Another energy and momentum SR problem

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Homework Help Overview

The problem involves a 100 keV photon scattering from a free electron, with the goal of finding the recoil velocity of the electron when the photon scattering angle is 180 degrees. The context is rooted in the principles of energy and momentum conservation, specifically within the framework of special relativity.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the setup of the conservation equations but express uncertainty about the correctness of their formulations. There are attempts to manipulate the equations to isolate the recoil velocity of the electron.

Discussion Status

Some participants have provided guidance on how to approach the problem, suggesting methods to combine equations. Others are questioning the correctness of their calculations and comparing their results to a reference answer from a textbook, which has led to further exploration of potential errors in their reasoning.

Contextual Notes

Participants note that the use of Compton's effect equation is not permitted, which adds a constraint to their problem-solving approach. There is also mention of an attachment containing a detailed solution that requires moderator approval before it can be reviewed by others.

zimo
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Homework Statement


A 100kev photon scatters from a free electron initially at res. Find the recoil velocity of the electron if the photon scattering angle is 180 degrees.


Homework Equations


Conservation of Energy, Conservation of Momentum
(It is not allowed to use Compton's effect equation!)


The Attempt at a Solution


I've written:
[tex] \eqalign{<br /> & {{hc} \over \lambda } + m_e c^2 = {{hc} \over {\lambda '}} + {{m_e c^2 } \over {\sqrt {1 - {{\bar v^2 } \over {c^2 }}} }} \cr[/tex]
[tex] & {h \over \lambda } = {{m_e \bar v} \over {\sqrt {1 - {{\bar v^2 } \over {c^2 }}} }} - {h \over {\lambda '}} \cr}[/tex]

but as I develop the equations on trying to solve the equations, somewhat it doesn't work out...
 
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zimo said:
I've written:
[tex] \eqalign{<br /> & {{hc} \over \lambda } + m_e c^2 = {{hc} \over {\lambda '}} + {{m_e c^2 } \over {\sqrt {1 - {{\bar v^2 } \over {c^2 }}} }} \cr[/tex]
[tex] & {h \over \lambda } = {{m_e \bar v} \over {\sqrt {1 - {{\bar v^2 } \over {c^2 }}} }} - {h \over {\lambda '}} \cr}[/tex]

but as I develop the equations on trying to solve the equations, somewhat it doesn't work out...

Hi zimo! :smile:

Well, you want v, but you don't want [tex]\lambda'[/tex] …

so just multiply the second equation by c, and add it to the first equation. :smile:
 
OK.

Did I write the equations correctly?
 
zimo said:
OK.

Did I write the equations correctly?

Looks ok to me.

Why are you asking? :confused:
 
because I get v=0.483c, in contrast to the final answer in the book, which is 0.319c.
 
zimo said:
because I get v=0.483c, in contrast to the final answer in the book, which is 0.319c.

Unless you show your full calculations, we can't see where you've gone wrong. :smile:
 
I attached my solution. (starting with the lines indicating I) and II) )
 

Attachments

  • 22.jpg
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can someone read my uploaded solution, please, and reveal my mistake?
 
zimo said:
can someone read my uploaded solution, please, and reveal my mistake?
Your attachment needs to be approved by a moderator before we can view it. Be patient, attachments are usually approved fairly quickly. If it hasn't been approved by this evening I'll see if I can get someone's attention.
 
  • #10
zimo said:
because I get v=0.483c, in contrast to the final answer in the book, which is 0.319c.

Hi zimo! :smile:
zimo said:
√(1 - u2)0.711 MeV = 0.511(u + 1) MeV

(1 - u) = 1.936(1 + u)

u = .483

That's with (.711/.511)2 = 1.936.

The book seems to have used .711/.511 = 1.391, giving the answer of .319

The book seems to be wrong. :frown:
 
  • #11
No.

I made an error during the strike-out of (u+1) on both sides, I needed to leave one (u+1) at the right side of the eqation.

Finally, I got 0.319 as requested.
 

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