Thickness of lead required for 200 keV x-rays?

[ireland01]

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

 

[mfb]

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.

 

[ireland01]

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

 

[mfb]

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