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Physics SL
Sample Extended Essays

Table of content
Introduction
Research Question
Theoretical Framework
Hypothesis
Methodology
Results tables
Data Analysis
Conclusion
Evaluation
Glossary
References

# How does activation voltage depend on the wavelength of LED radiation measured in the range 400 to 800 nm and how can this information be used to measure Planck’s constant?

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## Introduction

This essay focuses on using LEDs (Light Emitting Diodes) to explore the photoelectric effect, a physical phenomenon by which if the surface of an electrode is struck by light of sufficient frequency, an electron will absorb a photon and be ejected from that surface.

The scope of this experiment is to investigate the relationship between activation voltage, which is the minimum voltage required for the LED to light up, and the wavelength of various LED radiations, and, using this information, to get an experimental measurement of Planck’s constant (h). The experiment involves two methods: for the first, the activation voltage of the LEDs will be determined by performing the experiment in a dark room and looking at when the LEDs cease to emit light; the second entails setting up a circuit with a voltmeter connected in parallel and an ammeter connected in series to a variable resistor and a LED. The current will then be decreased until it is equal to 0 by increasing the resistance of the circuit, and the voltage across the LED when this happens (known as the activation voltage) will be recorded. This will then be extended to include a range of wavelengths spanning from Infrared to blue radiation.

I was quite sure that the topic of my essay would be related to Quantum Physics, a branch of Physics which I find interesting and peculiar because it concerns the extremely small. In the end, I opted for an experimental essay on the photoelectric effect because I am fascinated by its implications, which truly revolutionised the whole of classical Physics, and I would like to see it myself. Specifically, I chose to focus on a method using LEDs rather than a classical experiment using a cathode and an anode because I find that this method sheds more light on the photoelectric effect by looking at it from the perspective of the activation voltage rather than looking at well-known results involving the frequency of light. I believe this essay is relevant to me because it allows for an exploration of a field of Physics which is still unknown to me and deepens my knowledge of Physics by delving into a cutting-edge topic such as Quantum Physics.

## Research Question

How does activation voltage depend on the wavelength of LED radiation measured in the range 400 to 800 nm and how can this information be used to measure Planck’s constant?

## Theoretical Framework

Between 1886 and 1887 Heinrich Hertz discovered that an electric discharge between two electrodes has a higher chance of occurring if ultraviolet light falls on one of the electrodes. Lenard discovered that this is because it facilitates the discharge by causing electrons to be ejected from the cathode’s surface. The emission of electrons from a surface, due to light, is called the photoelectric effect.

Figure 1 - Shows A Classical Experiment Used To Investigate The Photoelectric Effect

A value V0 , the stopping voltage or potential, is reached when the photoelectric current drops to 0. This is because when this happens, electrons have the minimum energy needed to be ejected from the cathode, barely reach the collecting plate and then accelerate backwards towards the cathode. Hence, there is no current. The stopping voltage multiplied by the charge on an electron gives a measurement of the kinetic energy (Kmax) of the fastest electron ejected from the cathode-

Kmax = eV0          (1)

Albert Einstein found that the photoelectric effect revolutionised physics, because there are three main aspects of it which cannot be explained in terms of the classical wave theory of light-

• Although there exists a linear relationship between intensity of light and electrons emitted, according to wave theory, the kinetic energy of the electrons should increase as the light beam gains intensity, however the graph below shows what emerged from the photoelectric effect-
Figure 2 - shows that the activation voltage and hence the kinetic energy depends on the frequency and not the intensity. A greater frequency leads to a greater magnitude of the voltage.
• Wave theory requires that the photoelectric effect should occur at any frequency of light (assuming the light is intense enough to give the energy needed for the electrons to escape). However, the diagram below shows how frequencies less than f0 do not allow the photoelectric effect to occur-

Figure 3 - shows the relationship between activation voltage and frequency. It can be seen that they are proportional to each other. There is no voltage for frequencies below f0.

• If the energy absorbed by a photoelectron comes from a wave incident on the metal plate where the electron is found, classical theory predicts that an incident wave would have light energy uniformly distributed over its wave front. Therefore, there should be a time lag between when light strikes the metal surface and when the photoelectron is emitted, equivalent to the time taken by the electron to absorb the photon, but this time lag has been detected to be consistent with 0 s.

In 1905, Einstein proposed that electromagnetic or radiant energy, once radiated, was quantised into “concentrated bundles”, which we now call photons. Einstein assumed that this bundle of energy is initially located in a small volume of space, and that, as it moves away from the source with velocity c, it stays localised. Another assumption made by Einstein was that during the photoelectric effect one photon is completely absorbed by one electron in the cathode.

Planck, however, was the first to assume that the energy, E, of the photon (bundle) is related to its frequency, f, by the equation-

E = hf          (2)

When one electron is ejected from the surface of a metal, its kinetic energy K is given by-

K = hf - w          (3)

where hf gives the energy of the absorbed photon and w is the work needed to force the electron to be ejected from the metal. Work is required to overcome the attraction felt by the electron to the other atoms on the surface of the metal. Moreover, there are losses of kinetic energy to be considered. If there are no internal losses of energy involved in the ejection of the electron, that is, the electrons do not collide with other particles, losing some energy as thermal energy, then the kinetic energy is given by-

Kmax = hf - w0          (4)

where w0 is the work function of the metal, which is the minimum energy required for an electron to pass through the metal’s surface and escape the attractive forces binding it to the metal.

Equation (4) shows that if Kmax = 0, the electrons have just enough energy to be ejected from the metal, and we have-

hf0 = w0          (5)

This shows that a photon, having frequency f0, would have the minimum energy required for the electrons to be ejected. This is further evidence that if the frequency is reduced below f0, regardless of the illumination, the photons will not eject any electron.

Equation (4) can be rewritten using a substitution Kmax = eV0, which comes from equation (1). This gives-

eV0 = hf - w0          (6)

Einstein’s equation can be rearranged further, using the wave equation $$f=\frac{c}{\lambda}$$ , into-

$$eV_0=\frac{hc}{\lambda}-\frac{hc}{\lambda_0}$$          (7)

where λ0 is the minimum wavelength needed for a photon to eject an electron.

This is related to the threshold frequency, which is the minimum frequency of light for an electron to be ejected from a metal. After dividing by e, we get-

$$V_0=\frac{hc}{e}\frac{1}{\lambda}-\frac{hc}{e\lambda_0}$$          (8)

Which is in the form y = mx + c. A graph of V0 against $$\frac{1}{\lambda}$$ will therefore have a gradient equal to $$\frac{hc}{e}$$ and a y-intercept of $$-\frac{hc}{e\lambda_0}$$.

There are, however, other methods for conducting the photoelectric experiment, using semiconducting devices known as LEDs. LEDs are made from a thin layer of doped semiconductor material (in this experiment GaP) and when forward biased emit light at a specific wavelength, converting electrical energy into light energy. Lossev experimented on electro-luminescence effects and concluded that they represented the inverse of the photoelectric effect – a current producing light rather than light producing a current.

In 1951, Welker combined equiatomic quantities of an element from the third group and one element from the fifth group of the Periodic Table, creating compounds which showed a wide range of band gaps (Eg). The frequency f of these compounds is given by

$$f=\frac{E_g}{h}$$          (9)

Most of these frequencies correspond to wavelengths in the visible region of light. Today, most of these compounds are used as LEDs.

LEDs consist of a p-n junction diode. If no external voltage is applied, a voltage VD is generated in the area between the n- and p- type materials, known as a depletion layer. This layer prevents electrons from leaving their region and entering the other region.

Figure - 4 Shows A Diagram Of The p-n Junction Of An LED And The Depletion Zone (‘space charge region’)

If an external voltage V is applied, the layer between the n- and p- regions reduces to e(VDV).

If V VD the barrier’s thickness is almost 0 and electrons can flow from one side to the other. When electrons are injected into the depletion region, they eject a photon of energy

hf = Eg          (10)

For doped semiconductors, such as the materials used in my LEDs,

eVD = Eg          (11)

Assuming that all the energy of the electrons in the depletion region is converted into light, the frequency of this light is given by equation (10). Therefore, substituting eVD for Eg in equation (10)-

hf = eVD          (12)

Or, equivalently, the photoelectric effect can be demonstrated with LEDs because the energy of the ejected photons is related to the activation voltage by the equation

eVthreshold = Elight          (13)

where Vthreshold is the activation voltage and Elight = hf

This shows that VDVthreshold, which is the activation voltage. Planck’s constant can then be measured, since the wavelength of light emitted is known.

## Hypothesis

I think that as the independent variable, the wavelength of LED radiation, increases, the dependent variable, the activation voltage, will decrease, hence I predict the two are strongly negatively correlated. Their inverse proportionality can be inferred from the equation E = hf, where E is the radiant energy, f is the frequency of radiation and h is the constant of proportionality between the two which is nowadays known as Planck’s constant, which sets the scale for these kinds of experiments, concerned with low energies and phenomena at a microscopic scale. The radiant energy is equivalent to the kinetic energy; from equation (1), Kmax = eV0. Then, $$eV_0=\frac{hc}{\lambda}$$ which shows that activation voltage and wavelength have an inverse relationship.

However, these variables will not be inversely proportional. This is because the best-fit line will not pass through the origin since the work function of the LEDs will determine the y–intercept of the graph of V against $$\frac{1}{\lambda}$$ . I predict that the value for the y–intercept will be small and negative because looking at the y–intercept (the constants $$-\frac{hc}{e\lambda_0}$$), I can see that their orders of magnitude approximately cancel out, hence the y–intercept should be close to 1, indicating a very strong negative correlation between the two variables and excluding the possibility of an indirect proportionality.

I also predict that the activation voltage for the blue LED will be significantly higher because it has quite a small wavelength compared with the other three. Similarly, I also think that the infrared LED will have quite a small activation voltage because it has a substantially higher wavelength than the other LEDs.

## Methodology

### Variables

Independent variable- the wavelength of LED radiation, which changes for every different LED. The values for the wavelength of the LED radiation are obtained from the data sheets of Arduino, Shenzhen Jingyuanxing Photoelectric Co., Ltd and Kingbright, the companies from which the LEDs used in this experiment come from. The value for the LEDs used is “dominant wavelength”; this is because it is the wavelength which most probably is most similar to the wavelength of the light emitted by the LEDs in the experiment. The absolute uncertainties on the wavelength vary from LED to LED, and are as follows-

$$\Delta\lambda_{blue}=±5nm$$

$$\Delta\lambda_{red}=±5nm$$

$$\Delta_{green}=±10 nm$$

$$\Delta\lambda_{yellow}=±5nm$$

Dependent variable- the activation voltage, measured by eye and by using a voltmeter, in two different methods. The first method involves seeing when the LED stops emitting light and is enhanced by the use of a dark room. The second method involves measuring the activation voltage by looking at when the current through the LED reaches 0 (at this point the LED will have stopped emitting light). The absolute uncertainty on the voltmeter is V = ±0.001 V; the absolute uncertainty in the current measured by the ammeter is $$\Delta I=±0.05A$$.

Control variables

• Use the same LEDs throughout the trials, because different LEDs might have different uncertainties on their wavelength and give inconsistent results. The red, green and blue LEDs come from Shenzhen Jingyuanxing Photoelectric Co., Ltd.; the yellow LED comes from Kingbright Electronics.
• The materials of the LEDs have to be kept the same because on the material depends the amount of colour emitted by the LEDs, therefore a different intensity of colour can affect the activation voltage I perceive through my eyes. The LEDs were made out of Gallium Phosphide.
• Maintain the voltage constant in order to avoid changes in current which make the results inconsistent. The voltage will be kept at 8V. This not the activation voltage, but rather the voltage from the voltage source.
• Use the same 220Ω resistor throughout the trials of both methods because a resistor of different value may limit a greater or lesser amount of current through the circuit.

### Apparatus

• LEDs of various wavelengths: red, yellow, green, blue, infrared. LEDs of different kinds are needed because each one has a different activation voltage;
• D.C. power supply to supply 8V to the circuit. 8V have been chosen because they provide enough voltage to clearly see the light from the LED without incurring into the risk of being too high and damaging the LEDs;
• Ammeter used to measure the current through the diode. The ammeter was able to measure the current through the LED to various degrees: the degree in the range <2mA was chosen because it was the most appropriate for the experiment given the very low currents involved;
• Voltmeter used to measure the activation voltage. The voltmeter was able to measure the potential difference across the LED to various degrees: the degree in the range <10V was chosen because it was the most appropriate for the experiment given that the voltage of the sources was 8V;
• A 220Ω resistor used to limit the current through the LED, added to the circuit because otherwise the LED can be damaged or melt with a too high current passing through it;
• Potentiometer with a 10000Ω resistance which will be used to increase resistance until the current is equal to 0;
• Lab coats, used to set up the dark room even better, blocking light passing from under the doors.

### Diagram

Figure 5 - Shows An Electronic Schematic Of The Circuit Used In The Experiment

### Procedure

Method 1– obtaining the activation voltage in the dark

• Set up the circuit as shown in fig. 5 above
• Conduct the experiment in a very dark room. This experiment involves obtaining a value for the activation voltage by determining when the LED first stops emitting light, and this can be better seen in a dark room.
• Remain in a dark room for at least 10 minutes before starting the experiment, to get used to the dark.
• Vary the resistance across the circuit using the slider on the potentiometer. The resistance
must be increased until the current through the LED reaches 0 – i.e. when the LED does
not emit any light.
• When the LED stops emitting light, record the voltage across the LED as measured by the voltmeter.
• Conduct 5 trials for each LED.
• Between each trial turn off the power pack, then decrease the resistance to have a sizeable current through the LED. Then turn on the power supply and repeat steps 4–5

Method 2– obtaining the activation voltage using an ammeter and voltmeter in the light

• Set up the circuit as shown in the fig. 5 overleaf
• As in method 1, the resistance has to be increased until the current reaches 0.
• Assess when the current is 0 by using the ammeter. It is important to be looking at the ammeter from directly above because this avoids parallax errors in measurement.
• Repeat steps 5–6–7 from method 1
• Repeat steps 2–3.
Hazard
Risk
Control
Electricity
Getting a shock
Handle apparatus carefully. Turn off the power supply in between trials.
Heat from the wires and power supply
Getting burnt.
Turn off the power supply during trials so as to not let it heat up. Do not touch the wires throughout the experiment.
Light from the LED
Damage to the eye
Especially when in the dark room, do not stare at the LED closely nor directly. Stare at the LED only when the intensity of the light being emitted has been reduced.
Water, especially if in contact with electricity
Electrocution
Ensure that any sources of water are far away from the electrical equipment. If water is on the floor, dry the floor using paper towels.
Figure 6 - Risk Assessment

## Results tables

Activation voltage (V) [V0] ∆V0 = ±0.001 V

630
0.938
630
0.969
630
0.894
630
0.914
630
0.918

Average activation voltage= 0.93±0.04V

588
0.952
588
0.944
588
0.987
588
0.956
588
0.933

Average activation voltage= 0.95±0.03V

570
1.020
570
0.975
570
0.984
570
0.949
570
0.940

Average activation voltage= 0.97±0.04V

460
1.332
460
1.311
460
1.317
460
1.280
460
1.303

Average activation voltage= 1.31±0.03V

Figure 7 - Method 1 – In The Dark Room

Activation voltage (V) [V0] ∆V0 = ±0.001V

630
1.026
630
1.029
630
1.021
630
1.026
630
1.025

Average activation voltage= 1.025±0.004V

588
1.097
588
1.099
588
1.096
588
1.099
588
1.105

Average activation voltage= 1.099±0.004V

570
1.148
570
1.150
570
1.142
570
1.149
570
1.143

Average activation voltage= 1.146±0.003V

460
1.433
460
1.425
460
1.405
460
1.436
460
1.422

Average activation voltage= 1.42±0.02V

790
0.620
790
0.617
790
0.612
790
0.610
790
0.617

Average activation voltage= 0.615±0.005V

Figure 8 - Table on Method 2 – in the light

## Graphs

Figure 9 - Method 1 – In The Dark Room
Figure 10 - Method 2 – In The Light With The Addition Of The Infrared LED

## Data Analysis

### Error propagation and the experimental value of Planck’s constant

The absolute uncertainty on the voltage was found by considering the number of decimal places with which the voltmeter delivered recordings for the voltage across the LED, which was three. The uncertainty lies on the last digit, therefore ∆V = ±0.001 V.

The absolute uncertainty on the wavelength of each LED is variable, and was obtained by considering the range of dominant wavelength in the data sheet of the LEDs and dividing it by two, so for example for the red LED: $$\Delta\lambda=\frac{635-625}{2}=±5nm$$

The uncertainty in the current is I = ±0.05 A, obtained by taking the smallest measurement of the analogue ammeter (0.10 A) and dividing it by two.

The uncertainty on the gradient for experiment a was calculated using the formula-

$$\Delta m=\frac{m_{max}-m_{min}}{2}$$

where m ≡ $$\frac{hc}{e}$$

Substituting the values for method 1-

$$\Delta m=±100.10^{-9}m kgs^{-3}A^{-1}$$

Therefore-

m = (700 ± 100).10-9 m kgs-3 A-1

Similarly, for method 2-

$$\Delta m =±90.10^{-9}mkgs^{-3}A^{-1}$$

Thus-

m = (900 ± 90).10-9m kgs-3A-1

The uncertainty on the y–intercept is obtained in a similar way-

$$\Delta c=\frac{c_{max}-c_{min}}{2}$$

For method 1-

$$\Delta\frac{hc}{e\lambda _0}=±0.2kgm^2s^{-3}A^{-1}$$

Therefore-

$$\frac{hc}{e\lambda_0}=(-0.2±0.2)kgm^2s^{-3}A^{-1}$$

For method 2-

h = 4.10-34kgm2s-1

The uncertainty on Planck’s constant is calculated by considering the absolute uncertainties and absolute values for the speed of light in a vacuum, the electronic charge and for the gradient of the graphs, which corresponds to $$\frac{hc}{e}$$

The formula used for this is: $$\Delta h=h\sqrt{(\frac{\Delta e}{e})^2+(\frac{\Delta m}{m})^2}$$

The term $$(\frac{\Delta c}{c})^2$$ is not included because the value for the speed of light in a vacuum is exact, as it is defined in terms of the second (s) and has no uncertainty.

For method 1-

$$\Delta h =\pm0.9.10^{-34}kgm^2s^{-1}$$

For method 2-

$$\Delta h=\pm0.6.10^{-34}kgm^2s^{-6}$$

Hence, the experimental values obtained for Planck’s constant are-

h = (4 ± 0.9).10-34kgm2s-1

for the method in the dark, and

h = (5 ± 0.6).10-34kgm2s-1

for the method in the light.

## Conclusion

The first relationship which can be established from both the data and the graphs is that as the wavelength of the radiation emitted by the LED increases the activation voltage decreases. This is equivalent to saying that the activation voltage increases as the reciprocal of the wavelength increases. It can also be noticed that values for the activation voltage for Infrared light and blue light are quite smaller and bigger respectively compared with the values for red, yellow and green light. This is due to the fact that the wavelengths of these two LEDs were significantly higher and lower respectively when compared to the wavelengths of the other LEDs. Taking a closer look at the graphs shows that the activation voltage and the reciprocal of the wavelength are not directly proportional. This is because it is not possible to fit a straight line through all error bars and through the origin.

From the data analysis above, it can be seen that the values obtained for Planck’s constant for both methods are at more than two experimental uncertainties away from the accepted value. These are quite big errors, which show that although it is possible to combine the gradient of the graph of V0 against $$\frac{1}{\lambda}$$ with the constants e and c, it is not clear whether this method ensures great accuracy when determining Planck’s constant; this is further reinforced by the fact that the uncertainties on the gradients of both graphs are quite large, therefore the experimental value of Planck’s constant could be much closer or much farther from the accepted value.

## Evaluation

I believe that the background information I collected has been very useful in guiding me and in enhancing my understanding of the experiment. This is because not only was I able to fully derive relevant equations for my experiment, but I was also helped by them– for example, in predicting the relationship between my variables, and in knowing the physical meaning of the quantities in the y − intercept and slope of the graphs I used. I think that the high quality of the sources I used comes from their purpose, which is to provide guidance to students on the photoelectric effect, and their origin, since they come from people such as scientists who I consider to be authorities in their field, because they have published or have links to a lot of research on this topic.

The above analysis strongly suggests that my hypothesis was correct; I was able to correctly predict the relationship and correlation between my variables. However, I feel like to make an even stronger hypothesis I could have used a more convincing mathematical argument in support of the strong negative correlation between activation voltage and wavelength.

The procedure I followed also has its merits. First of all, having two different methods of determining activation voltage was an advantage because it allowed for comparison of the two methods and shed more light on the errors involved in this kind of experiment. Moreover, I believe that taking five trials for each LED improved my method because it allowed for a lot of data to be used to produce graphs and it reduced random errors. Other strong points of the method I used include the fact that it was very safe, and I didn’t find any risks for my safety throughout the experiment.

Nevertheless, the methods used suffered from a lot of weaknesses. First of all, something which must be considered is that there is some ambiguity in this experiment. It was chosen to take the activation voltage as the point at which the LED starts emitting light. However, the activation voltage can also be defined in terms of the I–V curve of a LED as when the current starts increasing exponentially.

It can be argued that both methods were unsuccessful, for various reasons; first of all, the ammeter and voltmeter were subject to systematic error, because sometimes even though there was no voltage applied to the circuit they were still measuring a current/voltage– this means my results are slightly skewed towards higher values. This error was compensated for by first recording the measurements of the voltage, and then adding on an amount equal to the absolute uncertainty on the voltmeter, which has eliminated this error from the experiment.

Other errors come from my own inability to fully concentrate: for example, in the dark room, the values for the activation voltage are certainly affected by some degree of random uncertainty, since it is not possible for my eye to always see the LED fade away at the same exact voltage. Unfortunately, it is very difficult to compensate for this type of random error. Furthermore, a limitation in this method is that, even in the dark room, some light was very faintly passing through the jackets used to cover the bottom of the doors, which impacts on the activation voltage in such a way that when there is light, I perceive the LED turning off faster than I would in complete darkness, hence this skewed my values to higher values.

For the method in the light, although great care was taken in trying to avoid parallax errors as far as possible, it is still subject to random uncertainty because estimating when the current is 0 using an analogue device entailed that the ‘hand’ of the ammeter would be in different positions; I believe that a digital ammeter would have probably been better for this experiment, because the measurement taken using an analogue ammeter is subject to both its inherent uncertainty as well as the human errors arising from my inability to see exactly when the ‘hand’ was indicating 0 current.

A further experiment to extend this method could involve two steps: firstly, the I–V characteristics of the LEDs used would have to be determined. This involves a graph which initially roughly approximates a straight line and then, for higher values of voltage, looks like an exponential graph. Then, a plot of ln(I) against V could be used to turn it into a straight-line graph. The x– intercept would then be used as the value for the activation voltage.

Something which could also be interesting to explore would be the work function of the LEDs. Several LEDs made out of different materials could be used in order to compare a range of work functions, for example to explore which materials are best for cathode ray tubes.

## Glossary

Activation voltage

• for this experiment, it is defined as the minimum voltage at which the LED will first start emitting light.

Anode

• a positively charged electrode through which current enters in an electrical device.

Band gap

• the energy difference between the top and the bottom bands in a semiconductor, where electrons reside

c

• the speed of light in a vacuum

Cathode

• a negatively charged electrode from which current leaves an electrical device.

Doped semiconductor

• a lattice of Germanium or Gallium to which has been added a small percentage of impurities

e

• the elementary charge

Electrode

• an electrical conductor kept in contact with a non-metallic part of a circuit

Forward biased

• an LED in which electrons are allowed to flow, thus creating a current

h

• Planck’s constant

Intensity

• power of radiation per unit area.

Kinetic energy

• the energy a body has due to its motion.

LED

• Light-Emitting Diode, an electrical device.

n-type semiconductor

• a semiconductor to which have been added impurities such as Antimony, which donate free electrons, making this semiconductor very conductive

p-n junction

• the joint interface created when a n-type semiconductor and a p-type semiconductor come into contact

Photoelectron

• an electron that has been ejected from the surface of a metal.

Potential difference

• the work done per unit charge in moving a point positive charge between two points.

Quantised energy

• energy that can take values in a form which is discrete rather than continuous.

• energy resulting from electromagnetic waves.

Semiconductor

• Materials with properties, notably electrical conductivity between those of a conductor and an insulator. Their conductivity can be changed by introducing impurities in their structure.

Threshold frequency

• the minimum frequency of light needed in order for an electron to be ejected from the surface of a metal.

Wavelength

• the length of a full wave.

Work function

the minimum amount of energy that must be supplied for an electron to escape the surface of a metal.

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