveling in theremaining section (of length b ). For each section, there aretwo possibilities: the photon passes through or does not.Let P( c , b ) be the probability that the photon passesthrough the first section. The probability of the photonpassing through the second is dependent on the latter: ifthe first one is not crossed the photon could never crossthe second one. In fact, this probability equals the prob-ability that the photon passes through the whole sample.Lets denote by P( c , b | c , b x > b ) to the probability that thephoton has crossed the second section since we know thatthe first section was crossed. The probability that the pho-ton passes through the sample, P( c , b + b ) , will then be P( c , b + b ) = P( c , b )P( c , b | c , b x > b ) according to thechain rule of probability. The fact that a photon has passedthrough a section does not change its propensity to be ab-sorbed in the future. This implies that P( c , b | c , b x > b ) mustbe equal to the probability of a photon passing a first sectionof length b , i.e. P( c , b | c , b x > b ) = P( c , b ) , then P( c , b + b ) = P( c , b )P( c , b ) (1)The only non-trivial continuous real function that satisfiesthe above equality for a fixed c is the exponential function P( c , b ) = a f ( c ) b . In fact, this is one of the many ways inwhich the exponential function can be characterized.We will analyze the dependence on concentration. Tomodel the process we will consider the following: the photonis absorbed when it collides with an absorbing particle that re-mains unmodified during that process. Thus, in order for thephoton not to be absorbed, no absorbing particle can be in theregion of space ( V ) where the photon would be obstructed.Then, the probability that it will not be absorbed is equal tothe probability that there are no absorbent particles in thatregion. If they are divided into two subsets built randomly,the sample concentration c may be expressed as the sum ofthe two concentrations ( c and c ) due to both subsets. Let P( c + c , b ) be the probability that there are no particles in V .No particles in V implies that there are no particles contribut-ing to c and no particles contributing to c in that region. Theprobabilities of these events are P( c , b ) and P( c , b ) , respec-tively. A good approximation is to consider that the absorbing P( c , b ) = P( c , b ) = particles are independent of each other and follow a uniformdistribution imposed by the homogeneity constraint. Thus,the two events are independent and the probability of themboth occurring equals the product of their probabilities P( c + c , b ) = P( c , b )P( c , b ) (2)which implies P( c , b ) = a f ( b ) c . Then, as P( c , b ) = a f ( c ) b = a f ( b ) c it follows that P( c , b ) = T ( c , b ) = a βbc which is the L-B law and β is constant. Because probabilitiesalways lie between 0 and 1, 0 ≤ a β ≤
1. In real systems0 < a β < Following the logic used to obtain Eq. 2, we can buildas many subsets as absorbent particles. If c represents theconcentration corresponding to having only one particle inthe sample, N represents the number of particles and A thecross section of the sample P( c , b ) = N (cid:214) i P( c , b ) = P( c , b ) N = P( c , b ) Abc P( c , b ) does not depend on b provided that A remains thesame, so the proof was completed. For the present derivation an alternative picture will beused. Again, it will be considered that the absorbing particlesmust be in V for the photon to be absorbed. In contrast to theprevious picture, a particle in V does not implies that the ab-sorption will necessarily occur, but there will be a probability P of it occurring.For a given absorbing particle, the probability ( P ) of beingin V is the the fraction of the total volume that V represents.Having N particles, the probability of having k of them in V follows the Bernoulli distribution P( k , N ; P ) = (cid:18) Nk (cid:19) P k ( − P ) N − k With k particles in V , the probability of m of them havingthe capacity to absorb the photon in case the encounter takesplace is also given by the Bernoulli distribution P( m , k ; P ) = (cid:18) km (cid:19) P m ( − P ) k − m Thus, the probability of having m particles capable of absorb-ing the photon is P( m , N ; P , P ) = N (cid:213) k = m (cid:18) Nk (cid:19) (cid:18) km (cid:19) P k ( −P ) N − k P m ( −P ) k − m Through algebraic transformations it can be simplified to P( m , N ; P × P ) = (cid:18) Nm (cid:19) (P P ) m ( − P P ) N − m
2e probability of the photon passing trough the sampleequals the above expression for m = P( c , b ) = P( , N ; P × P ) = ( − P P ) N = ( − P P ) Abc
By replacing probability with transmittance this expressionturns into the Lambert-Beer law.
This treatment can be simplified by analyzing the particlesone by one. For the photon not to be absorbed, each absorbingparticle must satisfy one of the two following conditions: beoutside of V (whose probability is 1 − P ), or be inside of V and not to absorb (of which the probability is P × ( − P ) as these are independent events). Because these are mutuallyexclusive the probability of one of them occurring is ( − P ) + P × ( − P ) = − P P The occurrence of the above events does not depend on theremaining particles, then for the whole sample ( − P P ) N The rest of the proof is exactly the same as the previous one.
The problem will be approached from a temporal ratherthan a spatial perspective. Consider the event of a photonbeing absorbed. Let τ be a non-negative random variabledenoting the waiting time until this (eventually) occurs, and f ( t ) the corresponding probability density function. The sur-vival function, S ( t ) , is the probability of the photon remainingunabsorbed until a time t . If we set t = S ( t ) = − ∫ t f ( t † ) dt † This implies that f ( t ) = − S (cid:48) ( t ) . The distribution of τ can alsobe characterized though the hazard function, h ( t ) , defined by h ( t ) = lim ∆ t → P( t ≤ τ < t + ∆ t | τ ≥ t ) ∆ t (3)The latter represents the instantaneous rate of occurrence ofthe event. The absorption of photons can be treated as anelemental chemical reaction between the photons and the ab-sorbent species, being of first-order in each of them providedthat light intensity is reasonably low. Then, as long as thephoton is inside of the solution, h ( t ) is proportional to thereaction rate with which the absorption can be modeled. Andthe concentration of photons is constant because it is due toa single photon. Then, h ( t ) = κc where κ is a proportionalityconstant. Notice that the numerator in Eq. 3 can be rewrittenas f ( t ) dt / S ( t ) . Thus, due to f ( t ) = − S (cid:48) ( t ) we have h ( t ) = − S (cid:48) ( t ) S ( t ) = − d ln S ( t ) dt then S ( t ) = e − ∫ t h ( t † ) dt † = e − ∫ t κcdt † = e − κct The time t is related to the distance traveled by a photon, b , and its speed, v x , through t = b / v x . This brings out theequality S ( t ) = T ( c , b ) for a large set of photons, then T ( b , c ) = e − κvx bc which is the L-B law. Consider a single photon. Let us imagine that once itsabsorption has taken place the photon is not altered and canbe absorbed again indefinitely. This is a false assumption, butit will help us to model the phenomenon as a Poisson process.Using the former is justified because we will only considerwhat happened in times prior to the first eventual absorption.As a photon travels through the solution its chance of beingabsorbed is the same for any point in time. If once againthe process is modeled as a chemical reaction, the numberof occurrences per unit time ( λ ) would be proportional tothe reaction rate. It can be written as λ = κc . The expectednumber of hypothetical absorption during a time t equals λt . Two other remarkable features are that the hypotheticalabsorptions are independent of each other, and that only oneabsorption can occur at a given time. This is enough to modelthe phenomenon as a Poisson process.The number of hypothetical absorptions can be representedwith a random variable ( X ) that follows the Poisson distribu-tion. The latter is a discrete probability distribution that canbe used for modeling the number of times ( k ) an event (inthis case absorption) takes place in a fixed time interval. Thecorresponding probability mass function is P( X = k ) = (cid:40) ( λt ) k e − λt k ! for k ∈ { , , , . . . } k =
0) up to atime t is P( X = ) = ( λt ) e − λt = e − κct (4)The proof concludes in the same way as that proposed in §2.4. Consider a set of n layers at positions b , b , . . . b n where b j > b i if j > i and let b =
0. From Eq. 1 we have that P( c , b k ) = P( c , b k − )P( c , b k − b k − ) . It can be rewritten as P( c , b k − b k − ) = − ∆ P k P( c , b k − ) where ∆ P k : = P( c , b k − ) − P( c , b k ) >
0. From this and Eq. 1the probability of a photon passing through the whole sampleis the product of the probability of passing through each layerof width b k − b k − , n (cid:214) k = P( c , b k − b k − ) = n (cid:214) k = (cid:18) − ∆ P k P( c , b k − ) (cid:19) e x can be expanded in Maclaurin series as 1 + x + O( x ) ,for small ∆ P k we have1 − ∆ P k P( c , b k − ) ∼ e − ∆ P k P( c , bk − ) then P( c , b ) = lim n →∞ n (cid:214) k = e − ∆ P k P( c , bk − ) = lim n →∞ e ln [ (cid:206) nk = e − ∆ P k P( c , bk − ) ] = lim n →∞ e − (cid:205) nk = (cid:18) ln e P( c , bk − ) ∆ P k (cid:19) = e − ∫ P ln (cid:18) e P† (cid:19) d P † = e − ∫ P P† d P † that can be rewritten in terms of P and P : = P ( c , ) as e − ∫ P P d P = e − ∫ PP P † dP † (5)A change corresponding to a proportion P( c , b k − b k − ) can beattributed to the existence of ∆ N = A c ( b k − b k − ) absorb-ing particles. Thus, there must be a function of ∆ N , f ( ∆ N ) ,that equals 1 − ∆ P k /P( c , b k − ) . We know about f ( ∆ N ) thatlim ∆ N → + f ( ∆ N ) =
1. Expanding in Taylor series aroundzero and for small ∆ N ,lim ∆ N → + f ( ∆ N ) = lim ∆ N → + + d f ( ∆ N ) d ∆ N (cid:12)(cid:12)(cid:12)(cid:12) ∆ N = ∆ N = lim ∆ N → + ( e α ) ∆ N where α : = f (cid:48) ( ) . For the whole sample, performing stepsanalogous to the previous oneslim n →∞ n (cid:214) k = f ( ∆ N ) = lim n →∞ n (cid:214) k = ( e α ) ∆ N = e ∫ N αdN = e ∫ b αAcdb † Thus, e − ∫ PP P dP = e ∫ b αAcdb † Taking logarithms of both sides the staring point of the stan-dard calculus-based derivations is obtained, from where theL-B law can be derived.
Let ϕ be the volume density of photons in the sample andlet v = [ v x , v y , v z ] T be their velocity field. Notice that ϕ depends on the position and, eventually, of time. The flux ofphotons in the sample is defined by J = [ J x , J y , J z ] T : = ϕ v .Let Q represents the number of photons absorbed per unitvolume per unit time. According to the continuity equationin its differential form ∂ ϕ ∂ t + ∇ · J + Q = Q = kcϕ where k is the reaction rateconstant. Being that the incident radiation is collimated wewill consider that light moves along the x -axis of a cartesiancoordinate system, which implies that ∇ · J = ∂ J x ∂ x . In steadystate ∂ ϕ / ∂ t =
0, then ∂ J x ∂ x = − kcϕ = − kc ( v x ϕ ) v x = − kc J x v x If J x is multiplied by the product between the frequency of theradiation ( ν ) and the Planck’s constant ( h ), the x − componentof the energy flux is obtained. Thus, multiplying by hν andconsidering that the derivative is a linear map hν ∂ J x ∂ x = ∂ ( hν J x ) ∂ x = − kc J x hνv x The radiant power can be obtained by integrating the en-ergy flux with respect to the transversal area. To write theabove equation in terms of the radiant power we do ∬ S ∂ ( J x hν ) ∂ x dydz = ∬ S − kcv x J x hνdydz and then we employ the Leibniz rule d ( ∬ S J x hνdydz ) dx = − kcv x ∬ S J x hνdydzdPdx = − kcv x P from where the usual steps of the standard derivation can befollowed. Many proofs were proposed in the previous section. Theirrelative advantages depend on the course in question. Theproof proposed in §2.2 is difficult to beat in terms of brevity.Originally this proof was to be discussed in an extended form.However, in a subsequent literature search conducted forthis paper, I noticed that it shares many similarities with theworks of Berberan-Santos and Daniels . Nevertheless, theproof was described here because it could be useful for somesince it is much simpler mathematically. Berberan-Santosimplicitly treated the process of including or not including agiven particle in V as a Bernoulli trial and used the BinomialLaw and the well-known equation e x = lim n →∞ (cid:16) + xn (cid:17) n His proof has been considered complex . Daniels’ solutionmakes use of geometric series which for many students is nottrivial. The simplicity of the proposed derivation can make itmore accessible while maintaining the same rigor and picturewhich is nicely described in the early works.The proof proposed in §2.1 is almost as simple and briefas that of §2.2. It makes use of conditional probability whendealing with the dependence on the optical path. An alter-native approach is to consider that the sample is split intoindependent layers and that for each of them there is a photon(all of them with the same frequency) trying to pass throughwith some probability. The probability of a photon (with thatfrequency) passing through the whole sample equals the prod-uct of those probabilities. This seems to be the idea behind theapproach of Bare . Making this clear, his derivation becomesanother simple alternative.4n the derivations proposed above (except those of §§2.5,2.6 and 2.7), and in those from the early works, it was con-sidered that a photon is absorbed whenever it encounters anabsorbing particle. The absorption properties of each parti-cle are attributed to a wavelength-dependent cross-sectionnot directly related to the size of the particle. This is a validapproach, though perhaps not the easiest to imagine. Forexample, it may be counterintuitive when comparing soluteand solvent in a colored diluted aqueous solution under visi-ble light. The proofs found in §2.3 employ an alternative andprobably more intuitive picture, so they can be consideredvaluable alternatives. They generalize the proof given in §2.2,the latter corresponds to the particular case where P =
1. Itcan be further generalized with relative simplicity to considermore than one type of absorbent particles. This can be doneby using the Poisson binomial distribution, but this derivationwill be omitted to conserve space.The derivation proposed in §2.4 makes use of fundamen-tal concepts of survival analysis and chemical kinetics. Thehazard function was used solely to include the dependenceon concentration. For obtaining only the dependence on theoptical path, we could consider the following. In §2.1 it wasmentioned that the fact that a photon has passed through agiven layer does not change its propensity to be absorbedin the future. This suggest a memoryless process, that is P( τ > t ) = P( τ > t + t | τ > t ) . Using the chain rule P( τ > t ) = P( τ > t + t )/P( τ > t ) . Then, due to the defi-nition of the survival function: S ( t ) S ( t ) = S ( t + t ) whichimplies that S is an exponential function of time, therefore,of the optical path.The proof provided in §2.5 may seem challenging, but theidea behind it is pretty simple. For a given photon, the absorp-tion probability density is constant in time. This implies that,if the photon could be adsorbed indefinitely, the variable rep-resenting the times at which the absorptions occur follows auniform distribution. The count of (hypothetical) absorptionsper some period is Poisson distributed. The exponential char-acter arises because the inter-successive-absorptions timesare exponentially distributed, and the expected value betweensuccessive absorptions equals the expected time for the firstabsorption.The conventional calculus-based proofs require some clari-fications to address the issues raised by Bare. This is avoidedin the proof of §2.6 at the expense of making it long. Thisderivation maintains the approach used in the traditionalproofs consisting of looking at losses caused by successivelayers. It is meant to illustrate a viable procedure but not tobe used in the classroom because simpler alternatives exist.It is noteworthy that this proof could be made much brieferby using product integrals , but it is not something studentsusually know about.Among the derivations proposed in this paper, only thatof §2.7 does not refer explicitly to probability. However, itis implicitly implied through the inclusion of the reactionrate . This derivation is based on the continuity equationwhich is used in many fields that are important to chemists and physicists. In this work, seven derivations of the Lambert-Beer lawwere proposed. It is most likely that only one derivation willbe shown in detail at the classroom, however, it is enrichingto know different approaches to achieve a more completeknowledge on the subject.
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