At the end of the XIX century, Heinrich Hertz discovered a phenomenon that allows electricity to be extracted from light energy. It is called the photoelectric effect. In this article, we will consider this effect in more detail, and also answer the question of what is the red border of the photoelectric effect.
Concept of photo effect
Before answering the question of what is the red border of the photoelectric effect, it is necessary to get acquainted with this physical phenomenon. How can this phenomenon be explained? The definition of the photoelectric effect can be given as follows: this is the process of formation of electrons free of atomic nuclei as a result of irradiation of matter with light.
This effect was discovered in 1887 by Hertz. In 1888, the Russian physicist, Alexander Stoletov, conducted a series of experiments in which he showed that this process occurs instantly after light hits the capacitor plate. The scientist also established the first law of the photoelectric effect: an increase in light intensity leads to a linear increase in the current in the circuit.
In 1905, Einstein published an article explaining the photoelectric effect by the quantum nature of the interaction of light with matter.
Einstein equation for photoelectric effect
A modern understanding of the photoelectric effect can be described in the form of the following model: falling on a substance, a photon meets an electron of some atom, the electron completely absorbs the photon, taking all the energy from it. If this energy is greater than some value, then the electron goes beyond the region of attraction of the atomic nucleus and becomes free.
The described process is expressed by the following equality of energies:
h * v = A + E k
Here h is the Planck constant, v is the frequency of the photon, A is the energy that you need to spend to “tear” an electron from the atom, E k is the kinetic energy of a free electron.
What is the red border of the photoelectric effect?
Let's pay attention to the Einstein equation given in the previous paragraph. It follows from this that if the photon energy is less than the electron work function, then no photoelectric effect will occur. This means that the phenomenon is observed only if:
h * v ≥ A.
The frequency v 0 corresponding to the value A is called the red border of the photoelectric effect. The wavelength corresponding to it is calculated by the formula:
λ 0 = c / v 0 or λ 0 = c * h / A.
Here c is the speed of light. Since the phenomenon under consideration is observed only for light wavelengths shorter than λ 0 , it becomes clear why this value is called the “red border” (red is the upper limit along the wavelength for the spectrum visible to a person).
If we talk about λ 0 for metals, then irradiation with red cannot lead to the appearance of a photoelectric effect for any of them, since the energy of “red” photons is too low to “tear” an electron from an atom. The highest value of λ 0 have alkali metals. For them, the red border is in the region of green and yellow (λ 0 ≈520-580 nm).
The solution to the problem of determining the type of element
Having figured out what the red border of the photoelectric effect is, we will solve one interesting problem for consolidating the knowledge gained. This will help to better understand this physical phenomenon.
In some laboratory, they decided to use the photoelectric effect to determine the type of chemical element. Prior to the experiment, it was found that this is the alkali metal of the first group of the table of D. I. Mendeleev. By changing the wavelength of light with which the metal was irradiated, it was determined that the photoelectric effect begins to be observed at 525 nm. What element did you work with in the laboratory?
We write out the corresponding formula for the red border of the photoelectric effect:
λ 0 = c * h / A
Where do we get:
A = c * h / λ 0
Substituting the corresponding constants and the value of λ 0 in the expression, we obtain the value of the electron work function for an unknown alkali metal:
A = c * h / λ 0 = 3 * 10 8 * 4.13567 * 10 -15 / (525 * 10 -9 ) = 2.333 eV
Note that the value of the Planck constant was substituted in units of eV * s.
The electron work function A is a unique characteristic for each chemical element. You can see it in the corresponding table. So, alkali metals are characterized by such values in eV:
- Li 2.93;
- Na 2.36;
- K 2.29;
- Cs 2.14;
- Rb 2.26.
These data show that the value of A found by us corresponds to sodium.