Bethe-Bloch : energy loss to a single electron?

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SUMMARY

The Bethe-Bloch equation quantifies the energy loss of charged particles as they traverse matter, specifically calculating the energy deposited per unit path length in materials like Silicon. It is established that the energy loss calculated is primarily transferred to a single electron, which subsequently induces further ionization at a rate of 3.65 eV per electron in Silicon. For minimally ionizing particles, this results in an energy transfer of 3.8 MeV per cm or 380 eV per micron. Accurate calculations of energy transfer should ideally be performed at the Angstrom scale, although this is often limited by computational processing times.

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  • Understanding of the Bethe-Bloch equation and its components
  • Familiarity with ionization processes in semiconductors, particularly Silicon
  • Knowledge of energy transfer mechanisms in particle physics
  • Basic principles of momentum transfer and Coulomb scattering
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  • Study the derivation and applications of the Bethe-Bloch equation in particle physics
  • Explore ionization rates in various materials beyond Silicon
  • Learn about delta rays and their significance in energy transfer processes
  • Investigate computational methods for modeling energy loss at the atomic scale
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nlanky85
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Hi guys,

the Bethe bloch (BB) equation, in combination with the material density, gives the energy deposited in the material per unit path length. One of the terms in the BB equation is the maximum energy transferred to an electron by the incident particle. So i am assuming, the energy loss calculated by the BB equation, is assumed to be transferred to a single electron and not numerous electrons? If the energy is transferred to a single electron, then this electron will cause further ionization, at a rate of one electron every 3.65 eV for Silicon.

For example, a minimally ionizing particle will deposit 3.8 MeV per cm, or 380 eV per micron in Silicon. Does this mean a minimally ionizing particle will transfer 3.8 MeV to a single electron in 1 cm of Si, then this electron will cause further ionization? Similarly, if i was working in microns, does this mean that on average 380 eV will be transferred to a single electron per micron in silicon? I know atom spacing’s are approximately in the Angstroms range. For the most accurate calculation of the energy transferred to a single electron, should dEdX be calculated every Angstrom, as opposed to every micron or cm ? Is this not done due to increased processing times for computers?

Thanks.
 
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The Bethe-Bloch equation is discussed in depth in Section 27.2 of

http://pdg.ihep.su/2009/reviews/rpp2009-rev-passage-particles-matter.pdf

The incident ionizing particle "collides" with many electrons in the silicon during its passage through matter. Because the mass m of recoil atomic electrons is so light, the momentum transfer is very small; p2 = 2mErecoil where Erecoil = ½mv2. The Bethe-Bloch equation Eq. 27.3, plotted in Fig. 27.2, is derived using the well-known Coulomb-scattering equations of the incident ionizing particle on the atomic electrons. The energy loss distribution of the recoiling electrons is based on the impact parameter (distance of closest approach) of the incident particle and atomic electron, and is best discussed in books like Ritson " Techniques of High Energy Physics" (Interscience) pages 11-24**. The "I" in the denominator of the Bethe Bloch equation is the mean ionization energy, or mean energy transferred from the incident ionizing particle to the recoil atomic electron in a single collision. Infrequently, the collision transfers a large energy to the recoil electron, this is called a delta ray. See Section 2.2.5 in above pdf. The recoil electrons in turn create ionization and subsequent conduction electrons (or holes) in semiconductor (germanium or silicon) depletion regions.

[added] **See also chapter 13 in Jackson Classical Electrodynamics (2nd edition) or

http://www.google.com/url?sa=t&sour...aJt4WQRjfsa2c2X3A&sig2=glnhffBd7l6-1ZsL8iN58w

Bob S
 
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