Ionizing Photons: Photoelectric Effects Compared

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In summary, the photoelectric effect in the ionization of atoms differs from the photoelectric effect that makes mirrors reflect because the ionizing wavelengths have more energy, as given by the equation E=hf, which completely liberates the electrons. In contrast, in reflection, the incoming photons do not move electrons to a higher energy state, and can even have very low energies such as radio frequencies. This can be explained classically by considering a classical electromagnetic wave incident on a conducting material.
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Elliot Fox
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This is my first time on this forum and my knowledge may be a bit basic but I was wondering how does the photoelectric effect in the ionization of atoms differ from the photoelectric effect that makes mirrors reflect. They both have incoming photons that give their energy to electrons which move to a higher energy state and yet one ionizes but the other reflects. Is it because in ionizing the ionizing wavelengths has more energy as E=hf so they completely liberate the electrons whereas other wavelengths lack that energy? Thanks.
 
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Elliot Fox said:
They both have incoming photons that give their energy to electrons which move to a higher energy state...

This is not correct. In reflection, the incoming photon does not move electrons in the reflecting material to a higher energy state. It is perfectly possible to build mirrors that reflect very low energy photons (like radio frequency) that have energies far too low to move electrons to a higher energy state. Reflection can be understood completely classically by considering a classical EM wave incident on a conducting material. The Feynman Lectures on Physics, Vol2 Chapter 33 has a good explanation of refection.
 

Related to Ionizing Photons: Photoelectric Effects Compared

1. What are ionizing photons?

Ionizing photons are high-energy photons that have enough energy to remove electrons from atoms or molecules. This process is known as ionization and can lead to the creation of free radicals, which can be harmful to living organisms.

2. How do ionizing photons cause the photoelectric effect?

The photoelectric effect occurs when ionizing photons strike a material, causing electrons to be ejected from the surface of the material. This is due to the high energy of the photons, which can overcome the binding energy of the electrons in the material.

3. What is the difference between the photoelectric effect and the Compton effect?

The photoelectric effect involves the ejection of electrons from the surface of a material, while the Compton effect involves the scattering of photons off of electrons within a material. Additionally, the photoelectric effect does not involve a change in the wavelength of the photons, while the Compton effect does.

4. How does the energy of ionizing photons affect the photoelectric effect?

The energy of ionizing photons is directly related to the likelihood of the photoelectric effect occurring. Higher energy photons have a greater chance of ejecting electrons from a material, while lower energy photons may not have enough energy to cause ionization.

5. What are the applications of ionizing photons and the photoelectric effect?

The photoelectric effect and ionizing photons have many practical applications, including in medical imaging, nuclear power, and radiation therapy. They are also used in devices such as photodiodes and solar cells. However, excessive exposure to ionizing photons can be harmful and precautions must be taken to minimize exposure.

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