What is the maximum energy an electron can have after scattering a 40keV photon?

In summary, when a photon with an energy of 40keV scatters from an electron initially at rest, the maximum energy the electron can have is its rest energy plus the kinetic energy from the photon. This is derived from the conservation of energy and momentum equations, which show that the change in wavelength of the photon is largest at an angle of pi.
  • #1
bmb2009
90
0

Homework Statement



A photon carrying energy of 40keV scatters from an electron initially at rest. what is the maximum energy the electron can have?


Homework Equations





The Attempt at a Solution



I tried using conservation of energy

Initial: 40keV (energy of photon) + mc^2 (rest energy of electron)
Final: To have the most energy all the energy of the photon should be transfer to kinetic energy of the electron correct?

so wouldn't the final energy just be the rest energy of the electron + the kinetic energy of the electron? Not sure where I am missing an idea..thanks for the help
 
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  • #2
There's something else that needs to be conserved besides energy.
 
  • #3
TSny said:
There's something else that needs to be conserved besides energy.

I know momentumn does too but how do i combine these two to get the total final energy?
 
  • #4
If you're allowed to use the well-known http://hyperphysics.phy-astr.gsu.edu/%E2%80%8Chbase/quantum/comptint.html then you will not need to go through the algebra of combining the energy and momentum equations.

If you are required to derive the result starting from conservation of energy and momentum, then what would each conservation equation look like?

Note that "scattering" implies that there will be a photon present after the "collision" as well as before.
 
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  • #5
Yes i derived that equation but i still don't see the connection to the energy of the electron... I assume that to maximize the energy of the electron the change in wavelength should be largest at an angle of pi correct? that would yield 2h/mc=delta lambda but now how do i find the energy of the electron??
 
  • #6
ahhh nevermind i figured it out.. thanks for the guidance..it's appreciated!
 
  • #7
ok, good!
 

What is Basic Compton Scattering?

Basic Compton Scattering is a physical phenomenon in which a photon (an elementary particle of light) collides with a charged particle, usually an electron, resulting in the transfer of energy and momentum. This process is important in understanding the behavior of light and matter in the microscopic world.

How does Basic Compton Scattering occur?

Basic Compton Scattering occurs when a photon collides with an electron. The electron absorbs some of the energy of the photon and is then ejected in a different direction. The photon's energy decreases, causing its wavelength to increase. This phenomenon is known as the Compton Effect.

What is the significance of Basic Compton Scattering?

Basic Compton Scattering is significant because it provides evidence for the dual nature of light, which can act as both a wave and a particle. It also helps in understanding the properties of electrons and their interactions with photons. Furthermore, it has practical applications in fields like medical imaging and materials science.

What are the factors that affect Basic Compton Scattering?

The factors that affect Basic Compton Scattering include the energy of the incident photon, the mass and movement of the target electron, and the angle at which the photon and electron collide. The scattering process is also influenced by the properties of the material through which the photon travels.

How is Basic Compton Scattering used in medical imaging?

Basic Compton Scattering is used in medical imaging techniques such as Computed Tomography (CT) scans. In this process, high-energy X-rays (photons) are directed towards the body, and the scattered photons are detected and used to create images of the internal tissues and organs. This allows for the non-invasive examination of the body's internal structures and is an important tool in diagnosing diseases and injuries.

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