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This is a research paper that is apart of the international baccalaureate about Measuring of the thickness of a thin film of graph in physics course.

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Physics Investigation
Measuring of the thickness of a thin film of graphite
Research Question

The objective of this investigation was to calculate the thickness of a thin film of a conducting lead of

in a pencil indirectly by measuring resistance and resistivity, and to calculate how the thickness of the

film is dependent on the resistance to calculate the number of carbon atoms in the lead layer.


Introduction

When I visited calligraphy museum, I was admired by the artworks made by pencils. The pencil has

been with us since our childhood. I feel that it has a secret behind this efficient magical work and a lot

of secrets. When we sharpen it or sometimes we use it with a shattered lead, we enjoy what we write

with it. During physics class we have studied that graphite are semi-conductor’s material so I decided

to relate this concept to what I studied. Therefore, I interlinked this concept with resistance and I was

curious to investigate the relationship of resistance and volume; specifically, the width and the length

to find the thickness.


Pencils are such an essential part of our life that has a lot of application also the thin film specifically

were it is used in sensors, diffusion barrels, and many filed in optics however the thin film has to

miniscule thickness that should be made on a surface of an object this means that they are indirect

substrate and should be applied on a material. However, this measurement is not possible way because

applying the thin film on a paper for instance will make the caliper to measure the dimensions of the

paper not the film. So the application of this investigation is to show the cheaper and easier way to

measure the thickness of a thin lead by a caliper since microscopy is expensive.


Background Information

The lead is made of graphite that means the line drawn into the paper will also conduct to form a film.

However, there are unknown variable that has to be interpreted using derived equations, such as

resistivity which is the specific electrical resistance of a surface that from it we can derive our

requirement measurement which is the thickness.

, From ohms’ law formula: R = V/I we can expand the formula to be R = ρL/A and substitute resistivity

(ρ) to be determined by the following equation: ρ = R A/L.


The area of a cross section of a lead is spherical so the formula will be: ρ = R π r2 / L


Where:


ρ: Resistivity constant of the material (Ω.m)

L: Length of the wire (m)

A: Cross sectional area (m2)

R: Resistance (Ω)


For the resistance that is calculated by changing one variable for instance length will lead in changing

another variable that is width to be alter.


R = pL/Wt


To illustrate the equation that the length acts as a directly proportional to the resister whereas the

length increase the resistance increase, while the width is inversely proportion to the resistance, and

by this equation all the variables will be known that helps inn finding the thickness of film.


It can be rewritten as: t = pL/WR


Also to eliminate the errors a graph between resistance and length could be plotted. The more the

linear the ideal the calculation was since the plot can also be used to calculate thickness by dividing

the resistivity value by the slope multiplied by the width will give more precise data than taking the

average for different lengths. This is because in plotting a table you can eliminate the points that are

of limits that decrease accuracy and precision.


The following formula shows the relation: Slope = p/wt so, t = p/slope.w

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Uploaded on
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Number of pages
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Written in
2019/2020
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SUMMARY

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