X & G Photon scattering

  • #1
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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Answers and Replies

  • #3
sorry, I edited while you wrote :( , staff mentor, could you or somebody else check the equation as I edited?
  • #5
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
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