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Bmax in Bohr's calculation for energy loss of charged particles (classical case)

  1. Nov 3, 2014 #1
    Dear all,

    I have inquiries about the Bohr's calculation for energy loss of charged particles. These inquires are from the "Techniques for Nuclear and Particle Physics Experiments" (W.Leo).

    On page 23 of this book, it is written that:

    • the the interaction time is t = b/(γv), with b is the distance of atomic electron from the trajectory of incident particle and v is the velocity of incident particle.
    • orbital period of electron is τ = 1/ν, with ν is the orbital frequency of atomic electron
    Why the interaction time of incident particle and atomic electron is t = b/v ? Does it means that it is considered as head-on collision ? It so, how about the electric force and Lorentz force that can be happen since incident particle and electron are both charged ?

    It is written that t = b/(γv) ≤ τ = 1/ν, and the reason is just for the interaction between incident charged particle and atomic electron can happen, isn't it ?

    Thank you so much.
     
  2. jcsd
  3. Nov 3, 2014 #2
    The calculus is done in the so-called impulsive approximation.
    The upper limit to the impact parameter b is given by the fact that the interaction duration must be much less than the electron orbital period, since otherwise the electron can not be considered at rest during the interaction.
    The classical momentum transfer in a collision goes as Dp = Ze^2/bv (1).
    The electrostatic force at the distance of b is F(b) = Ze^2/b^2 (2).
    From the impulse of the force you get Dp = F(b)*T (3), whereT is the interaction time.
    Comparing (1), (2), (3) it is easily seen that "it is as if" the interaction lasts T = 2b/v.
    In relativistic case, all is the same except that v becomes gamma*v, where gamma is the lorentz factor.
    Hope it helps.
     
  4. Nov 3, 2014 #3
    Dear Fek,

    Based on the impulse approximation, T = 2b/v. but the time for interaction that I mentioned is t = b/v. they are not the same !

    Please help. I am also thinking.
    Thanks.
     
  5. Nov 3, 2014 #4

    ChrisVer

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    Gold Member

    In fact I wonder how Fek_ got this 2 from the equations he wrote :)
    So let's write everything more readable...
    During the interaction, you have some momentum change: [itex]\Delta p = \frac{Z e^2}{bv}[/itex]
    and the force at [itex]r=b[/itex] is [itex]F(b)= \frac{Ze^2}{b^2} \Rightarrow Dp= F(b) \times T[/itex]
    Or:
    [itex] \frac{Z e^2}{bv}= \frac{Ze^2}{b^2} T \Rightarrow T= b/v [/itex]

    As noted, in the relativistic case again, [itex]T=\frac{b}{\gamma v}[/itex]
     
  6. Nov 3, 2014 #5
    I think T = 2b/v is right.
    Since the electric energy is E = Ze^2/b=mv^2/2 and T = mv=2E/v, thus, T = 2Ze^2/bv.
     
  7. Nov 5, 2014 #6
    Well, I wrote that very quickly and a factor 2 went lost :).
    The exact result for momentum transfer is Δp = (Ze2)/(2π ε0 bv) in SI units.
    The force at the closest distance b is F(b) = (Ze2)/(4π ε0 b2).
    So, when one takes Δp = F(b)τ, one gets the correct factor 2 for τ = 2b/v.
    Though, remind that the semi-classical Bohr's calculation, like this, is just a crude approximation. The Bethe-Bloch quantum mechanical approach is the proper way. It is not so difficult as one might think, at least for the "close" collisions". Try it :).
     
  8. Nov 5, 2014 #7
    Dear Fek,

    But can I know the reason why t = b/v ≤ τ = 1/ν ? I dont understand the physics meanings.
     
  9. Nov 5, 2014 #8
    Can you explain it for me?
    This will be helpful for me. because i have a report tomorrow.
     
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