Is This a Compton Scattering Example of a Gamma Ray Photon?

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The discussion confirms that the scenario described is an example of the Compton effect, where a photon collides elastically with a stationary electron, losing half its energy. The calculations indicate that the wavelength change results in a frequency of 4.04 * 10^21 Hz, classifying the photon as a gamma ray. Participants emphasize the importance of accuracy in calculations, particularly regarding the wavelength values. The behavior observed aligns with the expected characteristics of gamma rays, including energy loss and wavelength increase. Overall, the initial calculations are validated, and further assistance is offered for additional queries.
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A photon of frequency \nu collides elastically with an electron at rest. The phton loses half its nergy in the collision an its final path makes a 60 degree angle with the original path. What is \nu ? what sort of photon is this?

Well this is obviously the compton effect

\lambda_{1} - \lambda = \frac{h}{mc} (1 - Cos \phi ) = 1.21 * 10^12

Then Also the photon lost half it's energy so

\frac{h \nu}{2} = \frac{h \nu '}{2}
and figure that 2 \lambda = \lambda '

and then lambda = 1.21 * 10^12?

then \nu = 4.04 * 10^21 Hz

which is a gamma ray?

am i right in this? i have my doubts about the lambdas

Also i have another thread i need some help would greatly appreciate with both! Thank you!

https://www.physicsforums.com/showthread.php?t=51474
 
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Your thinking is correct...
stunner5000pt said:
and then lambda = 1.21 * 10^12?
...but watch those exponents. I assume you meant to write 1.21E-12.
then \nu = 4.04 * 10^21 Hz
redo that calculation.

which is a gamma ray?
Yep.
 


Yes, you are correct in your calculations. This would be a gamma ray photon, with a frequency of 4.04 * 10^21 Hz. The Compton effect is the scattering of a photon off an electron, resulting in a change in the photon's wavelength and energy. In this case, the photon has lost half its energy and its wavelength has increased. This is consistent with the behavior of gamma rays.

As for your other thread, I will take a look and see if I can offer any assistance. It's always a good idea to seek help and clarification when needed. Keep up the good work!
 
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