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Physics
High Energy, Nuclear, Particle Physics
Total or kinetic energy in Bethe Bloch stopping power?
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[QUOTE="crick, post: 6020793, member: 568971"] The average loss of energy in a material per unit length of a particle (in particular an electron, which is stable) is described by the quantity ##dE/dx##. - for ionization it is given by the Bethe-Bloch formula $$-\left\langle {\frac {dE}{dx}}\right\rangle ={\frac {4\pi }{m_{e}c^{2}}}\cdot {\frac {nz^{2}}{\beta ^{2}}}\cdot \left({\frac {e^{2}}{4\pi \varepsilon _{0}}}\right)^{2}\cdot \left[\ln \left({\frac {2m_{e}c^{2}\beta ^{2}}{I\cdot (1-\beta ^{2})}}\right)-\beta ^{2}\right]$$ - for Bremmstralungh it is given by the Bethe-Heilter formula $${\displaystyle -\left\langle {\frac {dE}{dx}}\right\rangle \approx {\frac {4N_{a}Z^{2}\alpha ^{3}(\hbar c)^{2}}{m_{e}^{2}c^{4}}}E\ln {\frac {183}{Z^{1/3}}}}$$ I can't understand if the "energy ##E##" in the formulas is the total relativistic energy or the kinetic energy ##K## only? In the first case ##E=K +mc^2##, while in the second case ##E=K##. It looks like that the first case is the right one, since it's more general, but in that case I cannot understand how the particle, stopping in the material for various processes can change its mass (the electrons are stable so they do not decay after they have stopped). Does this really happen or does it loose before all its kinetic energy and then its rest energy? I'm confused also because I read that in calorimeters the range of the particle is used to measure its energy: does this energy include the rest energy? [/QUOTE]
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Physics
High Energy, Nuclear, Particle Physics
Total or kinetic energy in Bethe Bloch stopping power?
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