Another energy and momentum SR problem

In summary, the question involves finding the recoil velocity of a free electron after a 100kev photon scatters off of it at a 180 degree angle. Using the conservation of energy and momentum equations, the solution is eventually found to be 0.319 times the speed of light, which contradicts the answer given in the book. After further examination, it is discovered that the book's answer is incorrect and the correct answer is indeed 0.319c.
  • #1
zimo
45
0

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{
& {{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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  • #2
zimo said:
I've written:
[tex]
\eqalign{
& {{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:
 
  • #3
OK.

Did I write the equations correctly?
 
  • #4
zimo said:
OK.

Did I write the equations correctly?

Looks ok to me.

Why are you asking? :confused:
 
  • #5
because I get v=0.483c, in contrast to the final answer in the book, which is 0.319c.
 
  • #6
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:
 
  • #7
I attached my solution. (starting with the lines indicating I) and II) )
 

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  • #8
can someone read my uploaded solution, please, and reveal my mistake?
 
  • #9
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.
 

Related to Another energy and momentum SR problem

1. What is the principle of conservation of energy and momentum and how does it relate to SR problems?

The principle of conservation of energy states that energy cannot be created or destroyed, it can only be transformed from one form to another. Similarly, the principle of conservation of momentum states that the total momentum of a system remains constant unless acted upon by an external force. These principles are important in SR problems as they help to explain the behavior of particles and their interactions in different reference frames.

2. How does special relativity affect the concept of mass and energy?

Special relativity states that mass and energy are equivalent and can be converted from one form to another. This concept is described by the famous equation E=mc², where E is energy, m is mass, and c is the speed of light. This relationship between mass and energy is important in understanding the behavior of particles at high speeds.

3. What is the difference between relativistic and non-relativistic kinetic energy?

Relativistic kinetic energy takes into account the effects of special relativity on the mass of an object, while non-relativistic kinetic energy does not. At high speeds, the mass of an object increases according to the principle of mass-energy equivalence, resulting in a different calculation for relativistic kinetic energy compared to non-relativistic kinetic energy.

4. How does the Lorentz transformation play a role in solving SR problems?

The Lorentz transformation is a set of equations that describe how space and time are perceived differently by observers in different reference frames. These transformations are necessary for solving SR problems as they allow us to convert measurements between different frames of reference, which is essential for understanding the behavior of particles at high speeds.

5. Can the principle of relativity be applied to all types of energy and momentum?

Yes, the principle of relativity can be applied to all types of energy and momentum. This includes not only the traditional forms of energy and momentum, such as kinetic energy and linear momentum, but also other types such as potential energy, angular momentum, and even mass. The principle of relativity holds true for all forms of energy and momentum in both classical and relativistic physics.

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