# Photoelectric effect and Compton scattering

• AllRelative
In summary, a photon with an energy of 13600 eV ejects an electron from a hydrogen atom at rest in the direction it was traveling. The electron requires 13.6 eV to be ejected. The speed of the photoelectron, momentum, and energy of the recoiling photon can be calculated using equations for energy of a photon, photoelectric effect, momentum of a photon, and Compton effect. However, the question is unclear about whether the electron is ejected due to the photoelectric effect or a collision with the photon. It is best to treat it as a conservation of energy and momentum problem without worrying about the specific mechanism. In any case, the photon transfers all its energy to the electron and there is no recoil
AllRelative

## Homework Statement

A photon with of 13600eV energy interacts with a hydrogen atom at rest and ejects the electron (photoelectrically) in the direction in which the photon was travelling. If 13.6 eV is required to eject the electron, find the speed of the photoelectron and the momentum and energy of the recoiling photon.

## Homework Equations

Energy of a Photo: E = h*f

Photo electric effect: KEmax = hf - phi

Momentum for a photon: p = h / lamda

Compton Effect: lamda ' - lamda = h / Me*c (1 - Cos theta)

## The Attempt at a Solution

Now I'm confused. In class we learned about the photoelectric effect and the Compton Scattering separately. I understand and can do problems about both concepts. What I don't understand here is the ejection of the electron.

Is the electron ejected due to the photoelectric effect or due to a collision with the photon?

I know that Phi is 13.6eV. This energy will be lost due to pulling the electron out of the atom.

Is the Kinetic energy of the ejected electron coming from the collision with the photon.Thanks for the help!

I know that Phi is 13.6eV. This energy will be lost due to pulling the electron out of the atom.
Is the Kinetic energy of the ejected electron coming from the collision with the photon.
... it is not useful to think of the interaction as a "collision".
It is just an interaction ... electrons and photons are not like little balls that have surfaces that can strike each other.

Is the electron ejected due to the photoelectric effect or due to a collision with the photon?
... the problem statement specifies "photoelectrically", but also says there is a recoil photon. In the photoelectric effect the incoming photon is completely absorbed by the electron so there is no "recoil photon" so I see your confusion.
http://hyperphysics.phy-astr.gsu.edu/hbase/mod2.html

I think you need to get a clarification from the person who wrote the problem.

In the absence of a clarification, I'd combine the information ... treat it as a conservation of energy and momentum problem without worrying about what name to use.
You have some initial state, something happens, and you get some final state. You need to relate energy to momentum and the initial state to the final state.

Thanks for the reply! Sadly this question is for an assignment I have to hand in tomorrow and I won't be able to get clarification on the problem.

I understand that the photon is absorbed. The question just confused me. The projected electron takes the energy of the incoming photon minus the work function of 13.6 eV. This will give it its Kinetic Energy. I can therefore find it's velocity, momentum and total energy.

Thanks for the reply! Sadly this question is for an assignment I have to hand in tomorrow and I won't be able to get clarification on the problem.
In the absence of a clarification, I'd combine the information ... treat it as a conservation of energy and momentum problem without worrying about what name to use.
You have some initial state, something happens, and you get some final state. You need to relate energy to momentum and the initial state to the final state.

The projected electron takes the energy of the incoming photon minus the work function of 13.6 eV. This will give it its Kinetic Energy. I can therefore find it's velocity, momentum and total energy.
... you could take it like that, but then: how will you answer the second part of the problem - the bit about the recoil photon?

Simon Bridge said:
In the absence of a clarification, I'd combine the information ... treat it as a conservation of energy and momentum problem without worrying about what name to use.
You have some initial state, something happens, and you get some final state. You need to relate energy to momentum and the initial state to the final state.

... you could take it like that, but then: how will you answer the second part of the problem - the bit about the recoil photon?

I just read the section on the photoelectric effect in my book and it says clearly that the photon transfers all its energy to the electron. Therefore there is no photon to speak of...

So your answer to the second question is that there is no recoil photon?
Fair enough - let me know how it went.

AllRelative

## 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 or other electromagnetic radiation. This effect was first observed by Heinrich Hertz in 1887 and was later explained by Albert Einstein in 1905 through his theory of quantum mechanics.

## 2. How does the photoelectric effect work?

The photoelectric effect occurs when photons of light with enough energy strike a material and transfer their energy to electrons, causing them to be ejected from the material. The energy of the photons must be greater than the work function of the material (the minimum amount of energy needed to remove an electron) for the photoelectric effect to occur.

## 3. What is Compton scattering?

Compton scattering is the process in which an incoming photon of high energy collides with an electron and transfers some of its energy to the electron. As a result, the scattered photon has a longer wavelength (lower energy) than the original photon. This phenomenon was first observed by Arthur Compton in 1923 and provided important evidence for the particle nature of light.

## 4. How does Compton scattering affect the energy of the scattered photon?

The energy of the scattered photon after a Compton scattering event can be calculated using the Compton scattering formula. This formula takes into account the initial energy and direction of the photon, as well as the energy and angle of the scattered photon. The energy of the scattered photon will always be less than the energy of the initial photon, as some energy is transferred to the electron during the scattering process.

## 5. What are the practical applications of the photoelectric effect and Compton scattering?

The photoelectric effect has many practical applications, including solar cells, photodiodes, and photomultiplier tubes. It is also used in the production of x-rays in medical imaging. Compton scattering is used in various fields, such as medical imaging and materials science, to study the structure and composition of materials. It is also important in understanding the behavior of photons and electrons in quantum mechanics.

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