- #1

- 17

- 0

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

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- Thread starter ireland01
- Start date

- #1

- 17

- 0

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

mfb

Mentor

- 35,720

- 12,305

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

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

mfb

Mentor

- 35,720

- 12,305

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

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