Thickness of lead required for 200 keV x-rays?

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Homework Help Overview

The discussion revolves around determining the thickness of lead required to attenuate 200 keV x-rays, focusing on the mass attenuation coefficient and its application in calculating necessary thickness for specific radiation levels.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of thickness based on the mass attenuation coefficient and question the origin of specific factors used in the calculations, such as the factor of 9.21 and its relation to radiation lengths and half-value layers (HVL).

Discussion Status

There is an ongoing exploration of different approaches to calculate the required thickness, with some participants suggesting that the calculations for 4.5 mm may be more accurate. The discussion includes verification of coefficients and factors used in the calculations, indicating a productive exchange of ideas without a clear consensus.

Contextual Notes

Participants reference specific values and formulas, such as the mass attenuation coefficient and the relationship between HVL and intensity reduction, while noting variations in the calculated thickness based on different assumptions.

ireland01
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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
 
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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.
 
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:
4.5mm looks better, and the new prefactor 7*ln(2) looks better as well.
 

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