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?

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.

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?

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.

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 (K_max) of the fastest electron ejected from the cathode-

 

K_max = 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-

• 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-

• 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-

 

K_max = 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 K_max = 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 K_max = 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 (E_g). 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 V_D 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.

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

 

If V ≈ V_D 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 = E_g          (10)

 

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

 

eV_D = E_g          (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 eV_D for Eg in equation (10)-

 

hf = eV_D          (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

 

eV_threshold = E_light          (13)

 

where V_threshold is the activation voltage and E_light = hf

 

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

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), K_max = 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.

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.

• 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.

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.

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.