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Thickness of lead required for 200 keV x-rays?

  1. Apr 18, 2013 #1
    is this correct?

    mass attenuation coeff. = 8.636 cm^2/g

    8.636 * (density of lead: 11.3 g/cm^3) = 975.868 per cm

    thickness (mm):

    t= 9.21 / 975.868

    = 0.009 cm

    = 0.09 mm
     
  2. jcsd
  3. Apr 18, 2013 #2

    mfb

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    Staff: Mentor

    Where does the factor of 9.21 come from?
    Assuming your coefficient is right, and you want 9.21 radiation lengths, the answer looks right.
     
  4. Apr 18, 2013 #3
    I found that factor online.It said to get intensity to < 1% but by my calculations, for 200 KeV beam using formula:
    HVL = 0.693/u

    need 7 HVL layers to get to 0.8% which is 4.48 mm.
    where u = 10.15 cm^-1
     
    Last edited: Apr 18, 2013
  5. Apr 18, 2013 #4

    mfb

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    Staff: Mentor

    4.5mm looks better, and the new prefactor 7*ln(2) looks better as well.
     
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