Photoelectricity is electricity generated by the interaction of light with matter. The photoelectric effect is the phenomenon that occurs when light impacts materials and causes electrons to be ejected from the material. This also means that the light is producing electricity, which is nothing more than charges (in this case, electrons) in movement.
We speak of photoelectricity because ‘photo’ is the Greek word for light. When light with high energy impacts a material and causes electrons to be ejected from it, the ejected electrons are called photoelectrons.
Figure 1. The emission of a photoelectron from a metal surface. Source: Oğulcan Tezcan, Vaia.
In today’s technology, the photoelectric effect is used in a variety of applications such as light metres that adjust the automatic iris on different types of cameras, solar cells, photomultipliers, and night vision devices.
The development of the concept of the photoelectric effect
The photoelectric effect was first observed by Heinrich Hertz in 1887, who used a coil with a spark gap in the receiver of the device he employed for his measurements. He did his observations in a dark box in order to see the spark more clearly, but when he noticed that the maximum spark length decreased inside the box, he had to choose a different approach.
He then used a coil for an inductor and a Leyden jar, which is a glass jar with layers of metal foil both on the exterior and the interior. It was used as an early kind of capacitor. Between the sender of electromagnetic waves and the receiver, a glass screen absorbed UV light, which helped the electrons to jump over the gap.
Figure 2. Hertz’s experiment setup. Source: Oğulcan Tezcan, Vaia.
Although Hertz was the one to observe the photoelectric effect, he wasn’t the one to explain it. This was done by Albert Einstein.
The initial experiments couldn’t explain why the emitted electrons were affected by using higher light frequencies and not by the brightness of the light. Einstein was the one to explain that the kinetic energy of a photoelectron is directly proportional to the light’s frequency. He claimed that light is acting as a particle (photon) and that its energy is equal to the multiplication of the Planck constant h and the light frequency f.
\[E_{photon} = hf\]
Robert Millikan disagreed with the proposal and stated that no electrons would be released if the experiment was done correctly. He conducted his own experiment in a vacuum to ensure that nothing in the atmosphere would interfere with or assist in the creation of a current.
However, Millikan ended up demonstrating that the photoelectric effect was true after all, which helped Einstein to win the 1921 Nobel Prize for his theory of the photoelectric effect. Millikan himself won the 1923 Nobel Prize for his work to determine the charge of an electron and confirm Einstein’s theory.
An example of the photoelectric effect
Let’s look at the following example to help us understand the concept of the photoelectric effect.
A photoelectron with a kinetic energy of 5.0eV is emitted from an iron plate. Determine the energy and frequency for the photon that impacted the plate.
To do this, we need to know what the work function is. A specific value of energy is required to release a photoelectron from the metal plate. The work function Φ is the least amount of energy that a photon must transmit in order to release the photoelectron.
In the case of an iron plate, the work function is 4.5eV:
\(\Phi = 4.5 e.V\)
This is the energy used to release the photoelectron. As the photoelectron has the kinetic energy of 5.0 eV, the energy of the photon must be the sum of both, as the energy is conserved.
\(E_{photon} = 4.5 [eV] + 5[eV] = 9.5 eV\)
One electron volt (eV) is equal to 1.6 ⋅ 10-19 joules. Converting the energy into joules, we get:
\(E_{photon} = 9.5 \cdot 1.6 \cdot 10^{-19} = 1.52 \cdot 10^{-18} J\)
To find the frequency of the photon, we need to replace E with hf.
\(hf = 1.52 \cdot 10^{-18} J\)
We know that the h is Planck constant, which is equal to 6.63 ⋅ 10-34 m2⋅kg/s. Dividing the energy by the Planck constant, we can determine the frequency of the light.
\(f = \frac{1.58 \cdot 10^{-18} [J]}{6.63 \cdot 10^{-34} [J/Hz]} = 2.383 \cdot 10^{15} Hz\)
How did photoelectricity affect the wave theory of light?
If the wave theory light were correct, then light impacting a metal surface would lead to the wave spreading equally over the surface and to each free electron on that surface gaining a little energy with every incoming wave. With the light continuing to impact the surface, eventually, every electron would gain enough energy to be released from the metal. For low-frequency lights, it would take more time, but the releasing of the electrons would eventually happen.
However, we now know that no electrons are emitted if the wave is below the threshold frequency (or work function energy), and the wave theory of light cannot explain that threshold frequency.
Photoelectricity - Key takeaways
- Photoelectricity is electricity produced as a result of the action of light on a material.
- The photoelectric effect is the phenomenon that occurs when electrons are ejected from a material by light impacting it. This also means that the light is producing electricity (photoelectricity).
- The particle form and basic fundamental unit of light is called a photon.
- The work function is the least energy that a photon must transmit in order to release the photoelectron from the metal plate.
- The equation used to calculate a photon’s energy is E = hf, where h is the Planck constant, which equals 6.63 ⋅ 10-34 m2⋅kg/s, and f is the frequency of the light.
Explanations 
Textbooks 
Exams 
Magazine 
