Physics and Tecnology of Radiations

[Radiation]

Homework Statement

this is the problem:

Y ≈ [ 3×10^-4 Z (T/mec^2) ] / [ 1 + 3×10^−4 Z (T/mec2) ]

this fraction gives us the approach energy lost by an electron from stopping radiation.
For beta particles emitted by 204Tl calculate the average value of energy lost by electron under the form of stopping radiation.

Homework Equations

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The Attempt at a Solution

Well, they give us this website to check the beta radiation from that component http://ie.lbl.gov/toi/nuclide.asp?iZA=810204" ( in the end of the page ). I suppose that i must do the average between the 29% particles with 347.5 Kev and the 97% with 763.72 Kev of energy. then i also know how to calculate de kinetic energy like T = E - mc^2
okay. know, radiation energy is like onde portion of the lost energy. the other part is colisional right? i can't just figure out what portion of energy is lost by radiation coz that Z ( atomic number) is not given, since is it suppose to be for any material.
I am a little stucked up, can anyone help me?

( sorry for my bad english, i hope that i made myself clear enough for you to understand :x )Thanks : )

 

[NateD]

The equation you gave is the stopping power of the material, which is the amount of energy per unit path length (typically MeV/cm) that is lost by the particle. This is not the energy lost by the particle in total. The total energy lost by the particle is equal to the kinetic energy it had when entering the material minus the kinetic energy it has when leaving the material (assuming no energy is lost as heat).For beta particles emitted by 204Tl, you can use the energy distributions given in the link you provided to calculate the average energy lost by the particle when traversing a certain path length. To do this, you will need to calculate the average energy of the particles and then multiply that by the stopping power equation you provided to get the average energy lost.