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

In summary, the NIST XCOM online tool is a valuable resource for calculating X and Gamma shielding. It provides comprehensive data on the atomic and nuclear properties of elements and compounds, allowing users to accurately estimate the amount of shielding material needed to protect against these types of radiation. The tool also offers a user-friendly interface and various customization options to cater to different scenarios and materials. Overall, the NIST XCOM online tool is a reliable and efficient solution for calculating shielding requirements in a wide range of applications.
  • #1
Javier Lopez
75
3
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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  • #2
Just multiply the two to get scattering per length.
 
  • #3
sorry, I edited while you wrote :( , staff mentor, could you or somebody else check the equation as I edited?
 
  • #5
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).
 
  • #6
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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  • #7
Javier Lopez said:
the equation should be less strange in latex (I do not know how if possible here)
Just surround it by ## for inline formulas and $$ for full-size formulas.
$$\frac{1}{4.2\cdot 10^{-2} \frac{cm^2}{g}}$$
 
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Likes Javier Lopez
  • #8
Done, we are doing good job
 

What is "X & G Photon scattering"?

"X & G Photon scattering" is a scientific phenomenon where photons, which are particles of light, interact with matter. This interaction can result in the scattering or redirection of photons in different directions.

How does "X & G Photon scattering" occur?

"X & G Photon scattering" can occur through various mechanisms such as Compton scattering, Rayleigh scattering, or Raman scattering. These mechanisms involve the absorption and re-emission of photons by atoms or molecules in a material.

What are the applications of "X & G Photon scattering"?

"X & G Photon scattering" has a wide range of applications in fields such as materials science, astronomy, and medical imaging. It is used to study the structure and properties of materials, determine the composition of distant objects, and create detailed images of the human body.

Can "X & G Photon scattering" be used for energy production?

No, "X & G Photon scattering" is not a viable method for energy production. While it involves the interaction of photons, it does not generate or release any significant amount of energy.

What are the implications of "X & G Photon scattering" in the study of the universe?

"X & G Photon scattering" plays a crucial role in our understanding of the universe. It allows us to observe and analyze the light from distant objects, providing valuable information about their composition, temperature, and movement. It also helps us understand the behavior of light in extreme environments such as black holes and supernovae.

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