Beta decay - derive an expression relating source angle to energy

Click For Summary

Homework Help Overview

The discussion revolves around deriving an expression that relates the energy of beta particles emitted from a source to the angle of the apparatus setup. The subject area includes concepts from particle physics and relativistic mechanics.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to combine equations related to momentum and radius to derive an expression for energy. Some participants question the clarity of the apparatus setup and the relationship between momentum and energy in the provided equations.

Discussion Status

Participants are exploring the relationships between the equations provided and discussing the complexity of the expressions. Some guidance has been offered regarding the nature of the energy equation, but there is no explicit consensus on the next steps or simplifications.

Contextual Notes

There is a request for a visual representation of the apparatus, indicating that the setup may not be fully understood by all participants. The original poster expresses uncertainty about simplifying the derived expression and understanding the components of the energy equation.

Exulus
Messages
50
Reaction score
0
Hi guys,
Need a spot of help as i can't seem to find where to go next :(
Ok, so there is a source which emits beta particles (currently unknown whether they are positrons or electrons). It can move at an angle +/- 90 degrees to the GM tube. I need to find an expression for the energy of the particles related to the angle the apparatus is at. Here are the equations given to me:
[tex]E = m_{0}c^2 [\sqrt{1+(\frac{p}{m_{0}c})^2} - 1][/tex] (1)
[tex]r = \frac{R}{\tan(\frac{\Theta}{2})}[/tex] (2)

Where

r = the radius of the electron/positron circle.
R = the radius of the circle used for the source/GM tube setup.

I also have the following equation:

[tex]F = \frac{mv^2}{r} = qvB[/tex] (3)

Combining (2) and (3) gives me:

[tex]\frac{R}{\tan(\frac{\Theta}{2})} = \frac{mv}{qB}[/tex]

I can't see where to go next...any ideas? Cheers :)
 
Physics news on Phys.org
I can't picture the setup of the apparatus. Could you provide a sketch?
Until then, have you noticed that momentum (mv) is in both formulas (the formulas at the top and bottom of your post)?
 
oops, sure :)

http://img427.imageshack.us/img427/6121/scan1lu.jpg

I didnt notice that momentum was in both, silly mistake. Ok so if i do substitute in momentum it still seems like a bit of a hefty equation..i don't know if i can simplify it much?

Out of interest, could anyone break down the energy equation for me or hint towards what part is what? Apparently its both kinetic and potential energies in one but i can't really see how they got that. Thanks :)
 
Last edited by a moderator:
Yes it is hefty. This is where physics gets fun. Just take a deep breath, get all your values together and thank god you don't need to use a slide-rule anymore.

And the first formula is the kinetic energy formula for objects traveling at relativistic speeds. it has conveniently been put in terms of the object's momentum rather than speed.
 
ok I've now got this:

[tex]E = m_{0}c^2 [\sqrt{1+(\frac{RqB}{m_{0}c\tan{\frac{\Theta}{2}}})^2} - 1][/tex]

I take it that can't be simplified any further. Ack this is going to be hard to work out errors with!

Thanks for your help!
 
I've seen worse.
 

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
Replies
9
Views
2K
Replies
1
Views
3K
  • · Replies 11 ·
Replies
11
Views
9K
Replies
2
Views
2K
Replies
6
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 17 ·
Replies
17
Views
5K
  • · Replies 4 ·
Replies
4
Views
5K