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Bethe formula

  1. Aug 4, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate the energy loss loss ##\Delta T## for protons, deuterons and ##\alpha##-particles between ##10## to ##200##MeV when they're passing through a 2mm thick plastic scintillator. Suppose ##Z/A=0.56##, ##I=65eV## and ##\rho = 1.10##g/cm^3.

    2. Relevant equations
    Bethe formula
    ##\frac{dE}{dx}=\left( \frac{e^2}{4 \pi \epsilon_0} \right)^2 \frac{4 \pi z^2 N_0 Z\rho}{mc^2\beta^2 A} \left[ \ln \left( \frac{ 2mc^2 \beta^2}{I}\right) - \ln (1-\beta^2)-\beta^2\right]##
    ##N_0## is Avogadro's constant, ##v = \beta c##, ##ze## the electric charge of the particle, ##Z, A## and ##\rho## the atomic number, atomic weight, and density of the stopping material and ##m## the electron mass.

    Relativistic energy
    ##T = mc^2-\frac{mc^2}{\sqrt{1-v^2/c^2}} = mc^2-\frac{mc^2}{\sqrt{1-\beta^2}}##

    Mass of proton ##m_p = 938.28##MeV/c^2
    Mass of electron ##m = 0.511003##MeV/c^2
    Electron charge ##1.60217662 \cdot 10^{-19}## coulomb
    Avogadro's constant ##6.0221409\cdot 10^{23}##
    permittivity of free space ##8.85418782\cdot 10^{-12}## F/m

    3. The attempt at a solution
    I started by just trying to compute ##\Delta T## for ##T = 10 MeV## for the proton.
    The ##\beta## factor after some algebra is
    ##\beta = \sqrt{1-\left(\frac{1}{\frac{T}{mc^2}+1}\right)^2}##

    For the rest of the values I converted everything to SI units and then input the formula into matlab. However I end up with an error in the order of ##10^2## if I do this and I think it may be since I'm using the formula with the wrong units. The ##Z/A## value in particular I have no unit of and I don't know if I should convert that to kg or not if it's in atomic mass units.

    Here's the matlab code for the calculation:
    Code (Text):

    e=1.60217662e-19; %electron charge
    e0 = 8.85418782e-12; %permitivity of free space
    ZA = 0.56; %unit??
    I=65e-6; %MeV
    rho = 1.10*1000; %converted to kg/m^3
    N0 = 6.0221409e+23; %avogadro
    d=0.002; %distance in meter
    z=1; %proton
    mc2 = 0.511003; %MeV electron
    mc2p = 938.28; %MeV proton
    T = 10; %MeV
    beta=sqrt(1-1./(T./mc2p+1).^2);
    k = (e^2/(4*pi*e0))^2*4*pi*z^2*N0*rho*ZA./(mc2*e*1e6); %convert to joule/meter
    temp = 1/beta^2*(log(2*mc2*beta.^2./I)-log(1-beta.^2)-beta.^2);
    dT = d*k.*temp/e/10^6 %convert to MeV
     
    This outputs ##0.0105## MeV while the graph provided in the answer says I should get something close to ##9## MeV.

    I realise this is quite a long post so I'm mostly hoping that someone is familiar with the formula to know which units I should be using since I suspect that is my error.
     
  2. jcsd
  3. Aug 4, 2016 #2

    haruspex

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    Isn't it just the ratio protons/(protons+neutrons)?
    I must be missing something here. mc2 is 0.5MeV, right? I don't see anything in the quoted equation that corresponds to the e*1e6.
     
  4. Aug 4, 2016 #3
    I don't think so, it should be the relative atomic mass https://en.wikipedia.org/wiki/Bethe_formula
    And unless ##A## got the unit ##kg## the units doesn't work out since
    ##\left(C^2/(F/m)\right)^2\cdot kg/m^3/J = \frac{C^4 \cdot kg}{F^2\cdot m \cdot J} = \frac{J\cdot kg}{m}##

    I converted to joule here. ##1 MeV = e \cdot 10^6##
     
  5. Aug 4, 2016 #4

    TSny

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    I'm thinking ##A## is the atomic mass in g/mole. The mole here will cancel with the mole in ##N_0## and the grams will cancel with the grams in ##\rho##. But if this is so, then Z/A should have been specified as 0.56 mole/g, rather than as a dimensionless quantity.
     
  6. Aug 4, 2016 #5

    Charles Link

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    I think I may have found an algebraic error. In your expression for ## \beta ## it should be ## T/(mc^2) -1 ## in that denominator instead of ## T/(mc^2)+1 ##.
     
  7. Aug 4, 2016 #6

    TSny

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    I think the expression for ##\beta## is OK. However, the original expression for ##T## was given as
    The two terms on the right should be interchanged.
     
  8. Aug 4, 2016 #7

    TSny

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    Incand, note that the rate of loss of energy, |dE/dx|, increases as the particle slows down. So, you might not be able to treat dE/dx as a constant while the particle is traveling the 2 mm distance through the plastic.

    EDIT: Is there any way you could post the graph you referred to in the OP?
     
  9. Aug 5, 2016 #8
    mole/g would make sense, I see if I can get some better values out of that.

    I redrew the answer figure. It's not exact but pretty close. It is a loglog plot. I get lines similar lines if I do the same calculation as above but varying beta but with the wrong factors infront.

    That's true TSny, I guess I may have to integrate then. Still at least for high energy particle the energy loss is low enough that this shouldn't affect the answer that much so maybe it's easier to work on the 200 MeV side first.
     

    Attached Files:

  10. Aug 5, 2016 #9
    Okey It seems you were correct about the unit of ##Z/A##. Using that I get reasonable answers

    Code (Text):

    e=1.60217662e-19; %electron charge
    e0 = 8.85418782e-12; %permitivity of free space
    ZA = 0.56; %unit??
    I=65e-6; %MeV
    rho = 1.10*10^6; %converted to g/m^3
    N0 = 6.0221409e+23; %avogadro
    d=0.002; %distance in meter
    z=1; %proton
    mc2 = 0.511003; %MeV electron
    mc2p = 938.28; %MeV proton
    T = 200; %MeV
    beta=sqrt(1-1./(T./mc2p+1).^2);
    k = (e^2/(4*pi*e0))^2*4*N0*pi*z^2*rho*ZA./(mc2*e*1e6); %convert to joule/meter
    temp = 1/beta^2*(log(2*mc2*beta.^2./I)-log(1-beta.^2)-beta.^2);
    dT = d*k.*temp/e/10^6 %convert to MeV
     
    This outputs about ##1##MeV. while if I change to ##T=10##MeV I get ##\Delta T \approx 11## MeV so I'm in the right order. I'm guessing it's just I need to integrate to get the last part. For ##200## MeV, it already works well
     
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