Undergrad How can I use the NIST XCOM online tool to calculate X and Gamma shielding?

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The NIST XCOM online tool can be used to calculate X and gamma shielding by providing a graph and table of scattering values. For example, to calculate shielding from 1e19 gamma rays of 4 MeV using 10 cm of lead, a scattering value of 4.2e-2 cm2/g and lead's density of 11.34 g/cc are utilized. The scattering length is derived from the total mass attenuation coefficient, resulting in a calculated length of 21 cm. The remaining rays after passing through the lead is calculated using the formula N=N0*exp(-thick/scattering length), yielding that 62% of the radiation passes through the lead. The discussion also emphasizes the importance of correctly identifying the terms used in the calculations, particularly distinguishing between scattering lengths and mass attenuation coefficients.
Javier Lopez
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In order to calculate X and Gamma shielding of I should like use the NIST XCOM online at https://physics.nist.gov/PhysRefData/Xcom/html/xcom1.html
In the Xcom tool it is obtained a graph and table of scattering in cm2/g.

In order to calculate as example the shielding from 1e19 gamma rays of 4MeV using 10cm of lead.
Using the tool I obtained a scattering of 4.2e-2 cm2/g and the density of Pb solid is 11.34 g/cc
I suppose scattering length= 1/(4.2e-2 cm2/g * 11.34 g/cc)=21cm
Then the remainder rays should be: N=N0*exp(- lead thick/scattering length)=6.2e18 so 62% of generated radiation passes through lead
 
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Just multiply the two to get scattering per length.
 
sorry, I edited while you wrote :( , staff mentor, could you or somebody else check the equation as I edited?
 
Javier,
your equation is good but the terms you used are Strange. 4.2e-2 cm2/g is not a "scattering length" but a total mass attenuation coefficient (due to Scattering, Photoelectric Absorption, Pair Production).
 
Hello PSR1919121, you are right, the equation should be less strange in latex (I do not know how if possible here):
$$\lambda =\frac{1}{4.2*10^{-2}*\frac{cm^2}{g}*11.34\frac{g}{cm^3}}=21cm$$

Then I write the formula here:
$$N=N_0*e^{-\frac{thick}{\lambda }}=N_0*e^{-thick*SC*\rho}$$
SC is the scattering parameter (cm2/g) and rho is the density (g/cm3), thick is in cm
(It could be used SI using kg and metres instead of g and cm but unfortunately SC is written in cgs)
 

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Javier Lopez said:
the equation should be less strange in latex (I do not know how if possible here)
Just surround it by ##[/color] for inline formulas and $[/color]$ for full-size formulas.
$$\frac{1}{4.2\cdot 10^{-2} \frac{cm^2}{g}}$$
 
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Done, we are doing good job