Have you ever gone to a restaurant and looked at the red open neon sign to make sure you could go in? If so, you’ve come across the photoelectric effect in action. The bright neon signs you see displayed to attract our attention are powered by neon, a noble gas, that glows specific colors when electricity is passed through it. This electricity provides energy for electron excitation resulting in light production.
Figure 1: An example of a neon sign you might see when passing by a restaurant, shop, or even a store. Daniela Lin, Study Smarter Originals.
So, without further ado, let's dive into the photoelectric effect!
First, we’ll look over what the photoelectric effect is.
Next, we’ll cover Einstein’s contributions and experiments regarding the photoelectric effect.
Then we’ll go over the photoelectric effect equation.
Finally, we’ll look at some examples of the photoelectric effect.
Photoelectric Effect Definition
Let's start by looking at the definition of photoelectric effect.
The photoelectric effect is the emission of electrons when light is shone on metal.
The emission or ejection of electrons is caused by metallic material absorbing electromagnetic radiation, like light.
Figure 2: Photoelectric effect illustrated. Daniela Lin, Study Smarter Originals.
Photoelectric Effect Experiments
Scientists, astronomers, mathematicians, and even philosophers have debated whether light is a wave or a particle. Wave theory proponents believed that light traveled as a wave due to Italian physicist Francesco Grimaldi pointing out that light diffracted like a wave in 1665.
By contrast, in 1704 Isaac Newton formulated his corpuscular theory, which was the beginning of the particle theory of light. Newton proposed this theory after he conducted experiments in optics. Newton argued that the geometric nature of refracted and reflected beams of light could only be explained if light were made up of particles. Newton noted that only particles travel in straight lines while waves do not. As a result of experimentation in optics, Newton proposed that light is made up of small particles called “corpuscles” that travel in straight lines.
Scientists such as Christian Huygens in turn argued that light had several wave-like characteristics such as polarization, diffraction, and color spectrum, so it couldn’t possibly be a particle. While the debate over what light was still raging on, in 1887 Heinrich Hertz first observed what we know today as the photoelectric effect when he increased sensitivity in his early radio device, by shining light onto the two metal spheres that worked as a receiver.
Just as science started to accept the wave theory due to Huygens’s influence and Newton’s lack of proof, the introduction of the photoelectric effect led to another problem. This arose because the photoelectric effect could not be explained only by the wave-like properties of light!
When we shine light of a sufficient frequency onto a metal surface electrons are ejected. However, an increase in the intensity of the light had no effect on the energy of electrons released but only increased the number of electrons ejected, which could not be explained by the wave theory of light.
The reason the photoelectric effect defies the wave theory of light is due to the fact that as intensity increases a wave’s energy also increases, which should then permit the ejection of electrons from a metal's surface for any wavelength of light as long as the intensity was high enough. However, this was not the case with the photoelectric effect.
As scientists observed, when a beam of light was below a certain frequency even the most intense beam of light could not eject any electrons. In contrast, if the beam of light was above a certain frequency threshold even the faintest beam of light could eject electrons.
For more information regarding light theory please reference “Newton’s and Huygens’ Theories of Light.”
Einstein and the Photoelectric Effect
In 1905, Albert Einstein settled the century-old long debate with the quantum theory of light.
The quantum theory of light states that light exists as tiny particles called photons while simultaneously exhibiting wave-like properties.
In other words, Einstein explained that light had what we call today a wave-particle duality.
But you’re probably wondering how did he do this when all the other preceding scientists were confused? Well, the answer lies in preceding scientists' experimental work. Einstein only did thought experiments, being a theoretician, but he also utilized the discoveries of other scientists to make conjectures to solve the mysteries of what was light.
Five years before Einstein’s theory, Max Planck, in 1900, theorized that energy isn’t continuous but is instead quantized.
When something is quantized it means that it can’t just have any value instead it’s limited to certain energy values or discrete states.
Planck's theory of the quantization of energy proposed that energy could only be emitted or absorbed in discrete energy packets. In the case of Planck’s theory, these quantized energy packets, these quanta of action, were subsequently called Planck’s constant, h.
After reading Planck’s findings, Einstein theorized that light must also be quantized. He called these particles of light photons. Einstein related the energy in one photon to its frequency through Planck's constant, E = hν.
Einstein’s theory helped explain the photoelectric effect because it demonstrated that it only takes one photon that exceeds a minimum frequency threshold to eject an electron from a given metal's surface.
Figure 3: The Photoelectric effect related to intensity and frequency. Daniela Lin, Study Smarter Originals.
As shown in figure 3, we only need 1 photon to meet the minimum threshold frequency to eject an electron from a metal surface. But if none of the photons meet the minimum threshold frequency, then there’s not going to be enough energy to eject an electron no matter the intensity of light. This means that if the light frequency is too low, then there will be no electrons ejected from the metal meaning there’s no electric current.
In contrast, if there’s low-intensity light but enough energy to where it’s above the minimum threshold frequency some electrons are ejected leading to a small current.
Lastly, if there’s high-intensity light but still enough energy to where it’s above the minimum threshold frequency then there will be more ejected electrons than in the previous scenario and thus a high current.
It’s important to note that the intensity of light only matters after the minimum threshold frequency for the incident light is met.
The minimum threshold frequency is the smallest amount of frequency required of light for an electron to be ejected from a metal surface.
For more information regarding light and their respective wavelengths and energies check out our “Electromagnetic Spectrum” article.
Photoelectric Effect equation
Now that we understand how the photoelectric effect was developed and how it works, we can move on to understanding how to calculate it.
The energy of a photon can be given by:
$$E=h\nu$$
where, Planck's constant is, h = 6.63 X 10-34 J·s, and the frequency of the photon is, ν in Hertz (Hz).
In other words, to calculate the energy of a photon we use Planck’s constant and the frequency of a photon. Another important thing to know is since light is both a particle and a wave how do we relate frequency to wavelength?
The answer is:
The Wave equation
\( c = \lambda\nu \)
where, the wavelength is, λ, in meters, the frequency of the photon is, ν, in Hertz (Hz), and the speed of light is c = 3.00 X 108 m/s
This means that the frequency and wavelength are inversely proportional to each other.
So how do we relate this to the photoelectric effect equation? Since we know that the photoelectric effect occurs when light above a certain frequency threshold is shone upon a metallic surface, this results in photoelectrons being ejected.
This suggests that we want to find the maximum kinetic energy of the ejected electrons. Kinetic energy is the energy something has due to it being in motion.
The law of the conservation of energy tells us that "energy can't be created or destroyed only converted" which signifies that the total energy of a photon (E) equals the maximum kinetic energy of an ejected electron (\(E_k\)) and the energy required to eject an electron from a metallic surface (Φ).
\(E= E_k + \phi \)
The part of the energy required to eject an electron from a metallic surface (Φ) is called the work function.
Einstein thought that a photon would go through the metallic material and transfer its energy to an electron. As the electron moves through the metal and gets ejected, the kinetic energy decreases by the energy required to eject an electron from a metallic surface, symbolized by, Φ.
Since the energy of photon is $$E=h\nu$$ then:
Then
\(E= h\nu= E_k + \phi \)
Solving for \(E_k\) or the electron's maximum kinetic energy we get the photoelectric effect equation:
\(E_k= h\nu - \phi \)
Photoelectric Effect examples
After going through the photoelectric effect equations, we should learn how to apply them to some examples, which are provided below.
What is the energy in joules of a photon with a wavelength of 550-nm, green light?
Since we're looking for the energy in Joules but we're only given the wavelength in nanometers, nm, we need to convert nanometers to meters, which is the dimension of length for the Joule:
Figure 4: Formula to find energy in joules. Daniela Lin, Study Smarter Originals.
h= \(6.63\times10^{-34}\space{J.s}\)
c= \(3.00\times10^{8}\space{m/s}\)
λ= convert it from nm to m
Figure 5: Conversion of nanometers (nm) to meters (m). Daniela Lin, Study Smarter Originals.
Figure 6: Formula to find energy in joules solved. Daniela Lin, Study Smarter Originals.
E = \(3.62\times10^{-19}\space{J}\)
The same green light with a wavelength of 550-nm, from above, is shone on a zinc metal surface, leading to the emission of photoelectrons. The zinc photoelectron has a work function of 4.31 eV. Find the maximum kinetic energy of the ejected photoelectron, \(E_k \).
To solve for the electron's maximum kinetic energy or \(E_k\) we use the photoelectric effect equation:
\(E_k= h\nu - \phi \)
where: \(E= h\nu \). Thus,from the problem above:
\(h\nu= \) \(3.62\times10^{-19}\space{J}\)
and
\( \phi \) = 4.31 eV
Thus,
\(E_k \) = \(3.62\times10^{-19}\space{J}\) - 4.32 eV
We need to convert Joules to electron volts (eV) which means that we need the conversion factor that relates 1 ev to Joules. Then given:
1 eV= \(1.602\times10^{-19}\space{J}\)
we find that,
\(3.62\times10^{-19}\space{J}\) x \( \frac {1 eV} {1.602\times10^{-19} J} \) = 2.26 eV
And plugging in this value into the equation for, \(E_k\), we get:
\(E_k \)= 2.26 eV - 4.31 eV = -2.05 eV
Photoelectric Effect - Key takeaways
The photoelectric effect results from the emission of electrons when light exceeding a metal-specific frequency threshold is shone on metal.
The quantum theory of light states that light exists as tiny particles called photons while simultaneously exhibiting wave-like properties. In other words, Einstein explained that light had what we call today a wave-particle duality.
Einstein’s theory helped explain the photoelectric effect because it only takes 1 photon of a minimum threshold frequency to eject electrons.
It’s important to note that the intensity of light only matters when the minimum threshold frequency for the incident light energy is met.
References
- Libretexts. (2022, April 21). 2.3: Photoelectric effect. Chemistry LibreTexts.
- Elert, G. (n.d.). Photoelectric effect. The Physics Hypertextbook.
- Libretexts. (2021, December 26). 6.2: Quantization: Planck, Einstein, energy, and photons. Chemistry LibreTexts.
- Ling, S. J., Sanny, J., & Moebs, W. (2016). Chapter 6.2/Photoelectric Effect . In University Physics Volume 3. OpenStax.
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