Thickness of lead required for 200 keV x-rays?

  • Thread starter ireland01
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
35,720
12,305
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.
 
  • #3
17
0
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:
  • #4
35,720
12,305
4.5mm looks better, and the new prefactor 7*ln(2) looks better as well.
 

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