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IB Higher Level Maths Exploration

Derivation of the Beer-Lambert Law

Chisato Tsuji

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Contents

Introduction………………………………………………………………………………...3

Derivation……………………………………………………………………………….…4

Reflection…………………………………………………………………………….……12

Conclusion………………………………………………………………………………...13

Bibliography……………………………………………………………………………....14

Glossary……………………………………………………………………………….…..14

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Introduction

Beer-Lambert law is a law used to calculate the concentration of a substance in a solution.

The first time that I came across it was when I was doing some work experience at a

laboratory looking at ways to isolate proteins embedded within the cell membrane and allow

the movement of molecules in and out of the cell. There is a machine called NanoDrop 1

which can determine the concentration of protein and DNA (which stores the genetic

information of organisms) in a microlitre of solution. The PhD student that I was following

briefly referenced this law when explaining how the machine works and said that details into

the principles behind it would be covered in a standard university course for biology, which is

what I am hoping to pursue. I was intrigued that the concentration of proteins and DNA,

which allows fundamental life processes to happen, can be measured using a single microlitre

of solution and attempted to research the mathematics behind the law. However, I was not

able to find any satisfactory explanation as to how the law works. Therefore, I decided to

explore and work out for myself how the law was derived.

In my research I also found out that in addition to the concentration of proteins and DNAs,

the Beer-Lambert law is used for finding out the concentration of a coloured substance using

a colorimeter. Its use can include the measurement of concentration of chlorophylls,

haemoglobin and food dyes as well as iodine solution to determine the concentration of starch

in a solution in a quantitative way which is more objective than the alternative method of

seeing its relative concentration by performing a titration and comparing colour changes.2

The Beer-Lambert law was originally two separate laws regarding absorbance of a solution.

Pierre Bouguer was a mathematician and physicist who related the thickness of a medium

with its absorbance in a paper in 1729 titled “Essai d'optique sur la gradation de la lumière

(Optical Treatise on the Gradation of Light)” which was later developed by Johann Heinrich

Lambert in 1760. August Beer was a German mathematician and natural scientist. In 1852, he

published a paper in which he related the absorbance of a solution with its concentration. The

Beer-Lambert’s law unites these two factors together to form one equation;

𝑨 = 𝒄𝝐𝒃

where: 𝐴 = absorbance

𝑐 = concentration

𝜖 = molar absorptivity (how much light is absorbed by 1 mole of a molecule)

𝑏 = length of the path that the light travels

1 Thermo Scientific, NanoDrop 2000/2000c Spectrophotometer, V1.0 User Manual, Thermo Fisher Scientific

Inc., USA, 2009 2 Adds, J. et al., Tools, Techniques and Assessment in Biology, Thomas Nelson & Sons Ltd., UK, 1999

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My derivation attempts go from first principles to see exactly how each variable relates

within the equation and how it can be represented mathematically.

Derivation

Factors which affect absorbance

The Beer-Lambert law is dependent on a beam of light being shone through a solution

containing solutes which absorb a fraction of the light as it goes through the solution. As the

solution absorbs some of the light, the intensity of the light that will come out of the other

side of the solution will be decreased in intensity.

This can be clearly represented using the following diagram (Figure 1) where 𝐼𝑒is the initial

intensity of light, 𝑏 is the total length of the path through which light travels and 𝐼𝑏 is the

intensity of light after it has travelled through the solution. In real usage, it is possible to

ignore the absorbance of light by the material of the container and the solution in which the

solute is dissolved in by calibrating the calorimeter. Therefore, these can be ignored in this

derivation as well.

Figure 1

The extent to which the intensity of light will decrease after it has gone through a solution

will depend on three factors. These three factors, suggested by the three variables of the right

hand side of the Beer-Lambert law equation (𝒄𝝐𝒃), are;

A) the number of molecules in the solution (derived as 𝒄)

B) the amount of light each molecule absorbs (derived as 𝝐)

C) the length of the path that the light travels (derived as 𝒃)

The effect of the factors can be visually illustrated by the following diagrams (Figure 2).

𝐼𝑒 𝐼𝑏

𝑏

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Figure 2: visual illustration of factors affecting absorbance

Factor A) The number of molecules in the solution that absorb the light (ie concentration of

solutes in a solution)

If there are more molecules in a solution of the same volume, the solution will absorbc more

light as there are more molecules to absorb the light.

Factor B) How much light each molecule absorbs

If the molecules themselves absorb more light, the soution will absorb more light.

Factor C) The length of the path that the light travels through.

If the light travels through a solution for a longer distance, more light will be absorbed,

assuming that the concentration is the same.

Therefore, the equation of the absorbance of light by the solute in a solution must be linked to

these three factors.

In a solution, the molecules of the solutes will move around. Therefore, in order to work out

the accurate absorbance of light at a specific moment in time, it is more acurate to consider

the whole block of solution as split into smaller blocks where the molecules will be spread

out evenly throughout the solution. I will use the notation 𝛿𝑦 to represent the thickness of

one block.

By using the value 𝛿𝑦, I can now consider how factor A, the number of molecules in the

solution, and factor B, how much light each molecule absorbs, are related to the total

absorbance as I have specified that the length of the path that the light travels, factor C, as 𝛿𝑦.

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A) the number of molecules in the solution (concentration)

I have said that the absorbance of light is dependent on the number of molecules there are in

the solution. If the number of molecules in a given area of one unit/cm2 is 𝑛, then the number

of molecules in a block of thickness 𝛿𝑦 can be shown by the following expression where 𝑠

signifies the cross sectional area of the block of solution which stays constant throughout.

The number of molecules in the block = 𝑛 × 𝑠 × 𝛿𝑦

B) how much light each molecule absorbs

If I assign the cross sectional area of one molecule as 𝜎 as illustrated in Figure 3, then the

fractional proportion of the area that absorbs light compared to the whole area, 𝑠, specified

above, can be represented as 𝜎

𝑠.

Figure 3

I can now link factor A and factor B together; the fractional proportion of the light that is

absorbed in a block with thickness 𝛿𝑦 is equal to fractional proportion of the area that absorbs

the light in a block with thickness 𝛿𝑦. ie;

Proportion of light absorbed = proportion of area of one molecule × number of molecules

Proportion of light absorbed = (𝑛 × 𝑠 × 𝛿𝑦) ×𝜎

𝑠

= 𝜎 × 𝑛 × 𝛿𝑦 (equation 1)

Cross sectional area of one molecule absorbing the light = 𝜎

Cross sectional area of the block= 𝑠

Proportion of cross sectional area absorbing light = 𝜎

𝑠

𝛿𝑦

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Absorbance

In equation 1, I worked out the fractional proportion of light absorbed. Therefore, in order to

substitute into the equation, I can work out the proportion of the light absorbed in terms of

light intensity.

The absorbance of light cannot be directly measured but it is easily worked out as it is the

difference in value between the intensity of light as it enters the solution and the intensity of

light as it leaves the solution, which can be written as 𝛿𝐼, as the difference in intensity will be

very small if the value of 𝛿𝑦 is small. As the intensity of light is decreasing, the value of 𝛿𝐼 is

negative and so absorbance can be represented as −𝛿𝐼. (Figure 4)

Figure 4

Therefore, the proportion of light absorbed would be the absorbance compared to the

intensity of light at the beginning which I specified earlier as 𝐼𝑒. ie; −𝛿𝐼

𝐼𝑒

Substituting this into equation 1;

−𝛿𝐼

𝐼𝑒= 𝜎 × 𝑛 × 𝛿𝑦

𝛿𝑦

𝐼𝑒

𝐼𝑒 − 𝛿𝐼

𝛿𝐼

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Applying the equation to the whole sample

I have succeeded in relating the absorbance of a solution with the factors that affect

absorbance. However, the equation is for a very small block of the whole solution. Therefore,

I need to use this equation to make it apply to the whole block of solution.

There will be many blocks of thickness 𝛿𝑦, in the solution which will have a constant value

of 𝜎𝑛 assuming that there is only one solute in the solution and that the solutes are evenly

dispersed in the solvent. However, for each block within the solution, the value of 𝐼𝑒 will be

different; if the block lies closest to the source of light, then the value of 𝐼𝑒 will be the

maximum, whereas if the block lies closer to the other side of the solution, then the value of

𝐼𝑒 will be smaller as light will have been absorbed already by the blocks preceding it. It is

clear that the value of 𝐼𝑒 is dependent on the distance from the source of light. This distance

can be represented with the notation 𝑦 which also link it to the notation 𝛿𝑦 that I used for the

thickness of the block, as 𝛿𝑦 is essentially a small difference in distance (Figure 5).

Therefore, the expression −𝛿𝐼

𝐼𝑒 can be better represented as −

𝛿𝐼

𝐼𝑦 where 𝐼𝑦 means the intensity

of light at distance 𝑦 from the source.:

Figure 5

In order to work out the absorbance of the whole solution and how this relates to the three

factors, I can add up what happens for every single small block. This can be expressed using

mathematical notation, but as each side of the equation has different variables (on the left

hand side I have the proportion of absorbance in terms of intensity and on the right hand side

I have the proportion of absorbance in terms of the thickness of the block) it is easier to sum

the respective sides of the equation separately. The resulting expression from these

summations can be equated later as each are expressing the proportion of absorbance.

Therefore,

𝐼𝑒

𝑦

𝐼𝑦

𝛿𝑦

𝑏

𝐼𝑦 − 𝛿𝐼

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On the left hand side, the expression for the

proportion of absorbance by the small block

was −𝛿𝐼

𝐼𝑦.

On the right hand side, the expression for

the proportion of light being absorbed

which depends on the three factors was

𝜎𝑛𝛿𝑦.

In order to extrapolate what happens within one small block to what happens throughout the

block of solution as a whole, I can sum the effects of one block together for however many

blocks there are. The mathematical expression for a summation is ∑ 𝑥. Therefore;

∑ −𝛿𝐼

𝐼𝑦 ∑ 𝜎𝑛𝛿𝑦

Limits can be added to the above equation.

The variable on this side is the intensity, 𝐼.

This value of 𝐼 starts with the value of 𝐼𝑒 as

this is the intensity of light before it enters the

solution. The intensity of light after it has

gone through the solution, where it is being

detected, is 𝐼𝑏. Therefore, the limits of 𝐼 are

𝐼𝑒 ≤ 𝐼 ≤ 𝐼𝑏, where the value of 𝐼 decreases.

The value of 𝛿𝐼 can theoretically signify any

difference in 𝐼 which is very small. However,

the approximation of this summation will

improve in accuracy as the value of 𝛿𝐼

decreases which is dependent on the value of

𝛿𝑦 on the right hand side decreasing.

Therefore, the more detailed expression using

mathematical notation is;

lim𝛿𝐼→0

∑ −𝛿𝐼

𝐼𝑦

𝐼=𝐼𝑏

𝐼=𝐼𝑒

The variable I am focusing upon on this

side is the value of 𝑦. This value of 𝑦 has a

minimum of 0 and a maximum of 𝑏, as 𝑦

cannot lie beyond the length of the block of

solution. Therefore, the limits of 𝑦 are, 0 ≤

𝑦 ≤ 𝑏.

The value of 𝛿𝑦 can theoretically signify

any difference in 𝑦 which is very small.

However, as the value of 𝛿𝑦 gets smaller,

the approximation for the number and areas

of molecules that absorb the light within

that very thin block will improve.

Therefore, the more detailed expression

using mathematical notation is;

lim𝛿𝑦→0

∑ 𝜎𝑛𝛿𝑦

𝑦=𝑏

𝑦=0

By definition, any summation that tends to zero is equal to an integral as the values in this

case are continuous (meaning that the values can be measured and has no exact numerical

values), ie;

lim𝛿𝑥→0

∑ 𝑓(𝑥)𝛿𝑥

𝑥=∞

𝑥=−∞

= ∫ 𝑓(𝑥)𝑑𝑥∞

−∞

Therefore the above expressions are the equivalent of the following expressions;

∫ − 𝑑𝐼

𝐼𝑦

𝐼𝑏

𝐼𝑒

∫ 𝜎𝑛𝑑𝑦𝑏

0

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These expressions can be clarified for integration;

∫ − 𝑑𝐼

𝐼𝑦

𝐼𝑏

𝐼𝑒

= − ∫ 1

𝐼𝑦

𝐼𝑏

𝐼𝑒

× 𝑑𝐼

∫ 𝜎𝑛𝑑𝑦𝑏

0

= 𝜎𝑛 ∫ 1 × 𝑑𝑦𝑏

0

These can be integrated with their respective methods;

Integrating any value of 1

𝑥 gives ln 𝑥.

Therefore;

− ∫ 1

𝐼𝑦

𝐼𝑏

𝐼𝑒

× 𝑑𝐼 = −[ln 𝐼𝑦]𝐼𝑏

𝐼𝑒

For integration, add 1 to the power and

divide by the power.

Therefore;

𝜎𝑛 ∫ 1 × 𝑑𝑦𝑏

0

= 𝜎𝑛 × [𝑦]𝑏0

Substituting in the limits;

−[ln 𝐼𝑦]𝐼𝑏

𝐼𝑒= −(ln 𝐼𝑏 − ln 𝐼𝑒)

= ln 𝐼𝑒 − ln 𝐼𝑏

By the rules of logs which states that

log(𝑎) − log(𝑏) = log (𝑎

𝑏), this equals to;

ln𝐼𝑒

𝐼𝑏

𝜎𝑛 × [𝑦]𝑏0

= 𝜎𝑛[𝑏 − 0]

= 𝜎𝑛𝑏

Having thus finished the integration for both sides, these can now be equated;

ln𝐼𝑒

𝐼𝑏= 𝜎𝑛𝑏

However, both sides of the equation can be converted into values with units that are more

conventional to use and are used in the Beer-Lambert law.

These are constants and

can be taken out of the

integral

This can be

expressed

as 1 × 𝑑𝑦

The negative

sign can be

taken out of the

integral

This can be

expressed as 1

𝐼𝑦× 𝑑𝐼

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At the moment, the expression I am using for

the proportion of the absorbance is ln𝐼𝑒

𝐼𝑏.

However, absorbance is conventionally

expressed in terms of log base 10 instead of

the natural log.

By making use of the rules of logs which

states that log𝑏 𝑎 = log𝑐 𝑎

log𝑐 𝑏, I can convert from

ln 𝑥 (= log𝑒 𝑥) to log 𝑥 (= log10 𝑥) by

inserting the values 𝑎 = 𝑥, 𝑏 = 𝑒, 𝑐 = 10 into

the rule;

ln x = loge x = log10 x

log10 e

= log10 x × 1

log10 e

= log10 x × 2.303

= 2.303 log x

Substituting in 𝐼𝑒

𝐼𝑏 for 𝑥,

ln𝐼𝑒

𝐼𝑏= 2.303 log

𝐼𝑒

𝐼𝑏

At the moment, the unit is molecules/cm3.

However, it is more conventional to express

concentration with moles/litre as molecules

are very small and the number of molecules

in a solution can get very large.

A mole is a unit, where one mole of a

substance equals to 6.02 x 1023 molecules

of the substance. The number is known as

Avogadro’s constant3. Therefore, to convert

a value from molecules/cm3 to

molecules/litre, the volume can first by

multiplying by 1000 as 1 litre = 1000cm3

and to convert the number of molecules into

moles the value must be divided the

Avogadro’s constant. ie;

𝑐 (moles/litre) = 𝑛 ×1000

6.02 × 1023 = 𝑛

6.02 × 1020

In order to substitute into the expression, n

must be made the subject;

𝑛 = 6.02 × 1020 × 𝑐

This can then be substituted in;

𝜎𝑛𝑏 = 𝑐 × 6.02 × 1020 × 𝜎 × 𝑏

Therefore,

ln𝐼𝑒

𝐼𝑏= 𝜎𝑛𝑏

2.303 log𝐼𝑒

𝐼𝑏 = 𝑐 × 6.02 × 1020 × 𝜎 × 𝑏

log𝐼𝑒

𝐼𝑏=

𝑐 × 6.02 × 1020 × 𝜎 × 𝑏

2.303

For clarification, this can be simplified further by using the notation 𝐴 for log𝐼𝑒

𝐼𝑏 as it is a

measure of absorbance. The part 6.02 × 1020×𝜎

2.303 is a constant specific to the one solute present

in a solution. This constant is known as the molar absorption coefficient and is usually given

the notation 휀.

3 Chemistry Data Booklet, International Baccalaureate Organisation, 2014

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Therefore, the whole equation can be simplified to;

𝐴 = 𝑐휀𝑏

This completes the derivation of the Beer-Lambert law.

In order to work out the concentration of a substance, which is the most frequent use for this

law, the above equation can be rearranged to make 𝑐 the subject of the equation:

𝑐 =𝐴

휀𝑏

Reflection

Whilst I was deriving this law, I had to assume that the molecules in the solution were evenly

dispersed in the solution so that each small block contained the same number of molecules. I

was also aware that if the molecule is not regularly shaped, the cross sectional area of the

molecule which absorbs the light can vary depending on its orientation. For example, if a

molecule was a perfect sphere, the cross sectional area of the molecule which absorbs light

will be equal for any orientation. However, the shape of the molecule is more likely to be

irregular which means that the orientation of the molecule will lead to different cross

sectional areas which absorbs the light; for example, if the shape of the molecule was like a

cuboid where its length is longer than the width and the base that form one surface, then the

area which absorbs light will be smaller if the cuboid lies with the smaller face towards the

light source than if it lies with the larger, rectangular face towards the light source (Figure 6).

These could introduce some inaccuracies in the prediction of the absorbance of a solution.

This law will also not be applicable if the solution fluoresced as it will affect the

measurement of concentration as the solution will absorb and emit light and the amount of

light detected will not necessarily be representative of the number of molecules there are

within that solution.

Figure 6

During my work experience, this law was being used to find out the concentration of a

specific protein in the solution. The protein being investigated was a product of several

methods starting with its biological manufacture using yeast, such as the disruption of the cell

membrane and using a centrifuge to separate out different components of the cell and its

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isolation was difficult. One disadvantage in using this law is the assumption that the solution

is pure and only contains the solute of interest, which is not necessarily guaranteed.

Therefore, although this law is useful in estimating the relative success in the isolation of the

protein in terms of its percentage produce, it is important to note that it can only produce an

estimate due to factors such as contamination by other solutes. In addition, the absorbance of

a protein is particularly difficult to dictate due to the different shapes that one protein can

exist in depending on the environment, such as temperature and pH. Therefore, the Beer-

Lambert law must be used in consideration of these limitations.

There is another difficulty with this law; looking back at the section where I outlined the

factors which affect the absorbance of light, I realised that this law would not be the best for

comparing the absorbance of solutes which absorb different wavelengths of light. Therefore,

this law requires some prior information regarding the solute such as the wavelength of light

that it absorbs and how much of that wavelength one molecule absorbs, in order to be able to

work out its concentration by measuring its absorbance. Upon realising this, I tried looking

up some literature values to be used with the Beer-Lambert law and came across the

following table;

Near-ultraviolet absorption bands of some amino acids and nucleotides4

Substance pH max(nm) (liter/cm*mole)

Phenylalanine 6 257 200

Tyrosine 6 275 1,300

Tryptophan 6 280 5,000

Adenosine-5'-phosphate 7 259 15,000

Cytidine-5'-phoosphate 7 271 9,000

Uridine-5'-phosphate 7 262 10,000

Guanosine-5'-phosphate 7 252 14,000

Reference to values such as those found in this table – pH, the wavelength of light ( and the

value of 휀, would improve the accuracy of the Beer-Lambert law and eradicates some of the

problems I have mentioned.

Conclusion

Doing this exploration has enabled me to fully comprehend the factors that each part of the

equation represents. This will enable me to consider the limitations of this law if I come to

use it later when I go to university.

4 http://life.nthu.edu.tw/~labcjw/BioPhyChem/UV/UV_absorption.htm (accessed 20/11/15)

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Bibliography

Adds, J. et al., Tools, Techniques and Assessment in Biology, Thomas Nelson & Sons Ltd.,

UK, 1999

Chemistry Data Booklet, International Baccalaureate Organisation, 2014

Thermo Scientific, NanoDrop 2000/2000c Spectrophotometer, V1.0 User Manual, Thermo

Fisher Scientific Inc., USA, 2009

http://life.nthu.edu.tw/~labcjw/BioPhyChem/UV/UV_absorption.htm (accessed 20/11/15)

Glossary of notations used:

𝐼𝑒 = initial intensity of light as it enters the solution

𝑏 = total length of the solution (distance of the path that the light travels)

𝐼𝑏 = intensity of light after travelling through the whole solution

𝑦 = distance from the source of light

𝛿𝑦 = infinitesimal thickness of one block

𝑛 = number of molecules

𝑠 = cross sectional area of the block of solution

𝜎 = cross sectional area of one molecule

𝛿𝐼 = difference in intensity of light after it has gone through a block of solution with

thickness 𝛿𝑦

𝐼𝑦 = intensity of light at distance 𝑦 from the source

c = concentration in moles/litre

𝐴 = absorbance

휀 = molar absorptivity coefficient