Photoelectric Effect In Solar Panels
- Page ID
- 418927
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I.D.1.a - The interaction of photons with molecules and atoms provides information about energy levels.
- X.E.1. - Rearrangement of common mathematical expressions can either isolate variables or identify relationships that enhance understanding of the system being described.
Context:
Climate change is one of the biggest issues putting the well-being of humanity at risk as global warming and abnormal weather patterns are disrupting our ways of life (National Climate Assessment, 2018). As noted in NASA’s climate change efforts, human activity leading to pollution and greenhouse gas emissions are significant historical factors contributing to the state of our environment today, but now we face many challenges in our efforts to adapt and tackle this problem (Shaftel, 2022). Along with industry, transportation, and agriculture, our use of electric power has been linked with the rise in human-made emissions of green-house gasses that have become trapped in the atmosphere, thus slowing down heat loss into space while increasing average temperatures (Environmental Protection Agency, 2022).
Figure 1: Solar Panels on Roofs
Many are now searching for clean, sustainable sources of energy to power our electric grid including wind, hydro, geothermal and solar power (Energy Information Administration, 2022). Rather than traditional, carbon producing sources such as fossil fuels, solar panels are notably becoming common sources of electricity in the residential and commercial sectors. It’s a common sight to see this technology on the roofs of city buildings or covered parking, but what is it that allows us to capture energy from the sun?
Background Chemistry:
The Photoelectric Effect describes the process that occurs when a light shone on metal causes the ejection of an electron from the metal’s surface. The energy required to eject an electron is based on the threshold frequency (ν) and Planck’s constant (h). The threshold frequency is the minimum frequency of light that is required to eject an electron from the surface of the metal; regardless of the intensity of the light, if the frequency is too low, no electrons will be emitted. The following relationship summarizes the relationship between energy (E) and frequency:
Energy Required to Remove an Electron
E = hν
E = Energy required to remove an electron
h = Planck’s Constant \((6.626 \text{x} 10^{-34} \ m^2 kg/s) \)
ν = Frequency
Based on the relationship between frequency and wavelength, the equation can also be rewritten in the form below, where energy is inversely proportional to wavelength:
E = \(h\frac{c}{λ}\)
c = Speed of Light (2.998 x \(10^8\) m/s)
λ = Wavelength
The photoelectric effect helps to describe the wave-particle duality of light as the equation reveals that there is no relationship between the intensity of light and the energy of an emitted electron. Increasing light intensity will lead to more electrons being emitted, but it has no effect on the kinetic energies of those emitted electrons. Additionally, the concept of threshold frequency displays the quantized energy of photoelectrons as light below the threshold frequency simply won’t cause electrons to be ejected. However, at frequencies above the threshold frequency of light, kinetic energy is proportional to the frequency of light irradiating the metal.
Figure 2: Diagram of the Photoelectric Effect
Image Credit: Wikimedia Commons
The kinetic energy of an emitted electron is based on the minimum energy required to remove an electron and the work function (Φ). The work function is the energy lost by an ejected electron as it is used to remove the electron from the surface of the metal. Thus, the kinetic energy of an emitted photoelectron can be described by the following equation:
KE = hν - Φ
Applications of the Photoelectric Effect:
One real-world application of the photoelectric effect is in solar panels; solar panels harness energy from the sun to create energy that can power solar heating, solar electricity, and solar lighting. To convert sunlight into usable energy, photovoltaic cells (solar cells) are used; photovoltaic technology utilizes the principles of the photoelectric effect to capture free electrons and convert their movement into the current.
Photovoltaic cells are made up of a semiconductor plate that has an electric field that is positive on one side of the semiconductor and negative on the other side; these semiconductors are usually made from silicon. When light strikes the semiconductor material of the photovoltaic cells, electrons are knocked out from the semiconductor and become loose; these electrons are captured by conductors that form an electrical circuit. An electric current is thus generated through the flow of emitted electrons; this current can power circuits and generate electricity.
Silicon is a common material used in the semiconductor surfaces of solar cells; the work function of silicon is 4.85 eV. When sunlight strikes the surface of a solar panel, an electron is ejected and then captured by the conductor to produce an electric current.
A solar panel has been irradiated with sunlight with a wavelength of 374 nm. Calculate the kinetic energy of an electron in the split second between when it is emitted from a silicon semiconductor and when it is captured by the conductor material.
Solution
Step 1: Convert work function from eV to Joules
4.85 eV x \(\frac{ 1 Joule}{6.242 \text{x} 10^{18} eV} = 7.7705 \text{x} 10^{-19} J\)
Step 2: Convert from nanometers to meters
\( 125 \text{nm x} \frac{1 m}{ 10^ 9 nm} = 1.25 \text{x} 10^{-7}m\)
Step 3: Calculated the energy supplied to remove the electron
E = \(h\frac{c}{λ}\)= \(\frac{(6.626 \text{x} 10^{-34}m^2/kg s)(2.998 \text{x} 10^8 m/s)}{1.25 \text{x} 10^{-7} m} = 1.5892 \text{x} 10^{-18} J \)
Step 4: Calculate Kinetic Energy
KE = \(h\frac{c}{λ} -Φ\)= \(1.5892 \text{x} 10^{-18}J - 7.7705 \text{x} 10^{-19} J = 8.1213 \text{x} 10^{-19}J\)
When sunlight strikes the surface of solar panels, the metal ejects an electron which is captured by the conductor material to produce a current, allowing us to harness the energy. Calculate how much energy the solar panel requires to eject an electron if it has a de Broglie wavelength of 7.21 × 10-9 m when irradiated with a light beam of 374 nm from the sun.
Solution
\(𝛌_{dB}\) = 7.21 × \(10^{-9}\) m
𝛌 = 374 nm
\(𝛌_{dB}\) = \(\frac{h}{mv}\)
7.21 × \(10^{-9}\) m = \( \frac{6.626 \text{×} 10^{-34} J/s} {9.11 \text{×} 10^{-31} kg ⋅ v}\)
v = \(\frac{6.626 × 10^{-34} J/s} {9.11 × 10^{-31} kg ⋅ 7.21 × 10^{-9} m}\)
v = 100,878.308119 m/s
E = \(h\frac{c}{𝛌}\)
E = \(\frac{6.626 \text{×} 10^{-34} J/s ⋅ 2.998 \text{×} 10^8 m/s} {374 nm ⋅ 1 m/1 \text{×} 10^9 nm}\)
E = 5.311429 × \(10^{-19}\) J
E = KE + Φ
Φ = E - ½m\(v^2\)
Φ = (5.311429 × \(10^{-19}\) J) - ½ ⋅ (9.11 × \(10^{-31}\) kg) ⋅ (100,878.308119 m/s)\(^2\)
Φ = 5.265075 × \(10^{-19}\) J
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Image Credit:
Figure 1: https://www.flickr.com/photos/400774...00/49534575181
Figure 2: https://commons.wikimedia.org/wiki/F..._-_diagram.svg