Photoelectric effect problem

In summary, the threshold wavelength for photoelectric emission from a sodium surface is 683 nm. The energy of the ejected electrons is K = 1.1E-19 J and the energy of incident photons is 2.91E-19 J. With an intensity of 2.0 W/m^2 and 1% of incident photons producing photoelectrons, it is estimated that 3.6E13 photoelectrons are emitted per second from a sodium surface of area 2*10^-4 m^2 when illuminated by light of wavelength 500 nm. This differs from the book's answer of 1.0E13 due to dividing by the KE of the ejected electrons instead of the energy of the incident photons
  • #1
kasse
384
1

Homework Statement



The wavelength treshold for photoelectric emission from a sodium surface is 683 nm. Calculate the energy of the electrons which are ejected when a sodium surface is illuminated by light of wavelength 500 nm. If the intensity of the light is 2.0 W/m^2 and if 1 per cent of incident photons produce photoelectrons, estimate the number of electrons emitted per second from a sodium surface of area 2*10^-4 m^2 when it is illuminated by light of the wavelength 500 nm.

2. The attempt at a solution

The treshold wavelength corresponds to a treshold frequency of 6E14 Hz, which in turn corresponds to a electron binding energy of 2.91E-19 J.

This means that the energy of the electrons which are ejected is K = 1.1E-19J.

Next, we observe that since the surface area is 2E-4 m^2, the total effect is 4E-4 W, and hence 4E-4 J of energy hit the surface each second.

Since 1 per cent of this energy produces photoelectrons, the total photoelectron energy per second is 4E-6 J.

Since each photoelectron has energy 1.1E-19, my estimated number of photoelectrons is 4E-6/1.1E-19 = 3.6E13 photoelectrons.

My book says 1.0E13. What have I done wrong?
 
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  • #2
kasse said:
Since each photoelectron has energy 1.1E-19, my estimated number of photoelectrons is 4E-6/1.1E-19 = 3.6E13 photoelectrons.

My book says 1.0E13. What have I done wrong?
You divided by the KE of the ejected electrons instead of the energy of the incident photons.
 

1. What is the photoelectric effect?

The photoelectric effect is the phenomenon where electrons are emitted from a material when it is exposed to light of a certain frequency. This was first observed by Heinrich Hertz in 1887 and was later explained by Albert Einstein in 1905.

2. How does the photoelectric effect work?

When light of a certain frequency, called the threshold frequency, hits a material, it transfers its energy to the electrons in the material. If the energy of the light is greater than the binding energy of the electron in the material, the electron will be ejected from the material.

3. What are the applications of the photoelectric effect?

The photoelectric effect has many practical applications, such as in photovoltaic cells, which convert light energy into electrical energy. It is also used in photoelectric sensors, solar panels, and in the production of x-rays.

4. What is the significance of the photoelectric effect in the development of quantum mechanics?

The photoelectric effect played a crucial role in the development of quantum mechanics as it provided evidence for the existence of discrete packets of energy, known as photons. This helped to overturn the classical wave theory of light and laid the foundation for the quantum theory of matter and energy.

5. Are there any limitations or challenges of the photoelectric effect?

One limitation of the photoelectric effect is that it only occurs in materials with a certain minimum frequency, which is called the threshold frequency. This makes it difficult to use in some applications, as not all light sources have the necessary frequency. Additionally, the intensity of the light does not affect the number of electrons emitted, which can be a challenge in some experiments.

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